# Paris-Caen Program: Artifaille for Palindromization and Its 23 Bacouettes

David Pigeon

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# Paris-Caen Program: Artifaille for Palindromization and Its 23 Bacouettes

David Pigeon*

January 29, 2025

**Keywords:** base decomposition, Paris-Caen program, artifaille, palindromization, mirror numbers, Norman numbers, Celt numbers, Breton numbers, Québécois numbers, Sicilian numbers, dukes, elingard, riboutins, bacouettes.

## Abstract

This text presents a palindromization algorithm applied to natural integers based on their base decomposition. We designate as "Norman numbers" those for which the algorithm converges. We develop a *artifaille*<fnref n="1" /> possible for studying this algorithm. Based on computer simulations, several bacouettes are formulated. The set of these bacouettes constitutes the Paris-Caen Program.

## Contents

<table><tr><td><strong>Introduction</strong></td><td><strong>3</strong></td></tr><tr><td>Norman Palindromic Legend</td><td>3</td></tr><tr><td>The Lower Normandy of Rollor</td><td>3</td></tr><tr><td>The Upper Normandy of Nollon</td><td>4</td></tr><tr><td>Quelques explications pour les non-normands</td><td>6</td></tr><tr><td>Two Simple Examples</td><td>6</td></tr><tr><td>Context</td><td>9</td></tr><tr><td><em>A tantôt</em> for more Norman legends</td><td>9</td></tr><tr><td><strong>1 Decomposition in Arithmetic Base</strong></td><td><strong>11</strong></td></tr><tr><td>1.1 Sets N<sub>a</sub><sup>b</sup></td><td>11</td></tr><tr><td>1.2 The Rollon</td><td>12</td></tr><tr><td>1.3 Basic Tools</td><td>12</td></tr><tr><td>1.4 Theorems</td><td>14</td></tr><tr><td>1.5 Addition in base b</td><td>20</td></tr><tr><td>1.6 Artifaille in Python</td><td>22</td></tr><tr><td><strong>2 Artifaille for Palindromization</strong></td><td><strong>28</strong></td></tr><tr><td>2.1 Reduction Operation</td><td>28</td></tr><tr><td>2.2 Mirror Operation</td><td>32</td></tr><tr><td>2.3 Palindromic Operation</td><td>34</td></tr><tr><td>2.4 Artifaille in Python</td><td>37</td></tr><tr><td><strong>3 Artifaille of Norman</strong></td><td><strong>40</strong></td></tr><tr><td>3.1 Norman Operation</td><td>40</td></tr><tr><td>3.2 Norman Sequence in Base b</td><td>40</td></tr><tr><td>3.3 Norman Integers in Base b</td><td>41</td></tr><tr><td>3.4 The Case of the Celts in Base b</td><td>42</td></tr></table>

<footnote>*https://sites.google.com/view/davidpigeon/</footnote>

<footnote n="1">¹A set of tools for an artisan.</footnote>
<table><tbody><tr><td>3.5</td><td>The Case of the Quebecers in Base b</td><td>44</td></tr><tr><td>3.6</td><td>The Case of the Sicilians in Base b</td><td>45</td></tr><tr><td>3.7</td><td>Artifaille in Python</td><td>45</td></tr><tr><td><b>4</b></td><td><b>Group Strategies</b></td><td><b>49</b></td></tr><tr><td>4.1</td><td>Depth of the Dream</td><td>49</td></tr><tr><td>4.2</td><td>Elingard Operation</td><td>49</td></tr><tr><td>4.3</td><td>Duc and Duc Elingard operations</td><td>52</td></tr><tr><td>4.4</td><td>Set of quasi-elingard ducs</td><td>55</td></tr><tr><td>4.5</td><td>Routine operations</td><td>56</td></tr><tr><td>4.6</td><td>Fundamental Result</td><td>58</td></tr><tr><td>4.7</td><td>From Nollon to Rollor</td><td>62</td></tr><tr><td>4.8</td><td>Applications to base 2</td><td>63</td></tr><tr><td>4.9</td><td>Artifaille in Python</td><td>64</td></tr><tr><td><b>5</b></td><td><b>The Bacouettes</b></td><td><b>72</b></td></tr><tr><td>5.1</td><td>Study of Palindromization</td><td>72</td></tr><tr><td>5.2</td><td>Study of the degree of the Norman sequence</td><td>73</td></tr><tr><td>5.3</td><td>Study of the length of a triskelion</td><td>73</td></tr><tr><td>5.4</td><td>Study of the ribouts</td><td>74</td></tr><tr><td>5.5</td><td>Study of B<sub>1,b</sub> &subset; N<sub>b</sub></td><td>75</td></tr><tr><td>5.6</td><td>Study on the density</td><td>76</td></tr><tr><td>5.7</td><td>Study of N<sub>b</sub> = N<sub>&bull;</sub> for small b</td><td>76</td></tr><tr><td>5.8</td><td>Study of N<sub>b</sub> &ne; N<sub>&bull;</sub></td><td>78</td></tr><tr><td><b>6</b></td><td><b>Extensions and Variants</b></td><td><b>80</b></td></tr><tr><td>6.1</td><td>Changing the Reduction Operation</td><td>80</td></tr><tr><td>6.2</td><td>Changing the Palindromization Operation</td><td>80</td></tr><tr><td>6.3</td><td>Euclidean Division in R<sub>+</sub><sup>*</sup></td><td>81</td></tr><tr><td>6.4</td><td>Euclidean Division in a Euclidean Ring</td><td>81</td></tr><tr><td><b>A</b></td><td><b>Computational Studies</b></td><td><b>82</b></td></tr><tr><td>A.1</td><td>Case B<sub>d,b</sub> &subset; N<sub>b</sub></td><td>82</td></tr><tr><td>A.2</td><td>Case of Bases with Triskelion</td><td>84</td></tr><tr><td>A.3</td><td>Celt Numbers Less Than 10<sup>4</sup></td><td>90</td></tr><tr><td>A.4</td><td>Breton Numbers and Multicelt Numbers</td><td>93</td></tr><tr><td>A.5</td><td>Sicilian Numbers in Base 66</td><td>94</td></tr><tr><td>A.6</td><td>Celt and Sicilian Numbers of Degree 1</td><td>96</td></tr><tr><td><b>B</b></td><td><b>Artifaille in Python</b></td><td><b>98</b></td></tr></tbody></table>
# Introduction

## Norman Palindromic Legend

This all began in the **Year Zero**. On the **First Day** (of year one), the **God Rollon** and his gods emerged from the ice.

On the **Second Day**, the God Rollon created **Normandy**.

*Quant Rou a Roen arriva*
*Qui de north homes amena*
*Cil furent Normant apelé*
*Por ceu que de north furent né*
*(...) E de Normanz est apelee*
*Normendie que il unt poplee.*

When Rollon arrived in Rouen,
Bringing men from the North,
These were called Normans
Because they were born in the North
(...) And Normandy was given this name
By the Normans who populated it.

Wace<fnref n="2" />, *Roman de Rou<fnref n="3" />*

Normandy, feeling lonely, Rollon created, on the **Third Day**, around Normandy, the celt lands (particularly the Britains), and in more distant regions, Sicily (also called the Normandy of the sun), and at the other end of the world, Quebec (also called distant Normandy).

On the **Fourth Day**, Normandy gave birth to two children: **Rollor** and **Nollon**, the palindromic demigods. They took the form of leopards<fnref n="4" />:

<figure><img src="image_2.png" /></figure>

On the **Fifth Day**, Rollon ordered his sons to populate this Norman land. The two sons disagreed on how to populate Normandy.

So, on the **Sixth Day**, Rollon divided Normandy in two: **Lower Normandy** for Rollor and **Upper Normandy** for Nollon.

After working for six days, Rollon rested and watched his sons populate Normandy until the end of time.

In the following, we recall the games proposed by Rollor and Nollon to populate Lower Normandy and Upper Normandy.

### The Lower Normandy of Rollor

Rollor settled not far from the sea, in a town called Cathim<fnref n="5" />. The Lower Normandy playground was created. Here is the game Rollor proposed to his father Rollon. Rollor decided to create nothing the first year. The game would begin in the second year ($n := 2$). In that second year, he created the base $b := 2$. He placed it in a region not far from Normandy, called Lutèce<fnref n="6" />.

At the beginning of each year $n$ (starting from the second year), Rollor would create a new base $b := n$. Thus, at the start of year $n$, there would be $n - 1$ bases already created. Each existing base

<footnote n="2">²Norman poet of the 12th century.</footnote>

<footnote n="3">³Many other texts in Norman can be found on the website of the Federation of Associations for the Norman Language E (FALE): https://www.fale-normandie.fr/.</footnote>

<footnote n="4">⁴Source Wikipedia: Norman_Flag</footnote>

<footnote n="5">⁵Caen.</footnote>

<footnote n="6">⁶Paris.</footnote>
had the possibility, each year, to leave Lutèce for the great outdoors of Lower Normandy, celt lands, or Quebec. Rollor gave them the ability to travel through time (to the past or the future) during their dreams.

For this, each base $b$ had to decompose the year $n$ (in its base $b$) into a finite sequence of small blocks of natural numbers less than $b-1$. During its dream, the base $b$ had the possibility to travel to another year $n$ in two different ways:

⊗ If the first block is equal to the last block, it could remove them, thus obtaining a sequence with two fewer blocks, corresponding to another year $n'$ smaller than year $n$;

⊗ Otherwise, it could add to its sequence of blocks the mirror year of that sequence and thus obtain a new sequence of blocks, corresponding to a year $n'$ greater than year $n$, aiming to achieve a year increasingly palindromic.

It could continue with the sequences of blocks it obtained. Three possibilities then arose:

⊕ If a base $b$ managed to eliminate all its blocks during its successive dreams and obtained what Rollor called the Rollon year • (the blockless year, named in homage to his father), it would become Lower Norman until the end of year $n$. It could then spend the end of year $n$ in Lower Normandy, enjoying Lower Norman pleasures, joining the happy bases like it;

⊕ If it returned to a sequence it had already encountered, it could orbit around Lower Normandy and spend the end of the year in one of the celt lands. A special case occurred if it returned to itself in its dreams: it would go directly to Brittany, which wasn't bad either;

⊕ If it was in none of these cases, it would be sent to infinity, which the inhabitants of Lutèce called Quebec, at the other end of the world, in the cold.

Unfortunately, if it couldn't figure out where it should go, it would stay in Lutèce, where space was scarce, as the population of Lutèce increased each year by one additional base $b := n$.

In this first draft of the game, Rollon noticed that his son had neglected the possibility that some bases would not have enough time to decide where to travel in their dreams. Rollor therefore changed the rules and decided that these bases, despite all their efforts, would be sent to the Sicilian sun.

Once they arrived in their Norman or celt lands, Rollor gave these bases $b$ the possibility of avoiding a return to Lutèce for year $n+1$. To do this, they had to, in their respective lands, dream and transform for year $n+1$ into the light year (they leave or remain in Lower Normandy) or into a triskelion (they leave or remain in celt lands). All others would be required to return to Lutèce before the end of year $n$.

Thus, great travels and great movements of bases occurred each year. Lower Normandy thus populated itself with native bases. They harvested and cultivated the delights created by Rollon.

"How provident is nature!
She grows apples in Normandy
Knowing the natives of this province drink only cider."

Henry Bonaventure Monnier (1799 - 1877)

This game was proposed to Rollon by Rollor for Lower Normandy. Rollon accepted it.

## The Upper Normandy of Nollon

Jealous of his brother, Nollon settled in a city called Roen<fnref n="7" />. The second playground was thus created. Out of pride, Nollon decided to create an infinite number of small beings: the natural numbers, arranged

<footnote n="7">⁷Rouen.</footnote>
from smallest to largest. He placed these numbers in a zone not far from Lutèce, Versalias<fnref n="8" />. Yearning for vast spaces, Nollon created a game for them starting in the second year.

Each year *b*, a number *n* had the possibility of leaving Versalias to breathe the fresh air of Normandy, celt lands, or Quebec. To do this, Nollon copied Rollor's idea and gave them, according to year *b*, the possibility, during their dreams, to decompose into a finite sequence of small blocks of natural numbers less than *b* − 1. During their dreams, they had the possibility of transforming into another number in two different ways:

⊗ If the first block was equal to the last block, they could remove them, thus obtaining a sequence with two fewer blocks, corresponding to another natural number *n*′ smaller than themselves;

⊗ Otherwise, they could add to their sequence of blocks the mirror number of that sequence, aiming to achieve a number increasingly palindromic.

They could continue thus with the sequences of blocks obtained. Three possibilities then arose:

⊕ If a number *n* managed to eliminate all its blocks during its successive dreams and obtained what Nollon called the Rollon number • (the blockless number, named in homage to his father as his brother Rollor had done), it would become Upper Normand until the end of year *b*. It could then spend the end of year *b* in Upper Normandy, enjoying Upper Normand pleasures and joining the happy numbers like it;

⊕ If it returned to a sequence it had already encountered, it could orbit around Normandy and spend the end of the year in one of the celt lands. A special case occurred if it returned to itself in its dreams: it would go directly to Brittany, which wasn't bad either;

⊕ If it was in none of these cases, it would be sent to Quebec.

Once again, if it couldn't figure out where to go, it would stay in Lutèce, where space was scarce.

Having taken into account the remark Rollon had made to his brother Rollor, Nollon added a rule for the numbers that, despite all their efforts, didn't have enough time to decide where to travel. He decided they would be sent to the Sicilian sun.

Once they arrived in their Norman or celt lands, Nollon gave these numbers *b* the possibility of avoiding a return to Lutèce for year *b* + 1. To do this, they had to, in their respective lands, dream and transform, for year *b* + 1, into a light number (they leave or remain in Upper Normandy) or into a cycle (they leave or remain in celt lands). All others would be required to return to Lutèce before the end of year *b*.

Thus, great travels and great movements of numbers occurred each year. Upper Normandy thus populated itself with native numbers. They harvested and cultivated the delights created by Rollon.

This game was proposed to Rollon by Nollon for Upper Normandy. Rollon accepted it.

Thus bases and numbers began to do mathematics.

The Paris-Caen program was created.

As they have said in Normandy since the Middle Ages:

Hora!

<footnote n="8">⁸Versailles.</footnote>
D auns men prechi

In my speech

Coumbyin de feis m'assyisaunt lenreit aù hâot de la tablle

How many times, sitting there at the end of the table,

Coume hiyi coume à c'sei, tenaunt ma taête à pouégnie

Like yesterday, like tonight, holding my head in my hands,

J'érais vouli ichin dire dé dequei d'indisablle

I wanted to say here something unspeakable,

Et je restais en jé n'sais. Je viyais touot s'alouégni

But I remained stuck, I saw everything slipping away,

Dauns mon loceis normaund pique ch'est ainchin qu'je prêche

In my Norman language, because that's how I express myself,

J'guette d'eun bord et de l'àote et touot me viynt coume i deit

I watch from one side to the other, and everything comes to me as it should,

Touot m'arrive en débord, touot s'mêle et touot se vouêche

Everything comes to me in overflow, everything tangles, everything stirs,

Dauns le loceis de tcheu nous, jé n'sis brin à l'étreit

In the language of our home, I don't feel constrained.

Côtis-Capel (1915 - 1986)<fnref n="9" />, Raz-bannes

## Quelques explications pour les non-normands

As we will see, we define a method allowing an integer $n$ to become Norman in base $b$ by applying two possible operations to the decompositions of $n$ in base $b$: reduction and palindromization. But as Rollor and Nollon suggest, there are two ways to think about this.

⊙ For Rollor, the game fixes a year $n$ (that is, a natural number) and observes the bases $b$ between 2 and $n$. In other words, for a fixed $n$ greater than 2 (the game begins in the second year), we consider the bases:

$$ \{2, \dots, n\}. $$

He does not consider bases greater than $n$ because Rollor understood that these bases would have no game to play, as their decompositions would have only one block. Indeed, any integer $n$ decomposes in base $b > n$ with a single block.

⊙ For Nollon, the game fixes a year $b$ (that is, a natural number greater than 2, called a base) and observes the integers $n$ from 0 to infinity. In other words, for a fixed base $b$ greater than 2 (the game also begins in the second year), we consider the integers:

$$ N := \{0, 1, 2, \dots\}. $$

## Two Simple Examples

Let us start with the integer $n_0 := 149\,823$ in base $b := 10$. It decomposes into blocks:

1 4 9 8 2 3

(1) The first and last block of $n_0$ are different. We therefore add to $n_0$ its "mirror," that is, the blocks in reverse order $[3 \ 2 \ 8 \ 9 \ 4 \ 1]$. We obtain:

$$ n_1 := \begin{array}{c} + \\ | \\ \hline 1 & 4 & 9 & 8 & 2 & 3 \\ 3 & 2 & 8 & 9 & 4 & 1 \\ | \\ \hline 4 & 7 & 8 & 7 & 6 & 4 \end{array} $$

<footnote n="9">⁹Norman writer and priest.</footnote>
(2) The number $n_1$ has the same first and last block. We simplify it by removing these two extreme digits:

$$n_2 := \begin{bmatrix} 7 & 8 & 7 & 6 \end{bmatrix}.$$

(3) The extreme blocks of $n_2$ are different. We then add to it its mirror $\begin{bmatrix} 6 & 7 & 8 & 7 \end{bmatrix}$:

$$n_3 := \begin{bmatrix} & 7 & 8 & 7 & 6 \\ + & 6 & 7 & 8 & 7 \\ \hline 1 & 4 & 6 & 6 & 3 \end{bmatrix}.$$

(4) Similarly, the extreme blocks of $n_3$ are different. We add to it its mirror $\begin{bmatrix} 3 & 6 & 6 & 4 & 1 \end{bmatrix}$:

$$n_4 := \begin{bmatrix} 1 & 4 & 6 & 6 & 3 \\ + & 3 & 6 & 6 & 4 \\ \hline 5 & 1 & 3 & 0 & 4 \end{bmatrix}.$$

(5) The extreme digits of $n_4$ are still different. We add to it its mirror $\begin{bmatrix} 4 & 0 & 3 & 1 & 5 \end{bmatrix}$:

$$n_5 := \begin{bmatrix} 5 & 1 & 3 & 0 & 4 \\ + & 4 & 0 & 3 & 5 \\ \hline \color{red}{9} & \color{red}{1} & \color{red}{6} & \color{red}{1} & \color{red}{9} \end{bmatrix}.$$

(6) The number $n_5$ is palindromic *id est* it reads the same backward and forward. We simplify it by removing all its blocks. We obtain a "blockless number," denoted:

$$n_6 := \bullet.$$

The algorithm seems to always end with the symbol $\bullet$, regardless of the initial natural number $n$. However, this is not true if the base $b$ changes. To simplify notation, the number $n_0 := 149\,823$ is written in base 10 as:

$$(1 \mid 4 \mid 9 \mid 8 \mid 2 \mid 3)_{10}.$$

To obtain this representation, note that:

$$149\,823 = \color{red}{1} \times 10^5 + \color{red}{4} \times 10^4 + \color{red}{9} \times 10^3 + \color{red}{8} \times 10^2 + \color{red}{2} \times 10^1 + \color{red}{3} \times 10^0.$$

The blocks of $n_0$ will now be called **digits**.

Let us now consider the number $n_0 := 143\,712$ with $b := 11$. As:

$$\begin{align*} 143\,712 &= 13\,064 \times 11 + 8 \\ &= (1\,187 \times 11 + 7) \times 11 + 8 = 1\,187 \times 11^2 + 7 \times 11 + 8 \\ &= (107 \times 11 + 10) \times 11^2 + 7 \times 11 + 8 = 107 \times 11^3 + 10 \times 11^2 + 7 \times 11 + 8 \\ &= (9 \times 11 + 8) \times 11^3 + 10 \times 11^2 + 7 \times 11 + 8 \\ &= \color{red}{9} \times 11^4 + \color{red}{8} \times 11^3 + \color{red}{10} \times 11^2 + \color{red}{7} \times 11^1 + \color{red}{8} \times 11^0 \end{align*}$$

we write it as:

$$n_0 = (9 \mid 8 \mid 10 \mid 7 \mid 8)_{11}.$$

Let us then apply the algorithm to this number.
(1) The first and last digits of $n_0$ are different. Thus, we add to $n_0$ its "mirror," $(8|7|10|8|9)_{11}$ (the number formed by reversing the order of the digits of $n_0$ in base 11). We get:

$$n_1 := (9|8|10|7|8)_{11} + (8|7|10|8|9)_{11} = (1|7|5|10|5|6)_{11}.$$

Addition proceeds as in base 10, except that a carry only occurs if the sum of digits exceeds 11 instead of 10. Let's revisit the above sum; we have (the 1s in red are the carries):

$$\begin{array}{r@{\ }r@{\ }r@{\ }r@{\ }r@{\ }r}
 & & \color{red}{1} & \color{red}{1} & \color{red}{1} & \color{red}{1} \\
 & & 9 & 8 & 10 & 7 & 8 \\
+ & & 8 & 7 & 10 & 8 & 9 \\
\hline
n_1 & := & 1 & 7 & 5 & 10 & 6
\end{array}$$

(2) The extreme digits of $n_1$ are different. We thus add its mirror $(6|5|10|5|7|1)_{11}$:

$$n_2 := (1|7|5|10|5|6)_{11} + (6|5|10|5|7|1)_{11} = (8|2|5|5|1|7)_{11}.$$

(3) The extreme digits of $n_2$ are different. We thus add its mirror $(7|1|5|5|2|8)_{11}$:

$$n_3 := (8|2|5|5|1|7)_{11} + (7|1|5|5|2|8)_{11} = (1|4|3|10|10|4|4)_{11}.$$

(4) The extreme digits of $n_3$ are different. We thus add its mirror $(4|4|10|10|3|4|1)_{11}$:

$$n_4 := (1|4|3|10|10|4|4)_{11} + (4|4|10|10|3|4|1)_{11} = (5|9|3|10|2|8|5)_{11}.$$

(5) The number $n_4$ has the same first and last digits. It is therefore simplified by removing these two extreme digits:

$$n_5 := (9|3|10|2|8)_{11}.$$

(6) The extreme digits of $n_5$ are different. We thus add its mirror $(8|2|10|3|9)_{11}$:

$$n_6 := (9|3|10|2|8)_{11} + (8|2|10|3|9)_{11} = (1|6|6|9|6|6)_{11}.$$

(7) The extreme digits of $n_6$ are different. We thus add its mirror $(6|6|9|6|6|1)_{11}$:

$$n_7 := (1|6|6|9|6|6)_{11} + (6|6|9|6|6|1)_{11} = (8|2|5|5|1|7)_{11} = n_2.$$

We then notice that:

$$n_7 = n_2$$

Thus, the algorithm will not end on • for $n_0$; it concludes on a cycle of length 5.

We explore computationally in this text, bases that contain cycles like base 11 and bases that appear not to contain any, such as base 10. Several conjectures are presented. In the following sections, we will explore these conjectures. We will define the key notions in any given base and present *bacouettes*<fnref n="10" /> supported by computer simulations. Python programs are provided in the text to test this algorithm and validate the bacouettes.

We conclude the text with extensions and variations.

<footnote n="10">¹⁰Bacouetter or "faire la bacouette" means to chat. A **bacouette** in this text is an open problem, a question, an exercise, or a statement for discussion. In the final section, we will gather all our bacouettes. If this text inspires some to engage in real mathematical discussions rather than just behind a screen, that will bring me *bonhour*.</footnote>
Context

The title highlights five particular aspects.

⊗ I imagined this conjecture in 2018, when I was teaching in Paris (Lutèce) while residing in Caen (Cathim). I spent many hours reflecting on this algorithm during my train commutes between Caen and Paris, with only a vague sense of the challenges involved. The idea of writing this text in 2024 was inspired by my students, who showed interest in this mathematical game. It was at the start of the 2024-2025 school year that I created this narrative.

⊗ Like the famous "Paris-Brest," a dessert created by pastry chefs for the mythical Paris-Brest-Paris cycling race, the title "Lutèce-Cathim" or "Paris-Caen" for the contemporary naturally imposed itself. Being Norman<sup>11</sup>, this name is a nod to this round crust<sup>12</sup>.

⊗ The title echoes the legendary Collatz conjecture, named after the American university where Collatz popularized his famous algorithm. I borrow many ideas and notations.

⊗ The word "Program" refers to the famous Langlands Program of arithmetic geometers. Here, at a much lower level, there is an opportunity to play with well-chosen arithmetic bases. Everyone has their own program: their base (arithmetic) and their Normandy (geometry).

⊗ Finally, the number 23 refers to the famous 23 problems of Hilbert. In reality, I did not have 23 bacouettes, but being so close to that number, I created new ones.

Enough! Let us move on to Norman mathematics and the bacouettes of our Paris-Caen Program.

A tantôt for more Norman legends

Two other Norman legends will soon be published:

★ **Pommeau Program: artifaille for calendonization.** Like a good Norman meal, it begins with good pommeau and camembert. This text will study the sharing of a *calendo*<sup>13</sup>. We will thus define the notion of multi-regular polygons, graphs, and the cyclotomic polynomial associated with such a polygon. We will examine the case of graphs with loops, leading to beautiful fractals. We will learn to draw beautiful calendos depending on our appetite. Many bacouettes are again provided, forming the Pommeau Program.

★ **Bourdelot Program: artifaille for normandization.** Like a Bourdelot enjoyed with good local apples<sup>14</sup>, it is unthinkable to use butter<sup>15</sup> for a Bourdelot on a non-Norman apple. Searching for the right apple is an adventure that will take us back in time through Norman legends. Apples in Normandy are like prime numbers in mathematics: they have always been there<sup>16</sup> and have structured the Norman language due to their importance. In this text, we study good apples (mathematical bases) that yield beautiful Norman languages, the normandization (beautiful mathematics). Many bacouettes are provided, forming the Bourdelot Program.

<footnote n="11"><sup>11</sup>In truth, I am Breton on my mother's side and Norman on my father's side.</footnote>

<footnote n="12"><sup>12</sup>Round crust with a hole.</footnote>

<footnote n="13"><sup>13</sup>Camembert.</footnote>

<footnote n="14"><sup>14</sup>Friends of the *Verger de sauvegarde de Maltot* (south of Caen) preserve more than 800 varieties of apples and pears carefully selected by our ancestors for their flavors.</footnote>

<footnote n="15"><sup>15</sup>Norman butter is by definition always salted; otherwise, it is churned cream.</footnote>

<footnote n="16"><sup>16</sup>Apples originate from Central and Western Asia (Kazakhstan).</footnote>
"O you all!
Brave Normans of the Channel Islands,
Quivering to speak as your fathers spoke,
Even as you have your sons taught English,
You who take away the old French names of your streets...
Know that your "patois" is venerable;
Your "patois" is sacred;
For it is from it, like a flower from its root,
That the French language emerged, which tomorrow will be the language of Europe.
Your "patois," your fathers from Normandy died to spread it
In England, Sicily, Judea, London, Naples, and even at the tomb of Christ.
For they knew that losing one's language is losing one's nationality,
And that by bringing their idiom, they brought their homeland with them.
Yes, your "patois" is venerable,
For the first poet who spoke it was the first of French poets."

François-Victor Hugo (1828 - 1873)<fnref n="17" />, *La Normandie inconnue*

<footnote n="17">¹⁷French writer and son of Victor Hugo.</footnote>
## 1 Decomposition in Arithmetic Base

In this section, we define the standard procedure for base decomposition.

### 1.1 Sets $\mathbb{N}_a^b$

We denote $\mathbb{N}$, the set of natural numbers id est:

$$ \mathbb{N} := \{0, 1, 2, 3, \dots\}. \qquad (1) $$

For all $a, b \in \mathbb{N}$ such that $a \le b$, we denote:

$$ \mathbb{N}_a := \mathbb{N} \cap [a, +\infty[ = \{k \in \mathbb{N}, k \ge a\} = \{a, a+1, \dots\} \qquad (2) $$

$$ [[a, b]] := \mathbb{N} \cap [a, b] = \{k \in \mathbb{N}, a \le k \le b\} = \{a, a+1, \dots, b-1, b\} \qquad (3) $$

#### Example 1.1.1

For all $a \in \mathbb{N}$, we have:

$$ \mathbb{N} = \mathbb{N}_0, \quad \mathbb{N}_* := \mathbb{N} - \{0\} = \mathbb{N}_1, \quad [[a, a]] = \{a\}. $$

and:

$$ [[0, a]] = \mathbb{N} \cap [0, a] = \{0, 1, \dots, a-1, a\}. $$

We extend notation (3) by defining for all $a, b \in \mathbb{N}$ such that $a > b$:

$$ [[a, b]] := \emptyset. \qquad (4) $$

#### Definition 1.1.2

A base is an element of $\mathbb{N}_2$, id est, an integer greater than or equal to 2.

Unless otherwise specified, we fix a base $b \in \mathbb{N}_2$ throughout this text.

We define the following notations.

★ For all $d \in \mathbb{N}$, we denote:

$$ \mathcal{B}_{d,b} := [[0, b^{d+1} - 1]] = \{0, 1, \dots, b^{d+1} - 1\} \qquad (5) $$

★ We define recursively for all $d \in \mathbb{N}$:

$$ \mathcal{B}_{(0),b} := \mathcal{B}_{0,b} \qquad (6) $$

$$ \mathcal{B}_{(d+1),b} := \mathcal{B}_{d+1,b} - \mathcal{B}_{d,b} = \{b^{d+1}, b^{d+1} + 1, \dots, b^{d+2} - 1\} \qquad (7) $$

We simply denote:

$$ \mathcal{B}_b := \mathcal{B}_{0,b} = \mathcal{B}_{(0),b} = [[0, b-1]] = \{0, 1, \dots, b-1\}. \qquad (8) $$

Then, for all $d \in \mathbb{N}$, we have:

$$ \mathcal{B}_b = \mathcal{B}_{0,b} \subset \dots \subset \mathcal{B}_{d,b} \subset \mathcal{B}_{d+1,b} \subset \dots \subset \mathbb{N} \qquad (9) $$

$$ \mathbb{N} = \bigcup_{d \in \mathbb{N}} \mathcal{B}_{d,b} = \bigcup_{d \in \mathbb{N}} \mathcal{B}_{(d),b} \qquad (10) $$

$$ \mathrm{card} (\mathcal{B}_{d,b}) = b^{d+1} \qquad (11) $$

$$ \mathrm{card} (\mathcal{B}_{(d),b}) = b^{d+1} - b^d \qquad (12) $$

★ For any subset $\mathcal{A}_b$ of $\mathbb{N}$ depending on $b$ and any $d \in \mathbb{N}$, we denote:

$$ \mathcal{A}_{d,b} := \mathcal{A}_b \cap \mathcal{B}_{d,b} \qquad (13) $$

$$ \mathcal{A}_{(d),b} := \mathcal{A}_b \cap \mathcal{B}_{(d),b} \qquad (14) $$
## 1.2 The Rollon

### Definition 1.2.1: The Rollon

We define the number $\bullet$ as being the number without digits, called the **Rollon**.

It can be seen as the number obtained by removing all digits from a given number. We assume that this number $\bullet$ is unique in the sense that, regardless of the number from which all digits are removed, we always obtain the number $\bullet$.

★ For any subset $I$ of $\mathbb{N}$, we denote:

$$I_{\bullet} := I \cup \{\bullet\} \qquad (15)$$

In particular, for all $a, b \in \mathbb{N}$ such that $a \le b$, we have:

$$\mathbb{N}_{\bullet} = \mathbb{N} \cup \{\bullet\} \qquad (16)$$

$$(\mathbb{N}_a)_{\bullet} = \{\bullet, a, a+1, \dots, \} \qquad (17)$$

$$[[0, b]]_{\bullet} = \{\bullet, 0, 1, \dots, b-1, b\} \qquad (18)$$

$$[[a, b]]_{\bullet} = \{\bullet, a, a+1, \dots, b-1, b\} \qquad (19)$$

We will abusively continue to call the elements of $\mathbb{N}_{\bullet}$ **integers** or **numbers**.

★ The usual order relation on $\mathbb{N}$ is extended to $\mathbb{N}_{\bullet}$ by defining for every integer $n \in \mathbb{N}$:

$$\bullet < n \qquad (20)$$

Thus, for any subset $I$ of $\mathbb{N}$, we have:

$$\min I_{\bullet} = \bullet. \qquad (21)$$

★ Addition between an integer and $\bullet$ is not defined; we simply denote:

$$\bullet + \bullet = \bullet \qquad (22)$$

This relation will be useful for the palindromic operation.

## 1.3 Basic Tools

In this subsection, we define four tools useful for the remainder of the text.

