# AN ANALYSIS OF THE COLLATZ CONJECTURE: SUPPLEMENTAL FORMAL STATEMENTS

EMANUEL J. THOMPSON

ABSTRACT. Let $n$ be a natural number belonging to a specific group based on its final digit. By investigating the behavior of $n$ under a series of transformations derived from the standard Collatz operation and implementing these group-based transformations, this structured dynamic analysis can treat $n$ as a unique arbitrary element within the natural numbers. The study explores how the dynamics of the Collatz Conjecture across different number groups provide a potential pathway for further logical analysis.

## INTRODUCTION

This paper examines the Collatz Conjecture through the mapping of numbers through group assignments in reference $|1|$ and their respective transformation rules, using proof structures discussed in Schütte's *Proof Theory* [2] to create a deterministic approach. The Collatz map, assigns $Col(N) := 3N + 1$ for odd $N$ and $Col(N) := N/2$ for even $N$. This study explores how these transformations contribute to understanding the conjecture's dynamics. To support the probabilistic methods of reference $|3|$

### 1. GROUP ASSIGNMENT AND LINEAR FORMULAS

For every natural number $n$, there exists a unique group $G_x(n)$ to which $n$ belongs.

$$ \forall n \exists! G_x(n) $$

Or by the finite group set theorem:

$$ \exists G_1, G_2, \dots, G_k (\forall n \exists G_i \text{ such that } G_x(n) = G_i) $$

Each number is uniquely assigned to a group based on its transformation, which conveniently corresponds to its final digit.

$$ G_A(n) = \{n \mid n \equiv 0 \pmod{10} \lor n \equiv 5 \pmod{10}\} $$

$$ G_B(n) = \{n \mid n \equiv 1 \pmod{10} \lor n \equiv 6 \pmod{10}\} $$

$$ G_C(n) = \{n \mid n \equiv 2 \pmod{10} \lor n \equiv 7 \pmod{10}\} $$

$$ G_D(n) = \{n \mid n \equiv 3 \pmod{10} \lor n \equiv 8 \pmod{10}\} $$

$$ G_E(n) = \{n \mid n \equiv 4 \pmod{10} \lor n \equiv 9 \pmod{10}\} $$

When $n$ represents the universal index, each group $G_x(n)$ is described by a linear function, where the value of $n$ determines the specific sequence within that group.

<footnote>Date: July 31, 2024.</footnote>
Each group follows a linear formula of the form:

$$G_x(n) = 5n + c$$

where c is a constant specific to the group:

* Group A has $c = 0$

* Group B has $c = 1$

* Group C has $c = 2$

* Group D has $c = 3$

* Group E has $c = 4$

Representing each group in this form as follows:

### Group A:

$$\text{For } n = 0, \quad G_E(0) = 5(0) + 0 = 0$$

$$\text{For } n = 1, \quad G_E(1) = 5(1) + 0 = 5$$

$$\text{For } n = 2, \quad G_E(2) = 5(2) + 0 = 10$$

$\vdots$

$$G_E(n) = 5n$$

### Group B:

$$\text{For } n = 0, \quad G_E(0) = 5(0) + 1 = 1$$

$$\text{For } n = 1, \quad G_E(1) = 5(1) + 1 = 6$$

$$\text{For } n = 2, \quad G_E(2) = 5(2) + 1 = 11$$

$\vdots$

$$G_E(n) = 5n + 1$$

### Group C:

$$\text{For } n = 0, \quad G_E(0) = 5(0) + 2 = 2$$

$$\text{For } n = 1, \quad G_E(1) = 5(1) + 2 = 7$$

$$\text{For } n = 2, \quad G_E(2) = 5(2) + 2 = 12$$

$\vdots$

$$G_E(n) = 5n + 2$$

### Group D:

$$\text{For } n = 0, \quad G_E(0) = 5(0) + 3 = 3$$

$$\text{For } n = 1, \quad G_E(1) = 5(1) + 3 = 8$$

$$\text{For } n = 2, \quad G_E(2) = 5(2) + 3 = 13$$

$\vdots$

$$G_E(n) = 5n + 3$$
Group E:

$$
\text{For } n = 0, \quad G_E(0) = 5(0) + 4 = 4
$$

$$
\text{For } n = 1, \quad G_E(1) = 5(1) + 4 = 9
$$

$$
\text{For } n = 2, \quad G_E(2) = 5(2) + 4 = 14
$$

∴

$$
G_E(n) = 5n + 4
$$

Where *n* represents the universal index of some given number. Using this com-
mon index approach allows a method of tracking the progression of numbers as
they undergo transformations within defined parameters.

<table><caption>Table 1.1: Number Groups with their applicable subsections.</caption>
  <thead>
    <tr>
      <th></th>
      <th>0</th>
      <th>1</th>
      <th>2</th>
      <th>3</th>
      <th>4</th>
      <th>5</th>
      <th>6</th>
      <th>...</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <th>A</th>
      <td>0</td>
      <td>5</td>
      <td>10</td>
      <td>15</td>
      <td>20</td>
      <td>25</td>
      <td>30</td>
      <td>...</td>
    </tr>
    <tr>
      <th>&delta;<sub>A</sub></th>
      <td>10</td>
      <td>40</td>
      <td>70</td>
      <td>100</td>
      <td>130</td>
      <td>160</td>
      <td>190</td>
      <td>...</td>
    </tr>
    <tr>
      <th>B</th>
      <td>1</td>
      <td>6</td>
      <td>11</td>
      <td>16</td>
      <td>21</td>
      <td>26</td>
      <td>31</td>
      <td>...</td>
    </tr>
    <tr>
      <th>&delta;<sub>B</sub></th>
      <td>16</td>
      <td>46</td>
      <td>76</td>
      <td>106</td>
      <td>136</td>
      <td>166</td>
      <td>196</td>
      <td>...</td>
    </tr>
    <tr>
      <th>C</th>
      <td>2</td>
      <td>7</td>
      <td>12</td>
      <td>17</td>
      <td>22</td>
      <td>27</td>
      <td>32</td>
      <td>...</td>
    </tr>
    <tr>
      <th>&delta;<sub>C</sub></th>
      <td>22</td>
      <td>52</td>
      <td>82</td>
      <td>112</td>
      <td>142</td>
      <td>172</td>
      <td>202</td>
      <td>...</td>
    </tr>
    <tr>
      <th>D</th>
      <td>3</td>
      <td>8</td>
      <td>13</td>
      <td>18</td>
      <td>23</td>
      <td>28</td>
      <td>33</td>
      <td>...</td>
    </tr>
    <tr>
      <th>&delta;<sub>D</sub></th>
      <td>28</td>
      <td>58</td>
      <td>88</td>
      <td>118</td>
      <td>148</td>
      <td>178</td>
      <td>208</td>
      <td>...</td>
    </tr>
    <tr>
      <th>E</th>
      <td>4</td>
      <td>9</td>
      <td>14</td>
      <td>19</td>
      <td>24</td>
      <td>29</td>
      <td>34</td>
      <td>...</td>
    </tr>
    <tr>
      <th>&delta;<sub>E</sub></th>
      <td>34</td>
      <td>64</td>
      <td>94</td>
      <td>124</td>
      <td>154</td>
      <td>184</td>
      <td>214</td>
      <td>...</td>
    </tr>
  </tbody>
</table>

2. DERIVATION OF TRANSFORMATION RULES FROM THE COLLATZ CONJECTURE

2.1. Group-Based Transformation Propositions.
Group A (Numbers ending in 0 or 5):

$$
\forall n \in \mathbb{N}, \text{ if } n \equiv 0 \pmod{10} \text{ or } n \equiv 5 \pmod{10}, \\
\text{then: } \begin{cases} n \text{ is odd} & \Rightarrow T(n) = 3n + 1 \text{ ends in 6, moving to Group B,} \\ n \text{ is even} & \Rightarrow T(n) = \frac{n}{2} \text{ stays in Group A.} \end{cases}
$$

