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Continued Fractions With Numerator

$(v + 1 + n)(v + 2 + n)(v + 3 + n)(v + 4 + n)$

David Naccache, Ofer Yifrach-Stav

► To cite this version:

David Naccache, Ofer Yifrach-Stav. Continued Fractions With Numerator $(v+1+n)(v+2+n)(v+3+n)(v+4+n)$. 2023. hal-04092469

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https://hal.science/hal-04092469

Preprint submitted on 9 May 2023

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Continued Fractions With Numerator

$$ (v + 1 + n)(v + 2 + n)(v + 3 + n)(v + 4 + n) $$

David Naccache¹ and Ofer Yifrach-Stav¹

DIÉNS, ÉNS, CNRS, PSL University, Paris, France
45 rue d'Ulm, 75230, Paris CEDEX 05, France
ofer.friedman@ens.fr, david.naccache@ens.fr

**Abstract.** Let $a_{n,v} = (v + 1 + n)(v + 2 + n)(v + 3 + n)(v + 4 + n)$ for $v \in \{0, -1\}$ and let $b_{n,v,c,t} = c((3 + v + n)^2 - t)$ for $t \in \{0, 1\}$. We consider the continued fractions:

$$ Q(v, c, t) = \mathcal{K}_{n=1}^{\infty} \left( \frac{a_{n,v}}{b_{n,v,c,t}} \right) $$

Denoting $d = \sqrt{c^2 + 4}$ We observe that:

$$ Q(v, c, t) = \frac{t}{2} (d-c) (5v+8) + \frac{(t-1)(v+3)\left(-3c^3 + 3\sqrt{d^3} + c(v-13)\right)}{c^2(v-2) - (v-3)^2} $$

Because we do not provide a proper peer-reviewed proof of the relations given here we do not claim them to be theorems, they were however intensively machine-checked.

# 1 Introduction

Let $a_n$ and $b_n$ be two nonzero polynomials and consider the continued fraction:

$$ Q = \mathcal{K}_{n=1}^{\infty} \left( \frac{a_n}{b_n} \right) $$

Trivially, if $a_k = 0$ for some $k$, the nested summation stops and we immediately get $Q$.

In particular $\forall b_n$, $a_{n,v} = (v + 1 + n)(v + 2 + n)(v + 3 + n)(v + 4 + n)$ and $v \le -6$:

$$ Q(v) = \mathcal{K}_{n=1}^{\infty} \left( \frac{a_{n,v}}{b_n} \right) = \mathcal{K}_{n=1}^{-v-5} \left( \frac{a_{n,v}}{b_n} \right) $$

For $-5 \le v \le -2$, $Q(v) = 0$. It is thus interesting to examine what happens for $v \in \{0, -1\}$.

Because we do not provide a proper peer-reviewed proof of the relations given here we do not claim them to be theorems, they were however intensively machine-checked.
As announced in the abstract, defining $b_{n,v,c,t} = c((3+v+n)^2-t)$, denoting $d = \sqrt{c^2+4}$ and

$$Q(v, c, t) = \mathcal{K}_{n=1}^{\infty} \left( \frac{a_{n,v}}{b_{n,v,c,t}} \right)$$

it appears that for $v \in \{0, -1\}$ and $t \in \{0, 1\}$:

$$Q(v, c, t) = \frac{t}{2}(d-c)(5v+8) + \frac{(t-1)(v+3)(-3c^3 + 3\sqrt{d^3} + c(v-13))}{c^2(v-2) - (v-3)^2}$$

With:

$$\ell(v,t) = \lim_{c \to \infty} cQ(v,c,t) = \frac{(t-1)(3+v)(5+v)}{v-2} + t(5v+8)$$

As checked by the following code:

1 Union[Flatten[
2 Table[Expand[((-1 + t) (-3 c^3 + 3 Sqrt[(4 + c^2)^3] +
3 c (-13 + v)) (3 + v))/((-3 + v)^2 + c^2 (-2 + v)) +
4 1/2 (-c + Sqrt[4 + c^2]) t (8 + 5 v)] ==
5 Expand[RootApproximant[
6 N[ContinuedFractionK[(1 + X + v)*(2 + X + v)*(3 + X + v)*(4 +
7 X + v), c ((3 + v + X)^2 - t), {X, 1, 1000}], 20]]], {t, 0,
8 1}, {v, -1, 0}, {c, 1, 10}]]][1]]

A better readable split-form is given in Table 1.

## 2 Generating specific radicals

Inspired by $[1]$, one might want to look for $Q(v, c, t)$ featuring only specific radicals. We illustrate how to generate such relations with:

$$Q(0, c, 0) = \mathcal{K}_{n=1}^{\infty} \left( \frac{(1+n)(2+n)(3+n)(4+n)}{c(3+n)^2} \right) = \frac{-39c - 9c^3 + 9\sqrt{(4+c^2)^3}}{9+2c^2}$$

If $\tau(c^2 + 4)$ is a square<fnref n="1" /> then $Q(0, c, 0)$ contains the radical $\sqrt{\tau}$.