### Power and Composition.

#### Definition 1.3.1

Let $I$ be a subset of $\mathbb{N}_{\bullet}$ and $f$ a function from $I$ to $I$. For any $k \in \mathbb{N}$, we denote $f^k$ the $k$-fold composition of $f$, defined by the following recurrence:

$$f^0 := \mathrm{Id}_I \qquad (23)$$

$$\forall k \in \mathbb{N}, \quad f^{k+1} := f^k \circ f. \qquad (24)$$

For all $k \in \mathbb{N}$, we have:

$$\mathrm{Id}_I^k = \mathrm{Id}_I. \qquad (25)$$

### Stable Subset.
**Definition 1.3.2**

Let $I$ be a subset of $\mathbb{N}_{\bullet}$ and $f$ a function from $\mathbb{N}_{\bullet}$ to $\mathbb{N}_{\bullet}$. The subset $I$ is said to be **stable** under $f$ if $f(I) \subset I$, id est, for all $x \in I$, we have $f(x) \in I$.

In this case, for all $k \in \mathbb{N}$, we have $f^k(I) \subset I$, id est, $I$ is stable under $f^k$.

★ **Equivalence Relations.** Various types of equivalence relations associated with a function $f$ can be defined.

**Definition 1.3.3**

Let $I$ and $J$ be subsets of $\mathbb{N}_{\bullet}$, $f$ a function from $I$ to $J$, and $g$ a function from $I$ to $I$.

(i) We denote $\triangleleft[f, I, J]$ the equivalence relation on $I$ defined for all $m, n \in I$ by $\{m, n\} \triangleleft [f, I, J]$ if $f(m) = f(n)$.

The equivalence class of an element $n \in I$ is denoted $[f, I, J] \triangleright \{n\}$:

$$[f, I, J] \triangleright \{n\} := \{m \in I, \{m, n\} \triangleleft [f, I, J]\} = \{m \in I, f(m) = f(n)\}. \quad (26)$$

(ii) Let $k \in \mathbb{N}_1$. We denote $\triangleleft[k, g, I]$ the equivalence relation on $I$ defined for all $m, n \in I$ by $\{m, n\} \triangleleft [k, g, I]$ if $g_k^k(m) = g_k^l(n)$.

The equivalence class of an element $n \in I$ is denoted $[k, g, I] \triangleright \{n\}$:

$$[f, I] \triangleright \{n\} := \{m \in I, \{m, n\} \triangleleft [k, g, I]\} = \{m \in I, g_k^k(m) = g_k^l(n)\}. \quad (27)$$

(iii) We denote $\triangleleft[\infty, g, I]$ the equivalence relation on $I$ defined for all $m, n \in I$ by $\{m, n\} \triangleleft [\infty, g, I]$ if there exist $k, l \in \mathbb{N}$ such that $g_k^k(m) = g_l^l(n)$.

The equivalence class of an element $n \in I$ is denoted $[\infty, g, I] \triangleright \{n\}$:

$$[\infty, g, I] \triangleright \{n\} := \{m \in I, \{m, n\} \triangleleft [\infty, g, I]\} = \{m \in I, \exists k, l \in \mathbb{N}, g_k^k(m) = g_l^l(n)\}. \quad (28)$$

Thus, for all $m, n \in I$ and $k \in \mathbb{N}_1$:

$$\{m, n\} \triangleleft [g, I] \implies \{m, n\} \triangleleft [k, g, I] \implies \{m, n\} \triangleleft [\infty, g, I] \quad (29)$$

★ **Densities.** We conclude this subsection with the density of a subset of $\mathbb{N}_{\bullet}$.

**Definition 1.3.4**

Let $I$ and $J$ be subsets of $\mathbb{N}_{\bullet}$.

(i) The **sequence of J-densities of I** is the sequence defined for all $n \in \mathbb{N}$ by:

$$\delta_{(n,J)}(I) := \frac{\text{card}(I \cap J \cap [0, n])}{\text{card}(J \cap [0, n])} \in [0, 1].$$

(ii) Suppose the sequence $(\delta_{(n,J)}(I))_{n \in \mathbb{N}}$ converges. The **J-density of I** is the real number defined by:

$$\delta_J(I) := \lim_{n \to \infty} \delta_{(n,J)}(I).$$
A particular case is when $J := \mathbb{N}_k$ with $k \in \mathbb{N}_\bullet$, and $I$ is a subset of $\mathbb{N}_k$. For all $n \in \mathbb{N}_k$, we denote:

$$ \delta_{(n,k)}(I) := \delta_{(n,\mathbb{N}_k)}(I) := \frac{\mathrm{card}(I \cap \llbracket k, n \rrbracket)}{\mathrm{card}(\llbracket k, n \rrbracket)} \in [0,1]. \quad (30) $$

Suppose the sequence $(\delta_{(n,k)}(I))_{n \in \mathbb{N}_k}$ converges. We denote:

$$ \delta_k(I) := \delta_{\mathbb{N}_k}(I) := \lim_{n \to \infty} \delta_{(n,k)}(I). \quad (31) $$

For instance, for any $k \in \mathbb{N}$ and $n \in \mathbb{N}_k$:

$$ \delta_{(n,k)}(\llbracket k, n \rrbracket) = \frac{n-k+1}{n-k+1} = 1 \quad (32) $$

and hence $\mathbb{N}_k$ has density 1 in $\mathbb{N}_k$.

## 1.4 Theorems

We have the usual theorem on base $b$ decomposition.

### Theorem-Definition 1.4.1

Let $n \in \mathbb{N}_1$. There exists a unique integer $d \in \mathbb{N}$ and a unique sequence $(a_0, \dots, a_d) \in (\mathcal{B}_b)^{d+1}$ such that $a_d \neq 0$ and:

$$ n = \sum_{k=0}^{d} a_k b^k. $$

(i) This sequence $(a_0, \dots, a_d)$ is called the **base $b$ decomposition of $n$**, and we write:

$$ (a_d | \dots | a_0)_{d,b} := n. $$

The integers $a_0, \dots, a_d$ are called the **digits of $n$ in base $b$**.

(ii) The integer $d$ is called the **degree of $n$ in base $b$**. We denote it as $d_b(n)$.

*Proof. Existence.* Let us denote:

$$ \Gamma := \{ n \in \mathbb{N}_1, \exists d \in \mathbb{N}, \exists (a_0, \dots, a_d) \in (\mathcal{B}_b)^{d+1}, a_d \neq 0 \wedge n = \sum_{k=0}^{d} a_k b^k \} \subset \mathbb{N}_1. $$

We reason by contradiction and assume the opposite, that is:

$$ (*) : \Gamma \not\subseteq \mathbb{N}_1 $$

that is, $\mathbb{N}_1 - \Gamma$ is a non-empty subset of $\mathbb{N}_1$. Let us define:

$$ n_0 := \min(\mathbb{N}_1 - \Gamma). $$

Since $1 \in \Gamma$ (because $1 = 1 \times b^0$ and $0 < 1 < 2 \le b$), we have $n_0 \ge 2$ that is, $n_0 - 1 \ge 1$ and $n_0 - 1 \in \Gamma$. Therefore, there exists $d \in \mathbb{N}$ and $(a_0, \dots, a_d) \in (\mathcal{B}_b)^{d+1}$ such that $a_d \neq 0$ and:

$$ n_0 - 1 = \sum_{k=0}^{d} a_k b^k. $$

Let us show that necessarily $a_k = b - 1$ for all $k \in \llbracket 0, d \rrbracket$. We reason by contradiction and assume the opposite, that is:

(**) : there exists a smallest $k_0 \in \llbracket 0, d \rrbracket$ such that $a_{k_0} \le b - 2$ and for all $k \in \llbracket 0, k_0 - 1 \rrbracket$, we have $a_k = b - 1$ (if $k_0 := 0$, $\llbracket 0, k_0 - 1 \rrbracket = \emptyset$).
We then have by telescoping:

$$
\begin{align*}
n_0 &= (n_0 - 1) + 1 \\
&= \sum_{k=0}^{d} a_k b^k + b^0 \\
&= \sum_{k=0}^{k_0-1} (b-1)b^k + \sum_{k=k_0}^{d} a_k b^k + b^0 \\
&= \sum_{k=0}^{k_0-1} b^{k+1} - \sum_{k=0}^{k_0-1} b^k + \sum_{k=k_0}^{d} a_k b^k + b^0 \\
&= b^{k_0} + a_{k_0} b^{k_0} + \sum_{k=k_0+1}^{d} a_k b^k \\
&= (a_{k_0} + 1)b^{k_0} + \sum_{k=k_0+1}^{d} a_k b^k
\end{align*}
$$

Since $a_{k_0} + 1 \le b - 2 + 1 = b - 1$, we thus have a base $b$ decomposition of $n_0$, that is, $n_0 \in \Gamma$ (which contradicts the assumption $(*)$). Therefore, for all $k \in [0, d]$, $a_k = b-1$. Thus, we have by telescoping:

$$
\begin{align*}
n_0 &= (n_0 - 1) + 1 \\
&= \sum_{k=0}^{d} a_k b^k + 1 \\
&= \sum_{k=0}^{d} (b-1)b^k + 1 \\
&= b^{d+1} - 1 + 1 \\
&= b^{d+1}
\end{align*}
$$

that is, $n_0 \in \Gamma$. This contradicts $(*)$. Thus, $\Gamma = \mathbb{N}_1$.

**Uniqueness.** Let $n \in \mathbb{N}_1$, $d, d' \in \mathbb{N}$, $(a_0, \dots, a_d) \in (\mathcal{B}_b)^{d+1}$ and $(a'_0, \dots, a'_{d'}) \in (\mathcal{B}_b)^{d'+1}$ such that $a_d, a'_{d'} \neq 0$ and:

$$n = \sum_{k=0}^{d} a_k b^k = \sum_{k=0}^{d'} a'_k b^k.$$

By symmetry of the two decompositions, we can assume that $d \le d'$. We reason by contradiction and assume that there exists $k_0 \in [0, d]$ such that:

$$(* * *) : a_{k_0} \neq a'_{k_0} \wedge [\forall k \in [0, k_0 - 1], a_k = a'_k]$$

Then we have:

$$
\begin{align*}
0 &= \sum_{k=0}^{d} a_k b^k - \sum_{k=0}^{d'} a'_k b^k \\
&= \sum_{k=k_0}^{d} (a_k - a'_k)b^k - \sum_{k=d+1}^{d'} a'_k b^k
\end{align*}
$$

Thus, we have:

$$
\begin{align*}
|a_{k_0} - a'_{k_0}| b^{k_0} &= \left| \sum_{k=k_0+1}^{d} (a_k - a'_k)b^k - \sum_{k=d+1}^{d'} a'_k b^k \right| \\
&= b^{k_0+1} \left| \sum_{k=k_0+1}^{d} (a_k - a'_k)b^{k-k_0-1} - \sum_{k=d+1}^{d'} a'_k b^{k-k_0-1} \right|
\end{align*}
$$
that is, simplifying by $b^{k_0}$:

$$|a_{k_0} - a'_{k_0}| = b \left| \sum_{k=k_0+1}^{d} (a_k - a'_k) b^{k-k_0-1} - \sum_{k=d+1}^{d'} a'_k b^{k-k_0-1} \right| = 0[b].$$

Since $a_{k_0}, a'_{k_0} \in \mathcal{B}_b$, by the pigeonhole principle, we have $|a_{k_0} - a'_{k_0}| \in \mathcal{B}_b$. Thus, since $|a_{k_0} - a'_{k_0}| = 0[b]$, we conclude that $a_{k_0} = a'_{k_0}$. This contradicts $(***)$. Therefore, for all $k \in [0, d]$, $a_k = a'_k$.

It remains to show that $d = d'$ that is, $d \ge d'$ (since $d \le d'$). We reason by contradiction and assume that $d \le d' - 1$. Since $a'_{d'} \ne 0$, we have:

$$0 = n - n = \sum_{k=0}^{d} a_k b^k - \sum_{k=0}^{d'} a'_k b^k = \sum_{k=d+1}^{d'} a'_k b^k \ge a'_{d'} b^{d'} > 0$$

which is a contradiction. Therefore, $d = d'$. Hence, the uniqueness is proven. □

### Example 1.4.2

(1) We saw

in the introduction that:

$$149\,823 = \mathbf{1} \times 10^5 + \mathbf{4} \times 10^4 + \mathbf{9} \times 10^3 + \mathbf{8} \times 10^2 + \mathbf{2} \times 10^1 + \mathbf{3} \times 10^0 = (1 \mid 4 \mid 9 \mid 8 \mid 2 \mid 3)_{5,10}$$

$$143\,712 = \mathbf{9} \times 11^4 + \mathbf{8} \times 11^3 + \mathbf{10} \times 11^2 + \mathbf{7} \times 11^1 + \mathbf{8} \times 11^0 = (9 \mid 8 \mid 10 \mid 7 \mid 8)_{4,11}$$

(2) One can take very large bases. For example, with $b := 2024$, we have:

$$123\,456\,789 = \mathbf{30} \times 2024^2 + \mathbf{276} \times 2024^1 + \mathbf{885} \times 2024^0 = (30 \mid 276 \mid 885)_{2,2024}$$

Thus, for all $d \in \mathbb{N}$:

$$\mathcal{B}_{d,b} = \{n \in \mathbb{N}, d_b(n) \le d\} \qquad (33)$$

$$\mathcal{B}_{(d),b} = \{n \in \mathbb{N}, d_b(n) = d\}. \qquad (34)$$

We extend this notation.

<figure><img src="image_3.png" /></figure>

★ In the same way that 124 and 00124 represent the same number in base 10, we can add as many 0s as we want to the left of the decomposition. For all $k \in \mathbb{N}$, we define:

$$(0 \mid \cdots \mid 0 \mid a_d \mid \cdots \mid a_0)_{d+k,b} := (a_d \mid \cdots \mid a_0)_{d,b} \qquad (35)$$

in particular:

$$(0 \mid a_d \mid \cdots \mid a_0)_{d+1,b} = (a_d \mid \cdots \mid a_0)_{d,b}$$

$$(0 \mid 0 \mid a_d \mid \cdots \mid a_0)_{d+2,b} = (a_d \mid \cdots \mid a_0)_{d,b}$$

<figure><img src="image_4.png" /></figure>

★ We define for 0:

$$(0)_b := 0 \qquad (36)$$

$$d_b(0) := 0 \qquad (37)$$

<figure><img src="image_5.png" /></figure>

★ We define for the number $\bullet$:

$$(\bullet)_b := \bullet \qquad (38)$$

$$d_b(\bullet) := -\infty. \qquad (39)$$
★ For $n := (a_d | \cdots | a_0)_{d,b}$, we denote for all $k \in \mathbb{N}$, $[n]_{k,b}$ the digit in position $k$ of $n$ in its base $b$ decomposition, i.e.:

$$ [n]_{k,b} := \begin{cases} a_k & \text{if } k \in \llbracket 0, d \rrbracket \\ 0 & \text{otherwise} \end{cases} \qquad (40) $$

Thus, for all $k \in \mathbb{N}$, we have:

$$ n = ([n]_{d,b} | \cdots | [n]_{0,b})_{d,b} = ([n]_{d+k,b} | \cdots | [n]_{0,b})_{d+k,d}. \qquad (41) $$

For all $m, n \in \mathbb{N}$, we have:

$$ m = n \iff \forall k \in \mathbb{N}, [m]_{k,b} = [n]_{k,b}. \qquad (42) $$

In particular, for all $n \in \mathbb{N}$, we have:

$$ n = 0 \iff \forall k \in \mathbb{N}, [n]_{k,b} = 0. \qquad (43) $$

★ Let $n \in \mathbb{N}$. Since the sequence $([n]_{k,b})_{k \in \mathbb{N}}$ becomes zero from a certain rank, we write:

$$ \sum_{k \ge 0} [n]_{k,b} b^k = n. \qquad (44) $$

(The advantage is that the degree of $n$ does not need to be specified if it is not useful for understanding).

★ We finish with a final notation. For all $m, n \in \mathcal{B}_b$ and all $k, l \in \mathbb{N}_1$, we write:

$$ ((n)_k)_{k-1,b} = (n | \cdots | n)_{k-1,b} \\ ((m)_k | a_d | \cdots | a_0 | (n)_l)_{d+k+l,b} = (m | \cdots | m | a_d | \cdots | a_0 | n | \cdots | n)_{d+k+l,b} $$

We could have proven the previous theorem of base $b$ decomposition by performing successive Euclidean divisions. As we have demonstrated without it, let us deduce it from the previous theorem.

### Corollary 1.4.3: Euclidean Division Theorem

Let $n \in \mathbb{N}$. Then there exist two natural integers $q_b(n), r_b(n)$ called the **quotient and remainder of the Euclidean division of $n$ by $b$** such that:

$$ n = q_b(n)b + r_b(n) \\ r_b(n) \in \mathcal{B}_b. $$

*Proof. Existence.* There are two cases.

(1) Case $n = 0$. We have $0 = 0b + 0$. Hence, existence is shown by setting:

$$ q_b(0) := 0, \quad r_b(0) := 0. $$

(2) Case $n \neq 0$ id est $n \in \mathbb{N}_1$. By the previous theorem, there exists a unique integer $d \in \mathbb{N}$ and a unique sequence $(a_0, \dots, a_d) \in (\mathcal{B}_b)^{d+1}$ such that $a_d \neq 0$ and:

$$ n = \sum_{k=0}^{d} a_k b^k = \left( \sum_{k=1}^{d} a_k b^{k-1} \right) b + a_0. $$

Thus, existence is shown by setting:

$$ q_b(n) := \sum_{k=1}^{d} a_k b^{k-1}, \quad r_b(n) := a_0. $$
**Uniqueness.** Let $n, q, q' \in \mathbb{N}$ and $r, r' \in \mathcal{B}_b$ such that:

$$n = qb + r = q'b + r'$$

Then we have:

$$(*) : b(q - q') = r' - r$$

and thus $b$ divides $r' - r$. By the pigeonhole principle, $|r' - r| \in \mathcal{B}_b$, we have $r' - r = 0 \text{ id est } r' = r$. Thus, by $(*)$, we have $b(q - q') = 0 \text{ id est } q = q'$ because $b$ is nonzero. Hence, uniqueness is shown. $\square$

Therefore, for all $m, n \in \mathbb{N}$, we have:

$$m = n \iff (r_b(m), q_b(m)) = (r_b(n), q_b(n)). \qquad (45)$$

### Example 1.4.4

(1) Since $31 = 7 \times 4 + 3$, we have:

$$(r_4(31), q_4(31)) = (3, 7).$$

(2) For all $n \in \mathbb{N}$, we have:

$$n = 0 \iff (r_b(n), q_b(n)) = (0, 0).$$

We then deduce a recursive expression for the digits of an integer in base $b$.

### Corollary 1.4.5

Let $n \in \mathbb{N}$. Then for all $k \in \mathbb{N}$:

$$[n]_{k,b} = r_b(q_b^k(n)).$$

*Proof.* We prove the result by strong induction on $n \in \mathbb{N}$:

$$(\mathcal{P}_n) : \forall k \in \mathbb{N}, \ [n]_{k,b} = r_b(q_b^k(n)).$$

(i) We prove by induction on $k \in \mathbb{N}$:

$$(\mathcal{Q}_n) : q_b^k(0) = 0.$$

(I) We have $q_b^0(0) = \text{Id}_{\mathbb{N}}(0) = 0$. Thus $(\mathcal{Q}_0)$ is true.

(II) Let $k \in \mathbb{N}$. Assume $(\mathcal{Q}_k)$ is true. We have

$$q_b^{k+1}(0) = q_b(q_b^k(0)) = q_b(0) = 0.$$

Thus $(\mathcal{Q}_{k+1})$ is true.

(III) By induction, $(\mathcal{Q}_k)$ is true for all $k \in \mathbb{N}$.

Thus we have:

$$\forall k \in \mathbb{N}, \ [0]_{k,b} = 0 = r_b(0) = r_b(q_b^k(0))$$

i.e. $(\mathcal{P}_0)$ is true.

(ii) Let $n \in \mathbb{N}$. Assume that $(\mathcal{P}_m)$ is true for all $m \in [0, n]$. Let $n' := q_b(n+1)$. Since $n+1 > 0$, we have $n' \le n$. Since $(\mathcal{P}_{n'})$ is true, we have for all $k \in \mathbb{N}$:

$$r_b(q_b^{k+1}(n+1)) = r_b(q_b^k(q_b(n+1))) = r_b(q_b^k(n')) = [n']_{k,b}.$$

Thus we have:

$$\sum_{k \ge 0} [n+1]_{k,b} b^k = n+1$$
$$
\begin{align*}
&= n'b + r_b(n+1) \\
&= \left( \sum_{k \ge 0} [n']_{k,b} b^k \right) b + r_b(n+1) \\
&= \left( \sum_{k \ge 0} r_b(q_b^{k+1}(n+1)) b^k \right) b + r_b(q_b^0(n+1)) \\
&= \sum_{k \ge 0} r_b(q_b^{k+1}(n+1)) b^{k+1} + r_b(q_b^0(n+1)) \\
&= \sum_{k \ge 0} r_b(q_b^k(n+1)) b^k
\end{align*}
$$

By uniqueness of the base $b$ decomposition of $n+1$, we have for all $k \in \mathbb{N}$:

$$
[n+1]_{k,b} = r_b(q_b^k(n+1)).
$$

Thus ($\mathscr{P}_{n+1}$) is true.

(iii) By strong induction, ($\mathscr{P}_n$) is true for all $n \in \mathbb{N}$.

☐

We finish this section with two objects associated with a base $b$ decomposition.

**Definition 1.4.6**

Let $n \in \mathbb{N}_{\bullet}$.

(i) The polynomial associated with $n$ in base $b$ is:

$$
P_{n,b} := \begin{cases} \displaystyle \sum_{k=0}^{d_b(n)} [n]_k X^k & \text{if } n \in \mathbb{N} \\ \bullet & \text{if } n := \bullet. \end{cases}
$$

(ii) The valuation of $n$ in base $b$ is:

$$
\nu_b(n) := \begin{cases} \max\{k \in \mathbb{N}, b^k|n\} & \text{if } n \in \mathbb{N}_1 \\ +\infty & \text{if } n \in \{\bullet, 0\}. \end{cases}
$$

**Example 1.4.7**

Let $n := 2528 = (2 \mid 1 \mid 3 \mid 2 \mid 0 \mid 0)_{5,4}$. Since $d_4(n) = 5$, we have:

$$
P_{n,4} = 2X^5 + X^4 + 3X^3 + 2X^2
$$

and:

$$
\nu_4(n) = \max \{k \in \mathbb{N}, 4^k | n\} = \max \{0, 1, 2\} = 2.
$$

Let:

$$
m := \frac{n}{4^{\nu_4(n)}} = \frac{n}{4^2} = 158.
$$

We have:

$$
m = (2 \mid 1 \mid 3 \mid 2)_{3,4} \quad , \quad P_{m,5} = 2X^3 + X^2 + 3X + 2.
$$

Then:

$$
P_{n,4}(4) = n, \quad P_{n,4} = X^2 P_{m,4} = X^{\nu_4(n)} P_{m,4}.
$$

We generalize the results seen in the example.
### Lemma 1.4.8

Let $n \in \mathbb{N}$.

(i) We have $P_{n,b}(b) = n$.

(ii) Let:

$$m := \frac{n}{b^{\nu_b(n)}}.$$

We have:

$$P_{n,b} = X^{\nu_b(n)} P_{m,b}.$$

*Proof.* (i) We have:

$$P_{n,b}(b) = \sum_{k=0}^{d_b(n)} [n]_k b^k = n.$$

(ii) Since:

$$m = \frac{n}{b^{\nu_b(n)}} = \frac{\sum_{k=0}^{d_b(n)} [n]_k b^k}{b^{\nu_b(n)}} = \frac{\sum_{k=\nu_b(n)}^{d_b(n)} [n]_k b^k}{b^{\nu_b(n)}} = \sum_{k=\nu_b(n)}^{d_b(n)} [n]_k b^{k-\nu_b(n)}$$

we have:

$$P_{n,b} = \sum_{k=0}^{d_b(n)} [n]_k X^k \\ = \sum_{k=\nu_b(n)}^{d_b(n)} [n]_k X^k \\ = X^{\nu_b(n)} \sum_{k=\nu_b(n)}^{d_b(n)} [n]_k X^{k-\nu_b(n)} \\ = X^{\nu_b(n)} P_{m,b}$$

□

## 1.5 Addition in base b

We now look at the addition of two integers decomposed in base $b$. The addition is similar to the one in base 10: we need to add the digits in the same position with a carry, the only difference is that the result should be considered modulo $b$ and not modulo 10.

Let $m, n \in \mathbb{N}$. To compute the base $b$ decomposition of $m+n$ from those of $m$ and $n$, we can use the auxiliary sequence $(\sigma(m,n)_k)_{k \in \mathbb{N}}$ defined for all $k \in \mathbb{N}_1$ by the following recurrence relation:

$$\sigma(m,n)_{0,b} := [m]_{0,b} + [n]_{0,b} \qquad (46)$$

$$\sigma(m,n)_{k,b} := [m]_{k,b} + [n]_{k,b} + q_b(\sigma(m,n)_{k-1,b}) \qquad (47)$$

Thus, for all $k \in \mathbb{N}$, we have:

$$[m+n]_{k,b} = r_b(\sigma(m,n)_{k,b}). \qquad (48)$$
## Example 1.5.1

Let $b := 4$ and:

$$m := 2998 = (2 \mid 3 \mid 2 \mid 3 \mid 2 \mid 2)_{5,4}, \quad n := 2603 = (2 \mid 2 \mid 0 \mid 2 \mid 2 \mid 3)_{5,4}.$$

We have the following table:

<table><thead><tr><th>k</th><th>6</th><th>5</th><th>4</th><th>3</th><th>2</th><th>1</th><th>0</th></tr></thead><tbody><tr><td>[m]<sub>k,4</sub></td><td>0</td><td>2</td><td>3</td><td>2</td><td>3</td><td>1</td><td>2</td></tr><tr><td>[n]<sub>k,4</sub></td><td>0</td><td>2</td><td>2</td><td>0</td><td>2</td><td>2</td><td>3</td></tr><tr><td>[m]<sub>k,4</sub> + [n]<sub>k,4</sub></td><td>0</td><td>4</td><td>5</td><td>2</td><td>5</td><td>3</td><td>5</td></tr><tr><td>σ(m, n)<sub>k,4</sub></td><td>1</td><td>5</td><td>5</td><td>3</td><td>6</td><td>4</td><td>5</td></tr><tr><td>q<sub>4</sub>(σ(m, n)<sub>k,4</sub>)</td><td>0</td><td>1</td><td>1</td><td>0</td><td>1</td><td>1</td><td>1</td></tr><tr><td>[m + n]<sub>k,4</sub> = r<sub>b</sub>(σ(m, n)<sub>k,4</sub>)</td><td>1</td><td>1</td><td>1</td><td>3</td><td>2</td><td>0</td><td>1</td></tr></tbody></table>

id est

$$(2 \mid 3 \mid 2 \mid 3 \mid 2 \mid 2)_{5,4} + (2 \mid 2 \mid 0 \mid 2 \mid 2 \mid 3)_{5,4} = (1 \mid 1 \mid 1 \mid 3 \mid 2 \mid 0 \mid 1)_{6,4}$$

or in base 10:

$$2998 + 2603 = 5601.$$

We then deduce the following definition.

## Definition 1.5.2: Carry Sequence

The sequence $(q_b(\sigma(m, n)_{k,b}))_{k \in \mathbb{N}}$ is called the **carry sequence in the addition of $m$ and $n$ in base $b$**.

We then have, by setting for all $k \in \mathbb{N}$, $u_k := [m]_{k,b} + [n]_{k,b}$:

$$\sigma(m, n)_{0,b} = u_0$$

$$\sigma(m, n)_{1,b} = u_1 + q_b(u_0)$$

$$\sigma(m, n)_{2,b} = u_2 + q_b(u_1 + q_b(u_2))$$

$$\sigma(m, n)_{3,b} = u_3 + q_b(u_2 + q_b(u_1 + q_b(u_2)))$$

...

and thus:

$$[m + n]_{0,b} = r_b(\sigma(m, n)_{0,b}) = r_b(u_0)$$

$$[m + n]_{1,b} = r_b(\sigma(m, n)_{1,b}) = r_b(u_1 + q_b(u_0))$$

$$[m + n]_{2,b} = r_b(\sigma(m, n)_{2,b}) = r_b(u_2 + q_b(u_1 + q_b(u_2)))$$

$$[m + n]_{3,b} = r_b(\sigma(m, n)_{3,b}) = r_b(u_3 + q_b(u_2 + q_b(u_1 + q_b(u_2))))$$

...

We see that there is no simple form for the digits of $m+n$. However, we have the following property.

## Lemma 1.5.3

For all $k \in \mathbb{N}$, $\sigma(m, n)_{k,b} \in [0, 2b - 1]$ and in particular, $q_b(\sigma(m, n)_{k,b}) \in \{0, 1\}$.

*Proof.* We prove the result by induction on $k \in \mathbb{N}$:

$$(\mathcal{P}_k) : \sigma(m, n)_{k,b} \in [0, 2b - 1].$$
(i) Since $[m]_{0,b}, [n]_{0,b} \in \mathcal{B}_b$, we have $\sigma(m,n)_{0,b} \in [0, 2b-2] \subset [0, 2b-1]$. Hence, $(\mathcal{P}_0)$ is true.

(ii) Let $k \in \mathbb{N}$. Suppose $(\mathcal{P}_k)$ is true. Since $[m]_{k,b}, [n]_{k,b} \in \mathcal{B}_b$ and $q_b(\sigma(m,n)_{k,b}) \in \{0,1\}$, we have:

$$ \sigma(m, n)_{k+1, b} = [m]_{k+1, b} + [n]_{k+1, b} + q_b(\sigma(m, n)_{k,b}) \le 2(b-1) + 1 = 2b-1. $$

Hence, $(\mathcal{P}_{k+1})$ is true.

(iii) By induction, $(\mathcal{P}_k)$ is true for all $k \in \mathbb{N}$.

<figure><img src="image_6.png" /></figure>

## 1.6 Artifaille in Python

We import two modules in Python:

```python
import math as ma
from typing import List, Tuple, Union, Dict
```

We denote a decomposition in base b in Python by:

$$ (B, b) := ([B[0], \dots, B[d]], b) := (B[d] | \dots | B[0])_{d,b} \quad (49) $$

and:

$$ [] := \bullet. \quad (50) $$

For example:

$$ ([4, 4, 3, 5, 5, 1], 6) = (1 | 5 | 5 | 3 | 4 | 4)_{5,6} = 15472 \quad (51) $$

$$ (\emptyset, 3) = (\bullet)_{3} = \bullet \quad (52) $$

We define two new types `IntB` and `Base`:

```python
Base = int
IntB = Tuple List[int], Base
```

♣ The program `int_to_base(n, b)` takes as input an integer `n` and a base `b`. It returns the decomposition `N = (B, b)` of type `IntB` of the integer `n` in base `b`.

### Python 1.6.1

```python
def int_to_base(n: int, b: Base) -> IntB:
    if n == []:
        return ([], b)
    elif n == 0:
        return ([0], b)
    else:
        B = []
        k = n
        while k > 0:
            B.append(k % b)
            k /= b
        return (B, b)
```
### Example 1.6.2

```python
>>> int_to_base([], 3)
([], 3)

>>> int_to_base(0, 4)
([0], 4)

>>> int_to_base(15472, 6)
([4, 4, 3, 5, 5, 1], 6)
```

The program `base_to_int(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns the integer `n` such that `int_to_base(n, b) = N`.

### Python 1.6.3

```python
def base_to_int(N: IntB) -> int:
    B, b = N
    if B == []:
        return []
    else:
        s = 0
        p = 1
        for c in B:
            s += c * p
            p *= b
        return s
```

### Example 1.6.4

```python
>>> base_to_int([[], 3])
[]
>>> base_to_int([0, 4])
0
>>> base_to_int([4, 4, 3, 5, 5, 1], 6)
15472
```

The program `deg_base(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns the degree of `N` of type `int` or `float` (`ma.inf` is of type `float`).

### Python 1.6.5

```python
def deg_base(N: IntB) -> int | float:
    B, _ = N
    if B == []:
        return -ma.inf
    elif len(B) == 1:
        return 0
    else:
        d = len(B) - 1
        while B[d] == 0:
```
```
d -= 1
return d
```

### Example 1.6.6

```python
>>> type(ma.inf)
<class 'float'>
>>> deg_base([], 5)
-inf
>>> deg_base([0, 5])
0
>>> deg_base([3, 4, 2, 0], 5)
2
>>> deg_base([3, 4, 2, 1], 5)
3
```

The program `val_base(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns the valuation of `N` of type `int` or `float`.