Group B (Numbers ending in 1 or 6):

$$
\forall n \in \mathbb{N}, \text{ if } n \equiv 1 \pmod{10} \text{ or } n \equiv 6 \pmod{10}, \\
\text{then: } \begin{cases} n \text{ is odd} & \Rightarrow T(n) = 3n + 1 \text{ ends in 4, moving to Group E,} \\ n \text{ is even} & \Rightarrow T(n) = \frac{n}{2} \text{ ends in 3 or 8, moving to Group D.} \end{cases}
$$

Group C (Numbers ending in 2 or 7):

$$
\forall n \in \mathbb{N}, \text{ if } n \equiv 2 \pmod{10} \text{ or } n \equiv 7 \pmod{10}, \\
\text{then: } \begin{cases} n \text{ is odd} & \Rightarrow T(n) = 3n + 1 \text{ ends in 2, staying in Group C,} \\ n \text{ is even} & \Rightarrow T(n) = \frac{n}{2} \text{ ends in 1 or 6, moving to Group B.} \end{cases}
$$

Group D (Numbers ending in 3 or 8):
$$ \forall n \in \mathbb{N}, \text{ if } n \equiv 3 \pmod{10} \text{ or } n \equiv 8 \pmod{10}, $$

$$ \text{then: } \begin{cases} n \text{ is odd} & \Rightarrow T(n) = 3n + 1 \text{ ends in 0, moving to Group A,} \\ n \text{ is even} & \Rightarrow T(n) = \frac{n}{2} \text{ ends in 4 or 9, moving to Group E.} \end{cases} $$

**Group E (Numbers ending in 4 or 9):**

$$ \forall n \in \mathbb{N}, \text{ if } n \equiv 4 \pmod{10} \text{ or } n \equiv 9 \pmod{10}, $$

$$ \text{then: } \begin{cases} n \text{ is odd} & \Rightarrow T(n) = 3n + 1 \text{ ends in 8, staying in Group E,} \\ n \text{ is even} & \Rightarrow T(n) = \frac{n}{2} \text{ ends in 2 or 7, moving to Group C or Group B.} \end{cases} $$

The Collatz Conjecture states that for any natural number $n$:

* If $n$ is even, divide it by 2.

* If $n$ is odd, multiply it by 3 and add 1.

This rule can be extended to analyze the behavior of numbers based on their final digits, leading to the following group-based transformations.

## 2.2. Group-Based Transformation Rules with Examples.

* **Group A (Numbers ending in 0 or 5):**

Applying $3n + 1$ to any odd number in Group A will result in a number ending in 6, moving it to Group B if $n$ is even, applying $n/2$ will maintain the resultant within Group A.

- **Example:** $n = 5, 3 \times 5 + 1 = 16$ which is ends in 6 and belongs to group B.

- **Generalization:** For any $n$ in Group A, if has the form $n = 10m + 0$ (even) or $n = 10m + 5$ (odd) for some integer $m$, then applying $n/2$ results in $5m + 0$ (for even) and applying $3n + 1$ results in $30m + 16$ (for odd). with an odd number always ending in 6 moving to group B.

* **Group B (Numbers ending in 1 or 6):**

Applying $3n + 1$ to any odd number in Group B will result in a number ending in 4, moving it to Group E. If $n$ is even applying $n/2$ will move the resultant in 3 or 8 moving to Group D.

- **Example:** $n = 11, 3 \times 11 + 1 = 34$, which ends in 4 and moves to Group E.