Note that there are several direct ways to efficiently generate specific radicals, e.g. $Q(0, c, 0)$ values related to the golden ratio are $Q(0, |q_5(i)|, 0)$ where:

$$q_5(i) = \begin{cases} -1 & \text{if } i \le 1 \\ -3q_5(i-1) - q_5(i-2) & \text{otherwise.} \end{cases}$$

This yields identities such as, e.g. $41Q(0, 4, 0) = 88Q(0, 1, 0) - 348$.

<footnote n="1">¹ i.e. a solution of the generalized Pell equation $\tau(c^2 + 4) = y^2$.</footnote>
<table><caption>Table 1: The conjectured relations.</caption><thead><tr><th>v</th><th>t</th><th>a<sub>n,v</sub></th><th>b<sub>n,v,c,t</sub></th><th>Q(v, c, t)</th><th>ℓ(v, t)</th></tr></thead><tbody><tr><td>0</td><td>0</td><td>(1 + n) × ... × (4 + n)</td><td>c(3 + n)<sup>2</sup></td><td><sup>3(-3c<sup>3</sup> + 3√(c<sup>2</sup> + 4)<sup>3</sup> - 13c)</sup> / (-2c<sup>2</sup> - 9)</td><td><sup>15</sup> / 2</td></tr><tr><td>0</td><td>1</td><td>(1 + n) × ... × (4 + n)</td><td>c((3 + n)<sup>2</sup> - 1)</td><td>4(√c<sup>2</sup> + 4 - c)</td><td>8</td></tr><tr><td>-1</td><td>0</td><td>n × ... × (3 + n)</td><td>c(2 + n)<sup>2</sup></td><td><sup>2(-3c<sup>3</sup> + 3√(c<sup>2</sup> + 4)<sup>3</sup> - 14c)</sup> / (-3c<sup>2</sup> - 16)</td><td><sup>8</sup> / 3</td></tr><tr><td>-1</td><td>1</td><td>n × ... × (3 + n)</td><td>c((2 + n)<sup>2</sup> - 1)</td><td><sup>3</sup> / 2(√c<sup>2</sup> + 4 - c)</td><td>3</td></tr></tbody></table>

Similarly, to provoke the appearance of a $\sqrt{2}$ use:

$$ q_2(i) = \begin{cases} 12i + 2 & \text{if } i \le 1 \\ 6q_2(i - 1) - q_2(i - 2) & \text{otherwise.} \end{cases} $$

Another possibility is to generate the fast increasing sequences:

$$ c_{k,i} = \left\lfloor \left( k + \frac{\sqrt{k^2 + 4}}{2} \right)^{2i+1} \right\rfloor $$

Where $\forall i \in \mathbb{N}$ each k value generates in $Q(0, c_{k,i}, 0)$ a radical $\sqrt{\tau}$ where $\tau$ is the square-free part of $(k^2 + 4)/4^{k+1 \mod 2}$ (OEIS A013946).

The exact same observations are also valid for $\{v, t\} = \{0, 1\}, \{-1, 0\}, \{-1, 1\}$.

### 3 Further investigations

The above indicates that it might be interesting to study thoroughly the behavior continued fractions having a $a_n$ of the form:

$$ a_{n,v,\kappa} = (v + \kappa n)(v + 1 + \kappa n)(v + 2 + \kappa n)(v + 3 + \kappa n) $$

One such algebraic continued fraction is given in [1], hence it is very likely that more could be discovered by automatic or algebraic exploration.

### References

1. H.-C. Chan. Golden ratio and a ramanujan-type integral. *Axioms*, 2(1):58–66, 2013.
<table><caption>Table 2: Radicals $\sqrt{\tau}$ appearing for all $i \in \mathbb{N}$ in $Q(0, c_{k,i}, 0)$ for $1 \le k \le 15$.</caption><thead><tr><th>k</th><th>causes the appearance in Q(0, c<sub>k,i</sub>, 0) of the radical √τ</th></tr></thead><tbody><tr><td>1, 4, 11</td><td>√5</td></tr><tr><td>2, 14</td><td>√2</td></tr><tr><td>3</td><td>√13</td></tr><tr><td>5</td><td>√29</td></tr><tr><td>6</td><td>√10</td></tr><tr><td>7</td><td>√53</td></tr><tr><td>8</td><td>√17</td></tr><tr><td>9</td><td>√85</td></tr><tr><td>10</td><td>√26</td></tr><tr><td>12</td><td>√37</td></tr><tr><td>13</td><td>√173</td></tr><tr><td>15</td><td>√229</td></tr></tbody></table>