### Python 1.6.7

```python
def val_base(N: IntB) -> int | float:
    B, _ = N
    if B == [] or B == [0]:
        return ma.inf
    else:
        v = 0
        while B[v] == 0:
            v += 1
        return v
```

### Example 1.6.8

```python
>>> val_base([], 5)
inf
>>> val_base([0, 5])
inf
>>> val_base([0, 0, 3, 4, 2, 0], 5)
2
>>> val_base([3, 4, 2, 1], 5)
0
```

The program `plus_1(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns the integer `N + 1` of type `IntB` written in the same base `b`.
Python 1.6.9

```python
def plus_1(N: IntB) -> IntB:
    B, b = N
    c = 1
    for i in range(len(B)):
        B[i] += c
        if B[i] < b:
            return (B, b)
        B[i] = 0
    B.append(1)
    return (B, b)
```

Example 1.6.10

```python
>>> plus_1([1, 3], 4)
([2, 3], 4)
>>> plus_1([3, 3, 3], 4)
([0, 0, 0, 1], 4)
```

♣ The program `compare_base(N, M)` takes as input two integers `N` and `M` of type `IntB` written in the same base. It returns `True` if `N` ≤ `M` and `False` otherwise. One could write this program simply by stating:

`compare_base_bis(N, M) := base_to_int(N) <= base_to_int(M).`

To limit the number of comparison operations, we will prefer this program:

Python 1.6.11

```python
def compare_base(N: IntB, M: IntB) -> bool: # N <= M ?
    B, C = N[0], M[0]
    if len(B) != len(C):
        return len(B) < len(C)
    for i in range(len(B) - 1, -1, -1):
        if B[i] > C[i]:
            return False
        elif B[i] < C[i]:
            return True
    return True
```

Example 1.6.12

```python
>>> compare_base([0, 5], 6), ((), 6))
False
>>> compare_base([1, 3, 2], 4), ([3, 1, 2], 4))
True
>>> compare_base([1, 1, 2, 1], 4), ([3, 1, 2], 4))
True
```

♣ The program `add_base(N, M)` takes as input two integers `N` and `M` of type `IntB` written in the same base `b` and with the same length of decomposition, i.e. `len(N[0]) = len(M[0])`. It returns the
sum N+M of type IntB in the same base b. A simple method is to sum in base 10, but it is more costly in terms of time, i.e.:

```python
add_base(M, N) := int_to_base(base_to_int(M) + base_to_int(N), N[1]).
```

To limit the number of addition operations, we will prefer this program:

## Python 1.6.13

```python
def add_base(N: IntB, M: IntB) -> IntB: # len(N[0]) = len(M[0])
    R, b, B, C, s = [], N[1], N[0], M[0], 0
    for i in range(len(B)):
        s = B[i] + C[i] + s
        R = R + [s % b]
        s = s // b
    return (R + (s * [1]), b)
```

## Example 1.6.14

```python
>>> add_base([], 3), ([], 3)
([], 3)
>>> add_base([0], 4), ([] , 4)
([0], 4)
>>> add_base([3, 3, 5], 6), ([3, 1, 2], 6))
([0, 5, 1, 1], 6)
```

We conclude this subsection with two programs to transform an integer or a list of integers in a base into LaTeX code.

♣ The program `latex_intB(N)` takes as input an integer N of type IntB written in the same base b. It returns the LaTeX code of (N as a str type).

## Python 1.6.15

```python
def latex_intB(N: IntB) -> str:
    B, b = N
    s = '\\ |\\ '.join([str(B[-i-1]) for i in range(len(B))])
    return '\\left(' + s + '\\right)_{'} + str(len(B) - 1) + ', ' + str(b) + '}'
```

## Example 1.6.16

```python
>>> latex_intB([2, 14, 1, 12], 24))
'\\left(12\ |\ 1\ |\ 14\ |\ 2\right)_{3, 24}'
>>> print(latex_intB([2, 14, 1, 12], 24))
\left(12\ |\ 1\ |\ 14\ |\ 2\right)_{3, 24}
```

♣ The program `latex_list_intB(L, b)` takes as input a list of lists of integers L and a base b. It returns the LaTeX code of the list L as a str type, separated by a comma.
## Python 1.6.17

```python
def latex_list_intB(L: List List[int]], b, Base) -> str:
    return ', '.join([latex_intB((x, b)) for x in L])
```

## Example 1.6.18

```python
>>> L = [[2, 14, 1, 12], [1, 12], [2, 14, 1], [2, 14, 1, 12, 0]]
>>> latex_list_IntB(L, 24)
'\left(12\ |\ 1\ |\ 14\ |\ 2\right)_{3, 24},\ \left(12\ |\ 1\right)_{1, 24},\ \left(0\ |\ 12\ |\ 1\ |\ 14\ |\ 2\right)_{4, 24}'
>>> print(latex_list_IntB(L, 24))
\left(12\ |\ 1\ |\ 14\ |\ 2\right)_{3, 24},\ \left(12\ |\ 1\right)_{1, 24},
\ \left(1\ |\ 14\ |\ 2\right)_{2, 24},\ \left(0\ |\ 12\ |\ 1\ |\ 14\ |\ 2\right)_{4, 24}
```
## 2 Artifaille for Palindromization

By adding an integer $n$ with its mirror, we hope to make this sum more "palindromic" than $n$ and increase the chances of reductions. In this section, we provide the definitions and initial properties that help understand this intuition.

### 2.1 Reduction Operation

The reduction operation seen in the introduction involves, if an integer has the same first and last "digit" in base $b$, removing these two extreme digits.

**Definition 2.1.1**

Let $n \in \mathbb{N}$ of degree $d$ in base $b$. The **reduction rank in base $b$ of $n$** is defined as:

$$rg_b(n) := \max \{r \in \mathbb{N}, \forall k \in [0, r-1], k \le d-k \wedge [n]_{k,b} = [n]_{d-k,b}\}.$$

In the case where $n \in \mathbb{N}$ of degree $d$ in base $b$ is such that $[n]_{0,b} \neq [n]_{d,b}$, we have:

$$rg_b(n) = \max \{0\} = 0. \qquad (53)$$

**Example 2.1.2**

We have:

$$\begin{align*}
\mathrm{rg}_4 ((1)_{0,4}) &= 1 \\
\mathrm{rg}_4 ((2 | 2)_{1,4}) &= 1 \\
\mathrm{rg}_4 ((2 | 1)_{1,4}) &= 0 \\
\mathrm{rg}_4 ((2 | 1 | 3 | 1 | 2)_{4,4}) &= 3 \\
\mathrm{rg}_4 ((2 | 2 | 2 | 0 | 2 | 2)_{5,4}) &= 2
\end{align*}$$

**Definition 2.1.3**

(i) (a) Let $r \in \mathbb{N}_1$ and $n \in \mathbb{N}$ of degree $d \in \mathbb{N}_{2r-2}$ in base $b$. We say that $n$ is **reducible of rank $r$ in base $b$** if $rg_b(n) = r$. We denote by $\mathcal{R}_b^r$ the set of numbers reducible of rank $r$ in base $b$.

(b) Let $n \in \mathbb{N}$. We say that $n$ is **reducible in base $b$** if there exists $r \in \mathbb{N}_1$ such that $n \in \mathcal{R}_b^r$. We denote by $\mathcal{R}_b$ the set of numbers reducible in base $b$.

(ii) Let $n \in \mathcal{R}_b^r$ of degree $d \in \mathbb{N}$. The **reduction of $n$ in base $b$** is:

$$\rho_b(n) := \begin{cases} \bullet & \text{if } d = 2r-2 \\ ([n]_{d-r,b} | \cdots | [n]_{r,b})_{d-2r,b} & \text{otherwise.} \end{cases}$$

We denote:

$$\mathcal{R}_b^0 := \mathbb{N}_\bullet - \mathcal{R}_b \qquad (54)$$

Thus, we have:

$$\mathcal{R}_b = \bigsqcup_{r \in \mathbb{N}_1} \mathcal{R}_b^r \qquad (55)$$

$$\mathbb{N}_\bullet = \bigsqcup_{r \in \mathbb{N}} \mathcal{R}_b^r \qquad (56)$$
In particular, for any number $n$ of degree 0, we have $n \in \mathcal{R}_b$. We assume that $\bullet$ is not reducible:

$$ \bullet \in \mathcal{R}_b^0. \qquad (57) $$

For any $n \in \mathcal{R}_b^r$ (with $r \in \mathbb{N}_1$), we have:

$$ n = ([n]_{d,b} | \cdots | [n]_{d-r+1,b} | [n]_{d-r,b} | \cdots | [n]_{r,b} | [n]_{r-1,b} | \cdots | [n]_{0,b})_{d,b} \quad (58) $$

$$ = ([n]_{0,b} | \cdots | [n]_{r-1,b} | [n]_{d-r,b} | \cdots | [n]_{r,b} | [n]_{r-1,b} | \cdots | [n]_{0,b})_{d,b} \quad (59) $$

and thus:

$$ \rho_b(n) = ([n]_{d-r,b} | \cdots | [n]_{r,b})_{d,b}. \qquad (60) $$

### Definition 2.1.4

The reduction operation in base $b$ is the application $\rho_b$ defined for all $n \in \mathbb{N}_\bullet$:

$$ \rho_b(n) := \begin{cases} \rho_b(n) & \text{if } n \in \mathcal{R}_b \\ n & \text{otherwise.} \end{cases} $$

### Example 2.1.5

We have:

$$ \rho_b((3 \mid 1 \mid 2 \mid 4 \mid 3 \mid 2)_{5,6}) = (3 \mid 1 \mid 2 \mid 4 \mid 3 \mid 2)_{5,6} $$

$$ \rho_b((3 \mid 1 \mid 2 \mid 4 \mid 1 \mid 3)_{5,6}) = (2 \mid 4)_{1,6} $$

$$ \rho_b((3 \mid 2 \mid 3)_{2,5}) = \bullet $$

$$ \rho_b((3)_{0,5}) = \bullet $$

$$ \rho_b(\bullet) = \bullet $$

We have a particular case when the integer is reducible to the "maximum," meaning it reads the same forwards and backwards; we say it is palindromic.

### Definition 2.1.6

Let $n \in \mathbb{N}$. We say that $n$ is a **palindrome in base $b$** or **palindromic in base $b$** if $\rho_b(n) = \bullet$. We denote by $\mathcal{P}_b$ the set of palindromes in base $b$.

For all $n \in \mathbb{N}$ of degree $d := 2d' + \delta$ with $\delta \in \{0, 1\}$, we have equivalence between:

$$ n \in \mathcal{P}_b \iff \rho_b(n) = \bullet \iff n \in \mathcal{R}_b^{d'+1}. \qquad (61) $$

In the two following examples, we see the role of $d'$ and $\delta$ and the reason for $d' + 1$ in $\mathcal{R}_b^{d'+1}$.

### Example 2.1.7

(1) For $n := (2 \mid 3 \mid 2)_{2,4}$, we have $d_4(n) = 2 = 2 \times 1 + 0$ ($d' := 1$ and $\delta := 0$). We have:

$$ \mathrm{rg}_4(n) = 2 = d' + 1, \quad n \in \mathcal{R}_4^2 = \mathcal{R}_4^{d'+1}. $$

(2) For $n := (2 \mid 3 \mid 3 \mid 2)_{3,4}$, we have $d_4(n) = 3 = 2 \times 1 + 1$ ($d' := 1$ and $\delta := 1$). We have:

$$ \mathrm{rg}_4(n) = 2 = d' + 1, \quad n \in \mathcal{R}_4^2 = \mathcal{R}_4^{d'+1}. $$
For all $n \in \mathbb{N}$ of degree $d$ in base $b$, by setting $r := \text{rg}_b(n)$, we have:

$$ \rho_b(n) = \sum_{k=0}^{d} [n]_{k,b} b^{k-r} = \sum_{k=r}^{d-r} [n]_{k,b} b^{k-r} = \sum_{l=0}^{d-2r} [n]_{l+r,b} b^l \quad (62) $$

and thus:

$$ d_b(\rho_b(n)) = \begin{cases} -\infty & \text{if } n \in \mathcal{P}_b \\ d_b(n) - 2\text{rg}_b(n) & \text{otherwise.} \end{cases} \quad (63) $$

### Proposition 2.1.8

Let $r \in \mathbb{N}_1$ and $n \in \mathcal{R}_b^r$ such that $d_b(n) \ge 2r$. We have:

$$ \frac{n - b^{d-r+2} + b^d - b^r + 1}{b^r} \le \rho_b(n) \le \frac{n - b^{d_b(n)}}{b^r}. $$

The bounds are moreover optimal.

*Proof.* We have:

$$ \rho_b(n) = \sum_{k=r}^{d-r} [n]_{k,b} b^{k-r} = \frac{n - \sum_{k=0}^{r-1} ([n]_{d-k,b} b^{d-k} + [n]_{k,b} b^k)}{b^r}. $$

Let us prove the two inequalities separately.

(1) We have:

$$ \rho_b(n) = \frac{n - \sum_{k=0}^{r-1} ([n]_{d-k,b} b^{d-k} + [n]_{k,b} b^k)}{b^r} \le \frac{n - b^d - 1}{b^r}. $$

The upper bound is optimal because for $n := (1 | (0)_{r-1} | 1 | 0 | (0)_{r-1} | 1)_{2r+1,b}$ (thus $d := 2r+1$), we have:

$$ n = b^{2r+1} + b^r + 1 $$

and thus:

$$ \rho_b(n) = (1 | 0)_{1,b} = b = \frac{(b^{2r+1} + b^r + 1) - b^{2r+1} - 1}{b^r} = \frac{n - b^d - 1}{b^r}. $$

(2) We have:

$$ \begin{aligned} \rho_b(n) &= \frac{n - \sum_{k=0}^{r-1} ([n]_{d-k,b} b^{d-k} + [n]_{k,b} b^k)}{b^r} \\ &\ge \frac{n - \sum_{k=0}^{r-1} ((b-1)b^{d-k} + (b-1)b^k)}{b^r} \\ &= \frac{n - \sum_{k=0}^{r-1} (b^{d+1-k} - b^{d-k} + b^{k+1} - b^k)}{b^r} \\ &= \frac{n - b^{d-r+2} + b^d - b^r + 1}{b^r} \end{aligned} $$

The lower bound is optimal because for $n := ((b-1)_r | 1 | 0 | (b-1)_r)_{2r+1,b}$ (thus $d := 2r+1$), we have:

$$ \begin{aligned} n &= \sum_{k=0}^{r-1} (b-1)b^{2r+1-k} + 1 \times b^{r+1} + 0 \times b^r + \sum_{k=0}^{r-1} (b-1)b^k \\ &= b^{r+3} - b^{2r+1} + b^{r+1} + b^r - 1 \end{aligned} $$

and thus:

$$ \begin{aligned} \rho_b(n) &= (1 | 0)_{1,b} = b \\ &= \frac{(b^{r+3} - b^{2r+1} + b^{r+1} + b^r - 1) - b^{r+3} + b^{2r+1} - b^r + 1}{b^r} \\ &= \frac{n - b^{d-r+2} + b^d - b^r + 1}{b^r} \end{aligned} $$
☐

Let us finish this subsection with cardinality calculations. For this, let us denote for all elements $x, y \in \mathbb{N}$:

$$ \delta_{x,y} := \begin{cases} 1 & \text{if } x = y; \\ 0 & \text{if } x \neq y \end{cases} $$

called the **Kronecker symbol of x and y**.

**Proposition 2.1.9**

Let $d, r \in \mathbb{N}$. We have:

$$ \mathrm{card}(\mathcal{R}_{d,b}^r) = \begin{cases} 0 & \text{if } d \le 2r-3 \text{ or } d = 2r; \\ b^{r-1}(b-1) + \delta_{r,1}\delta_{d,0} & \text{if } d \in \{2r-2, 2r-1\}; \\ (b-1)^2 b^{d-r-1} & \text{if } d \ge 2r+1 \end{cases} $$

*Proof.* Let us address six cases separately.

(1) **Case $d \le 2r-3$.** Then the elements of $\mathcal{B}_{d,b}$ do not have enough digits in base $b$ to be reducible of rank $r$ in base $b$ id est $\mathcal{R}_{d,b}^r = \emptyset$.

(2) **Case $d = 2r$.** Reasoning by contradiction, suppose there exists $n \in \mathcal{R}_{2r,b}^r$. Then $d_b(\rho_b(n)) = 0$ and thus $\rho_b(n)$ is still reducible in base $b$. This is contradictory id est $\mathcal{R}_{2r,b}^r = \emptyset$.

(3) **Case $d=0$ and $r=1$.** Then we have $\mathcal{R}_{0,b}^1 = \mathcal{B}_b$ and thus

$$ \mathrm{card}(\mathcal{R}_{0,b}^1) = b = b^{1-1}(b-1) + \delta_{r,1}\delta_{d,0}. $$

(4) **Case $d = 2r-2$ and $r \neq 1$.** Then $n$ is palindromic in base $b$ and is of the type:

$$ n = ([n]_{0,b} | [n]_{1,b} | \cdots | [n]_{r-2,b} | [n]_{r-1,b} | [n]_{r-2,b} | \cdots | [n]_{1,b} | [n]_{0,b})_{2r-2,b}. $$

Choosing an element $n$ of $\mathcal{R}_{2r-2,b}^r$ thus amounts to choosing:

$$ \begin{gathered} [n]_{0,b} \in [1, b-1] \\ \forall k \in [1, r-1], [n]_{k,b} \in \mathcal{B}_b \end{gathered} $$

And thus:

$$ \mathrm{card}(\mathcal{R}_{2r-2,b}^r) = \mathrm{card}([1, b-1]) \prod_{k=1}^{r-1} \mathrm{card}(\mathcal{B}_b) = b^{r-1}(b-1). $$

(5) **Case $d = 2r-1$.** It is similar to the previous case; $n$ is palindromic in base $b$ and is thus of the type:

$$ n = ([n]_{0,b} | [n]_{1,b} | \cdots | [n]_{r-2,b} | [n]_{r-1,b} | [n]_{r-1,b} | [n]_{r-2,b} | \cdots | [n]_{1,b} | [n]_{0,b})_{2r-1,b}. $$

Choosing an element $n$ of $\mathcal{R}_{2r-1,b}^r$ thus amounts to choosing:

$$ \begin{gathered} [n]_{0,b} \in [1, b-1] \\ \forall k \in [1, r-1], [n]_{k,b} \in \mathcal{B}_b \end{gathered} $$

And thus:

$$ \mathrm{card}(\mathcal{R}_{2r-1,b}^r) = \mathrm{card}([1, b-1]) \prod_{k=1}^{r-1} \mathrm{card}(\mathcal{B}_b) = b^{r-1}(b-1). $$
(6) Case $d \ge 2r+1$. Then $n$ is of the type:

$$n = ([n]_{0,b} | \cdots | [n]_{r-1,b} | [n]_{d-r,b} | \cdots | [n]_{r,b} | [n]_{r-1,b} | \cdots | [n]_{0,b})_{2r-1,b}$$

Choosing an element $n$ of $\mathcal{R}_{d,b}^r$ thus amounts to choosing:

$$ \begin{gathered} [n]_{0,b} \in [1, b-1] \\ \forall k \in [1, r-1], [n]_{k,b} \in \mathcal{B}_b \\ [n]_{r,b} \in \mathcal{B}_b \wedge [n]_{d-r,b} \in \mathcal{B}_b - \{[n]_{r,b}\} \\ \forall k \in [r+1, d-r-1], [n]_{k,b} \in \mathcal{B}_b \end{gathered} $$

And thus:

$$ \begin{align*} \mathrm{card}(\mathcal{R}_{d,b}^r) &= \mathrm{card}([1, b-1]) \left( \prod_{k=1}^{r-1} \mathrm{card}(\mathcal{B}_b) \right) \mathrm{card}([1, b-1]) \mathrm{card}(\mathcal{B}_b) \prod_{k=r+1}^{d-r-1} \mathrm{card}(\mathcal{B}_b) \\ &= (b-1)b^{r-1}(b-1)bb^{d-2r-1} \\ &= (b-1)^2 b^{d-r-1}. \end{align*} $$

□

## 2.2 Mirror Operation

The mirror operation seen in the introduction consists of reversing the order of digits in the decomposition of an integer in base $b$. Although we will apply the mirror application to irreducible integers, this operation is defined for all elements of $\mathbb{N}_\bullet$.

### Definition 2.2.1

Let $n \in \mathbb{N}$ of degree $d \in \mathbb{N}$.

(i) The **mirror in base $b$ of $n$** is defined by:

$$\mu_b(n) := ([n]_{0,b} | \cdots | [n]_{d,b})_{d,b}.$$

(ii) For all $k \in [0, d]$, $[n]_{d-k,b}$ is called the **mirror digit of $[n]_{k,b}$ associated with $n$ in base $b$**.

### Example 2.2.2

We have:

$$ \begin{align*} \mu_b((3)_{0,4}) &= (3)_{0,4} \\ \mu_b((3 \mid 1 \mid 2 \mid 4 \mid 3 \mid 2)_{5,8}) &= (2 \mid 3 \mid 4 \mid 2 \mid 1 \mid 3)_{5,8} \end{align*} $$

Thus, for all $n \in \mathbb{N}_1$:

$$\mu_b(n) = ([n]_{0,b} | \cdots | [n]_{d_b(n),b})_{d_b(n),b} = ([n]_{\nu_b(n),b} | \cdots | [n]_{d_b(n),b})_{d_b(n)-\nu_b(n),b} \quad (64)$$

id est we have:

$$d_b(\mu_b(n)) = d_b(n) - \nu_b(n) \quad (65)$$

and thus:

$$\mu_b(n) = \sum_{k=0}^{d_b(n)-\nu_b(n)} [n]_{d_b(n)-k,b} b^k. \quad (66)$$
**Definition 2.2.3**

The mirror operation in base $b$ is the application $\mu_b$ defined for all $n \in \mathbb{N}$:

$$ \mu_b(n) := \begin{cases} \bullet & \text{if } n := \bullet \\ \mu_b(n) & \text{otherwise} \end{cases} $$

Thus, for all $n \in \mathbb{N}$:

$$ n \in \mathcal{P}_b \iff \mu_b(n) = n. \qquad (67) $$

The application $\mu_b$ is not injective because $b \neq 1$ and:

$$ \mu_b(b) = \mu_b((1 | 0)_{1,b}) = (0 | 1)_{1,b} = (1)_{0,b} = \mu_b((1)_{0,b}) = \mu_b(1). \qquad (68) $$

From this, we deduce the following result.

**Proposition 2.2.4**

(i) Let $m, n \in \mathbb{N}$. We have:

$$ \mu_b(m) = \mu_b(n) \iff b^{\nu_b(n)}m = b^{\nu_b(m)}n. $$

In particular, for all $k \in \mathbb{N}$:

$$ \mu_b(b^k n) = \mu_b(n). $$

(ii) For all $k \in \mathbb{N}$:

$$ \mu_b(b^k n) = \mu_b(n). $$

(iii) We have:

$$ b^{\nu_b(n)} \mu_b^2(n) = n. $$

(iv) Let $n \in \mathbb{N}_1$. We have:

$$ P_{\mu_b(n),b} = X^{d_b(n)} P_{n,b} \left( \frac{1}{X} \right). $$

*Proof.* (i) We have two cases.

(1) Case $m := 0$ or $n := 0$. By symmetry of $m$ and $n$, assume $m := 0$. Then we have:

$$ 0 = \mu_b(m) = \mu_b(n) \iff n = 0 = m. $$

(2) Case $m, n \neq 0$. Let:

$$ m' := \frac{m}{b^{\nu_b(m')}}, \quad n' := \frac{n}{b^{\nu_b(n)}}. $$

We must prove that:

$$ \mu_b(m) = \mu_b(n) \iff m' = n'. $$

Since $\mu_b(m') = \mu_b(m)$, $\mu_b(n') = \mu_b(n)$, and $\nu_b(m') = \nu_b(n') = 0$, we have:

$$ \begin{align*} \mu_b(m) &= \mu_b(n) \\ \iff \mu_b(m') &= \mu_b(n') \\ \iff d_b(\mu_b(m')) &= d_b(\mu_b(n')) \quad \forall k \in [0, d_b(\mu_b(m'))], \quad [\mu_b(m')]_{k,b} = [\mu_b(n')]_{k,b} \\ \iff d_b(m') &= d_b(n') \quad \forall k \in [0, d_b(m')], \quad [m']_{d_b(m')-k,b} = [n']_{d_b(n')-k,b} \\ \iff m' &= n' \end{align*} $$
(ii) We have:

$$
\begin{align*}
b^{\nu_b(n)} \mu_b^2(n) &= b^{\nu_b(n)} \mu_b \left( \sum_{k=0}^{d_b(n)-\nu_b(n)} [n]_{d_b(n)-k,b} b^k \right) \\
&= b^{\nu_b(n)} \left( \sum_{k=0}^{d_b(n)-\nu_b(n)} [n]_{k+\nu_b(n),b} b^k \right) \\
&= \sum_{k=0}^{d_b(n)-\nu_b(n)} [n]_{k+\nu_b(n),b} b^{k+\nu_b(n)} \\
&= \sum_{l=\nu_b(n)}^{d_b(n)} [n]_{l,b} b^l \\
&= n
\end{align*}
$$

(iii) We have:

$$
\begin{align*}
X^{d_b(n)} P_{n,b} \left( \frac{1}{X} \right) &= X^{d_b(n)} \left( \sum_{k=\nu_b(n)}^{d_b(n)} [n]_{k,b} \frac{1}{X^k} \right) \\
&= \sum_{k=\nu_b(n)}^{d_b(n)} [n]_{k,b} X^{d_b(n)-k} \\
&= \sum_{l=0}^{d_b(n)-\nu_b(n)} [n]_{d_b(n)-l,b} X^l \\
&= P_{\mu_b(n),b}.
\end{align*}
$$

<figure><img src="image_7.png" /></figure>

2.3 Palindromic Operation

The palindromic operation introduced in the introduction consists of adding an integer to its mirror in
base b. As with the mirror application, we will only apply it to irreducible integers, but we still define
it for every element of $\mathbb{N}_{\bullet}$.

**Definition 2.3.1**

The **palindromic operation in base b** is the application $\pi_b$ defined for every $n \in \mathbb{N}_\bullet$ by:

$$
\pi_b(n) := n + \mu_b(n).
$$

We call $\pi_b(n)$ the **pseudo-palindrome associated with $n$ in base $b$**.

By definition, the pseudo-palindrome of an integer is equal to the pseudo-palindrome of its mirror,
id est we have on $\mathbb{N}_{\bullet}$:

$$
\pi_b \circ \mu_b = \pi_b. \tag{69}
$$

**Example 2.3.2**

(1) Since $\bullet + \bullet = \bullet$, we have:

$$
\pi_b(\bullet) = \bullet + \mu_b(\bullet) = \bullet + \bullet = \bullet.
$$

(2) We have:

$$
\pi_b((0)_{0,6}) = (0)_{1,6} + (0)_{0,6} = (0)_{0,6}
$$
$$
\begin{align*}
\pi_b((3)_{0,6}) &= (3)_{1,6} + (3)_{0,6} = (1 \mid 1)_{1,6} \\
\pi_b((3 \mid 1 \mid 4)_{2,6}) &= (3 \mid 1 \mid 4)_{2,6} + (4 \mid 1 \mid 3)_{2,6} = (1 \mid 1 \mid 3 \mid 1)_{3,6} \\
\pi_b((2 \mid 1 \mid 3)_{2,6}) &= (2 \mid 1 \mid 3)_{2,6} + (3 \mid 1 \mid 2)_{2,6} = (5 \mid 2 \mid 5)_{2,6}
\end{align*}
$$

In the last example, we see that if the digit-by-digit sums between an integer and its mirror do not exceed $b-1$, then its pseudo-palindrome is a palindrome (this argument justifies the definition). We will see a more general argument in the following proposition. We also see from the three examples that the pseudo-palindrome of a non-zero integer is strictly greater than the integer, *id est* since for every $n \in \mathbb{N}_1$, $\mu_b(n) > 0$, we have:

$$ \pi_b(n) > n \tag{70} $$

Thus, the only fixed points of $\pi_b$ are $\bullet$ and $0$. And for every $n \in \mathbb{N}_1$, we have:

$$ n < \pi_b(n) < 2 \max\{n, \mu_b(n)\}. \tag{71} $$

In particular, $d_b(\pi_b(n)) \le d_b(n) + 1$ and for every $k \in \mathbb{N}$:

$$ d_b(\pi_b^k(n)) \le d_b(n) + k. \tag{72} $$

From this, we deduce the following results.

### Proposition 2.3.3

(i) Let $n \in \mathbb{N}$ with degree $d := 2d' + \delta$ in base $b$, where $\delta \in \{0, 1\}$. Suppose that for every $k \in [0, d']$:

$$ [n]_{k,b} + [n]_{d-k,b} \le b-1. $$

Then $\pi_b(n) \in \mathcal{P}_b$. In particular, we have $\rho_b(\pi_b(n)) = \bullet$.

(ii) Let $n, m \in \mathbb{N}_1$ with degree $d := 2d' + \delta$ in base $b$, where $\delta \in \{0, 1\}$. Suppose that:

$$ \forall k \in [0, d'], [m]_{k,b} + [m]_{d-k,b} = [n]_{k,b} + [n]_{d-k,b}. $$

Then $\pi_b(m) = \pi_b(n)$.

*Proof.* (i) We want to show that $\pi_b(n) \in \mathcal{P}_b$, *id est* $\mu_b(\pi_b(n)) = \pi_b(n)$. Let us define for every $k \in \mathbb{N}$:

$$ s_k := \sigma(n, \mu_b(n))_{k,b}. $$

We prove by finite induction on $k \in [0, d]$:

$$ (\mathcal{P}_k) : s_k = [n]_{k,b} + [n]_{d-k,b}. $$

(I) By definition, $s_0 = [n]_{0,b} + [\mu_b(n)]_{0,b}$. Hence, $(\mathcal{P}_0)$ holds.

(II) Let $k \in \mathbb{N}_1^d$. Suppose $(\mathcal{P}_{k-1})$ holds. Since $[n]_{k-1,b} + [n]_{d-k+1,b} \le b-1$, we have:

$$ q_b(s_{k-1}) = q_b([n]_{k-1,b} + [n]_{d-k+1,b}) = 0 $$

and thus:

$$ s_k = [n]_{k,b} + [\mu_b(n)]_{k,b} + q_b(s_{k-1}) = [n]_{k,b} + [n]_{d-k,b}. $$

So $(\mathcal{P}_k)$ holds.

(III) By finite induction, $(\mathcal{P}_k)$ holds for every $k \in [0, d]$.
Thus, for every $k \in [0, d]$:

$$
\begin{align*}
[\pi_b(n)]_{k,b} &= s_k \\
&= [n]_{k,b} + [n]_{d-k,b} \\
&= [n]_{d-k,b} + [n]_{k,b} \\
&= [\pi_b(n)]_{d-k,b}
\end{align*}
$$

(ii) Let us define for every $k \in \mathbb{N}$:

$$
s_k := \sigma(m, \mu_b(m))_{k,b} , \quad s'_k := \sigma(n, \mu_b(n))_{k,b}.
$$

We prove by finite induction on $k \in [0, d]$:

$$
(\mathcal{P}_k) : s_k = s'_k.
$$

(I) We have:

$$
\begin{align*}
s_0 &= [m]_{0,b} + [\pi_b(m)]_{0,b} = [m]_{0,b} + [m]_{d,b} \\
&= [n]_{0,b} + [n]_{d,b} = [n]_{0,b} + [\pi_b(n)]_{d,b} \\
&= s'_0
\end{align*}
$$

Thus, ($\mathcal{P}_0$) holds.

(II) Let $k \in \mathbb{N}_1^d$. Suppose ($\mathcal{P}_{k-1}$) holds. We have:

$$
\begin{align*}
s_k &= [m]_{k,b} + [\pi_b(m)]_{k,b} + q_b(s_{k-1}) \\
&= [n]_{k,b} + [\pi_b(n)]_{k,b} + q_b(s'_{k-1}) \\
&= s'_k
\end{align*}
$$

Thus, ($\mathcal{P}_k$) holds.

(III) By finite induction, ($\mathcal{P}_k$) holds for every $k \in [0, d]$.

Thus, for every $k \in \mathbb{N}$:

$$
\begin{align*}
[\pi_b(m)]_{k,b} &= [m + \mu_b(m)]_{k,b} \\
&= r_b(s_k) \\
&= r_b(s'_k) \\
&= [n + \mu_b(n)]_{k,b} \\
&= [\pi_b(n)]_{k,b}.
\end{align*}
$$

□

Example 2.3.4

We have:

$$
\begin{align*}
(1 \mid 1 \mid 3 \mid 1)_{3,6} &= \pi_6((3 \mid 1 \mid 4)_{2,6}) = \pi_6((4 \mid 1 \mid 3)_{2,6}) = \pi_6((2 \mid 1 \mid 5)_{2,6}) = \pi_6((5 \mid 1 \mid 2)_{2,6}) \\
(1 \mid 1 \mid 1 \mid 1)_{3,6} &= \pi_6((3 \mid 0 \mid 4)_{2,6}) = \pi_6((4 \mid 0 \mid 3)_{2,6}) = \pi_6((2 \mid 0 \mid 5)_{2,6}) \\
&= \pi_6((5 \mid 0 \mid 2)_{2,6})
\end{align*}
$$

It turns out that point (ii) of the previous proposition 2.3.3 is an equivalence, though we won't
need it. One might wonder if the property always holds when *n* and *m* are not of the same degree.
The answer is no, as we have for example:

$$
(1 | 1 | 1 | 1)_{3,6} = \pi_6((1 | 1 | 0 | 0)_{3,6}) = \pi_6((5 | 0 | 2)_{2,6}). \quad (73)
$$

We leave this for the "bacouettes" (see subsection 5.1).
## 2.4 Artifaille in Python

♠ The program `est_red(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns `True` if `N` is reducible and `False` otherwise.