- **Generalization:** For any $n$ in Group B, if it has the form $n = 10m+1$ (odd) or $n = 10m + 6$ (even), then applying $n/2$ results in $5m + 3$ (for even) with the resultant numbers ending in 3 or 8 and applying $3n+1$ results in $30m + 4$ (for odd), with the resultant number always ending in 4, moving to Group E.

* **Group C (Numbers ending in 2 or 7):**

Applying $3n + 1$ to any odd number in Group C will result in a number ending in 2, maintaining the resultant number in Group C. If $n$ is even, applying $n/2$ will move the resultant to group B.

- **Example:** $n = 7, 3 \times 7 + 1 = 22$, which ends in 2 and stays within Group C.

- **Generalization:** For any $n$ in Group C, if it has the form $n = 10m+2$ (even) or $n = 10m + 7$ (odd), then applying $n/2$ results in $5m + 1$ (for even) causing a resultant to end in 1 or 6 and applying $3n+1$ results in
$30m+22$ (for odd), with the resultant always ending in 2, maintaining it in Group C.

• **Group D (Numbers ending in 3 or 8):**

Applying $3n + 1$ to any odd number in Group D will result in a number ending in 0, moving the resultant to Group A. If $n$ is even, applying $n/2$ will result in a number ending in 4 or 9, moving the resultant to Group E.

- **Example:** $n = 3, 3 \times 3 + 1 = 10$, which ends in 0 and moves to Group A.

- **Generalization:** For any $n$ in Group D, if it has the form $n = 10m+3$ (odd) or $n = 10m+8$ (even), then applying $3n+1$ results in $30m+10$ (for odd) or $n/2$ results in $5m+4$ (for even), with the resultant number always ending in 0 (moving to Group A) or ending in 4 or 9 (moving to Group E).

• **Group E (Numbers ending in 4 or 9):**

Applying $3n + 1$ to any odd number in Group E will result in a number ending in 8, staying within Group E. If $n$ is even, applying $n/2$ will result in a number ending in 2 or 7, moving to Group C.

- **Example:** $n = 9, 3 \times 9 + 1 = 28$, which ends in 8 and stays within Group E.

- **Generalization:** For any $n$ in Group E, if it has the form $n = 10m+4$ (even) or $n = 10m+9$ (odd), then applying $3n+1$ results in $30m+28$ (for odd) or applying $n/2$ results in $5m+2$ (for even), with the resultant number ending in 8 for 0dd (moving to Group D) or ending in 2 or 7 (moving to Group C).

### 3. TRANSFORMATION PRESERVATION OF GROUP STRUCTURE

$$ \forall n (\exists G_x(n) \implies \exists G_y(T_x(n))) $$

Or for group transposition

$$ \forall n (\exists G_x(n) \wedge T_x(n) \implies G_x(T_x(n)) \vee G_y(T_x(n))) $$

Applying the transformation $T_x(n)$ to any number $n$ either keeps the number within its current group or predictably moves it to another group.

#### **Sequent Calculus by Group and Parity.**

**Group A:** $G_A(n) \vdash G_{AB}(T_A(n))$. **Odd statement:** $G_A(n) \vdash T_A(n)$

- $n$ is odd, therefore $T_A(n) = 3n$

- **Sequent:** $G_A(n) \vdash G_B(3n)$

**Even statement:** $G_A(n) \vdash T_A(n)$

- $n$ is even, therefore $T_A(n) = \frac{n}{2}$

- **Sequent:** $G_A(n) \vdash G_A(\frac{n}{2})$

**Group B:** $G_B(n) \vdash G_{DE}(T_B(n))$. **Odd statement:** $G_B(n) \vdash T_B(n)$

- $n$ is odd, therefore $T_B(n) = \frac{n-1}{2}$

- **Sequent:** $G_B(n) \vdash G_D(\frac{n-1}{2})$

**Even statement:** $G_B(n) \vdash T_B(n)$

- $n$ is even, therefore $T_A(n) = 3n$
• Sequent: $G_A(n) \vdash G_D(3n)$

Group C: $G_C(n) \vdash G_{BC}(T_C(n))$. Odd statement: $G_C(n) \vdash T_C(n)$

• $n$ is odd, therefore $T_C(n) = 3n + 1$

• Sequent: $G_C(n) \vdash G_C(3n + 1)$

Even statement: $G_C(n) \vdash T_C(n)$

• $n$ is even, therefore $T_C(n) = \frac{n}{2}$

• Sequent: $G_C(n) \vdash G_B(\frac{n}{2})$

Group D: $G_D(n) \vdash G_{AE}(T_D(n))$. Odd statement: $G_D(n) \vdash T_D(n)$

• $n$ is odd, therefore $T_D(n) = \frac{n-1}{2}$

• Sequent: $G_D(n) \vdash G_E(\frac{n-1}{2})$

Even statement: $G_D(n) \vdash T_D(n)$

• $n$ is even, therefore $T_D(n) = 3n + 2$

• Sequent: $G_D(n) \vdash G_A(3n + 2)$

Group E: $G_E(n) \vdash G_{CD}(T_E(n))$. Odd statement: $G_E(n) \vdash T_E(n)$

• $n$ is odd, therefore $T_E(n) = 3n + 2$

• Sequent: $G_C(n) \vdash G_D(3n + 2)$

Even statement: $G_E(n) \vdash T_E(n)$

• $n$ is even, therefore $T_E(n) = \frac{n}{2}$

• Sequent: $G_E(n) \vdash G_C(\frac{n}{2})$

Transformations are consistent and predictable, preserving the structure and allowing us to track a number's progress through groups.

### 4. DIRECTIONALITY AND COLOR PRESERVATION

**Directionality:**

$$D_x(n) = \begin{cases} \text{Up,} & n \equiv 0 \text{ or } 2 \pmod 8 \\ \text{Down,} & n \equiv 1 \text{ or } 3 \pmod 8 \\ \text{Left,} & n \equiv 4 \text{ or } 6 \pmod 8 \\ \text{Right,} & n \equiv 5 \text{ or } 7 \pmod 8 \end{cases}$$

**Color:**

$$C_x(n) = \begin{cases} \text{Red,} & n \equiv 0 \text{ or } 1 \pmod 4 \\ \text{Green,} & n \equiv 2 \text{ or } 3 \pmod 4 \end{cases}$$

For every natural number $n$, its directionality and color are uniquely determined and are preserved or predictably altered under transformations.

$$\forall n(D_x(n) \implies \exists D_y(T_x(n))) \text{ and } \forall n(\exists C_x(n) \implies \exists C_y(T_x(n)))$$

Since a number's behavior and sequential transformations are controlled, and it follows predictable paths under transformation.

By introducing directionality and color, additional layers of predictability are added under the Collatz transformation. These properties help describe how numbers evolve through the sequence.
## 5. CYCLE EXISTENCE AND CONVERGENCE

$$ \forall n \exists m (\exists G_x(n) \implies \exists Cycle(T^m(n))) $$

After a sufficient number of transformations, a number will enter a repeating cycle.

**Known Cycle (4-2-1):**

$$ \exists Cycle(4, 2, 1) : T_x(n) \in \{4, 2, 1\} \implies T_x(T_x(T_x(n))) = n $$

Every sequence generated repeatedly by applying $T_x(n)$ will eventually enter a known cycle.

## 6. STABILIZATION AND CYCLE RETURN

Once a sequence stabilizes under transformation, i.e. $T_x^m(n) = T_x^{m+1}(n)$, it must be part of the known cycle.

$$ \forall n (\exists m (T_x^m(n) = T_x^{m+1}(n) \implies Cycle(n))) $$

Any stabilized sequence is part of the known cycle.