### Python 2.4.1

```python
def est_red(N: IntB) -> bool:
    B, _ = N
    if B == []:
        return False
    return B[0] == B[-1]
```

### Example 2.4.2

```python
>>> est_red([1, 3, 2], 5)
False
>>> est_red([1, 3, 1], 5)
True
>>> est_red([], 5)
False
```

♠ The program `rg_red(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns the rank of `N` in base `b`.

### Python 2.4.3

```python
def rg_red(N: IntB) -> int:
    if len(N[0]) <= 2:
        return int est_red(N)
    elif est_red(N):
        return 1 + rg_red((N[0][1: -1], N[1]))
    else:
        return 0
```

### Example 2.4.4

```python
>>> rg_red([1, 3, 2], 5)
0
>>> rg_red([1, 3, 1], 5)
2
>>> rg_red([], 5)
0
```

♠ The program `est_pal(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns `True` if `N` is palindromic and `False` otherwise.
### Python 2.4.5

```python
def est_pal(N: IntB) -> bool:
    B, _ = N
    if B == []:
        return False
    return B == B[: : -1]
```

### Example 2.4.6

```python
>>> est_pal([1, 3, 1], 5)
True
>>> est_pal([1, 3, 2], 5)
False
>>> est_pal([], 5)
False
```

The program `rho_fct(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns $\rho_b(N)$ of type `IntB` in the same base `b`.

### Python 2.4.7

```python
def rho_fct(N: IntB) -> IntB:
    r = rg_red(N)
    return (N[0][r: len(N[0]) - r], N[1])
```

### Example 2.4.8

```python
>>> rho_fct([4, 3, 3, 1, 3], 5)
([4, 3, 3, 1, 3], 5)
>>> rho_fct([1, 3, 1], 5)
([1, 3, 1], 5)
```

The program `mu_fct(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns $\mu_b(N)$ of type `IntB` in the same base `b`.

### Python 2.4.9

```python
def mu_fct(N: IntB) -> IntB:
    return (N[0][ : : -1], N[1])
```

### Example 2.4.10

```python
>>> mu_fct([1, 3, 2], 5)
([2, 3, 1], 5)
>>> mu_fct([], 5)
([], 5)
```
♣ The program `pi_fct(N)` takes as input an integer *N* of type `IntB` written in base *b*. It returns $\pi_b(N)$ of type `IntB` in the same base *b*.

Python 2.4.11

```python
def pi_fct(N: IntB) -> IntB:
    B, b = N
    R, s = [], 0
    for k in range(len(B)):
        s = B[k] + B[-k - 1] + s
        R.append(s % b)
        s //:= b
    if s:
        R.append(1)
    return (R, b)
```

Example 2.4.12

```python
>>> pi_fct([4, 3, 3, 1, 3], 5)
([2, 0, 2, 0, 3, 1], 5)
>>> pi_fct([1, 3, 2], 5)
([3, 1, 4], 5)
>>> pi_fct([], 5)
[]
```
# 3 Artifaille of Norman

In this section, we provide the necessary explanation to understand the Norman sequence in base $b$ associated with an integer.

## 3.1 Norman Operation

The Norman operation consists of either reducing an integer if it is reducible or calculating the pseudo-palindrome otherwise.

**Definition 3.1.1**

The Norman operation in base $b$ is defined for all $n \in \mathbb{N}$ by:

$$ \gamma_b(n) := \begin{cases} \rho_b(n) & \text{if } n \in \mathcal{R}_b \\ \pi_b(n) & \text{if } n \in \mathcal{R}_b^0 \end{cases} $$

In particular, we have:

$$ \gamma_b(\bullet) = \pi_b(\bullet) = \bullet. \qquad (74) $$

**Lemma 3.1.2**

The only fixed point of $\gamma_b$ is $\bullet$.

*Proof.* We have $\gamma_b(\bullet) = \bullet$. Reasoning by contradiction, suppose there exists $n \in \mathbb{N}$ such that $\gamma_b(n) = n$. There are two cases:

(1) **Case** $n \in \mathcal{R}_b$. Thus, $n = \gamma_b(n) = \rho_b(n) < n$, which is contradictory.

(2) **Case** $n \in \mathcal{R}_b^0$. Thus, $n = \gamma_b(n) = \pi_b(n) > n$, which is contradictory.

Hence the result.

## 3.2 Norman Sequence in Base $b$

By iterating the Norman operation in base $b$ on an integer, we obtain its Norman sequence in base $b$.

**Definition 3.2.1**

The Norman sequence associated with $n$ in base $b$ is the sequence:

$$ \Gamma_b(n) := (\gamma_b^k(n))_{k \in \mathbb{N}}. $$

For any $N \in \mathbb{N}$, we define:

$$ \Gamma_{N,b}(n) := (\gamma_b^k(n))_{k \in \mathbb{N}[0,N]} = (\gamma_b^0(n), \dots, \gamma_b^N(n)). \qquad (75) $$

There are three possible cases for $\Gamma_b(n)$ with $n \in \mathbb{N}$:

(Case I): The sequence $\Gamma_b(n)$ is bounded and converges. It converges to $\bullet$ (the only fixed point of $\gamma_b$).

(Case II): The sequence $\Gamma_b(n)$ is bounded and diverges. It becomes periodic after a certain rank.

(Case III): The sequence $\Gamma_b(n)$ diverges to $+\infty$.

From this, we derive the following definitions.
### Definition 3.2.2

Let $n \in \mathbb{N}$.

(i) $n$ is called **Norman in base $b$** if there exists $k \in \mathbb{N}$ such that:

$$ \gamma_b^k(n) = \bullet. $$

The set of Norman integers in base $b$ is denoted by $\mathcal{N}_b$.

(ii) $n$ is called **celt in base $b$** if there exist distinct $k, l \in \mathbb{N}$ such that:

$$ \gamma_b^k(n) \neq \bullet \wedge \gamma_b^k(n) = \gamma_b^l(n). $$

The set of celt integers in base $b$ is denoted by $\mathcal{C}_b$.

(iii) $n$ is called **Quebecois in base $b$** if $n$ is neither Norman nor celt in base $b$. The set of Quebecois integers in base $b$ is denoted by $\mathcal{Q}_b$.

Thus:

$$ \mathbb{N}_{\bullet} = \mathcal{N}_b \cup \mathcal{C}_b \cup \mathcal{Q}_b. \qquad (76) $$

For statistical analysis via computing, it may be useful to examine three aspects:

(1) For a Norman integer in base $b$, the number of steps to "reach" the integer $\bullet$;

(2) For a celt integer in base $b$, the number of steps to "reach" the cycle and the length of the cycle;

(3) For a Quebecois integer in base $b$, the evolution of the degrees of the sequence $\Gamma_{b,N}(n)$ for sufficiently large $N$, i.e., the study of the sequence:

$$ (d_b(\gamma_b^k(n)))_{k \in [0,N]}. $$

To do this, we define the identity of a $\gamma_b$-sequence as the number of times we apply $\pi_b$ or $\rho_b$ to move from one element of the sequence to the next.

### Definition 3.2.3

Let $N \in \mathbb{N}_1$.

(i) A finite sequence of integers $\mathcal{A} := (a_0, \dots, a_N)$ of natural numbers is called a $\gamma_b$-sequence if there exists $n \in \mathbb{N}$ such that $\mathcal{A} = \Gamma_{N,b}(n)$.

(ii) For a $\gamma_b$-sequence $\mathcal{A} := \Gamma_{N,b}(n)$ with $n \in \mathbb{N}$, we define:

$$ \lambda_{\rho,b}(\mathcal{A}) := \mathrm{card}\{k \in \mathbb{N}_1^N, \gamma_b^k(n) = \rho_b(\gamma_b^{k-1}(n))\} $$

$$ \lambda_{\pi,b}(\mathcal{A}) := \mathrm{card}\{k \in \mathbb{N}_1^N, \gamma_b^k(n) = \pi_b(\gamma_b^{k-1}(n))\} $$

The identity of $\mathcal{A}$ in base $b$ is the pair:

$$ \mathrm{Id}_b(\mathcal{A}) := (\lambda_{\rho,b}(\mathcal{A}), \lambda_{\pi,b}(\mathcal{A})). $$

## 3.3 Norman Integers in Base $b$

In this subsection, we define terms related to Norman numbers in base $b$.
**Definition 3.3.1**

Let $n \in \mathcal{N}_b$.

(i) The depth of the dream of $n$ in base $b$ is defined as:

$$ \beta_b(n) := \min\{k \in \mathbb{N}, \gamma_b^k(n) = \bullet\}. $$

(ii) The identity of $n$ in base $b$ is:

$$ \mathrm{Id}_b(n) := \mathrm{Id}_b(\Gamma_{b,\beta_b(n)}(n)). $$

We also define:

$$ \lambda_{\rho,b}(n) := \lambda_{\rho,b}(\Gamma_{b,\beta_b(n)}(n)) $$

$$ \lambda_{\pi,b}(n) := \lambda_{\pi,b}(\Gamma_{b,\beta_b(n)}(n)). $$

Thus:

$$ \beta_b(n) = \lambda_{\rho,b}(n) + \lambda_{\pi,b}(n). \qquad (77) $$

**Example 3.3.2**

In the introduction, we showed that the integer $n := 149823$ is Norman in base 10. We have:

$$ \gamma_{10}(n) = \pi_{10}(149823) = 478764 $$

$$ \gamma_{10}^2(n) = \rho_{10}(478764) = 7876 $$

$$ \gamma_{10}^3(n) = \pi_{10}(7876) = 14663 $$

$$ \gamma_{10}^4(n) = \pi_{10}(14663) = 51304 $$

$$ \gamma_{10}^5(n) = \pi_{10}(51304) = 91619 $$

$$ \gamma_{10}^6(n) = \rho_{10}(91619) = \bullet $$

Thus:

$$ \lambda_{\rho,10}(n) = 2, \quad \lambda_{\pi,10}(n) = 4, \quad \beta_{10}(n) = 6, \quad \mathrm{Id}_b(n) = (2,4). $$

**3.4 The Case of the Celts in Base $b$**

In this subsection, we define the concepts related to celt numbers in base $b$. The Norman sequence in base $b$ of a celt number in base $b$ will be split into two subsequences: the dream voyage and the cyclic dream (the triskelion<fnref n="18" />).

**Definition 3.4.1**

Let $n \in \mathcal{C}_b$.

(i) The depth of the dream of $n$ in base $b$ is defined by:

$$ \beta_b(n) := \min\{k \in \mathbb{N}, \exists l \in \mathbb{N}_{k+1}, \gamma_b^k(n) = \gamma_b^l(n)\}. $$

(ii) The length of the triskelion of $n$ in base $b$ is defined by:

$$ \beta_b^\dagger(n) := \min\{k \in \mathbb{N}_1, \gamma_b^{\beta_b(n)+k}(n) = \gamma_b^{\beta_b(n)}(n)\}. $$

<footnote n="18">¹⁸ Symbol existing since the Neolithic, the triskelion is used by druidic circles at the end of the 19th century, it is the symbol of the Celts for this text.</footnote>
The **trisk of $n$ in base $b$** is defined by:

$$ \tau_b(n) := \min\{\gamma_b^k(n), k \in \mathbb{N}_{\beta_b(n)}\} $$

and the **triskelion of $n$ in base $b$** is defined by:

$$ \mathcal{T}_b(n) := \{\gamma_b^k(n), \beta_b(n) \le k \le \beta_b(n) + \beta_b^\dagger(n) - 1\}. $$

(iii) The **identity of $n$ in base $b$** is:

$$ \mathrm{Id}_b(n) := \left( \mathrm{Id}_b(\Gamma_{b,\beta_b(n)}(n)), \mathrm{Id}_b\left(\Gamma_{b,\beta_b^\dagger(n)}(\gamma_{\beta_b(n)} - 1)\right) \right). $$

We could place both Norman and celt numbers in base $b$ in the same framework, stating that a Norman number is a celt number that reaches a triskelion of length 1 reduced to $\bullet$.

Thus, the trisk of $n$ in base $b$ is a generator of the triskelion of $n$ in base $b$:

$$ \mathcal{T}_b(n) = \{\gamma_b^k(\tau_b(n)), 0 \le k \le \beta_b^\dagger(n) - 1\}. \qquad (78) $$

A special case occurs when the depth of the dream of a celt number $n$ in base $b$ is zero. The associated Norman sequence of $n$ in base $b$ is simply a triskelion. We call these Breton integers; such a triskelion is uniquely determined by its smallest element, i.e., its trisk in base $b$, leading to the following definitions.

### Example 3.4.2

(1) Let a sequence $\mathcal{A} := \Gamma_{N,b}(n)$ with $n \in \mathbb{N}$ and $N \in \mathbb{N}_1$. We say that the sequence $\mathcal{A}$ is **Breton in base $b$** if the sequence $\mathcal{A}$ is a sequence of distinct elements of $\mathbb{N}$ such that:

$$ \gamma_b^{N+1}(n) = n $$

i.e., if:

$$ \mathcal{A} = \mathcal{T}_b(n). $$

(a) In this case, the **length of $\mathcal{A}$ in base $b$** is:

$$ \beta_b^\dagger(\mathcal{A}) := N. $$

(b) In this case, the **trisk of $\mathcal{A}$ in base $b$** is:

$$ \tau_b(\mathcal{A}) := \min\{\gamma_b^k(n), k \in [0, N]\}. $$

(c) In this case, the **identity of $\mathcal{A}$ in base $b$** is $\mathrm{Id}_b(\mathcal{A})$.

(2) Let $n \in \mathcal{C}_b$. We say that $n$ is **Breton in base $b$** if $\beta_b(n) = 0$ i.e., there exists $N \in \mathbb{N}_1$ such that the sequence $\Gamma_{N,b}(n)$ is Breton in base $b$.

We denote $\mathcal{C}_b^\dagger$ as the set of Breton numbers in base $b$. Thus, the length of $n$ in base $b$ is:

$$ \beta_b^\dagger(n) = \beta_b^\dagger(\Gamma_{N,b}(n)) = N $$

the trisk of $n$ in base $b$ is:

$$ \tau_b(n) := \tau_b(\Gamma_{N,b}(n)). $$

and the identity of $n$ in base $b$ is:

$$ \mathrm{Id}_b(n) := \mathrm{Id}_b(\Gamma_{N,b}(n)). $$
Thus, we have:

$$ \mathscr{C}_b^\dagger \subset \mathscr{C}_b. \qquad (79) $$

For every celt number $n$ in base $b$, we will note:

$$ \mathrm{Id}_b^\dagger(n) := \mathrm{Id}_b(\tau_b(n)). \qquad (80) $$

the identity of its trisk in base $b$.

### Example 3.4.3

In the introduction, we showed that the integer $n := (9 \mid 8 \mid 10 \mid 7 \mid 8)_{4,11}$ is celt in base 11. We have:

$$ \gamma_{11}(n) = \pi_{11}((9 \mid 8 \mid 10 \mid 7 \mid 8)_{4,11}) = (1 \mid 7 \mid 5 \mid 10 \mid 5 \mid 6)_{5,11} $$

$$ \gamma_{11}^2(n) = \pi_{11}((1 \mid 7 \mid 5 \mid 10 \mid 5 \mid 6)_{5,11}) = (8 \mid 2 \mid 5 \mid 5 \mid 1 \mid 7)_{5,11} $$

$$ \gamma_{11}^3(n) = \pi_{11}((8 \mid 2 \mid 5 \mid 5 \mid 1 \mid 7)_{5,11}) = (1 \mid 4 \mid 3 \mid 10 \mid 10 \mid 4 \mid 4)_{6,11} $$

$$ \gamma_{11}^4(n) = \pi_{11}((1 \mid 4 \mid 3 \mid 10 \mid 10 \mid 4 \mid 4)_{6,11}) = (5 \mid 9 \mid 3 \mid 10 \mid 2 \mid 8 \mid 5)_{6,11} $$

$$ \gamma_{11}^5(n) = \rho_{11}((5 \mid 9 \mid 3 \mid 10 \mid 2 \mid 8 \mid 5)_{6,11}) = (9 \mid 3 \mid 10 \mid 2 \mid 8)_{4,11} $$

$$ \gamma_{11}^6(n) = \pi_{11}((9 \mid 3 \mid 10 \mid 2 \mid 8)_{4,11}) = (1 \mid 6 \mid 6 \mid 9 \mid 6 \mid 6)_{5,11} $$

$$ \gamma_{11}^7(n) = \pi_{11}((1 \mid 6 \mid 6 \mid 9 \mid 6 \mid 6)_{5,11}) = (8 \mid 2 \mid 5 \mid 5 \mid 1 \mid 7)_{5,11} = \gamma_{11}^2(n) $$

and thus:

$$ \beta_{11}(n) = 2, \quad \beta_{11}^{\dagger}(n) = 5, \quad \tau_b(n) = (9 \mid 3 \mid 10 \mid 2 \mid 8)_{4,11}, \quad \mathrm{Id}_b(n) = ((0,2), (1,4)). $$

## 3.5 The Case of the Quebecers in Base $b$

In this subsection, we define the concepts related to Quebecer numbers. We begin with a lemma.

### Lemma 3.5.1

Let $n \in \mathbb{N}$. Suppose that $n$ is neither Norman nor celt in base $b$. Then:

$$ \lim_{k \to \infty} \gamma_b^k(n) = \infty. $$

*Proof.* Let us show that for every $N \in \mathbb{N}$, there exists a $k \in \mathbb{N}$ such that:

$$ (*) : \forall l \in \mathbb{N}_k, \gamma_b^l(n) > N. $$

We proceed by contradiction. Suppose that there exists $N_0 \in \mathbb{N}$ such that for every $k \in \mathbb{N}$, there exists $l \in \mathbb{N}_k$ such that $\gamma_b^l(n) \le N_0$. We then construct the following sequence $(l_i)_{i \in \mathbb{N}}$ by recurrence:

$$ l_0 \in \mathbb{N}, \quad \gamma_b^{l_0}(n) \le N_0 $$

$$ l_{i+1} \in \mathbb{N}_{l_i+1}, \quad \gamma_b^{l_{i+1}}(n) \le N_0 $$

Thus, the infinite sequence $(\gamma_b^{l_i}(n))_{i \in \mathbb{N}}$ is a sequence of elements in $[0, N_0]$. By the pigeonhole principle, there exist distinct $i, j \in \mathbb{N}$ such that $\gamma_b^{l_i}(n) = \gamma_b^{l_j}(n)$. Therefore, $n$ is celt in base $b$, which contradicts the assumption on $n$. Hence, by taking the limit over $N \in \mathbb{N}$ in $(*)$, we obtain the result. $\square$
Quebecer numbers in base $b$ are those whose Norman sequence in base $b$ diverges to $+\infty$. For these numbers, it may then be interesting to calculate the sequence of the degrees of their Norman sequence in base $b$, i.e., the sequence:

$$ (d_b(\gamma_b^k(n)))_{k \in \mathbb{N}}. \qquad (81) $$

We then have the equivalences:

$$ n \in \mathcal{Q}_b \iff \lim_{k \in \mathbb{N}} \gamma_b^k(n) = +\infty \iff \lim_{k \in \mathbb{N}} d_b(\gamma_b^k(n)) = +\infty. \qquad (82) $$

I have not managed to find a Quebecer number in a given base $b$. In the appendix, we will explore possible Quebecer numbers. Given this difficulty, we define a final class of integers in base $b$ in the next section: Sicilian numbers.

## 3.6 The Case of the Sicilians in Base $b$

In this subsection, we define Sicilian numbers in base $b$, which are numbers for which we "still" cannot or "will never" be able to say whether they are Norman, celt, or Quebecer in a given base $b$.

### Definition 3.6.1

Let $n \in \mathbb{N}$.

(i) We say that $n$ is **Sicilian in base $b$ of order $N \in \mathbb{N}$** if the study of the sequence $\Gamma_{b,N}(n)$ does not allow us to say whether it is Norman or celt in base $b$. We denote $\mathcal{S}_b^N$ as the set of Sicilian integers in base $b$ of order $N$.

(ii) We say that $n$ is **Sicilian in base $b$** if the problem of determining whether $n \in \mathcal{N}_b$ or $n \in \mathcal{C}_b$ or $n \in \mathcal{Q}_b$ is undecidable. We denote $\mathcal{S}_b^\dagger$ as the set of Sicilian integers in base $b$.

As stated at the end of the previous subsection. I did not find any Quebec numbers in a given base $b$. Through a computational study, by looking at the sequence $(d_b(\gamma_b^k(n)))_{k \in \llbracket 0, N \rrbracket}$ associated with an integer $n \in \mathbb{N}$ for a sufficiently large integer $N$ (on the order of $10^5$), Sicilian numbers of order $N$ that could be potential Quebec numbers are found. But nothing prevents such numbers from having a sequence $(d_b(\gamma_b^k(n)))_{k \in \llbracket 0, N \rrbracket}$ that is visually increasing and then decreasing from a very large integer $N'$ before $N$ towards $\bullet$ or a triskelion.

I am even more uncertain whether Sicilian numbers exist in base $b$, i.e., numbers for which the convergence of their Norman sequences is undecidable.

## 3.7 Artifaille in Python

♣ The program `gamma_fct(N)` takes an integer $N$ of type `IntB` written in base $b$. It returns $\gamma_b(N)$ of type `IntB` in the same base $b$.

### Python 3.7.1

```python
def gamma_fct(N: IntB) -> IntB:
    if est_red(N):
        return rho_fct((N[0][1: -1], N[1]))
    else:
        return pi_fct(N)
```
### Example 3.7.2

```python
>>> gamma_fct([1, 3, 2], 5)
([3, 1, 4], 5)
```

```python
>>> gamma_fct([4, 3, 1, 3, 3, 4], 5)
([1, 3], 5)
```

```python
>>> gamma_fct((), 5)
((), 5)
```

♣ The program `suite_normande(N, VOL, ID, LIST, INF, INF_DEG)` takes an integer `N` of type `IntB` written in base `b`, three boolean parameters `VOL`, `ID`, and `LIST` defaulting to `False`, a triplet of integers `INF` defaulting to `(10 ** 3, 0, 10)`, and a triplet of integers `INF_DEG` defaulting to `(1, 10 ** 3, 100)`:

★ `N` = `(B, b)`: the integer of type `IntB` for which we wish to compute the Norman sequence in base `b`;

★ `VOL`: this parameter is used for the Norman numbers in base `b`. If it is `False`, the program will stop if there exists `k ∈ N` such that `γ_b^k(N) = []`, and if it is `True`, the program will stop if there exists `k ∈ N` such that `d_b(γ_b^k(N)) < d_b(N)`.

★ `ID`: this parameter is used to detail the identity of `N`. If it is `False`, the program will simply return the identity of `N`, and if it is `True`, it will also return a list of `0` and `1` (we put an `0` when we used the application `ρ_b` and `1` for the application `π_b`);

★ `LIST`: for displaying the details of the Norman sequence as a list of decompositions without specifying the base;

★ `INF`: it is a triplet `(a, b, c)` for Sicilian numbers, where `a` is the length of the computed Norman sequence, and `(b, c)` is a slice for displaying the list if `LIST = True`;

★ `INF_DEG`: it is a slice of the type `(a, b, c)` for displaying the sequence of degrees of the Norman sequence associated with `N` from `a` to `b-1` with a step of `c`.

It returns a tuple of type `Tuple[Union=bool, float, Dict]` containing the Norman sequence associated with `N` with all these details depending on the values of the three booleans (see the examples for more details).

### Python 3.7.3

```python
def suite_normande(N, VOL=False, ID=False, LIST=False,
                    INF=(10 ** 4, 0, 10),
                    INF_DEG = (1, 10 ** 4, 1000)) -> Tuple[Unionbool, float, Dict]:
    B, b = N
    S = [B]
    t_ID = []
    D = {'b': b}
    for i in range(1, INF[0]):
        if est_red((B, b)):
            B, _ = rho_fct((B[1: -1], b))
            t_ID.append(0)
        else:
            B, _ = pi_fct((B, b))
```
```python
t_ID.append(1)
if B in S:
    l = [S.index(B), i - S.index(B)]
    v_ID, c_ID = t_ID[: S.index(B)], t_ID[S.index(B):]
    D['id'] = {'V': (1[0], (v_ID.count(0), v_ID.count(1))) + ID * (v_ID,)}
    D['id'][ 'C'] = (1[1], (c_ID.count(0), c_ID.count(1))) + ID * (c_ID,)
    D['min_c'] = min(S[S.index(B):], key=lambda x: base_to_int((x, b)))
    if LIST:
        D.update({'V': S[: S.index(B)], 'C': S[S.index(B): ]})
    return (False, D)
elif (VOL and B == []) or (not VOL and len(B) < len(N[0])):
    S.append[])
    D['id'] = (i, ((t_ID.count(0), t_ID.count(1)))) + ID * (t_ID,)
    if LIST:
        D.update({ 'S': S})
    return (True, D)
else:
    S.append(B)
    D['id'] = (i, (t_ID.count(0), t_ID.count(1)))
    if LIST:
        D.update({ 'V': S[INF[1]: INF[2]], 'inf': INF})
    if INF_DEG:
        D['deg'] = (INF_DEG, [len(x) for x in S[slice(*INF_DEG)]])
    return (ma.inf, D)
```

### Example 3.7.4

(1) Take the Norman number in base 40:

$$ (0 \mid 22 \mid 36)_{2,40}. $$

We have:

```python
>>> suite_normande([36, 22, 0], 40))
(True, {'b': 40, 'id': (6, (1, 5))})

>>> suite_normande([36, 22, 0], 40), False, True, True)
(True, {'b': 24, 'id': (7, (1, 6), [1, 1, 1, 1, 1, 1, 0]), 'S': [[8, 23, 2],
[10, 22, 11], [21, 20, 22], [19, 17, 20, 1], [20, 13, 14, 21],
[17, 4, 4, 18, 1], [18, 22, 8, 22, 18], []}])
```

(2) Take the celt number in base 40:

$$ (1 \mid 22 \mid 36)_{2,40}. $$

We have:

```python
>>> suite_normande([36, 22, 1], 40))
(False, {'b': 40, 'id': {'V': (3, (0, 3)), 'C': (5, (1, 4))},
'min_c': [36, 4, 5, 37]})

>>> suite_normande([36, 22, 1], 40), False, True, True)
(False, {'b': 40, 'id': {'V': (3, (0, 3), [1, 1, 1]), 'C': (5, (1, 4),
[1, 1, 1, 1, 0])}, 'min_c': [36, 4, 5, 37], 'V': [[36, 22, 1], [37, 4, 38],
[35, 9, 35, 1]], 'C': [[36, 4, 5, 37], [33, 10, 9, 33, 1], [34, 3, 19, 3, 35],
[29, 7, 38, 6, 29, 1], [30, 36, 4, 5, 37, 30]]})
```
(3) Take the celt number in base 24:

$$ (3 | 23 | 8)_{2,24}. $$

We have:

```h
>>> suite_normande([8, 23, 3], 24), True, True, True)
(False, {'b': 24, 'id': {'V': (5, (1, 4), [1, 1, 1, 1, 0]), 'C': (9, (1, 8),
[1, 1, 1, 1, 1, 1, 1, 0])}, 'min_c': [21, 16, 22], 'V': [[8, 23, 3],
[11, 22, 12], [23, 20, 0, 1], [0, 21, 20, 0, 1], [1, 21, 16, 22, 1]]},
'C': [[21, 16, 22], [19, 9, 20, 1], [20, 5, 6, 21], [17, 12, 11, 17, 1],
[18, 5, 23, 5, 19], [13, 11, 22, 11, 13, 1], [14, 0, 10, 10, 1, 15],
[5, 2, 20, 20, 1, 5, 1], [6, 7, 21, 16, 22, 7, 6]])
```

(4) Take the Sicilian number in base 66:

$$ (64 | 34 | 0)_{2,66}. $$

We have:

```h
>>> suite_normande([0, 34, 64], 66)
(inf, {'b': 66, 'id': (999, (0, 999)), 'deg': ((1, 1000, 100),
[3, 53, 103, 153, 203, 253, 303, 353, 403, 453])})

>>> suite_normande([0, 34, 64], 66), True, True, True)
(inf, {'b': 66, 'id': (999, (0, 999)), 'V': [[0, 34, 64], [64, 2, 65],
[63, 5, 63, 1], [64, 2, 3, 65], [63, 6, 5, 63, 1], [64, 3, 11, 3, 65],
[63, 7, 22, 6, 63, 1], [64, 4, 29, 28, 4, 65], [63, 9, 57, 57, 8, 63, 1],
[64, 6, 0, 49, 0, 7, 65]], 'inf': (1000, 0, 10), 'deg': ((1, 1000, 100),
[3, 53, 103, 153, 203, 253, 303, 353, 403, 453]))
```

```h
>>> suite_normande([0, 34, 64], 66), True, True, True, (10 ** 3, 0, 10),
(1, 10 ** 3, 100))
(inf, {'b': 66, 'id': (999, (0, 999)), 'V': [[0, 34, 64], [64, 2, 65],
[63, 5, 63, 1], [64, 2, 3, 65], [63, 6, 5, 63, 1], [64, 3, 11, 3, 65],
[63, 7, 22, 6, 63, 1], [64, 4, 29, 28, 4, 65], [63, 9, 57, 57, 8, 63, 1],
[64, 6, 0, 49, 0, 7, 65]], 'inf': (1000, 0, 10), 'deg': ((1, 1000, 100),
[3, 53, 103, 153, 203, 253, 303, 353, 403, 453]))
```
# 4 Group Strategies

The integers to verify if they are Norman, celt, or Québécois in a certain base have started to perform more advanced mathematics. They realize that they can group themselves into small groups, and these same small groups form larger groups. The idea is that if an element of the group is Norman in a certain base, all the others will automatically be.

## 4.1 Depth of the Dream

In the Syracuse conjecture, the flight time of an integer is defined as the first time when the Syracuse sequence associated with that integer has an element strictly smaller than the starting integer. Here we can define the equivalent of the "flight time" associated with an integer. There are then two possibilities: either as the first moment when its Norman sequence has an integer strictly smaller than the starting integer, or as the first moment when its Norman sequence has an integer of strictly smaller degree than the starting integer.

**Definition 4.1.1**

Let $n \in \mathbb{N}$.

(i) (a) The **journey in base $b$ of $n$** is the set:

$$ \mathcal{V}_b(n) := \{k \in \mathbb{N}, \gamma_b^k(n) \geq n\} \subset \mathbb{N}. $$

(b) We say that $n$ has a **finite journey in base $b$** if $\mathcal{V}_b(n) \neq \mathbb{N}$, i.e., there exists $k \in \mathbb{N}_1$ such that:

$$ \gamma_b^k(n) < n. $$

We then denote:

$$ \omega_b(n) := \min\{k \in \mathbb{N}_1, \gamma_b^k(n) < n\} $$

and $\mathcal{V}_b$ is the set of integers having a finite journey in base $b$.

(ii) (a) The **degree journey in base $b$ of $n$** is the set:

$$ \mathcal{V}_b^{\dagger}(n) := \{k \in \mathbb{N}, d_b(\gamma_b^k(n)) \geq d_b(n)\} \subset \mathbb{N}. $$

(b) We say that $n$ has a **finite degree journey in base $b$** if $\mathcal{V}_b^{\dagger}(n) \neq \mathbb{N}$, i.e., there exists $k \in \mathbb{N}_1$ such that:

$$ d_b(\gamma_b^k(n)) < d_b(n). $$

We then denote:

$$ \omega_b^{\dagger}(n) := \min\{k \in \mathbb{N}_1, d_b(\gamma_b^k(n)) < d_b(n)\} $$

and $\mathcal{V}_b^{\dagger}$ is the set of integers having a finite degree journey in base $b$.

## 4.2 Elingard Operation

The concept of elingard<fnref n="19" /> for non-reducible numbers will be used for the following subsection.

<footnote n="19">¹⁹Slender, thin, and large.</footnote>
Definition 4.2.1

Let $n, m \in \mathcal{R}_b^0$.

(i) We say that $n$ is the **elingard in base $b$ associated with $m$** if there exists $k \in \mathbb{N}$ such that $n = \pi_b^k(m)$ and if:

$$ 
\begin{align*}
d_b(n) &= d_b(m) \\
d_b(\pi_b(n)) &= d_b(m) + 1 \\
\forall l \in [1, k+1], \pi_b^l(m) &\in \mathcal{R}_b^0
\end{align*}
$$

The integer $k$ satisfying these relations is called the **degraded version of $m$ in base $b$**, denoted $dg_b(m)$.

We define:

$$ \varepsilon_b(m) := n. $$

We then denote $\mathcal{E}_b$ as the set of integers that have an elingard in base $b$.

(ii) We say that $n$ is **elingard in base $b$** if $dg_b(n) = 0$, i.e., $d_b(\pi_b(n)) = d_b(n) + 1$. We then denote $\mathcal{E}_b^\dagger$ as the set of elingards in base $b$.

In particular, for every $m \in \mathcal{E}$:

$$ \varepsilon_b(m) \in \mathcal{R}_b^0. \qquad (83) $$

The condition $\pi_b^{k+1}(m) \in \mathcal{R}_b^0$, i.e., $\pi_b(n) \in \mathcal{R}_b^0$ in point (i), allows us to exclude numbers of the type:

$$ n := (b-1 | \cdots | 2)_{d,b} \qquad (84) $$

with $b \neq 3$ and $d \in \mathbb{N}_2^{20}$. Since $\pi_b(n) = (1 | \cdots | 1)_{d+1,b}$, $\pi_b(n)$ is reducible.

Therefore, for all $n \in \mathcal{E}_b$:

$$ dg_b(n) = \max \{k \in \mathbb{N}, d_b(\pi_b^k(n)) = d_b(n) \wedge d_b(\pi_b^{k+1}(n)) = d_b(n) + 1 \wedge \forall l \in [0, k+1], \pi_b^l(n) \notin \mathcal{R}_b \}. \qquad (85) $$

And for all $n \in \mathcal{E}_b^\dagger$:

$$ \varepsilon_b(n) = n. \qquad (86) $$

Example 4.2.2

(1) Take $n := 172 \in \mathcal{R}_{10}^0$. We have:

$$ \pi_{10}^2((1 | 7 | 2)_{2,10}) = \pi_b((4 | 4 | 3)_{2,10}) = (7 | 8 | 7)_{2,10} \in \mathcal{R}_{10} $$

i.e., $n \in \mathcal{R}_{10}^0 - \mathcal{E}_{10}$.

(2) Take $n := 192 \in \mathcal{R}_{10}^0$. We have:

$$ \pi_{10}^2((1 | 9 | 2)_{2,10}) = \pi_{10}((4 | 8 | 3)_{2,10}) = (8 | 6 | 7)_{2,10} $$

$$ \pi_{10}((8 | 6 | 7)_{2,10}) = (1 | 6 | 3 | 5)_{3,10} $$

i.e., $n \in \mathcal{E}_{10}$, so 867 is the elingard of $n$ in base 10 (thus, $867 \in \mathcal{E}_{10}^\dagger$) and:

$$ dg_{10}(n) = 2. $$

(3) Let $n := (2 | 2^k - 1 | 1)_{2,2^k}$ with $k \in \mathbb{N}_3$. We have:

$$ \pi_{2^k}(n) = (4 | 2^k - 2 | 3)_{2,2^k} = (2^{1+1} | 2^k - 2^1 | 2^{1+1} - 1)_{2,2^k} $$

<footnote>$^{20}$We excluded the case $b := 3$ because in this case $n := (2 | \cdots | 2)_{d,3}$ is not an elingard in base 3 because it is reducible in base 3.</footnote>
$$ \pi_{2k}^2(n) = (8 | 2^k - 4 | 7)_{2,2k} = (2^{2+1} | 2^k - 2^2 | 2^{2+1} - 1)_{2,2k} $$

Thus, by simple finite recurrence, we have for all $l \in [0, k-2]$:

$$ \pi_{2k}^l(n) = (2^{l+1} | 2^k - 2^l | 2^{l+1} - 1)_{2,2k} $$

In particular, we have:

$$ \pi_{2k}^{k-2}(n) = (2^{k-1} | 3.2^{k-2} | 2^{k-1} - 1)_{2,2k} $$

Thus, we have:

$$ \pi_{2k}^{k-1}(n) = (1 | 0 | 2^{k-1} | 2^k - 1)_{3,2k} \in \mathcal{R}_{2k}, \quad \mathrm{dg}_{2k}(n) = k-2. $$

### Definition 4.2.3

The elingard operation in base $b$ is the application $\varepsilon_b$ that associates to each $n \in \mathcal{E}_b$ the value $\varepsilon_b(n)$.

We have the following property.

### Proposition 4.2.4

Let $n \in \mathcal{E}_b$ of degree $d$ and $k \in \mathbb{N}_1$. Let $u := [n]_0 + [n]_d$. We have the equivalence between:

$$ \mathrm{dg}_b(n) = k \iff 2^{k-1}(u+1) + 1 \le b \le 2^k(u+1) + q_b(\sigma(\pi_b^k(n), \mu_b(\pi_b^k(n)))_{d,b}) - 1. $$

*Proof.* (i) Suppose $\mathrm{dg}_b(n) = k$. We will prove by finite induction on $l \in [1, k]$:

$$ (\mathcal{P}_l) : \pi_b^l(n) = (2^{l-1}(u+1) | \cdots | 2^{l-1}(u+1) - 1)_{d,b}. $$

(I) Since $\pi_b^1(n) \in \mathcal{R}_b^0$, we necessarily have $\pi_b^1(n) = (u+1 | \cdots | u)_{d,b}$ (note that $u+1 < b$ because otherwise $\mathrm{dg}_b(n) = k = 0$, and $[\pi_b^1(n)]_{d,b} = u+1$ because if $[\pi_b^1(n)]_{d,b} = u$ then $u \in \mathcal{R}_b$). Therefore, $(\mathcal{P}_1)$ is true.

(II) Let $l \in [1, k-1]$. Suppose $(\mathcal{P}_l)$ is true. Then we have:

$$ \begin{aligned} \pi_b^{l+1}(n) &= \pi_b((2^{l-1}(u+1) | \cdots | 2^{l-1}(u+1) - 1)_{d,b}) \\ &= (2^{l-1}(u+1) + 2^{l-1}(u+1) - 1 + 1) | \cdots | 2^{l-1}(u+1) + 2^{l-1}(u+1) - 1)_{d,b} \\ &= (2^l(u+1) | \cdots | 2^l(u+1) - 1)_{d,b} \end{aligned} $$

Thus $(\mathcal{P}_{l+1})$ is true.

(III) By finite induction, $(\mathcal{P}_l)$ is true for all $l \in [1, k]$.

Thus we particularly have:

$$ \pi_b^k(n) = (2^{k-1}(u+1) | \cdots | 2^{k-1}(u+1) - 1)_{d,b} $$

and hence $2^{k-1}(u+1) \le b-1$ i.e.:

$$ b \ge 2^{k-1}(u+1) + 1. $$

Since $\mathrm{dg}_b(n) = k$, we must have:

$$ 2^{k-1}(u+1) + 2^{k-1}(u+1) - 1 + q_b(\sigma(\pi_b^k(n), \mu_b(\pi_b^k(n)))_{d,b}) \ge b $$

i.e.:

$$ b \le 2^k(u+1) + q_b(\sigma(\pi_b^k(n), \mu_b(\pi_b^k(n)))_{d,b}) - 1. $$
(ii) Suppose $2^{k-1}(u+1) + 1 \le b \le 2^k(u+1) + q_b(\sigma(\pi_b^k(n), \mu_b(\pi_b^k(n))_{d,b}) - 1$. By the reasoning above and since $2^{k-1}(u+1) \le b-1$, we have:

$$ \pi_b^k(n) = (2^{k-1}(u+1) | \dots | 2^{k-1}(u+1) - 1)_{d,b} $$

and thus $dg_b(n) \ge k$. Since $2^k(u+1) + q_b(\sigma(\pi_b^k(n), \mu_b(\pi_b^k(n))_{d,b}) - 1 \ge b$, we have:

$$ \pi_b^{k+1}(n) = (1 | r_b(2^k(u+1) + q_b(\sigma(\pi_b^k(n), \mu_b(\pi_b^k(n))_{d,b}) - 1) | \dots | r_b(2^k(u+1) - 1))_{d+1,b} $$

Thus $dg_b(n) = k$.

□

## 4.3 Duc and Duc Elingard operations

The duc associated with an irreducible integer in base $b$ can be seen as the integer having the same pseudo-palindrome in base $b$ but with the left digits having the largest possible values relative to their mirror digits on the right. The core of this definition is point (ii) of proposition 2.3.3.

### Definition 4.3.1

Let $n \in \mathbb{N}_1$ of degree $d := d_b(n) = 2d' - \delta$ with $\delta \in \{0, 1\}$ and $d \in \mathbb{N}$. The duc associated with $n$ in base $b$ is the integer $\kappa_b(n)$ of degree less than $d$ defined for all $k \in [0, d' - 1]$ by:

$$ \begin{aligned} [\kappa_b(n)]_{d-k,b} &:= \min\{[n]_{k,b} + [n]_{d-k,b}, b-1\} \\ [\kappa_b(n)]_{k,b} &:= \max\{[n]_{k,b} + [n]_{d-k,b} - b + 1, 0\} \end{aligned} $$

and if $\delta = 0$:

$$ [\kappa_b(n)]_{d',b} := [n]_{d',b}. $$

We have for all $n \in \mathbb{N}$:

$$ \kappa_b(n) = n \iff n \in \mathcal{B}_b. \qquad (87) $$

In particular:

$$ \kappa_b(0) = 0. \qquad (88) $$

Let us begin with a lemma that justifies the notation.

### Lemma 4.3.2

Let $n \in \mathbb{N}$.

(i) Suppose that $[n]_{0,b} + [n]_{d(n),b} \ll [1, 2b - 3]$ and $d_b(n) \in \mathbb{N}_1$. Then $\kappa_b(n) \in \mathcal{R}_b^0$. In particular, if $n \in \mathcal{R}_b^0$ then $\kappa_b(n) \in \mathcal{R}_b^0$.

(ii) We have:

$$ d_b(\kappa_b(n)) = d_b(n). $$

(iii) We have:

$$ \pi_b(\kappa_b(n)) = \pi_b(n). $$

*Proof.* Let $d := d_b(n) = 2d' - \delta$ with $\delta \in \{0, 1\}$.

(i) We prove that $[\kappa_b(n)]_{d,b} \neq [\kappa_b(n)]_{0,b}$. Since:

$$ [\kappa_b(n)]_{0,b} = \max\{[n]_{0,b} + [n]_{d,b} - b + 1, 0\} \le \max\{2b - 3 - b + 1, 0\} = b - 2 $$

we have:

$$ [\kappa_b(n)]_{d,b} = \min\{[n]_{0,b} + [n]_{d,b}, b-1\} \ge b-1 > [\kappa_b(n)]_{0,b}. $$

Thus the result follows.
(ii) By definition, we have $d_b(\kappa_b(n)) \le d_b(n)$. There are two cases.

(1) **Case $n=0$.** We have $\kappa_b(0) = \kappa_b(0) = 0$ (thus they have the same degree).

(2) **Case $n \ne 0$.** Since $n > 0$, we have $[n]_d > 0$ and thus:

$$[\kappa_b(n)]_d = \min\{[n]_{0,b} + [n]_{d,b}, b-1\} \ge [n]_{0,b} + [n]_{d,b} \ge [n]_{d,b} > 0.$$

Thus the result also holds in this case.

(iii) As for all $x, y \in \mathbb{N}$:

$$\min\{x, y\} = \frac{x+y}{2} - \frac{|x-y|}{2}, \quad \max\{x, y\} = \frac{x+y}{2} + \frac{|x-y|}{2}$$

we have for all $k \in [0, d' - 1]$:

$$\begin{align*}
& [\kappa_b(n)]_{k,b} + [\kappa_b(n)]_{d-k,b} \\
&= \max\{[n]_{k,b} + [n]_{d-k,b} - b + 1, 0\} + \min\{[n]_{k,b} + [n]_{d-k,b}, b-1\} \\
&= \frac{[n]_{k,b} + [n]_{d-k,b} - b + 1}{2} + \frac{|[n]_{k,b} + [n]_{d-k,b} - b + 1|}{2} + \frac{[n]_{k,b} + [n]_{d-k,b} + b - 1}{2} - \frac{|[n]_{k,b} + [n]_{d-k,b} - b + 1|}{2} \\
&= [n]_{k,b} + [n]_{d-k,b}
\end{align*}$$

□

The converse of point (i) is false because for example:

$$(1 | 1)_{1,3} \in \mathcal{R}_3 \wedge \pi_3((1 | 1)_{1,3}) = (2 | 0)_{1,3} \in \mathcal{R}_3^0. \qquad (89)$$

In particular, we have for all:

$$[\pi_b, \mathbb{N}_\bullet] \triangleright \{\kappa_b(n)\} = [\pi_b, \mathbb{N}_\bullet] \triangleright \{n\}. \qquad (90)$$

Thus $\kappa_b(n)$ is a privileged representative of the class $[\pi_b, \mathbb{N}_\bullet] \triangleright \{n\}$ associated with $n$.

### Definition 4.3.3

The **duc operation in base $b$** is the application $\kappa_b$ that associates to $n \in \mathbb{N}$ the value $\kappa_b(n)$.

### Example 4.3.4

We have:

$$\kappa_6((3 \mid 3 \mid 4 \mid 4 \mid 2)_{4,6}) = (5 \mid 5 \mid 4 \mid 2 \mid 0)_{4,6}$$

$$\kappa_6((5 \mid 5 \mid 2 \mid 0)_{3,6}) = (5 \mid 5 \mid 2 \mid 0)_{3,6}$$

We now restrict ourselves to irreducible integers.

### Definition 4.3.5

Let $n \in \mathcal{R}_b^0$. We say that $n$ is a **duc in base $b$** if $\kappa_b(n) = n$. We denote by $\mathcal{D}_b$ the set of ducs in base $b$.

The map $\kappa_b$ induces a map from $\mathcal{R}_b^0$ to $\mathcal{D}_b$.

We define two auxiliary sets:

$$\mathcal{A}_b := \{(\min\{j, b-1\}, \max\{0, j-b+1\}), j \in [0, 2b-2]\} \qquad (91)$$
$$ \mathcal{A}'_b := \{(\min\{j, b-1\}, \max\{0, j-b+1\}), j \in [0, 2b-3]\} \quad (92) $$

We then have for all $d \in \mathbb{N}_1$ (this depends on the parity of $d$):

$$ \mathcal{D}_{(0),b} = \emptyset \quad (93) $$

$$ \mathcal{D}_{(2d),b} = \{(a_{2d} | \cdots | a_0)_{2d,b}, a_d \in \mathcal{B}_b \wedge (a_0, a_{2d}) \in \mathcal{A}'_b \wedge \forall i \in [1, d-1], (a_i, a_{2d-i}) \in \mathcal{A}_b\} \quad (94) $$

$$ \mathcal{D}_{(2d-1),b} = \{(a_{2d-1} | \cdots | a_0)_{2d-1,b}, (a_0, a_{2d-1}) \in \mathcal{A}'_b \wedge \forall i \in [1, d-1], (a_i, a_{2d-1-i}) \in \mathcal{A}_b\} \quad (95) $$

We can then calculate the cardinalities of $\mathcal{D}_{(d),b}$ by counting.

### Proposition 4.3.6

Let $d \in \mathbb{N}_1$. We have:

$$ \mathrm{card}(\mathcal{D}_{(d),b}) = 2(b-1)b^{r_2(d+1)}(2b-1)^{q_2(d)-1}. $$

*Proof.* First, note that:

$$ \mathrm{card}(\mathcal{A}_b) = 2b-1 $$

$$ \mathrm{card}(\mathcal{A}'_b) = 2b-2. $$

We have two cases.

(1) Case $d := 2d' (d' \in \mathbb{N}_1)$. We then have:

$$ \begin{aligned} \mathrm{card}(\mathcal{D}_{(d),b}) &= \mathrm{card}(\mathcal{B}_b) \mathrm{card}(\mathcal{A}'_b) \mathrm{card}(\mathcal{A}_b)^{d'-1} \\ &= 2(b-1)b(2b-1)^{d'-1} \\ &= 2(b-1)b^{r_2(d+1)}(2b-1)^{q_2(d)-1}. \end{aligned} $$

(2) Case $d := 2d' - 1 (d' \in \mathbb{N}_1)$. We then have:

$$ \begin{aligned} \mathrm{card}(\mathcal{D}_{(d),b}) &= \mathrm{card}(\mathcal{A}'_b) \mathrm{card}(\mathcal{A}_b)^{d'-1} \\ &= 2(b-1)(2b-1)^{d'-1} \\ &= 2(b-1)b^{r_2(d+1)}(2b-1)^{q_2(d)-1} \end{aligned} $$

□

### Example 4.3.7

From example 4.9.31, we have:

$$ \mathrm{card}(\mathcal{D}_{4,10}) = 3420 = 2 \times 9 \times 10^1 \times 19^1. $$

### Definition 4.3.8

(i) Let $n \in \mathcal{R}_b^0$. We say that $n$ is a **duc elingard in base $b$** if $n$ is elingard in base $b$ and a **duc in base $b$**. We denote by $\mathcal{D}_b^\dagger$ the set of duc elingards in base $b$.

(ii) The **duc elingard operation in base $b$** is the map $\kappa_b^\dagger$ defined for all $n \in \mathcal{E}_b$ by:

$$ \kappa_b^\dagger(n) := \kappa_b(\varepsilon_b(n)(n)). $$

Thus we have:

$$ \mathcal{D}_b^\dagger = \mathcal{D}_b \cap \mathcal{E}_b^\dagger. \quad (96) $$

Therefore, the map $\kappa_b^\dagger$ is a map from $\mathcal{E}_b$ to $\mathcal{D}_b^\dagger$, and the map $\kappa_b$ induces a map from $\mathcal{E}_b$ to $\mathcal{D}_b^\dagger$.
## 4.4 Set of quasi-elingard ducs

I have not found a simple criterion for the duc elingards in base $b$ that would allow listing them quickly in computing. This subsection is mainly for programming purposes. We will define a set containing the duc elingards with very few additional elements. This is the goal of this subsection. Let's begin with an example.

### Example 4.4.1

In example 4.9.31, we found the following cardinalities:

$$ \mathrm{card}(\mathcal{B}_{4,10}) = 90\,000 $$

$$ \mathrm{card}(\mathcal{R}_{4,10}^0) = 81\,000 $$

$$ \mathrm{card}(\mathcal{E}_{4,10}^\dagger) = 36\,500 $$

$$ \mathrm{card}(\mathcal{D}_{4,10}) = 3\,450 $$

$$ \mathrm{card}(\mathcal{D}_{4,10}^\dagger) = 1\,425 $$

The cardinality of the quasi-elingard ducs of degree 4 in base 10 is

$$ \mathrm{card}(\tilde{\mathcal{D}}_{4,10}) = 1\,520. $$

We notice that it is slightly higher than that of the duc elingards. Hence the interest of this new set.

We now define the set of quasi-elingard ducs.

### Definition 4.4.2

We define for all $d \in \mathbb{N}$, the set $\tilde{\mathcal{D}}_{(d),b}$ of quasi-elingard dukes of degree $d$ in base $b$. To do this, we define two auxiliary sets:

$$ \begin{aligned} \mathcal{A}_b &= \{ (\min\{j, b-1\}, \max\{0, j-b+1\}), j \in \llbracket 0, 2b-2 \rrbracket \} \\ \mathcal{A}_b'' &= \{ (\min\{j, b-1\}, \max\{0, j-b+1\}), j \in \llbracket b-1, 2b-3 \rrbracket \wedge j-b+1 \neq 2 \} \end{aligned} $$

Then, for every $d \in \mathbb{N}$ (depending on the parity of $d$):

$$ \begin{aligned} \tilde{\mathcal{D}}_{(0),b} &:= \emptyset \\ \tilde{\mathcal{D}}_{(2d),b} &:= \{ (a_{2d} | \cdots | a_0)_{2d,b}, a_d \in \mathcal{B}_b \wedge (a_0, a_{2d}) \in \mathcal{A}_b'' \wedge \forall i \in \llbracket 1, d-1 \rrbracket, (a_i, a_{2d-i}) \in \mathcal{A}_b \} \\ \tilde{\mathcal{D}}_{(2d-1),b} &:= \{ (a_{2d-1} | \cdots | a_0)_{2d-1,b}, (a_0, a_{2d-1}) \in \mathcal{A}_b'' \wedge \forall i \in \llbracket 1, d-1 \rrbracket, (a_i, a_{2d-i}) \in \mathcal{A}_b \} \end{aligned} $$

Another advantage of this set is that its cardinality is easy to calculate. By using the proof of proposition 4.3.6, we have for all $d \in \mathbb{N}_1$:

$$ \mathrm{card}(\tilde{\mathcal{D}}_{(d),b}) = (b-2)b^{r_2(d+1)}(2b-1)^{q_2(d)-1}. \qquad (97) $$

We also see that the elements $n$ of $\mathcal{D}_{(d),b}$ that were removed to construct $\tilde{\mathcal{D}}_{(d),b}$ are not elingard in base $b$ because they are such that $[n]_{0,b} + [n]_{d,b} < b-1$, and thus $d_b(\pi_b(n)) = d_b(n)$. Therefore, we have:

$$ \mathcal{D}_{(d),b}^\dagger \subset \tilde{\mathcal{D}}_{(d),b} \subset \mathcal{D}_{(d),b}. \qquad (98) $$

Thus, in particular, we have:

$$ \mathcal{D}_{(d),b}^\dagger = \mathcal{D}_{(d),b} \cap \mathcal{E}_{(d),b}^\dagger = \tilde{\mathcal{D}}_{(d),b} \cap \mathcal{E}_{(d),b}^\dagger. \qquad (99) $$
### Example 4.4.3

(1) From example 4.9.31, we have:

$$ \mathrm{card}(\tilde{\mathcal{D}}_{(4),10}) = 1520 = 8 \times 10^1 \times 19^1. $$

(2) We have:

$$ \tilde{\mathcal{D}}_{(1),b} = \{(a \mid a')_{1,b}, (a,a') \in \mathcal{A}'_b\} $$

$$ \tilde{\mathcal{D}}_{(2),b} = \{(a \mid a' \mid a'')_{2,b}, (a,a'') \in \mathcal{A}'_b, a' \in \mathcal{B}_d\} $$

(3) For $b := 5$, we have:

$$ \mathcal{A}_5 = \{(\min\{j, 4\}, \max\{0, j-4\}), j \in [0, 7]\} $$

$$ = \{(0, 0), (1, 0), (2, 0), (3, 0), (4, 0), (4, 1), (4, 2), (4, 3)\} $$

$$ \mathcal{A}'_5 = \{(\min\{j, 4\}, \max\{0, j-4\}), j \in [4, 8] \wedge j \neq 6\} $$

$$ = \{(4, 0), (4, 1), (4, 3)\} $$

Thus, we have:

$$ \tilde{\mathcal{D}}_{(1),5} = \{(4 \mid 0)_{1,b}, (4 \mid 1)_{1,b}, (4 \mid 3)_{1,b}, \} $$

$$ \tilde{\mathcal{D}}_{(2),5} = \{(4 \mid 0 \mid 0)_{2,b}, (4 \mid 1 \mid 0)_{2,b}, (4 \mid 2 \mid 0)_{2,b}, (4 \mid 3 \mid 0)_{2,b}, (4 \mid 4 \mid 0)_{2,b}, \\ (4 \mid 0 \mid 1)_{2,b}, (4 \mid 1 \mid 1)_{2,b}, (4 \mid 2 \mid 1)_{2,b}, (4 \mid 3 \mid 1)_{2,b}, (4 \mid 4 \mid 1)_{2,b}, \\ (4 \mid 0 \mid 3)_{2,b}, (4 \mid 1 \mid 3)_{2,b}, (4 \mid 2 \mid 3)_{2,b}, (4 \mid 3 \mid 3)_{2,b}, (4 \mid 4 \mid 3)_{2,b}\} $$

Since $\mathcal{D}^\dagger_{(2),5}$ is a subset of $\tilde{\mathcal{D}}_{(2),5}$, we just need to check those that are elingard in base 5. We have:

$$ \mathcal{D}^\dagger_{(2),5} = \{(4 \mid 3 \mid 0)_{2,b}(4 \mid 4 \mid 0)_{2,b}, (4 \mid 0 \mid 1)_{2,b}, (4 \mid 1 \mid 1)_{2,b}, (4 \mid 2 \mid 1)_{2,b}, \\ (4 \mid 3 \mid 1)_{2,b}, (4 \mid 4 \mid 1)_{2,b}, (4 \mid 0 \mid 3)_{2,b}, (4 \mid 1 \mid 3)_{2,b}, (4 \mid 2 \mid 3)_{2,b}, \\ (4 \mid 3 \mid 3)_{2,b}, (4 \mid 4 \mid 3)_{2,b}\} $$

## 4.5 Riboutine operations

The elingard duke associated with a non-reducible integer in base $b$ is the largest in a sense we defined in the previous subsection, but it is not as large as we would like: several elingard dukes can have the same value by $\gamma_b^k$ with the degree increasing by at most 1. The *riboutin*<fnref n="21" /> is better. Let's start with an example.

### Example 4.5.1

Let's consider the elingard integers in base 4:

$$ m := (3 \mid 2 \mid 0 \mid 1)_{3,4} = 225, \quad n := (3 \mid 3 \mid 2 \mid 1)_{3,4} = 245. $$

We have:

$$ \gamma_4(m) = \pi_4(m) = (1 \mid 0 \mid 2 \mid 3 \mid 0)_{4,4} $$

$$ \gamma_4(m) = \pi_4(n) = (1 \mid 1 \mid 2 \mid 2 \mid 0)_{4,4} $$

<footnote n="21">²¹Roitelet, a small passerine bird from our Norman countryside with a yellow head and an orange crest, the word likely comes from the Latin word *rex* (king).</footnote>
$$ \gamma_4^2(m) = \pi_4^2(m) = \pi_4^2(n) = (2 \mid 0 \mid 0 \mid 3 \mid 1)_{4,4} $$

Thus, we see that $\gamma_4^2(m) = \gamma_4^2(n)$. A priori, if everything goes well for $m$, it will go well for $n$.

We then deduce a subset of the elingards in base $b$: the riboutins in base $b$.

### Definition 4.5.2

Let $n \in \mathcal{D}_{(d),b}^{\dagger}$ (thus $d \in \mathbb{N}_1$) and $k \in \mathbb{N}$.

(i) The **tin of order $k$ in base $b$ of $n$** is:

$$ t_b^k(n) := \max \{l \in \mathbb{N}, d_b(\pi_b^l(n)) \le d_b(n) + k\}. $$

(ii) We say that $n$ is a **boutin of order $k$ in base $b$ or is boutin of order $k$ in base $b$** if:

$$ \forall l \in [1, t_b^k(n)], \pi_b^l(n) \in \mathcal{R}_b^0. $$

We denote by $\tilde{\mathfrak{R}}_{(d),b}^k$ the set of boutins of order $k$ and degree $d$ in base $b$.

(iii) Suppose that $n \in \tilde{\mathfrak{R}}_{(d),b}^k$.

(a) The **riboutin of order $k$ in base $b$ associated with $n$** is:

$$ \Psi_b^k(n) := \max \left\{ m \in \tilde{\mathfrak{R}}_{(d),b}^k, \pi_b^{t_b^k(m)}(m) = \pi_b^{t_b^k(n)}(n) \right\}. $$

(b) We say that $n$ is a **riboutin of order $k$ in base $b$ or is riboutin of order $k$ in base $b$** if $\Psi_b^k(n) = n$.

We denote by $\mathfrak{R}_{(d),b}^k$ the set of riboutins of order $k$ and degree $d$ in base $b$.

Note that in points (ii) and (iii), we can replace the function $\pi_b$ by $\gamma_b$, but not in point (i).
We define:

$$ \tilde{\mathfrak{R}}_b^k := \bigsqcup_{d \in \mathbb{N}_1} \tilde{\mathfrak{R}}_{(d),b}^k \qquad (100) $$

$$ \mathfrak{R}_b^k := \bigsqcup_{d \in \mathbb{N}_1} \mathfrak{R}_{(d),b}^k \qquad (101) $$

For all $n \in \mathcal{D}_{(d),b}^{\dagger}$, we have:

$$ t_b^0(n) = 0 \qquad (102) $$

and thus:

$$ \tilde{\mathfrak{R}}_{(d),b}^0 = \mathfrak{R}_{(d),b}^0 = \mathcal{D}_{(d),b}^{\dagger} \qquad (103) $$

We have the inclusion sequences for all $k \in \mathbb{N}$ and all $d \in \mathbb{N}_1$:

$$ \tilde{\mathfrak{R}}_{(d),b}^{k+1} \subset \tilde{\mathfrak{R}}_{(d),b}^k \cdots \subset \tilde{\mathfrak{R}}_{(d),b}^0 = \mathcal{D}_{(d),b}^{\dagger}. $$

We will see in our main result 4.6.1 that the case $k := 1$ is particularly interesting.
**Definition 4.5.3**

Let $k \in \mathbb{N}$. The **riboutin** operation of order $k$ in base $b$ is the application $\Psi_b^k$ which associates to $n \in \mathfrak{R}_b^k$ the value $\Psi_b^k(n)$.

**Example 4.5.4**

For $n := (4 | 4 | 0 | 2 | 3)_{4,5} \in \mathcal{D}_{(4),5}^{\dagger}$, we have:

$$ \pi_5(n) = (3 | 0 | 3 | 2 | 0 | 4)_{5,5} \in \mathfrak{R}_5^0 $$

$$ \pi_5^2(n) = (3 | 0 | 3 | 2 | 0 | 4)_{5,5} \in \mathfrak{R}_5^0 $$

and thus $t_5^2(n) = 2$. We can show that:

$$ \Psi_5^2(n) = \max \{m \in \tilde{\mathfrak{R}}_5^2, \pi_5^{t_5^2(m)}(m) = \pi_5^{t_5^2(n)}(n)\} $$

and thus find the largest riboutin of order 2 in base 5.

**4.6 Fundamental Result**

We have a result analogous to the one found in the Syracuse conjecture.

**Theorem 4.6.1**

Let $d \in \mathbb{N}_1$. Then the following equivalences hold:

(i) $\mathcal{B}_{d,b} \subset \mathcal{N}_b$;

(ii) $\mathcal{B}_{d,b} \subset \mathcal{Y}_b$;

(iii) $\mathcal{D}_{d,b} \subset \mathcal{Y}_b^{\dagger}$;

(iv) $\mathfrak{R}_{d,b}^0 = \mathcal{D}_{d,b}^{\dagger} \subset \mathcal{Y}_b^{\dagger}$.

(v) $\mathfrak{R}_{d,b}^1 \subset \mathcal{Y}_b^{\dagger}$.

*Proof.* We have the following inclusions:

(1) $\mathcal{N}_b - \{\bullet\} \subset \mathcal{Y}_b$

(2) $\Psi_{d,b}^1 \subset \mathcal{D}_{d,b}^{\dagger} \subset \mathcal{D}_{d,b}$

• (i) $\implies$ (ii). Assume point (i). By (1), we have:

$$ \mathcal{B}_{d,b} \subset \mathcal{N}_b - \{\bullet\} \subset \mathcal{Y}_b. $$

• (ii) $\implies$ (i). Assume point (ii). Let us prove by strong finite induction on $k \in \mathcal{B}_{d,b} = [0, b^d - 1]$ : $(\mathcal{P}_k) : k \in \mathcal{N}_b$.

(I) We have $\gamma_b((0)_b) = \rho_b((0)_b) = \bullet$. Hence, $0 \in \mathcal{N}_b \text{ id est } (\mathcal{P}_0)$ is true.

(II) Let $k \in [0, b^d - 2]$. Assume that $(\mathcal{P}_{k'})$ is true for all $k' \in [0, k]$. By point (ii), there exists $l \in \mathbb{N}_1$ such that $k' := \gamma_b^l(k + 1) < k + 1$. Since $k' \in [0, k]$, by $(\mathcal{P}_{k'})$, there exists $l' \in \mathbb{N}$ such that $\gamma_b^{l'}(k') = \bullet$. We have:

$$ \gamma_b^{l+l'}(k+1) = \bullet $$

id est $k+1 \in \mathcal{N}_b$. Thus, $(\mathcal{P}_{k+1})$ is true.
(III) By strong finite induction, ($\mathcal{P}_k$) is true for all $k \in \mathcal{B}_{d,b}$.

• (i) $\implies$ (iii). Assume point (i). Let $n \in \mathcal{D}_{d,b} \subset \mathcal{B}_{d,b}$. Hence $n \in \mathcal{N}_b$ id est there exists $k \in \mathbb{N}_1$ such that $\gamma_b^k(n) = \bullet$. Thus, we have:

$$-\infty = d_b(\bullet) = d_b(\gamma_b^k(n)) < d_b(n)$$

and so $n \in \mathcal{Y}_b^\dagger$.

• (iii) $\implies$ (iv). Assume point (iii). By (2), we have:

$$\mathcal{D}_{d,b}^{\dagger} \subset \mathcal{D}_{d,b} \subset \mathcal{Y}_{b}^{\dagger}.$$

• (iv) $\implies$ (v). Assume point (iv). By (2), we have:

$$\mathfrak{R}_{d,b}^{1} \subset \mathfrak{R}_{d,b}^{0} = \mathcal{D}_{d,b}^{\dagger} \subset \mathcal{Y}_{b}^{\dagger}.$$

• (v) $\implies$ (i). Assume point (v). Let us prove by strong finite induction on $k \in \mathcal{B}_{d,b} = [0, b^d - 1]$:

$$(\mathcal{P}_k) : k \in \mathcal{N}_b.$$

(I) We have $\gamma_b((0)_b) = \rho_b((0)_b) = \bullet$. Hence, $0 \in \mathcal{N}_b$ id est $(\mathcal{P}_0)$ is true.

(II) Let $k \in [0, b^d - 2]$ of degree $d' \in [0, d]$. Assume that $(\mathcal{P}_{k'})$ is true for all $k' \in [0, k]$. We have three cases.

(a) **Case $k+1 \in \mathcal{E}_{(d'),b}$**. Then there exists $l \in \mathbb{N}$ such that:

$$\gamma_b^l(k+1) = \pi_b^l(k+1) \in \mathcal{E}_{(d'),b}^{\dagger}$$

and thus:

$$m := \kappa_b^{\dagger}(\gamma_b^l(k+1)) \in \mathcal{D}_{(d'),b}^{\dagger} \subset \mathcal{B}_{d,b}.$$

We then have three subcases.

(1) **Case $m \in \mathfrak{R}_{(d'),b}^1$**. By point (v), we have $m \in \mathcal{Y}_b^\dagger$ id est there exists $l' \in \mathbb{N}_2$ ($l' \neq 1$ since $m \in \mathcal{E}_{(d'),b}^\dagger$) such that:

$$d_b(m') < d_b(m) = d_b(k+1) = d'$$

with:

$$m' := \gamma_b^{l'}(m) \in \mathcal{B}_{d'-1,b}.$$

Since $d_b(m') < d_b(k+1)$, we have $m' < k+1$ id est $m' \in [0, k]$. Thus, by $(\mathcal{P}_{m'})$, there exists $l'' \in \mathbb{N}$ such that $\gamma_b^{l''}(m') = \bullet$. Since $l' \in \mathbb{N}_2$, we have:

$$\begin{align*}
\gamma_b^{l+l'+l''}(k+1) &= \gamma_b^{l''}(\gamma_b^{l'-1}(\gamma_b(\gamma_b^l(k+1)))) \\
&= \gamma_b^{l''}(\gamma_b^{l'-1}(\pi_b(\gamma_b^l(k+1)))) \\
&= \gamma_b^{l''}(\gamma_b^{l'-1}(\pi_b(\kappa_b^{\dagger}(\gamma_b^l(k+1))))) \\
&= \gamma_b^{l''}(\gamma_b^{l'-1}(\pi_b(m))) \\
&= \gamma_b^{l''}(\gamma_b^l(m)) \\
&= \gamma_b^{l''}(m') \\
&= \bullet.
\end{align*}$$
(2) Case $m \in \tilde{\mathfrak{R}}_{(d'),b}^1 - \mathfrak{R}_{(d'),b}^1$. Since $m' := \Psi_b^1(n) \in \mathfrak{R}_{(d'),b}^1$, by case (1), $m' \in \mathcal{N}_b$ id est there exists $l' \in \mathbb{N}_{\max\{t_b^1(m), t_b^1(m')\}}$ such that $\gamma_b^{l'}(m') = \bullet$. Thus, we have:

$$
\begin{align*}
\gamma_b^{l+l'}(k+1) &= \gamma^{l'-1}(\gamma_b(\gamma_b^l(k+1))) \\
&= \gamma^{l'-1}(\pi_b(\gamma_b^l(k+1))) \\
&= \gamma^{l'-1}(\pi_b(\kappa_b^\dagger(\gamma_b^l(k+1)))) \\
&= \gamma^{l'-1}(\pi_b(m)) \\
&= \gamma^{l'}(m) \\
&= \gamma^{l'-t_b^1(m)}(\gamma_b^{t_b^1}(m)(m)) \\
&= \gamma^{l'-t_b^1(m')}(\gamma_b^{t_b^1(m')}(m')) \\
&= \gamma^{l'}(m') \\
&= \bullet
\end{align*}
$$

(3) Case $m \notin \tilde{\mathfrak{R}}_{(d'),b}^1$ id est there exists $l' \in \mathbb{N}_1$ such that $\gamma_b^{l'}(m) = \pi_b^{l'}(m) \in \mathcal{R}_b$ and $d_b(\pi_b^{l'}(m)) = d_b(m) + 1$. Let us define:

$$m' := \gamma_b^{l'+1}(m) = \rho_b(\pi_b^{l'}(m)).$$

We have:

$$d_b(m') \le d_b(m) - 1 = d_b(k + 1) - 1$$

in particular $m' < k + 1$ id est $k' \in [0, k]$. Thus, by $(\mathcal{P}_{m'})$, there exists $l'' \in \mathbb{N}$ such that $\gamma_b^{l''}(m') = \bullet$. We then have:

$$
\begin{align*}
\gamma_b^{l+l'+l''+1}(k+1) &= \gamma^{l''}(\gamma_b^{l'}(\gamma_b^l(k+1))) \\
&= \gamma^{l''}(\pi_b^{l'+1}(\gamma_b^l(k+1))) \\
&= \gamma^{l''}(\pi_b^{l'+1}(\kappa_b^\dagger(\gamma_b^l(k+1)))) \\
&= \gamma^{l''}(\pi_b^{l'+1}(m)) \\
&= \gamma^{l''}(m') \\
&= \bullet
\end{align*}
$$

(b) Case $k+1 \notin \mathcal{E}_{d',b}$. Then there exists $l \in \mathbb{N}$ such that:

$$
\begin{align*}
k' &:= \gamma_b^{l+1}(k+1) = \rho_b(\pi_b^l(k+1)) \\
d_b(\pi_b^l(k+1)) &= d_b(k+1)
\end{align*}
$$

Thus, we have:

$$d_b(k') < d_b(\pi_b^l(k+1)) = d_b(k+1)$$

in particular $k' < k + 1$ id est $k' \in [0, k]$. By $(\mathcal{P}_{k'})$, there exists $l' \in \mathbb{N}$ such that $\gamma_b^{l'}(k') = \bullet$. Thus, we have:

$$\gamma_b^{l+l'}(k+1) = \bullet$$

id est $k+1 \in \mathcal{N}_b$ in this case.

Thus, $(\mathcal{P}_{k+1})$ is true in all cases.

(III) By strong finite induction, $(\mathcal{P}_k)$ is true for all $k \in \mathcal{B}_{d,b}$.

<figure><img src="image_8.png" /></figure>

In other words, the equivalences are as follows:
(i) All numbers of degree less than $d$ in base $b$ are normands in base $b$;

(ii) All numbers of degree less than $d$ in base $b$ have a finite journey in base $b$;

(iii) All ducs in base $b$ of degree less than $d$ in base $b$ have a finite journey of degree in base $b$;

(iv) All ducs elingard in base $b$ of degree less than $d$ in base $b$ have a finite journey of degree in base $b$;

(v) All riboutins of order 1 in base $b$ of degree less than $d$ in base $b$ have a finite journey of degree in base $b$.

Taking the limit over $d \in \mathbb{N}$, we deduce the following corollary.

### Corollary 4.6.2

The following equivalences hold:

(i) $\mathcal{N}_b = \mathbb{N}_{\bullet}$;

(ii) $\mathbb{N} \subset \mathcal{Y}_b$.

(iii) $\mathcal{D}_b \subset \mathcal{Y}_b^{\dagger}$;

(iv) $\mathfrak{R}_b^0 = \mathcal{D}_b^{\dagger} \subset \mathcal{Y}_b^{\dagger}$;

(v) $\mathfrak{R}_b^1 \subset \mathcal{Y}_b^{\dagger}$.

*Proof.* Since:

$$ \mathbb{N} = \bigcup_{d \in \mathbb{N}} \mathcal{B}_{d,b} \quad , \quad \mathcal{D}_b = \bigcup_{d \in \mathbb{N}} \mathcal{D}_{d,b} \quad , \quad \mathcal{D}_b^{\dagger} = \bigcup_{d \in \mathbb{N}} \mathcal{D}_{d,b}^{\dagger} $$

we have the following equivalences by Theorem 4.6.1:

$$
\begin{align*}
(i) \quad \mathcal{N}_b = \mathbb{N}_{\bullet} &\iff \mathbb{N} \subset \mathcal{N}_b \iff \forall d \in \mathbb{N}, \mathcal{B}_{d,b} \subset \mathcal{N}_b \\
&\iff \forall d \in \mathbb{N}, \mathcal{B}_{d,b} \subset \mathcal{Y}_b \iff (ii) \quad : \mathbb{N} \subset \mathcal{Y}_b \\
&\iff \forall d \in \mathbb{N}, \mathcal{D}_{d,b} \subset \mathcal{Y}_b^{\dagger} \iff (iii) \quad : \mathcal{D}_b \subset \mathcal{Y}_b^{\dagger} \\
&\iff \forall d \in \mathbb{N}, \mathcal{D}_{d,b}^{\dagger} \subset \mathcal{Y}_b^{\dagger} \iff (iv) \quad : \mathcal{D}_b^{\dagger} \subset \mathcal{Y}_b^{\dagger} \\
&\iff \forall d \in \mathbb{N}, \mathfrak{R}_{d,b}^0 \subset \mathcal{Y}_b^{\dagger} \iff (v) \quad : \mathfrak{R}_b^0 \subset \mathcal{Y}_b^{\dagger} \\
&\iff \forall d \in \mathbb{N}, \mathfrak{R}_{d,b}^1 \subset \mathcal{Y}_b^{\dagger} \iff (v) \quad : \mathfrak{R}_b^1 \subset \mathcal{Y}_b^{\dagger}.
\end{align*}
$$

□

In other words, we have equivalence between:

(i) All integers are normands in base $b$;

(ii) All integers have a finite journey in base $b$;

(iii) All dukes in base $b$ have a finite journey in degree in base $b$;

(iv) All elingard dukes in base $b$ have a finite journey in degree in base $b$;

(v) All riboutins of order 1 in base $b$ have a finite journey in degree in base $b$.
One might wonder if we can add the equivalence with:

$$ \mathfrak{R}_b^2 \subset \mathcal{V}_b^\dagger. \qquad (104) $$

The answer is no; just take a triskelion in base $b$ of length 5. For example, the triskelion $\mathcal{T}_{11}(n)$ associated with the number $n := (9 | 3 | 10 | 2 | 8)_{4,11}$ Breton in base 11:

$$ \mathcal{T}_{11}(n) = \{(9 | 3 | 10 | 2 | 8)_{4,11}, (1 | 6 | 6 | 9 | 6 | 6)_{5,11}, (8 | 2 | 5 | 5 | 1 | 7)_{5,11}, (1 | 4 | 3 | 10 | 10 | 4 | 4)_{6,11}, (5 | 9 | 3 | 10 | 2 | 8 | 5)_{6,11}\}. $$

Let:

$$ m := \kappa_{11}(n) = (10 | 5 | 10 | 0 | 7)_{4,11} \in \mathcal{D}_{(4),11}^{\dagger} $$

We have $m \in \mathfrak{R}_b^1 - \mathfrak{R}_b^2$ because

$$ \gamma_{11}^4(m) = \gamma_{11}^5(n) = (5 | 9 | 3 | 10 | 2 | 8 | 5)_{6,11} \in \mathfrak{R}_{11}. $$

## 4.7 From Nollon to Rollor

Since the beginning of this text, we have essentially worked with a fixed base $b$. We can also, in the manner of Rollor, work with a fixed integer $n$ and vary the base $b$. We define for all $n \in \mathbb{N}$:

$$ \mathcal{M}_n := \{b \in \mathbb{N}_2, n \in \mathcal{N}_b\}. \qquad (105) $$

We begin with a lemma.

### Lemma 4.7.1

(i) For any base $b$, we have $\mathcal{B}_b \subset \mathcal{N}_b$.

(ii) We have:

$$ \mathbb{N}_n \subset \mathcal{M}_n. $$

*Proof.* (i) Let $n \in \mathcal{B}_b$. We have $n = (n)_b$ and thus $\rho_b(n) = \bullet$. Hence $\gamma_b(n) = \bullet$ *id est* $n \in \mathcal{N}_b$.

(ii) Since for all $n \in \mathbb{N}$, we have:

$$ \forall b \in \mathbb{N}_{n+1}, (n)_{0,b}. $$

Thus, we have $\mathbb{N}_{n+1} \subset \mathcal{M}_n$. Since $n = (1 | 0)_{1,n}$ and that:

$$ \gamma_n^2(n) = \gamma_n(\pi_n((1 | 0)_{1,n})) = \gamma_n((1 | 1)_{1,n}) = \rho_n((1 | 1)_{1,n}) = \bullet $$

we have $n \in \mathcal{M}_n$. Hence the result.

□

We have the following lemma which improves point (ii) of the previous lemma 4.7.1.

### Lemma 4.7.2

Let $n \in \mathbb{N}$. We have:

$$ \{b \in \mathbb{N}_2, r_b(n) + q_b(n) \leq b - 1\} \subset \mathcal{M}_n. $$

*Proof.* Let $b \in \mathbb{N}_2$ such that $b(b-1) \geq n$ (so $d_b(n) \leq 1$). We have three cases.

(1) Case $b > n$ *id est* $n \in \mathcal{B}_b$. Then we have $n = (n)_{0,b}$ and thus:

$$ \gamma_b(n) = \rho_b(n) = \bullet $$
(2) **Case** $b \le n \le b(b-1)$ **and** $n \in \mathcal{R}_b$. Since $n = (q_b(n) | r_b(n))_{1,b}$, we have $q_b(n) = r_b(n)$ id est $n = (r_b(n) | r_b(n))_{1,b}$. Thus, we have:

$$ \gamma_b(n) = \rho_b(n) = \bullet $$

(3) **Case** $b \le n \le b(b-1)$ **and** $n \notin \mathcal{R}_b$. We have:

$$ \gamma_b^2(n) = \gamma_b(\pi_b(n)) = \gamma_b((r_b(n) + q_b(n) | r_b(n) + q_b(n))_{1,b}) = \bullet $$

Thus, in all cases, we have $n \in \mathcal{N}_b$.

<figure><img src="image_9.png" /></figure>

Since for all $n \in \mathbb{N}$:

$$ \{b \in \mathbb{N}_2, n \le 2(b-2)\} \subset \{b \in \mathbb{N}_2, r_b(n) + q_b(n) \le b-1\} $$

we have:

$$ \{b \in \mathbb{N}_2, n \le 2(b-2)\} \subset \mathcal{M}_n. \qquad (106) $$

This viewpoint still seems delicate because there is no simple method for an integer $n$ to change the base of its decomposition: deducing the digits $[n]_{k,b}$ from the digits $[n]_{l,b'}$ for two distinct bases $b$ and $b'$.

We have an easily calculable density.

**Proposition 4.7.3**

For all $n \in \mathbb{N}$, we have:

$$ \delta_2(\mathcal{M}_n) = 1. $$

*Proof.* From point (ii) of lemma 4.7.1, we have:

$$ \mathbb{N}_n \subset \mathcal{M}_n $$

and thus for all $m \in \mathbb{N}_n$:

$$ \frac{m-n+1}{m-1} \le \delta_{(m,2)}(\mathcal{M}_n) \le 1 $$

Hence the result by taking the limit as $m$ tends to infinity:

$$ \delta_k(\mathcal{M}_n) = \lim_{n \to \infty} \delta_{(n,k)}(\mathcal{M}_n) = 1. $$

<figure><img src="image_10.png" /></figure>

**4.8 Applications to base 2**

The first simple result is that all integers are normands in base 2. This allows for a simple example to practice the notations and results from above.

**Theorem 4.8.1**

We have:

$$ \mathcal{N}_2 = \mathbb{N}_\bullet. $$

*Proof.* According to corollary 4.6.2, it is sufficient to show that all elingard dukes in base 2 have a finite journey in degree in base 2. Let's show that $\mathcal{B}_{2,2} \subset \mathcal{N}_2$ id est $\mathcal{D}_{2,2}^\dagger \subset \mathcal{V}_2^\dagger$ by theorem 4.6.1. We have:

$$ \mathcal{D}_{2,2}^\dagger = \{6 := (1 | 1 | 0)_{2,2}\}. $$
and

$$
\begin{align*}
\gamma_2^2(6) &= \gamma_2^2((1 \mid 1 \mid 0)_{2,2}) = \gamma_2(\pi_2((1 \mid 1 \mid 0)_{2,2})) = \gamma_2((1 \mid 0 \mid 0 \mid 1)_{3,2}) \\
&= \rho_2((1 \mid 0 \mid 0 \mid 1)_{3,2}) = (0 \mid 0)_{1,2}
\end{align*}
$$

Thus, $\mathcal{D}_{2,2}^{\dagger} \subset \mathcal{Y}_{2}^{\dagger}$.

Let $n \in \mathcal{D}_b^\dagger$ such that $d := d_2(n) \ge 3$. Then $n$ is written as:

$$
n := (1 \mid a \mid \cdots \mid a' \mid 0)_{d,2}
$$

with $a, a' \in \{0,1\}$ and $a \ge a'$. There exist $a'', \delta \in \{0,1\}$ (depending on whether there is a carry or not)
such that:

$$
\begin{align*}
\gamma_2(n) &= \pi_2(n) \\
&= n + \mu_2(n) \\
&= (1 \mid a \mid \cdots \mid a' \mid 0)_{2,d} + (0 \mid a' \mid \cdots \mid a \mid 1)_{2,d} \\
&= (1 \mid a'' \mid \cdots \mid r_2(a + a') \mid 1)_{2,d+\delta}
\end{align*}
$$

And thus:

$$
\gamma_2^2(n) = \rho_2(\pi_2(n)) = \rho_b((1 \mid a'' \mid \cdots \mid r_2(a+a') \mid 1)_{d+\delta,2}) = (a'' \mid \cdots \mid r_2(a+a'))_{d-2+\delta,2}
$$

and thus:

$$
d_2(\gamma_2^2(n)) = d - 2 + \delta < d = d_2(n).
$$

Thus, we have:

$$
\mathcal{D}_2^\dagger \subset \mathcal{Y}_2^\dagger.
$$

<figure><img src="image_11.png" /></figure>

4.9 Artifaille in Python

<figure><img src="image_12.png" /></figure>

♣ The program est_elingard(N) takes as input an integer N of type IntB written in base b. It returns True if N is elingard and False otherwise.

Python 4.9.1

def est_elingard(N: IntB) -> bool:
    if est_red(N):
        return False
    else:
        B, b = N
        d = len(B)
        B, b = pi_fct((list(B), b))
        return len(B) > d and not est_red((B, b))

Example 4.9.2

>>> est_elingard([1, 3, 2], 5)
False

>>> est_elingard([4, 3, 2], 5)
False

>>> est_elingard([4, 3, 1], 5)
True

♣ The program `elingard_fct(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns $\epsilon_b(N)$ of type `IntB` in the same base `b` if `N` is elingard and `False` otherwise.

### Python 4.9.3

```python
def elingard_fct(N: IntB) -> IntB:
    if est_red(N):
        return False
    else:
        d = len(N[0])
        while len(N[0]) == d:
            NO = list(N)
            N = pi_fct(N)
    if est_elingard(NO):
        return NO
    else:
        return False
```

### Example 4.9.4

```python
>>> elingard_fct([4, 3, 2], 5)
False
>>> elingard_fct([4, 3, 1], 5)
[[4, 3, 1], 5]
>>> elingard_fct([1, 3, 2], 5)
[[3, 1, 4], 5]
```

♣ The program `est_duc(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns `True` if `N` is duc and `False` otherwise.

### Python 4.9.5

```python
def est_duc(N: IntB) -> bool:
    return kappa_fct((list(N[0]), N[1])) == N
```

### Example 4.9.6

```python
>>> est_duc([1, 3, 2], 5)
False
>>> est_duc([1, 3, 4], 5)
True
```

♣ The program `est_duc_elingard(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns `True` if `N` is duc elingard and `False` otherwise.
Python 4.9.7

```python
def est_duc_elingard(N: IntB) -> bool:
    return est_elingard(N) and est_duc(N)
```

Example 4.9.8

```python
>>> est_duc_elingard([1, 3, 4], 5))
True
```

♣ The program `kappa_fct(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns $\kappa_b(N)$ of type `IntB` in the same base `b`.

Python 4.9.9

```python
def kappa_fct(N: IntB) -> IntB:
    B, b = N
    for k in range(len(B) // 2):
        a = B[k] + B[-k - 1]
        B[k], B[-k - 1] = max(0, a - b + 1), min(a, b - 1)
    return (B, b)
```

Example 4.9.10

```python
>>> kappa_fct([1, 1, 0, 8, 2, 9, 1, 5, 0, 0, 0], 10))
([1, 1, 0, 9, 3, 9, 0, 4, 0, 0, 0], 10)
```

♣ The program `kappaDag_fct(N)` takes as input an integer `N` of type `IntB` written in base `b`. It returns $\kappa_b^\dagger(N)$ of type `IntB` in the same base `b`.

Python 4.9.11

```python
def kappaDag_fct(N: IntB) -> IntB:
    B, b = N
d = len(B)
while True:
    if est_red((B, b)):
        return False
    B0, _ = pi_fct((B, b))
    if len(B0) > d:
        return kappa_fct((B, b))
    B = list(B0)
```

Example 4.9.12

```python
>>> kappaDag_fct([1, 1, 0, 8, 2, 9, 1, 5, 0, 0, 0], 10))
([9, 9, 9, 9, 9, 7, 3, 2, 1, 7, 7], 10)
>>> kappaDag_fct([1, 1, 0, 3, 5, 0, 0, 0], 10))
False
```

♣ The program `tin(N, k)` takes as input an integer `N` of type `IntB` written in base `b` and an integer `k`. It returns the integer $t_b^k(N)$.
Python 4.9.13

```python
def tin(N: IntB, k:int) -> int:
    B, b = N
    d = len(B)
    i = 0
    while len(B) <= d + k:
        B, _ = pi_fct((B, b))
        i += 1
    return i - 1
```

Example 4.9.14

```python
>>> tin([4, 2], 5), 0)
0
>>> tin([4, 2], 5), 1)
2
>>> tin([4, 2], 5), 2)
3
```

♣ The program `est_boutin(N, k)` takes as input an integer `N` of type `IntB` written in base `b` and an integer `k`. It returns the integer `True` if `N` is a boutin of order `k` in base `b`.

Python 4.9.15

```python
def est_boutin(N: IntB, k:int) -> bool:
    B, b = N
    d = len(B)
    while len(B) <= d + k:
        if est_red((B, b)):
            return False
        B, _ = pi_fct((B, b))
    return True
```

Example 4.9.16

```python
>>> est_boutin([4, 2], 5), 0)
True
>>> est_boutin([4, 2], 5), 1)
False
```

<figure><img src="image_13.png" /></figure>

The program `liste_B_db(d, b)` takes as input an integer `d` and a base `b`. It returns the list:

$$ \mathcal{B}_{(d),b}. $$

Python 4.9.17

```python
def liste_B_db(d: int, b: base) -> List List[int]]:
    L = [[i] for i in range(d != 0, b)]
    for _ in range(d):
```
L = [[i] + x for i in range(b) for x in L]
return L

### Example 4.9.18

>>> liste_B_db(2, 4)
[[0, 0, 1], [0, 0, 2], [0, 0, 3], [0, 1, 1], [0, 1, 2], [0, 1, 3], [0, 2, 1],
[0, 2, 2], [0, 2, 3], [0, 3, 1], [0, 3, 2], [0, 3, 3], [1, 0, 1], [1, 0, 2],
[1, 0, 3], [1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 2, 1], [1, 2, 2], [1, 2, 3],
[1, 3, 1], [1, 3, 2], [1, 3, 3], [2, 0, 1], [2, 0, 2], [2, 0, 3], [2, 1, 1],
[2, 1, 2], [2, 1, 3], [2, 2, 1], [2, 2, 2], [2, 2, 3], [2, 3, 1], [2, 3, 2],
[2, 3, 3], [3, 0, 1], [3, 0, 2], [3, 0, 3], [3, 1, 1], [3, 1, 2], [3, 1, 3],
[3, 2, 1], [3, 2, 2], [3, 2, 3], [3, 3, 1], [3, 3, 2], [3, 3, 3]]

<figure><img src="image_14.png" /></figure>

The program `liste_RoO(d, b)` takes as input an integer `d` and a base `b`. It returns the list:

$$ \mathcal{R}^0_{(d),b}. $$

### Python 4.9.19

def liste_RoO(d: int, b: base) -> List List[int]]:
    if d == 0:
        return []
    else:
        L = [[i] for i in range(1, b)]
        for _ in range(d - 1):
            L = [[i] + x for x in range(b) for x in L]
            L = [[i] + x for x in range(b) for x in L if i != x[-1]]
        return L

We could have done:

```python
liste_RoO(d, b) = [x for x in liste_B_db(d, b) if not est_red((x, b))]
```

but this method is slower.

### Example 4.9.20

>>> liste_RoO(2, 4)
[[0, 0, 1], [0, 0, 2], [0, 0, 3], [0, 1, 1], [0, 1, 2], [0, 1, 3], [0, 2, 1],
[0, 2, 2], [0, 2, 3], [0, 3, 1], [0, 3, 2], [0, 3, 3], [1, 0, 2], [1, 0, 3],
[1, 1, 2], [1, 1, 3], [1, 2, 2], [1, 2, 3], [1, 3, 2], [1, 3, 3], [2, 0, 1],
[2, 0, 3], [2, 1, 1], [2, 1, 3], [2, 2, 1], [2, 2, 3], [2, 3, 1], [2, 3, 3],
[3, 0, 1], [3, 0, 2], [3, 1, 1], [3, 1, 2], [3, 2, 1], [3, 2, 2], [3, 3, 1],
[3, 3, 2]]

<figure><img src="image_15.png" /></figure>

The program `liste_elingard(d, b)` takes an integer `d` and a base `b` as input. It returns the list:

$$ \mathcal{E}^{\dagger}_{(d),b}. $$
Python 4.9.21

```python
def liste_elingard(d: int, b: base) -> List List[int]]:
    return [x for x in liste_Ro0(d, b) if est_elingard((list(x), b))]
```

Example 4.9.22

```python
>>> liste_elingard(2, 4)
[[0, 2, 3], [0, 3, 3], [1, 0, 3], [1, 1, 3], [1, 2, 2], [1, 2, 3], [1, 3, 2],
[1, 3, 3], [2, 2, 1], [2, 3, 1], [3, 0, 1], [3, 1, 1], [3, 2, 1], [3, 3, 1]]
```

<figure><img src="image_16.png" /></figure>

The program `liste_duc(d, b)` takes an integer `d` and a base `b` as input. It returns the list:

$$ \mathcal{D}_{(d),b}. $$

Python 4.9.23

```python
def liste_duc(d: int, b: base) -> List List[int]]:
    return [x for x in liste_B_db(d, b) if est_duc((list(x), b))]
```

Example 4.9.24

```python
>>> liste_duc(2, 4)
[[0, 0, 1], [0, 0, 2], [0, 0, 3], [0, 1, 1], [0, 1, 2], [0, 1, 3], [0, 2, 1],
[0, 2, 2], [0, 2, 3], [0, 3, 1], [0, 3, 2], [0, 3, 3], [1, 0, 3], [1, 1, 3],
[1, 2, 3], [1, 3, 3], [2, 0, 3], [2, 1, 3], [2, 2, 3], [2, 3, 3], [3, 0, 3],
[3, 1, 3], [3, 2, 3], [3, 3, 3]]
```

<figure><img src="image_17.png" /></figure>

The program `liste_duc_quasi_elingard(d, b)` takes an integer `d` and a base `b` as input. It returns the list:

$$ \mathcal{D}^{\dagger}_{(d),b}. $$

Python 4.9.25

```python
def liste_duc_elingard(d: int, b: base) -> List List[int]]:
    return [x for x in liste_duc_quasi_elingard(d, b) if
             x[0] + x[-1] >= b or est_elingard((list(x), b))]
```

Example 4.9.26

```python
>>> liste_duc_elingard(2, 4)
[[0, 2, 3], [0, 3, 3], [1, 0, 3], [1, 1, 3], [1, 2, 3], [1, 3, 3]]
```

We could have done:

```python
liste_duc_elingard_bis(d, b) = [x for x in liste_Ro0(d, b) if est_duc_elingard((list(x), b))].
```

<figure><img src="image_18.png" /></figure>

The program `liste_duc_quasi_elingard(d, b)` takes an integer `d` and a base `b` as input. It returns the list:

$$ \tilde{\mathcal{D}}_{(d),b}. $$
Python 4.9.27

```python
def liste_duc_quasi_elingard(d: int, b: base) -> List List[int]:
    if d % 2 == 0:
        L = [[j] for j in range(b)]
    else:
        L = [[]]
    if d >= 1:
        A = [(max(0, j - b + 1), min(j, b - 1)) for j in range(2 * b - 1)]
        for _ in range((d + 1) // 2 - 1):
            L = [[a] + x + [b] for x in L for a, b in A]
            A = [(max(0, j - b + 1), min(j, b - 1))
                  for j in range(b - 1, 2 * b - 2) if j - b + 1 != 2]
            L = [[a] + x + [b] for x in L for a, b in A]
    return L
```

Example 4.9.28

```python
>>> liste_duc_quasi_elingard(2, 4)
[[0, 0, 3], [1, 0, 3], [3, 0, 3], [0, 1, 3], [1, 1, 3], [3, 1, 3], [0, 2, 3],
[1, 2, 3], [3, 2, 3], [0, 3, 3], [1, 3, 3], [3, 3, 3]]
```

♣ The program `liste_riboutin(d, b)` takes an integer `d` and a base `b` as input. It returns the list:

$$ \mathfrak{R}^k_{(d),b}. $$

Python 4.9.29

```python
def liste_riboutin(d: int, b: Base, k: int) -> List List[int]:
    L = dict()
    i = 0
    for y in liste_quasi_duc_elingard(d, b):
        x = list(y)
        while len(x) <= d + 1 + k:
            B = list(x)
            if est_red((x, b)):
                B = False
                break
            x, _ = pi_fct((x, b))
        if B != False:
            B = tuple(B)
            if B not in L or not compare_base((L[B], b), (B, b)) :
                i += 1
                L[B] = y
    return list(L.values())
```

Example 4.9.30

```python
>>> liste_riboutin(5, 4, 1)
[[0, 1, 0, 0, 3, 3], [0, 2, 0, 0, 3, 3], [0, 3, 0, 0, 3, 3], [0, 1, 0, 1, 3, 3],
[0, 2, 0, 1, 3, 3], [0, 3, 0, 1, 3, 3], [0, 1, 0, 2, 3, 3], [0, 2, 0, 2, 3, 3],
[0, 3, 0, 2, 3, 3], [1, 0, 0, 3, 2, 3], [0, 1, 0, 3, 3, 3], [0, 2, 0, 3, 3, 3],
[0, 3, 0, 3, 3, 3], [0, 0, 1, 3, 3, 3], [0, 1, 1, 3, 3, 3], [0, 2, 1, 3, 3, 3],
```
```
[0, 3, 1, 3, 3, 3], [0, 0, 2, 3, 3, 3], [0, 1, 2, 3, 3, 3], [0, 2, 2, 3, 3, 3],
[0, 3, 2, 3, 3, 3], [0, 0, 3, 3, 3, 3], [0, 1, 3, 3, 3, 3], [0, 2, 3, 3, 3, 3],
[0, 3, 3, 3, 3, 3]]

>>> liste_riboutin(5, 4, 2)
[[1, 0, 0, 3, 2, 3], [0, 3, 3, 3, 3, 3]]

>>> liste_riboutin(5, 4, 3)
[[0, 3, 3, 3, 3, 3]]

>>> liste_riboutin(5, 4, 4)
[]
```

We conclude this subsection with an example for large values of *b* and *d*. This helps to realize that we have significantly reduced the number of tests required to show that:

$$ \mathcal{B}_{(d),b} \subset \mathcal{N}_b. $$

### Example 4.9.31

```
>>> d, b = 4, 10
>>> len(liste_B_db(d, b))
90000

>>> len(liste_RoO(d, b))
81000

>>> len(liste_elingard(d, b))
36500

>>> len(liste_duc(d, b))
3420

>>> len(liste_duc_elingard(3, 5))
1425

>>> len(liste_duc_quasi_elingard(d, b))
1520

>>> len(liste_riboutin(d, b, 1))
365

>>> len(liste_riboutin(d, b, 2))
81
```

So to show that the 90,000 elements of $\mathcal{B}_{(4),10}$ are Normand in base 10, it is enough to show that the 365 riboutins of order 1 in base 10 of degree 4 are Normand in base 10.

The problem with the program `liste_riboutin(d, b)` is that it is very slow for very large values. I do not know if there is a faster way to calculate the list of riboutins of degree *d* in base *b*.
# 5 The Bacouettes

In this section, we present a list of bacouettes. The meaning of the word "bacouette" is vague; it generally refers to an exercise, conjecture, question, or problem in the mathematical sense. For example, the bacouette (BAC - 1) is a theorem to prove, the bacouette (BAC - 2) is a problem, the bacouette (BAC - 3) is a conjecture, and the bacouette (BAC - 4) is a question.

A bacouette is primarily meant to encourage discussion, sharing, and exchange: to engage in the bacouette.

## 5.1 Study of Palindromization

In proposition 2.3.3, we showed an implication for studying the injectivity of $\pi_b$. It turns out that the proposition is an equivalence.

### Bacouette 5.1.1

(BAC - 1) Let $n, m \in \mathbb{N}_1$ of degree $d$ in base $b$. Then the following equivalence holds:

(i) For all $k \in \{0, \dots, d\}$, we have:

$$[n]_{k,b} + [n]_{d-k,b} = [m]_{k,b} + [m]_{d-k,b};$$

(ii) $\pi_b(n) = \pi_b(m)$.

(BAC - 2) Study $\pi_b(n) = \pi_b(m)$ in the case of integers $n, m \in \mathbb{N}_1$ such that $|d_b(n) - d_b(m)| \le 1$.

For bacouette (BAC - 2), the following example shows that $n$ and $m$ do not necessarily have the same degree:

$$(1 | 1 | 3 | 1)_{3,6} = \pi_6((3 | 1 | 4)_{2,6}) = \pi_6((1 | 1 | 0 | 0)_{3,6}).$$

Some properties can be given for bacouette (BAC - 2). Suppose there exist $n, m \in \mathbb{N}_1$ such that $\pi_b(n) = \pi_b(m)$ and $d_b(n) = d_b(m) + 1$. Then, with $d := d_b(m)$, we have:

(i) $n$ is of the form:

$$n = (1 | \cdots | 0)_{d+1,b};$$

(ii) $\pi_b(n)$ is of the form:

$$\pi_b(n) = (1 | \cdots)_{d+1,b}.$$

We have defined palindromization for all integers $n \in \mathbb{N}$, which allows us to examine iterates of $\pi_b$ without performing reductions. This leads to the following natural bacouettes.

### Bacouette 5.1.2

(BAC - 3) Let $n \in \mathbb{N}$. There exists $k \in \mathbb{N}$ such that $\pi_b^k(n) \in \mathcal{R}_b$.

(BAC - 4) Let $n \in \mathbb{N}$. Does there exist $k \in \mathbb{N}$ such that $\pi_b^k(n) \in \mathcal{P}_b$?

By performing computer simulations, we can search for the maximum value of $k \in \mathbb{N}$ (depending on $b$ and independent of $n$) that exists in bacouette (BAC - 3).
**Definition 5.1.3**

(i) Let $n \in \mathbb{N}$. We say that $n$ is **iteratively reducible in base $b$** if there exists an integer (necessarily unique) $\mathfrak{p}_b(n) \in \mathbb{N}$ such that:

$$ \forall k \in [0, \mathfrak{p}_b(n) - 1], \pi_b^{\mathfrak{p}_b(n)}(n) \in \mathcal{R}_b^0 \wedge \pi_b^{\mathfrak{p}_b(n)}(n) \in \mathcal{R}_b. $$

(ii) We say that $b$ is an **iteratively reductive base** if the set:

$$ \{\mathfrak{p}_b(n), n \in \mathbb{N}\} $$

is bounded. In this case, we define:

$$ \mathfrak{p}_b := \max\{\mathfrak{p}_b(n), n \in \mathbb{N}\} $$

By revisiting the proof of theorem 4.8.1, we find:

$$ \mathfrak{p}_2 = 1. $$

We then have the following bacouette by testing up to $n := 10^9$.

**Bacouette 5.1.4**

(BAC - 5) The following table shows the values of $\mathfrak{p}_b$ for $b \in \mathbb{N}_3^{10}$:

<table><tr><td>b</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td></tr><tr><td>p<sub>b</sub></td><td>6</td><td>6</td><td>44</td><td>53</td><td>28</td><td>38</td><td>22</td><td>143</td></tr></table>

The two numbers less than $10^9$ in base 10 such that $\mathfrak{p}_{10}(n) = 143$ are:

$$ 899\,906\,080, 899\,951\,080. $$

We note that for $b \in \{3, 4\}$, we have $\mathfrak{p}_b = 6$. We will see the relevance of this bacouette in subsection 5.7 for these two bases.

Voici la traduction en anglais du texte LaTeX, avec les balises intactes et sans modification des programmes Python :

## 5.2 Study of the degree of the Norman sequence

We have the following bacouette.

**Bacouette 5.2.1**

(BAC - 6) Let $n \in \mathbb{N}$. Then for all $k \in \mathbb{N}_5$:

$$ d_b(\gamma_b^{2k-2}(n)) \le d_b(n) + k. $$

## 5.3 Study of the length of a triskelion

We have the following proposition on the length of a triskelion.
### Proposition 5.3.1

Let $n \in \mathscr{C}_b^\dagger$. Then:

$$ \beta_b^\dagger(n) \geq 4. $$

*Proof.* Let $n \in \mathscr{R}_{(d),b}^0$ be a trisk in base $b$ of a triskelion of length $l$ (thus $n$ is the smallest element of the triskelion and $n \in \mathscr{R}_b$). We have two cases.

(1) **Case** $d_b(\gamma_b^2(n)) \leq d_b(n) + 1$. Then we have $\gamma_b^3(n) = \pi_b^3(n)$ (otherwise $\gamma_b^3(n) = \rho_b(\pi_b^2(n))$ and thus $d_b(\gamma_b^3(n)) < d_b(n)$, which is contradictory to the minimality of $n$). Thus $l \geq 4$.

(2) **Case** $d_b(\gamma_b^2(n)) = d_b(n) + 2$. Then $\kappa_b(n)$ is of the form $(b-2 | b-1 | \cdots | a | 1)_{d,b}$ with $a \in \mathbb{N}_1$ (thus $d \in \mathbb{N}_3$). Therefore:

$$ \begin{aligned} \gamma_b^2(n) &= \pi_b(\pi_b(n)) = \pi_b(\pi_b(\kappa_b(n))) \\ &= \pi_b((1 | \cdots | b-1)_{d+1,b}) \\ &= (1 | \cdots | 0)_{d+2,b} \end{aligned} $$

and thus:

$$ \gamma_b^3(n) = \pi_b(\gamma_b^2(n)) = (a' | \cdots | 1)_{d+2,b}. $$

In particular, $l \geq 4$ in this case as well.

<figure><img src="image_19.png" /></figure>

Based on computational simulations, it seems that we can improve the result. The values of $\beta_b^\dagger(n)$ obtained by looking at the bases up to 100 are:

5, 6, 9, 18, 26, 51, 84, 117.

The value 5 occurs many times.

### Bacouette 5.3.2

(BAC - 7) Let $n \in \mathscr{C}_b^\dagger$. Then:

$$ \beta_b^\dagger(n) \geq 5. $$

That is, we have:

$$ \{\beta_b^\dagger(n), b \in \mathbb{N}_2 \wedge n \in \mathscr{C}_b^\dagger\} \subset \mathbb{N}_5. $$

(BAC - 8) We have:

$$ \mathrm{card} \{(b, n) \in \mathbb{N}_2 \times \mathbb{N}, n \in \mathscr{C}^\dagger \wedge \beta_b^\dagger(n) = 5\} = +\infty. $$

(BAC - 9) Do we have:

$$ \{\beta_b^\dagger(n), b \in \mathbb{N}_2 \wedge n \in \mathscr{C}_b^\dagger\} = \mathbb{N}_5? $$

## 5.4 Study of the riboutins

We have calculated in Proposition 2.1.9 the cardinality of $\mathscr{R}_{(d),b}^r$, and in Proposition 4.3.6 the cardinality of $\mathscr{D}_{(d),b}$. We then have this bacouette.
Bacouette 5.4.1

(BAC - 10) Let $d \in \mathbb{N}$. Calculate the cardinalities of:

$$ \mathcal{R}_{d,b}^r, \mathcal{R}_{(d),b}^r, \mathcal{R}_{d,b}, \mathcal{R}_{(d),b}, \mathcal{P}_{d,b}, \mathcal{P}_{(d),b}. \\ \mathcal{E}_{(d),b}, \mathcal{E}_{d,b}, \mathcal{E}_{(d),b}^\dagger, \mathcal{E}_{d,b}^\dagger, \mathcal{D}_{(d),b}^\dagger, \mathcal{D}_{d,b}^\dagger, \mathfrak{R}_{(d),b}^1, \mathfrak{R}_{d,b}^1. $$

From these cardinalities, we can derive many densities. Here is a list of possible cardinalities (I found them by hand with possible errors):

$$ \begin{align*} \mathrm{card}(\mathcal{E}_{(2d+1),2b}^\dagger) &= 2(2b)^{2d}(b-1)^2 + \frac{1}{2}(2b)^d(2b-1)^2 \left(\sum_{k=0}^{d-1}(2b)^k\right) \\ &= 2(2b)^{2d}(b-1)^2 + \frac{1}{2}(2b)^d(2b-1)^2 \frac{(2b)^d-1}{2b} \\ \mathrm{card}(\mathcal{E}_{(d),2b+1}^\dagger) &= 2(2b+1)^{d-1}(b^2-b+1) + b(2b-1)(2b+1)^{q_2(d-1)} \left(\sum_{k=0}^{q_2(d)-1}(2b+1)^k\right) \\ &= 2(2b+1)^{d-1}(b^2-b+1) + (2b-1)(2b+1)^{q_2(d-1)} \frac{(2b+1)^{q_2(d)}-1}{2} \\ \mathrm{card}(\mathcal{D}_{(2d+1),b}^\dagger) &= (b-3)(2b-1)^d + (b-1) \left(\sum_{k=0}^{d-1}(2b-1)^k\right) \\ &= (b-3)(2b-1)^d + \frac{(2b-1)^d-1}{2} \\ &= \frac{1}{2}((2b-5)(2b-1)^d-1) \\ \mathrm{card}(\mathcal{D}_{(2d),b}^\dagger) &= b(b-3)(2b-1)^{d-1} + b(b-1) \left(\sum_{k=0}^{d-2}(2b-1)^k\right) + q_2(b) \\ &= b(b-3)(2b-1)^{d-1} + b \frac{(2b-1)^{d-1}-1}{2} + q_2(b) \\ &= \frac{b}{2}((2b-5)(2b-1)^{d-1}-1) + q_2(b) \end{align*} $$

In the program of subsection 4.9, we gave a program to calculate the riboutins. This program calculates the riboutins of degree $d$ in base $b$ from the list of quasi-elingard ducs of degree $d$ in base $b$. One might wonder if there is a faster method to calculate them.

Bacouette 5.4.2

(BAC - 11) Find a mathematical criterion for riboutins of degree $d$ in base $b$ that allows calculating the set $\mathcal{D}_{d,b}^\dagger$ more easily.

5.5 Study of $\mathcal{B}_{1,b} \subset \mathcal{N}_b$

We showed in Lemma 4.7.1 that any number of degree 0 in base $b$ is Norman in base $b$. We might wonder if every number of degree 1 in base $b$ is Norman in base $b$, and if so, whether there exists a simple proof of the result.

To imagine having a simple proof, it would be necessary that for every base $b$, the integer:

$$ \Lambda_{1,b} := \max \lambda_b (\mathcal{B}_{1,b}) $$

is small, and even better, bounded independently of base $b$. Here are some values of $\Lambda_{1,b}$:

<table><tr><td>b</td><td>2</td><td>3</td><td>4</td><td>5</td><td>10</td><td>20</td><td>29</td><td>51</td><td>104</td><td>138</td><td>189</td></tr><tr><td>Λ<sub>1,b</sub></td><td>3</td><td>4</td><td>3</td><td>8</td><td>15</td><td>23</td><td>148</td><td>760</td><td>1158</td><td>2624</td><td>3875</td></tr></table>
It seems from the table that $\Lambda_{1,b}$ is not bounded by a constant independent of $b$. However, all is not lost because for "almost all" $b \in \mathbb{N}_2^{2000}$, we have computationally:

$$ \lambda_b((b-1 | b-2)_{1,b}) = \Lambda_{b,2} $$

that is, the maximum value of $\lambda_b$ on $\mathcal{B}_{1,b}$ is at:

$$ (b-1 | b-2)_{1,b}. $$

We conjecture the following two bacouettes.

### Bacouette 5.5.1

(BAC - 12) We have:

$$ \delta_2(\{b \in \mathbb{N}_2, \mathcal{B}_{2,b} \subset \mathcal{N}_b\}) = 1. $$

(BAC - 13) We have:

$$ \delta_2(\{b \in \mathbb{N}_2, \lambda_b((b-1 | b-2)_{2,b}) = \Lambda_{b,2}\}) = 1. $$

## 5.6 Study on the density

We showed in Proposition 4.7.3 that $\delta_2(\mathcal{M}_n) = 1$. We conjecture the following two bacouettes.

### Bacouette 5.6.1

(BAC - 14) For any base $b$, we have:

$$ \delta_{\bullet}(\mathcal{N}_b) = 1. $$

(BAC - 15) We have:

$$ \delta_2(\{b \in \mathbb{N}_2, \mathcal{N}_b \subset \mathcal{C}_b^{\dagger}\}) = 1. $$

## 5.7 Study of $\mathcal{N}_b = \mathbb{N}_{\bullet}$ for small $b$

It seems that $\mathcal{N}_b = \mathbb{N}_{\bullet}$ for $b \in \{3, 4\}$ can be shown as in the case $b := 2$ by simply studying cases, but this time using a computer program. Let's justify this.

In subsection 5.1, we studied the integer $\mathfrak{p}_b$ and showed that computationally for $n \le 10^9$, $\mathfrak{p}_b = 6$. We can continue the study. Let us define for all $r \in \mathbb{N}_1$ (the integer equal to $+\infty$ in the case where the set is empty):

$$ \mathfrak{p}_{r,b} := \max \left\{ k \in \mathbb{N}, \forall n \in \mathbb{N}, \sum_{l=0}^{k} \mathrm{rg}_b(\gamma_b^l(n)) \le r \right\} \quad (107) $$

We have the following table by studying the numbers up to $10^9$:

$$ \tilde{\mathfrak{p}}_{r,b} := \max \left\{ k \in \mathbb{N}, \forall n \in \mathbb{N}, \sum_{l=0}^{k} \mathrm{rg}_b(\gamma_b^l(n)) \le r \right\} \quad (108) $$

<table><thead><tr><th>r</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th><th>9</th><th>10</th><th>11</th><th>12</th><th>13</th></tr></thead><tbody><tr><td>\tilde{\mathfrak{p}}_{r,3}</td><td>6</td><td>11</td><td>14</td><td>14</td><td>14</td><td>14</td><td>14</td><td>14</td><td>14</td><td>14</td><td>14</td><td>14</td><td>14</td></tr><tr><td>\tilde{\mathfrak{p}}_{r,4}</td><td>6</td><td>10</td><td>11</td><td>11</td><td>11</td><td>11</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td>\tilde{\mathfrak{p}}_{r,10}</td><td>143</td><td>143</td><td>143</td><td>143</td><td>143</td><td>143</td><td>143</td><td>143</td><td>143</td><td>143</td><td>143</td><td>143</td><td>143</td></tr></tbody></table>
Bacouette 5.7.1

(BAC - 16) It can be shown using a computer program that $N_b = N_\bullet$ for $b \in \{3, 4\}$.

(BAC - 17) We have $p_{r,3} = 14$ and $p_{r,4} = 11$ for all $r \in N_3$ and $p_{r,10} = 143$ for all $r \in N$.

We have the following relation between the two bacouettes.

Proposition 5.7.2

The bacouettes (BAC - 6) and (BAC - 17) imply the bacouette (BAC - 16).

In the case $b := 3$, we notice that we can simplify the number of cases to study.

Example 5.7.3

Let $n \in \mathcal{D}_3^\dagger$. Then we have only two possibilities (with $d \in N_1$):

$$n = (2 | \dots | 0)_{d,3}, \quad n = (2 | \dots | 1)_{d,3}.$$

Let us show that the difficult cases of the first case reduce to the second case by replacing $d$ with $d + 1$.

Case $n = (2 | \dots | 0)_{d,3}$. We then have four cases:

$$(1.a) \qquad n = (2 | 1 | \dots | 0 | 0)_{d,3}$$

$$(1.b) \qquad n = (2 | 2 | \dots | 0 | 0)_{d,3}$$

$$(1.c) \qquad n = (2 | 2 | \dots | 1 | 0)_{d,3}$$

$$(1.d) \qquad n = (2 | 2 | \dots | 2 | 0)_{d,3}$$

For case (1.a), we notice that it has a finite degree travel because:

$$\gamma_3^2((2 | 1 | \dots | 0 | 0)_{d,3}) = \gamma_3((2 | a | \dots | 1 | 2)_{d,3}) = (a | \dots | 1)_{d-2,3}$$

For case (1.b), we have either:

$$\gamma_3^2((2 | 2 | \dots | 0 | 0)_{d,3}) = \gamma_3((2 | 2 | \dots | 2 | 2)_{d,3}) = (\dots)_{d-4,3}$$

$$(1.b') \qquad \gamma_3((2 | 2 | \dots | 0 | 0)_{d,3}) = (1 | \dots | 2)_{d+1,3}$$

For cases (1.c) and (1.d), we have (with $a \in \{0, 1, 2\}$):

$$\gamma_3((2 | 2 | \dots | a | 0)_{d,3}) = (1 | \dots | 2)_{d+1,3}.$$

Thus, to show cases (1.b'), (1.c), and (1.d), it suffices to show the case:

$$(2 | \dots | 1)_{d+1,3}.$$

Thus, to show that $\mathcal{D}_3^\dagger \subset \mathcal{V}_3^\dagger$, it suffices to show that for all $n := (2 | \dots | 1)_{d+1,3}$, there exists $k \in N_1$ such that:

$$d_3(\gamma_3^k(n)) < d.$$
## 5.8 Study of $N_b \neq N_\bullet$.

In this subsection, we give the bacouettes concerning bases $b$ where there are non-normand numbers.

We begin with bases $b$ containing celt numbers.

### Bacouette 5.8.1

(BAC - 18) The first value of $b$ with non-normand integers $id \ est \ N_b \neq N_\bullet$ is $b := 9$, more precisely, it is the first value of $b$ with Breton (or celt) numbers:

$$ \min\{b \in \mathbb{N}_2, N_b \neq N_\bullet\} = \min\{b \in \mathbb{N}_2, \mathcal{F}_b \neq \emptyset\} = 9. $$

(BAC - 19) The first even value with celt numbers is $b := 18 = 2 \times 9$.

We can then study the density of integers that are at least celt in a base. By running computer simulations, the curve seems to approach a constant.

<figure><img src="image_20.png" /></figure>

The curve is:

$$ n \in [0, 10^6] \mapsto \mathrm{card}(\{n \in \mathbb{N}, \exists b \in \mathbb{N}_2^{100}, n \in \mathscr{C}_b\}) \approx \frac{1}{200}. \quad (109) $$

We did not consider bases larger than 100 containing celt numbers for easier calculation and to allow simulations on larger integers. A priori, the exact curve should be slightly higher for large values $id \ est$ there are certainly integers between 1 and $10^6$ that are celt for a base $b > 100$.

### Bacouette 5.8.2

(BAC - 20) We have:

$$ \mathrm{card}(\{n \in \mathbb{N}, \exists b \in \mathbb{N}_2, n \in \mathscr{C}_b\}) \approx \frac{1}{200}. $$

According to the computer simulations in Appendix A, it seems that there are Québécois numbers only for certain carefully chosen bases.
Bacouette 5.8.3

(BAC - 21) The first four values of b with Québécois numbers of degree 1 are:

$$258 = 2^8 + 2, \quad 512 = 2^9, \quad 1024 = 2^{10}, \quad 4096 = 2^{12}.$$

(BAC - 22) The values of b with Québécois numbers are of the type:

$$b := 2^{b'}, \quad b := 2^{b'} + 2 \quad (b' \in \mathbb{N}_6).$$

We end this section with the Sicilian numbers.

Bacouette 5.8.4

(BAC - 23) Are there Sicilian numbers in a chosen base b?
# 6 Extensions and Variants

We finish the text with various extensions and variants. In this subsection, for a given integer $n_0$ and a fixed base $b$, we will denote:

$$ \forall k \in \mathbb{N}, n_k := \gamma_b^k(n_0). \qquad (110) $$

The associated Norman sequence for $n_0$ in base $b$ will then be written with an arrow:

$$ n_0 \to n_1 \to \dots \qquad (111) $$

## 6.1 Changing the Reduction Operation

Let's consider the number $n_0 := 305\,403$ in base 10. There are two possibilities. We either completely reduce the palindromic part 30 on the left and 03 on the right as:

$$ n_0 := \textcolor{red}{305\,403} \to n_1 := 54 \to n_2 := 54 + 45 = \textcolor{red}{99} \to n_3 := \textcolor{red}{\bullet} $$

or we reduce once (removing leading 0s):

$$ n_0 := \textcolor{red}{305\,403} \to n_1 := \textcolor{blue}{0540} = 540 \to n_2 := 540 + 045 = \textcolor{red}{585} \to n_3 := \textcolor{red}{8} \to n_4 := \textcolor{red}{\bullet}. $$

We have chosen the first option in this text. The other could certainly be justified. The reason I chose the first option is for palindromic numbers like $n_0 := 3075\,703$ where with the first option we immediately get:

$$ n_0 := \textcolor{red}{3075\,703} \to n_1 := \textcolor{red}{\bullet} $$

whereas with the second option, we get:

$$ \begin{align*} n_0 := \textcolor{red}{3075\,703} &\to n_1 := \textcolor{blue}{07570} = 7570 \to n_2 := 7570 + 757 = 8327 \to n_3 := 8327 + 7238 = 15565 \\ &\to n_4 := 15565 + 56551 = 72116 \to n_5 := 72116 + 61127 = 133243 \\ &\to n_6 := 133243 + 342331 = \textcolor{red}{475574} \to n_7 := \textcolor{red}{7557} \\ &\to n_8 := \textcolor{red}{55} \to n_9 := \textcolor{red}{\bullet}. \end{align*} $$

In both cases, the final result is the same. Testing the second option computationally, we realize that bases $b$ less than 100 with celt numbers are the same. One question remains, however, whether the problematic bases with celt or Quebecois numbers remain the same?

## 6.2 Changing the Palindromization Operation

When we add a number to its mirror, we can change the addition digit by digit and take what is called an internal composition law.

Let's take for example $n_0 := 149\,823$ in base 10. Since the extreme digits of $n_0$ are different, we "add" (by multiplication) its mirror:

$$ \begin{array}{c} \textcolor{red}{1} \quad \textcolor{red}{7} \quad \textcolor{red}{7} \\ \textcolor{red}{1} \quad \textcolor{red}{4} \quad \textcolor{red}{9} \quad \textcolor{red}{8} \quad \textcolor{red}{2} \quad \textcolor{red}{3} \\ \textcolor{red}{\otimes} \quad \textcolor{red}{3} \quad \textcolor{red}{2} \quad \textcolor{red}{8} \quad \textcolor{red}{9} \quad \textcolor{red}{4} \quad \textcolor{red}{1} \\ \hline \textcolor{red}{4} \quad \textcolor{red}{5} \quad \textcolor{red}{9} \quad \textcolor{red}{2} \quad \textcolor{red}{8} \quad \textcolor{red}{3} \end{array} $$

We notice that the carry is no longer necessarily 0 or 1, but is a number less than 9 (and $b-1$ in the general case). And so on, we have:

$$ \begin{array}{c} n_0 \to n_1 := 459\,283 \to \\ \text{\Large$\otimes$} \\ \hline n_2 := \textcolor{red}{1} \quad \textcolor{red}{6} \quad \textcolor{red}{2} \quad \textcolor{red}{0} \quad \textcolor{red}{2} \quad \textcolor{red}{1} \quad \textcolor{red}{1} \end{array} \quad \begin{array}{c} \textcolor{red}{1} \quad \textcolor{red}{4} \quad \textcolor{red}{2} \quad \textcolor{red}{2} \quad \textcolor{red}{4} \quad \textcolor{red}{1} \\ \textcolor{red}{4} \quad \textcolor{red}{5} \quad \textcolor{red}{9} \quad \textcolor{red}{2} \quad \textcolor{red}{8} \quad \textcolor{red}{3} \\ \textcolor{red}{3} \quad \textcolor{red}{8} \quad \textcolor{red}{2} \quad \textcolor{red}{9} \quad \textcolor{red}{5} \quad \textcolor{red}{4} \\ \hline \textcolor{red}{6} \quad \textcolor{red}{2} \quad \textcolor{red}{0} \quad \textcolor{red}{2} \quad \textcolor{red}{1} \quad \textcolor{red}{1} \end{array} \to n_3 := 62\,021 \to \dots $$
## 6.3 Euclidean Division in $\mathbb{R}_+^*$

We can extend the decompositions in rational or real bases to rationals or reals. For example, let's take $n_0 := 38.5$ in base $b := 2.1$. We have:

$$
\begin{align*}
n_0 &= 18 \times 2.1 + 0.7 \\
&= (8 \times 2.1 + 1.2) \times 2.1 + 0.7 = 8 \times 2.1^2 + 1.2 \times 2.1 + 0.7 \\
&= (3 \times 2.1 + 1.7) \times 2.1^2 + 1.2 \times 2.1 + 0.7 = 3 \times 2.1^3 + 1.7 \times 2.1^2 + 1.2 \times 2.1 + 0.7 \\
&= (1 \times 2.1 + 0.7) \times 2.1^3 + 1.7 \times 2.1^2 + 1.2 \times 2.1 + 0.7 \\
&= \mathcolor{red}{1} \times 2.1^4 + \mathcolor{red}{0.7} \times 2.1^3 + \mathcolor{red}{1.7} \times 2.1^2 + \mathcolor{red}{1.2} \times 2.1^1 + \mathcolor{red}{0.7} \times 2.1^0
\end{align*}
$$

which we will denote as:

$$n_0 = (1 \mid 0.7 \mid 1.7 \mid 1.2 \mid 0.7)_{4,2.1}.$$

We can then apply a similar algorithm. Since the extreme digits of $n_0$ are different, we add its mirror:

$$
\begin{array}{r@{\quad}r@{\quad}r@{\quad}r@{\quad}r@{\quad}r}
 & \multicolumn{1}{c}{\mathcolor{red}{1}} & \mathcolor{red}{1} & \mathcolor{red}{1} & & & \\
 & & \mathcolor{red}{1} & 0.7 & 1.7 & 1.2 & 0.7 \\
+ & & 0.7 & 1.2 & 1.7 & 0.7 & \mathcolor{red}{1} \\
\hline
n_1 & \mathcolor{red}{1} & 0.6 & 0.8 & 1.3 & 1.9 & 1.7
\end{array}
$$

And so on, we can study its Norman sequence in base $b := 2.1$.

## 6.4 Euclidean Division in a Euclidean Ring

We can work in any Euclidean ring. For example, the ring of polynomials $\mathbb{R}[X]$ over the field of real numbers. It is easy to see that the addition "digit" by "digit" is not interesting, but multiplication is of great interest.
# A Computational Studies

In this section, we study what happens for small bases using Python programs.

## A.1 Case $\mathcal{B}_{d,b} \subset \mathcal{N}_b$

In this first subsection, we study bases $b$ between 2 and 100 such that:

$$ \mathcal{B}_{d,b} \subset \mathcal{N}_b $$

with $d \in \mathbb{N}$ such that:

$$ b^d - 1 \geq 10^9. $$

The table has four columns:

* the first column is the base $b$ considered;

* the second column is the integer $d$ considered;

* the third column is the value of $b^d - 1$;

* the fourth column is the cardinal $\text{card}(\tilde{\mathcal{D}}_{d,b})$. This represents the number of quasi-elingard dukes in base $b$, which we have shown to have a finite journey in degree in base $b$ to demonstrate that $\mathcal{B}_{d,b} \subset \mathcal{N}_b$.

For example, for $b := 3$, we had to test 7812499 quasi-elingard dukes in base 3 with degree less than 19 to show that all integers between 1 and 1162261466 are norms in base 3.

<table><thead><tr><th>b</th><th>d</th><th>b<sup>d</sup> - 1</th><th>card(\tilde{\mathcal{D}}<sub>d,b</sub>)</th></tr></thead><tbody><tr><td>3</td><td>19</td><td>3<sup>19</sup> - 1 = 1162261466</td><td>7812499</td></tr><tr><td>4</td><td>17</td><td>4<sup>17</sup> - 1 = 17179869183</td><td>19216004</td></tr><tr><td>5</td><td>15</td><td>5<sup>15</sup> - 1 = 30517578124</td><td>17936135</td></tr><tr><td>6</td><td>13</td><td>6<sup>13</sup> - 1 = 13060694015</td><td>7440558</td></tr><tr><td>7</td><td>12</td><td>7<sup>12</sup> - 1 = 13841287200</td><td>4331754</td></tr><tr><td>8</td><td>12</td><td>8<sup>12</sup> - 1 = 68719476735</td><td>9980360</td></tr><tr><td>10</td><td>11</td><td>10<sup>11</sup> - 1 = 99999999999</td><td>15131720</td></tr><tr><td>12</td><td>9</td><td>12<sup>9</sup> - 1 = 5159780351</td><td>3637932</td></tr><tr><td>14</td><td>8</td><td>14<sup>8</sup> - 1 = 1475789055</td><td>807002</td></tr><tr><td>15</td><td>10</td><td>15<sup>10</sup> - 1 = 576650390624</td><td>16671630</td></tr><tr><td>16</td><td>8</td><td>16<sup>8</sup> - 1 = 4294967295</td><td>1400176</td></tr><tr><td>20</td><td>7</td><td>20<sup>7</sup> - 1 = 1279999999</td><td>1245698</td></tr><tr><td>21</td><td>8</td><td>21<sup>8</sup> - 1 = 37822859360</td><td>2243388</td></tr><tr><td>22</td><td>8</td><td>22<sup>8</sup> - 1 = 54875873535</td><td>2707034</td></tr><tr><td>23</td><td>8</td><td>23<sup>8</sup> - 1 = 78310985280</td><td>3239090</td></tr><tr><td>27</td><td>7</td><td>27<sup>7</sup> - 1 = 10460353202</td><td>2164455</td></tr><tr><td>29</td><td>7</td><td>29<sup>7</sup> - 1 = 17249876308</td><td>2877119</td></tr><tr><td>30</td><td>7</td><td>30<sup>7</sup> - 1 = 218699999999</td><td>3293160</td></tr></tbody></table>

<table><thead><tr><th>b</th><th>d</th><th>b<sup>d</sup> - 1</th><th>card(\tilde{\mathcal{D}}<sub>d,b</sub>)</th></tr></thead><tbody><tr><td>31</td><td>7</td><td>31<sup>7</sup> - 1 = 27512614110</td><td>3752767</td></tr><tr><td>32</td><td>6</td><td>32<sup>6</sup> - 1 = 1073741823</td><td>194624</td></tr><tr><td>33</td><td>6</td><td>33<sup>6</sup> - 1 = 1291467968</td><td>213510</td></tr><tr><td>34</td><td>6</td><td>34<sup>6</sup> - 1 = 1544804415</td><td>233580</td></tr><tr><td>35</td><td>6</td><td>35<sup>6</sup> - 1 = 1838265624</td><td>254870</td></tr><tr><td>36</td><td>6</td><td>36<sup>6</sup> - 1 = 2176782335</td><td>277416</td></tr><tr><td>37</td><td>6</td><td>37<sup>6</sup> - 1 = 2565726408</td><td>301254</td></tr><tr><td>39</td><td>6</td><td>39<sup>6</sup> - 1 = 3518743760</td><td>352950</td></tr><tr><td>41</td><td>6</td><td>41<sup>6</sup> - 1 = 4750104240</td><td>410246</td></tr><tr><td>42</td><td>6</td><td>42<sup>6</sup> - 1 = 5489031743</td><td>441084</td></tr><tr><td>43</td><td>6</td><td>43<sup>6</sup> - 1 = 6321363048</td><td>473430</td></tr><tr><td>44</td><td>6</td><td>44<sup>6</sup> - 1 = 7256313855</td><td>507320</td></tr><tr><td>48</td><td>6</td><td>48<sup>6</sup> - 1 = 12230590463</td><td>659040</td></tr><tr><td>50</td><td>6</td><td>50<sup>6</sup> - 1 = 156249999999</td><td>745100</td></tr><tr><td>52</td><td>6</td><td>52<sup>6</sup> - 1 = 19770609663</td><td>838344</td></tr><tr><td>53</td><td>6</td><td>53<sup>6</sup> - 1 = 22164361128</td><td>887750</td></tr><tr><td>54</td><td>6</td><td>54<sup>6</sup> - 1 = 24794911295</td><td>939060</td></tr><tr><td>55</td><td>6</td><td>55<sup>6</sup> - 1 = 27680640624</td><td>992310</td></tr></tbody></table>
<table><thead><tr><th>b</th><th>d</th><th>b<sup>d</sup> - 1</th><th>card(\tilde{\mathcal{D}}<sub>d,b</sub>)</th></tr></thead><tbody><tr><td>56</td><td>6</td><td>56<sup>6</sup> - 1 = 30 840 979 455</td><td>1 047 536</td></tr><tr><td>57</td><td>6</td><td>57<sup>6</sup> - 1 = 34 296 447 248</td><td>1 104 774</td></tr><tr><td>58</td><td>6</td><td>58<sup>6</sup> - 1 = 38 068 692 543</td><td>1 164 060</td></tr><tr><td>59</td><td>6</td><td>59<sup>6</sup> - 1 = 42 180 533 640</td><td>1 225 430</td></tr><tr><td>60</td><td>6</td><td>60<sup>6</sup> - 1 = 46 655 999 999</td><td>1 288 920</td></tr><tr><td>61</td><td>6</td><td>61<sup>6</sup> - 1 = 51 520 374 360</td><td>1 354 566</td></tr><tr><td>62</td><td>6</td><td>62<sup>6</sup> - 1 = 56 800 235 583</td><td>1 422 404</td></tr><tr><td>64</td><td>5</td><td>64<sup>5</sup> - 1 = 1 073 741 823</td><td>532 544</td></tr><tr><td>65</td><td>5</td><td>65<sup>5</sup> - 1 = 1 160 290 624</td><td>557 765</td></tr><tr><td>66</td><td>5</td><td>66<sup>5</sup> - 1 = 1 252 332 575</td><td>583 770</td></tr><tr><td>67</td><td>5</td><td>67<sup>5</sup> - 1 = 1 350 125 106</td><td>610 571</td></tr><tr><td>68</td><td>5</td><td>68<sup>5</sup> - 1 = 1 453 933 567</td><td>638 180</td></tr><tr><td>69</td><td>5</td><td>69<sup>5</sup> - 1 = 1 564 031 348</td><td>666 609</td></tr><tr><td>70</td><td>5</td><td>70<sup>5</sup> - 1 = 1 680 699 999</td><td>695 870</td></tr><tr><td>71</td><td>5</td><td>71<sup>5</sup> - 1 = 1 804 229 350</td><td>725 975</td></tr><tr><td>72</td><td>5</td><td>72<sup>5</sup> - 1 = 1 934 917 631</td><td>756 936</td></tr><tr><td>73</td><td>5</td><td>73<sup>5</sup> - 1 = 2 073 071 592</td><td>788 765</td></tr><tr><td>74</td><td>5</td><td>74<sup>5</sup> - 1 = 2 219 006 623</td><td>821 474</td></tr><tr><td>75</td><td>5</td><td>75<sup>5</sup> - 1 = 2 373 046 874</td><td>855 075</td></tr></tbody></table>

<table><thead><tr><th>b</th><th>d</th><th>b<sup>d</sup> - 1</th><th>card(\tilde{\mathcal{D}}<sub>d,b</sub>)</th></tr></thead><tbody><tr><td>76</td><td>5</td><td>76<sup>5</sup> - 1 = 2 535 525 375</td><td>889 580</td></tr><tr><td>77</td><td>5</td><td>77<sup>5</sup> - 1 = 2 706 784 156</td><td>925 001</td></tr><tr><td>78</td><td>5</td><td>78<sup>5</sup> - 1 = 2 887 174 367</td><td>961 350</td></tr><tr><td>79</td><td>5</td><td>79<sup>5</sup> - 1 = 3 077 056 398</td><td>998 639</td></tr><tr><td>81</td><td>5</td><td>81<sup>5</sup> - 1 = 3 486 784 400</td><td>1 076 085</td></tr><tr><td>82</td><td>5</td><td>82<sup>5</sup> - 1 = 3 707 398 431</td><td>1 116 266</td></tr><tr><td>84</td><td>5</td><td>84<sup>5</sup> - 1 = 4 182 119 423</td><td>1 199 604</td></tr><tr><td>85</td><td>5</td><td>85<sup>5</sup> - 1 = 4 437 053 124</td><td>1 242 785</td></tr><tr><td>87</td><td>5</td><td>87<sup>5</sup> - 1 = 4 984 209 206</td><td>1 332 231</td></tr><tr><td>88</td><td>5</td><td>88<sup>5</sup> - 1 = 5 277 319 167</td><td>1 378 520</td></tr><tr><td>90</td><td>5</td><td>90<sup>5</sup> - 1 = 5 904 899 999</td><td>1 474 290</td></tr><tr><td>91</td><td>5</td><td>91<sup>5</sup> - 1 = 6 240 321 450</td><td>1 523 795</td></tr><tr><td>93</td><td>5</td><td>93<sup>5</sup> - 1 = 6 956 883 692</td><td>1 626 105</td></tr><tr><td>94</td><td>5</td><td>94<sup>5</sup> - 1 = 7 339 040 223</td><td>1 678 934</td></tr><tr><td>95</td><td>5</td><td>95<sup>5</sup> - 1 = 7 737 809 374</td><td>1 732 895</td></tr><tr><td>96</td><td>5</td><td>96<sup>5</sup> - 1 = 8 153 726 975</td><td>1 788 000</td></tr><tr><td>98</td><td>5</td><td>98<sup>5</sup> - 1 = 9 039 207 967</td><td>1 901 690</td></tr><tr><td>99</td><td>5</td><td>99<sup>5</sup> - 1 = 9 509 900 498</td><td>1 960 299</td></tr><tr><td>100</td><td>5</td><td>100<sup>5</sup> - 1 = 9 999 999 999</td><td>3 999 700</td></tr></tbody></table>
## A.2 Case of Bases with Triskelion

In this subsection, we give the bases $b$ less than 100 that contain celt numbers. Here is how we present the results. For each base $b$, we give:

(1) $d, b^d - 1$, $\text{card}(\tilde{\mathcal{D}}_{d,b})$: we tested all integers between 1 and $b^d - 1$. This required testing $\text{card}(\tilde{\mathcal{D}}_{d,b})$ quasi-elingard dukes.

(2) $\mathfrak{X}$: for each triskelion $\mathcal{T}$ in base $b$, we denote $\mathfrak{b} := \tau_b(\mathcal{T}) = \min \mathcal{T}$ (the Breton number in base $b$ that is the minimum of the triskelion $\mathcal{T}$), we give its identity, the smallest elingard duke $n_b$ in base $b$ that is celt in base $b$ and such that $\mathfrak{b}$ is its trisk, and the number $N_b$ of elingard dukes with the same degree as $n_b$ having $\mathfrak{b}$ as their trisk.

$$ \spadesuit \ b = 9 $$

$$ d := 12, \ b^d - 1 = 9^{12} - 1 = 282\,429\,536\,480, \ \text{card}(\tilde{\mathcal{D}}_{d,b}) = 20\,765\,412 $$

$$ \mathfrak{X} \quad \mathfrak{b} = (8 \mid 2 \mid 8 \mid 6 \mid 2 \mid 0 \mid 2 \mid 7 \mid 0 \mid 2 \mid 7)_{10,9} = 29\,043\,268\,900 : $$

$$ \text{Id}_{\mathfrak{b}}(\mathfrak{b}) = (18, (2, 16)), \quad N_{\mathfrak{b}} = 38 $$

$$ n_{\mathfrak{b}} = (8 \mid 4 \mid 8 \mid 3 \mid 3 \mid 0 \mid 0 \mid 0 \mid 7)_{8,9} = 367\,954\,009 $$

$$ \spadesuit \ b = 11 $$

$$ d := 11, \ b^d - 1 = 11^{11} - 1 = 285\,311\,670\,610, \ \text{card}(\tilde{\mathcal{D}}_{d,b}) = 26\,955\,071 $$

$$ \mathfrak{X} \quad \mathfrak{b} = (9 \mid 3 \mid 10 \mid 2 \mid 8)_{4,11} = 137\,002 : $$

$$ \text{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 1 $$

$$ n_{\mathfrak{b}} = (10 \mid 10 \mid 10 \mid 8)_{3,11} = 14\,638 $$

$$ \mathfrak{X} \quad \mathfrak{b} = (9 \mid 3 \mid 5 \mid 1 \mid 7 \mid 0 \mid 8 \mid 2 \mid 5 \mid 2 \mid 8)_{10,11} = 241\,614\,467\,410 : $$

$$ \text{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 13 $$

$$ n_{\mathfrak{b}} = (10 \mid 5 \mid 10 \mid 2 \mid 0 \mid 0 \mid 0 \mid 0 \mid 8)_{9,11} = 24\,849\,686\,155 $$

$$ \mathfrak{X} \quad \mathfrak{b} = (9 \mid 3 \mid 10 \mid 2 \mid 8 \mid 5 \mid 5 \mid 9 \mid 3 \mid 10 \mid 2 \mid 8)_{11,11} = 2\,669\,791\,201\,404 : $$

$$ \text{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 34 $$

$$ n_{\mathfrak{b}} = (10 \mid 5 \mid 10 \mid 3 \mid 0 \mid 0 \mid 0 \mid 0 \mid 0 \mid 8)_{10,11} = 273\,366\,034\,796 $$

$$ \spadesuit \ b = 13 $$

$$ d := 10, \ b^d - 1 = 13^{10} - 1 = 137\,858\,491\,848, \ \text{card}(\tilde{\mathcal{D}}_{d,b}) = 8\,040\,370 $$

$$ \mathfrak{X} \quad \mathfrak{b} = (10 \mid 5 \mid 7 \mid 4 \mid 7 \mid 4 \mid 9)_{6,13} = 50\,334\,514 : $$

$$ \text{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 4 $$

$$ n_{\mathfrak{b}} = (12 \mid 7 \mid 2 \mid 0 \mid 10)_{4,13} = 358\,459 $$
$$ \mathbf{\star} \quad b = (12 \mid 2 \mid 11 \mid 12 \mid 2 \mid 11)_{5,13} = 4538870 : $$

$$ \begin{aligned} \mathrm{Id}_b(\mathfrak{b}) &= (18, (2, 16)) , \quad N_{\mathfrak{b}} = 4 \\ n_{\mathfrak{b}} &= (12 \mid 7 \mid 2 \mid 0 \mid 11)_{4,13} = 358460 \end{aligned} $$

$$ \mathbf{\star} \quad b = (10 \mid 5 \mid 7 \mid 7 \mid 4 \mid 9)_{5,13} = 3872358 : $$

$$ \begin{aligned} \mathrm{Id}_b(\mathfrak{b}) &= (5, (1, 4)) , \quad N_{\mathfrak{b}} = 4 \\ n_{\mathfrak{b}} &= (12 \mid 9 \mid 2 \mid 0 \mid 0 \mid 7)_{5,13} = 4716966 \end{aligned} $$

$$ \mathbf{\star} \quad b = (10 \mid 5 \mid 7 \mid 7 \mid 1 \mid 11 \mid 12 \mid 2 \mid 7 \mid 7 \mid 4 \mid 9)_{11,13} = 18690958382950 : $$

$$ \begin{aligned} \mathrm{Id}_b(\mathfrak{b}) &= (5, (1, 4)) , \quad N_{\mathfrak{b}} = 4 \\ n_{\mathfrak{b}} &= (12 \mid 7 \mid 1 \mid 3 \mid 7 \mid 0 \mid 0 \mid 0 \mid 0 \mid 10)_{9,13} = 133043935528 \end{aligned} $$

$$ \spadesuit \quad b = 17 $$

$$ d := 9 , \quad b^d - 1 = 17^9 - 1 = 118587876496 , \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 11340377 $$

$$ \mathbf{\star} \quad b = (14 \mid 5 \mid 11 \mid 8 \mid 11 \mid 4 \mid 13)_{6,17} = 345986546 : $$

$$ \begin{aligned} \mathrm{Id}_b(\mathfrak{b}) &= (5, (1, 4)) , \quad N_{\mathfrak{b}} = 44 \\ n_{\mathfrak{b}} &= (16 \mid 9 \mid 6 \mid 0 \mid 0 \mid 0 \mid 11)_{6,17} = 399480954 \end{aligned} $$

$$ \mathbf{\star} \quad b = (14 \mid 5 \mid 11 \mid 12 \mid 12 \mid 11 \mid 4 \mid 13)_{7,17} = 5882112162 : $$

$$ \begin{aligned} \mathrm{Id}_b(\mathfrak{b}) &= (5, (1, 4)) , \quad N_{\mathfrak{b}} = 8 \\ n_{\mathfrak{b}} &= (16 \mid 9 \mid 6 \mid 8 \mid 0 \mid 0 \mid 0 \mid 11)_{7,17} = 6791844210 \end{aligned} $$

$$ \spadesuit \quad b = 18 $$

$$ d := 9 , \quad b^d - 1 = 18^9 - 1 = 198359290367 , \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 15094530 $$

$$ \mathbf{\star} \quad b = (17 \mid 4 \mid 4 \mid 5 \mid 4 \mid 16)_{5,18} = 32567596 : $$

$$ \begin{aligned} \mathrm{Id}_b(\mathfrak{b}) &= (84, (6, 78)) , \quad N_{\mathfrak{b}} = 2 \\ n_{\mathfrak{b}} &= (17 \mid 14 \mid 8 \mid 0 \mid 16)_{4,18} = 1868848 \end{aligned} $$

$$ \spadesuit \quad b = 19 $$

$$ d := 9 , \quad b^d - 1 = 19^9 - 1 = 322687697778 , \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 19782819 $$

$$ \mathbf{\star} \quad b = (16 \mid 5 \mid 13 \mid 13 \mid 4 \mid 15)_{5,19} = 40363140 : $$

$$ \begin{aligned} \mathrm{Id}_b(\mathfrak{b}) &= (5, (1, 4)) , \quad N_{\mathfrak{b}} = 8 \\ n_{\mathfrak{b}} &= (18 \mid 8 \mid 7 \mid 0 \mid 15)_{4,19} = 2403192 \end{aligned} $$
$$ \spadesuit \ b = 24 $$

$$ d := 7, \ b^d - 1 = 24^7 - 1 = 4586471423, \ \text{card}(\tilde{\mathcal{D}}_{d,b}) = 1354224 $$

$$ \blacklozenge \quad \mathfrak{b} = (22 \mid 16 \mid 21)_{2,24} = 13077 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (9, (1, 8)), \quad N_{\mathfrak{b}} = 2 $$

$$ n_{\mathfrak{b}} = (23 \mid 16 \mid 20)_{2,24} = 13652 $$

$$ \spadesuit \ b = 25 $$

$$ d := 7, \ b^d - 1 = 25^7 - 1 = 6103515624, \ \text{card}(\tilde{\mathcal{D}}_{d,b}) = 1593175 $$

$$ \blacklozenge \quad \mathfrak{b} = (22 \mid 5 \mid 19 \mid 16 \mid 19 \mid 4 \mid 21)_{6,25} = 5427605746 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 16 $$

$$ n_{\mathfrak{b}} = (24 \mid 8 \mid 17 \mid 0 \mid 0 \mid 21)_{5,25} = 237765646 $$

$$ \blacklozenge \quad \mathfrak{b} = (22 \mid 5 \mid 19 \mid 4 \mid 4 \mid 19 \mid 4 \mid 21)_{7,25} = 135685230746 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 98 $$

$$ n_{\mathfrak{b}} = (24 \mid 11 \mid 17 \mid 6 \mid 0 \mid 0 \mid 20)_{6,25} = 5973531270 $$

$$ \spadesuit \ b = 26 $$

$$ d := 7, \ b^d - 1 = 26^7 - 1 = 8031810175, \ \text{card}(\tilde{\mathcal{D}}_{d,b}) = 1862432 $$

$$ \blacklozenge \quad \mathfrak{b} = (25 \mid 10 \mid 11 \mid 15 \mid 11 \mid 9 \mid 24)_{6,26} = 7847006230 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (51, (7, 44)), \quad N_{\mathfrak{b}} = 8 $$

$$ n_{\mathfrak{b}} = (25 \mid 19 \mid 10 \mid 2 \mid 0 \mid 0 \mid 24)_{6,26} = 7953245480 $$

$$ \spadesuit \ b = 28 $$

$$ d := 7, \ b^d - 1 = 28^7 - 1 = 13492928511, \ \text{card}(\tilde{\mathcal{D}}_{d,b}) = 2501800 $$

$$ \blacklozenge \quad \mathfrak{b} = (22 \mid 10 \mid 9 \mid 21)_{3,28} = 491057 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 1 $$

$$ n_{\mathfrak{b}} = (27 \mid 8 \mid 21)_{2,28} = 21413 $$

$$ \blacklozenge \quad \mathfrak{b} = (25 \mid 5 \mid 1 \mid 26 \mid 27 \mid 2 \mid 4 \mid 24)_{7,28} = 339766450152 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 4 $$

$$ n_{\mathfrak{b}} = (27 \mid 8 \mid 21 \mid 16 \mid 0 \mid 0 \mid 24)_{6,28} = 13161980184 $$

$$ \spadesuit \ b = 38 $$

$$ d := 6, \ b^d - 1 = 38^6 - 1 = 3010936383, \ \text{card}(\tilde{\mathcal{D}}_{d,b}) = 326420 $$
$$ \mathbf{\star} \quad b = (35 \mid 18 \mid 33 \mid 17 \mid 34)_{4,38} = 74015788 : $$

$$ \text{Id}_b(\mathfrak{b}) = (9, (1, 8)), \quad N_{\mathfrak{b}} = 2 $$
$$ n_{\mathfrak{b}} = (37 \mid 35 \mid 33 \mid 0 \mid 32)_{4,38} = 79118236 $$

$$ \spadesuit \quad b = 40 $$

$$ d := 6, \quad b^d - 1 = 40^6 - 1 = 4095999999, \quad \text{card}(\tilde{\mathcal{D}}_{d,b}) = 380880 $$

$$ \mathbf{\star} \quad \mathfrak{b} = (37 \mid 5 \mid 4 \mid 36)_{3,40} = 2376196 : $$

$$ \text{Id}_b(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 1 $$
$$ n_{\mathfrak{b}} = (39 \mid 4 \mid 36)_{2,40} = 62596 $$

$$ \spadesuit \quad b = 45 $$

$$ d := 6, \quad b^d - 1 = 45^6 - 1 = 8303765624, \quad \text{card}(\tilde{\mathcal{D}}_{d,b}) = 542790 $$

$$ \mathbf{\star} \quad \mathfrak{b} = (35 \mid 17 \mid 14 \mid 14 \mid 16 \mid 34)_{5,45} = 6529499854 : $$

$$ \text{Id}_b(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 12 $$
$$ n_{\mathfrak{b}} = (44 \mid 15 \mid 37 \mid 0 \mid 0 \mid 42)_{5,45} = 8184118542 $$

$$ \spadesuit \quad b = 46 $$

$$ d := 6, \quad b^d - 1 = 46^6 - 1 = 9474296895, \quad \text{card}(\tilde{\mathcal{D}}_{d,b}) = 579876 $$

$$ \mathbf{\star} \quad \mathfrak{b} = (45 \mid 6 \mid 44)_{2,46} = 95540 : $$

$$ \text{Id}_b(\mathfrak{b}) = (117, (10, 107)), \quad N_{\mathfrak{b}} = 1 $$
$$ n_{\mathfrak{b}} = (45 \mid 6 \mid 44)_{2,46} = 95540 $$

$$ \mathbf{\star} \quad \mathfrak{b} = (44 \mid 22 \mid 21 \mid 43)_{3,46} = 4330345 : $$

$$ \text{Id}_b(\mathfrak{b}) = (9, (1, 8)), \quad N_{\mathfrak{b}} = 1 $$
$$ n_{\mathfrak{b}} = (45 \mid 12 \mid 43)_{2,46} = 95815 $$

$$ \spadesuit \quad b = 47 $$

$$ d := 6, \quad b^d - 1 = 47^6 - 1 = 10779215328, \quad \text{card}(\tilde{\mathcal{D}}_{d,b}) = 618614 $$

$$ \mathbf{\star} \quad \mathfrak{b} = (40 \mid 41 \mid 39)_{2,47} = 90326 : $$

$$ \text{Id}_b(\mathfrak{b}) = (6, (1, 5)), \quad N_{\mathfrak{b}} = 2 $$
$$ n_{\mathfrak{b}} = (46 \mid 18 \mid 33)_{2,47} = 102493 $$
$$ \mathbf{\star} \quad \mathfrak{b} = (35 \mid 20 \mid 34 \mid 35 \mid 20 \mid 34)_{5,47} = 8128277136 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 6 $$
$$ n_{\mathfrak{b}} = (46 \mid 40 \mid 23 \mid 0 \mid 0 \mid 23)_{5,47} = 10747445514 $$

$$ \spadesuit \quad b = 49 $$

$d := 6, \quad b^d - 1 = 49^6 - 1 = 13841287200, \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 701190$

$$ \mathbf{\star} \quad \mathfrak{b} = (46 \mid 5 \mid 7 \mid 7 \mid 4 \mid 45)_{5,49} = 13023526050 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 28 $$
$$ n_{\mathfrak{b}} = (48 \mid 17 \mid 16 \mid 0 \mid 42)_{4,49} = 278748939 $$

$$ \spadesuit \quad b = 51 $$

$d := 6, \quad b^d - 1 = 51^6 - 1 = 17596287800, \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 790806$

$$ \mathbf{\star} \quad \mathfrak{b} = (44 \mid 12 \mid 11 \mid 43)_{3,51} = 5868460 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 2 $$
$$ n_{\mathfrak{b}} = (50 \mid 22 \mid 43)_{2,51} = 131215 $$

$$ \spadesuit \quad b = 63 $$

$d := 6, \quad b^d - 1 = 63^6 - 1 = 62523502208, \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 1492470$

$$ \mathbf{\star} \quad \mathfrak{b} = (47 \mid 27 \mid 46)_{2,63} = 188290 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 3 $$
$$ n_{\mathfrak{b}} = (62 \mid 27 \mid 31)_{2,63} = 247810 $$

$$ \spadesuit \quad b = 80 $$

$d := 5, \quad b^d - 1 = 80^5 - 1 = 3276799999, \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 1036880$

$$ \mathbf{\star} \quad \mathfrak{b} = (1 \mid 57 \mid 58)_{2,80} = 11018 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 1 $$
$$ n_{\mathfrak{b}} = (79 \mid 34 \mid 40)_{2,80} = 508360 $$

$$ \spadesuit \quad b = 83 $$

$d := 5, \quad b^d - 1 = 83^5 - 1 = 3939040642, \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 1157435$
$$ \mathbf{\vdash} \quad \mathfrak{b} = (64 \mid 33 \mid 3 \mid 3 \mid 32 \mid 63)_{5,83} = 253\,666\,464\,492 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 4 $$
$$ n_{\mathfrak{b}} = (82 \mid 52 \mid 37 \mid 0 \mid 63)_{4,83} = 3\,921\,570\,202 $$

$$ \spadesuit \quad b = 86 $$

$$ d := 5, \quad b^d - 1 = 86^5 - 1 = 4\,704\,270\,175, \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 1\,286\,990 $$

$$ \mathbf{\vdash} \quad \mathfrak{b} = (85 \mid 18 \mid 17 \mid 84)_{3,86} = 54\,199\,434 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (26, (4, 22)), \quad N_{\mathfrak{b}} = 1 $$
$$ n_{\mathfrak{b}} = (85 \mid 52 \mid 84)_{2,86} = 633\,216 $$

$$ \spadesuit \quad b = 89 $$

$$ d := 5, \quad b^d - 1 = 89^5 - 1 = 5\,584\,059\,448, \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 1\,425\,869 $$

$$ \mathbf{\vdash} \quad \mathfrak{b} = (71 \mid 31 \mid 30 \mid 70)_{3,89} = 50\,301\,090 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 2 $$
$$ n_{\mathfrak{b}} = (88 \mid 61 \mid 0 \mid 53)_{3,89} = 62\,520\,506 $$

$$ \spadesuit \quad b = 92 $$

$$ d := 5, \quad b^d - 1 = 92^5 - 1 = 6\,590\,815\,231, \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 1\,574\,396 $$

$$ \mathbf{\vdash} \quad \mathfrak{b} = (90 \mid 74 \mid 73 \mid 89)_{3,92} = 70\,715\,061 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (9, (1, 8)), \quad N_{\mathfrak{b}} = 1 $$
$$ n_{\mathfrak{b}} = (91 \mid 38 \mid 89)_{2,92} = 773\,809 $$

$$ \spadesuit \quad b = 97 $$

$$ d := 5, \quad b^d - 1 = 97^5 - 1 = 8\,587\,340\,256, \quad \mathrm{card}(\tilde{\mathcal{D}}_{d,b}) = 1\,844\,261 $$

$$ \mathbf{\vdash} \quad \mathfrak{b} = (73 \mid 41 \mid 72)_{2,97} = 690\,906 : $$

$$ \mathrm{Id}_{\mathfrak{b}}(\mathfrak{b}) = (5, (1, 4)), \quad N_{\mathfrak{b}} = 2 $$
$$ n_{\mathfrak{b}} = (96 \mid 41 \mid 49)_{2,97} = 907\,290 $$
### A.3 celt Numbers Less Than $10^4$

In this subsection, we give all the integers $n$ less than $10^4$ that are celt in a base $b$. The table contains five columns:

* the first column is the number of the integer (its ranking);

* the second column is the considered base $b$;

* the third column is the considered integer $n$ and its decomposition in base $b$;

* the fourth column is the trisk associated with $n$ in base $b$;

* the fifth column is the identity of the trisk associated with $n$ in base $b$.

We have highlighted trisk values that appear frequently with the same color (in red and blue).

<table><thead><tr><th>N°</th><th>b</th><th>n</th><th>τ<sub>b</sub>(n)</th><th>Id<sub>b</sub><sup>†</sup>(n)</th></tr></thead><tbody><tr><td>1</td><td>24</td><td>1076 = (1 | 20 | 20)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>2</td><td>24</td><td>1113 = (1 | 22 | 9)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>3</td><td>24</td><td>1126 = (1 | 22 | 22)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>4</td><td>24</td><td>1138 = (1 | 23 | 10)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>5</td><td>24</td><td>1651 = (2 | 20 | 19)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>6</td><td>24</td><td>1688 = (2 | 22 | 8)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>7</td><td>24</td><td>1701 = (2 | 22 | 21)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>8</td><td>24</td><td>1713 = (2 | 23 | 9)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>9</td><td>24</td><td>2226 = (3 | 20 | 18)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>10</td><td>24</td><td>2263 = (3 | 22 | 7)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>11</td><td>24</td><td>2276 = (3 | 22 | 20)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>12</td><td>24</td><td>2288 = (3 | 23 | 8)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>13</td><td>40</td><td>2516 = (1 | 22 | 36)<sub>2,40</sub></td><td>2376196 = (37 | 5 | 4 | 36)<sub>3,40</sub></td><td>(5, (1, 4))</td></tr><tr><td>14</td><td>24</td><td>2801 = (4 | 20 | 17)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>15</td><td>24</td><td>2838 = (4 | 22 | 6)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>16</td><td>24</td><td>2851 = (4 | 22 | 19)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>17</td><td>40</td><td>2857 = (1 | 31 | 17)<sub>2,40</sub></td><td>2376196 = (37 | 5 | 4 | 36)<sub>3,40</sub></td><td>(5, (1, 4))</td></tr><tr><td>18</td><td>24</td><td>2863 = (4 | 23 | 7)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>19</td><td>46</td><td>3355 = (1 | 26 | 43)<sub>2,46</sub></td><td>95540 = (45 | 6 | 44)<sub>2,46</sub></td><td>(117, (10, 107))</td></tr><tr><td>20</td><td>24</td><td>3376 = (5 | 20 | 16)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>21</td><td>24</td><td>3426 = (5 | 22 | 18)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>22</td><td>24</td><td>3438 = (5 | 23 | 6)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>23</td><td>24</td><td>3951 = (6 | 20 | 15)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>24</td><td>24</td><td>3988 = (6 | 22 | 4)<sub>2,24</sub></td><td>13077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr></tbody></table>
<table><thead><tr><th>N°</th><th>b</th><th>n</th><th>τ<sub>b</sub>(n)</th><th>Id<sup>†</sup><sub>b</sub>(n)</th></tr></thead><tbody><tr><td>25</td><td>24</td><td>4 001 = (6 | 22 | 17)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>26</td><td>24</td><td>4 013 = (6 | 23 | 5)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>27</td><td>40</td><td>4 115 = (2 | 22 | 35)<sub>2,40</sub></td><td>2 376 196 = (37 | 5 | 4 | 36)<sub>3,40</sub></td><td>(5, (1, 4))</td></tr><tr><td>28</td><td>40</td><td>4 456 = (2 | 31 | 16)<sub>2,40</sub></td><td>2 376 196 = (37 | 5 | 4 | 36)<sub>3,40</sub></td><td>(5, (1, 4))</td></tr><tr><td>29</td><td>24</td><td>4 526 = (7 | 20 | 14)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>30</td><td>24</td><td>4 563 = (7 | 22 | 3)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>31</td><td>24</td><td>4 576 = (7 | 22 | 16)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>32</td><td>24</td><td>4 588 = (7 | 23 | 4)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>33</td><td>24</td><td>5 101 = (8 | 20 | 13)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>34</td><td>24</td><td>5 138 = (8 | 22 | 2)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>35</td><td>51</td><td>5 145 = (1 | 49 | 45)<sub>2,51</sub></td><td>5 868 460 = (44 | 12 | 11 | 43)<sub>3,51</sub></td><td>(5, (1, 4))</td></tr><tr><td>36</td><td>24</td><td>5 151 = (8 | 22 | 15)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>37</td><td>24</td><td>5 163 = (8 | 23 | 3)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>38</td><td>46</td><td>5 470 = (2 | 26 | 42)<sub>2,46</sub></td><td>95 540 = (45 | 6 | 44)<sub>2,46</sub></td><td>(117, (10, 107))</td></tr><tr><td>39</td><td>24</td><td>5 676 = (9 | 20 | 12)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>40</td><td>24</td><td>5 713 = (9 | 22 | 1)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>41</td><td>40</td><td>5 714 = (3 | 22 | 34)<sub>2,40</sub></td><td>2 376 196 = (37 | 5 | 4 | 36)<sub>3,40</sub></td><td>(5, (1, 4))</td></tr><tr><td>42</td><td>24</td><td>5 726 = (9 | 22 | 14)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>43</td><td>24</td><td>5 738 = (9 | 23 | 2)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>44</td><td>40</td><td>6 055 = (3 | 31 | 15)<sub>2,40</sub></td><td>2 376 196 = (37 | 5 | 4 | 36)<sub>3,40</sub></td><td>(5, (1, 4))</td></tr><tr><td>45</td><td>24</td><td>6 251 = (10 | 20 | 11)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>46</td><td>24</td><td>6 288 = (10 | 22 | 0)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>47</td><td>24</td><td>6 301 = (10 | 22 | 13)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>48</td><td>24</td><td>6 313 = (10 | 23 | 1)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>49</td><td>24</td><td>6 826 = (11 | 20 | 10)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>50</td><td>63</td><td>6 849 = (1 | 45 | 45)<sub>2,63</sub></td><td>188 290 = (47 | 27 | 46)<sub>2,63</sub></td><td>(5, (1, 4))</td></tr><tr><td>51</td><td>24</td><td>6 876 = (11 | 22 | 12)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>52</td><td>24</td><td>6 888 = (11 | 23 | 0)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>53</td><td>40</td><td>7 313 = (4 | 22 | 33)<sub>2,40</sub></td><td>2 376 196 = (37 | 5 | 4 | 36)<sub>3,40</sub></td><td>(5, (1, 4))</td></tr><tr><td>54</td><td>24</td><td>7 401 = (12 | 20 | 9)<sub>2,24</sub></td><td>13 077 = (22 | 16 | 21)<sub>2,24</sub></td><td>(9, (1, 8))</td></tr><tr><td>55</td><td>63</td><td>7 432 = (1 | 54 | 61)<sub>2,63</sub></td><td>188 290 = (47 | 27 | 46)<sub>2,63</sub></td><td>(5, (1, 4))</td></tr></tbody></table>