## 7. MONOTONICITY OF TRANSFORMATION AND BOUNDEDNESS

$$ \forall n (\exists G_x(n) \implies T_x(n) \le n \lor T_x(n) > n) \land \exists N (\forall n < N \exists Cycle(T_x(n))) $$

Therefore a sequence cannot continue indefinitely without eventually reducing in magnitude or repeating, leading to the known cycle.

## 8. COLOR-BASED $n_{i+1}$ TRANSFORMS

### 8.1. Formal Definition.
Let $n_i$ represent an arbitrary element in the universal index. For each $n_i$, we define the following attributes:

* **Color $C(n_i)$:** A function that assigns a color to $n_i$. The color is either Red or Green based on specific properties of $n_i$ (e.g., modulo operation).

* **Directionality $D(n_i)$:** A function that assigns a direction to $n_i$. Directionality can be Up, Down, Left, or Right, and is determined by the current state of $n_i$ and its transformation.

* **Parity $P(n_i)$:** A function that determines whether $n_i$ is Even or Odd.

The transformations of $n_{i+1}$ are governed by the rules that depend on the values of $C(n_i)$, $D(n_i)$, and $P(n_i)$.

### 8.2. Red $n_{i+1}$ Transforms.

* Each number $n_i$ in the cycle has a color (Red or Green), direction (Up, Down, Left, or Right), and Parity (Even or Odd).

* If $n_i$ is even, Up, and Red: $n_{i+1}$ will be odd, remain in the Red group, and change directionality to Down.

* If $n_i$ is even, Left, and Red: $n_{i+1}$ will be odd, remain in the Red group, and change directionality to Right.

* If $n_i$ is odd, Down, and Red: $n_{i+1}$ will be even, move into the Green group, and change directionality to Up.
* If $n_i$ is odd, Right, and Red: $n_{i+1}$ will be even, move into the Green group, and change directionality to Left.

### 8.3. Green $n_{i+1}$ Transforms.

* If $n_i$ is even, Up, and Green: $n_{i+1}$ will be odd, remain in the Green group, and change directionality to Down.

* If $n_i$ is even, Left, and Green: $n_{i+1}$ will be odd, remain in the Green group, and change directionality to Right.

* If $n_i$ is odd, Down, and Green: $n_{i+1}$ will be even, move into the Red group, and change directionality to Left.

* If $n_i$ is odd, Right, and Green: $n_{i+1}$ will be even, move into the Red group, and change directionality to Up.

## 9. DIRECTIONAL CHANGES INDEPENDENT OF COLOR

When referencing $n_i$ as an element of the universal index, when the index $i$ is applied to the respective formulae, the resulting number will have a specific direction, color, and parity as follows:

* If $n_i$ is even, Up, and Red: $n_{i+1}$ will be odd, Red, and Right.

* If $n_i$ is odd, Down, and Red: $n_{i+1}$ will be even, Green, and Left.

* If $n_i$ is even, Up, and Green: $n_{i+1}$ will be odd, Green, and Right.

* If $n_i$ is odd, Down, and Green: $n_{i+1}$ will be even, Red, and Up.

* If $n_i$ is even, Left, and Red: $n_{i+1}$ will be odd, Red, and Down.

* If $n_i$ is odd, Right, and Red: $n_{i+1}$ will be even, Green, and Up.

* If $n_i$ is even, Left, and Green: $n_{i+1}$ will be odd, Green, and Down.

* If $n_i$ is odd, Right, and Green: $n_{i+1}$ will be even, Red, and Left.

## 10. THEOREMS AND LEMMAS

**Theorem 10.1.** Every number belongs to a specific group based on its final digit, and adding 1 to the number will move it to the next appropriate group.

*Proof.* Induction will be used to prove that every number $n$ belongs to a specific group based on its final digit, and that adding 1 to $n$ will move the number to the next group.

**Group Definitions:**

* Group A: Numbers ending in 0 or 5.

* Group B: Numbers ending in 1 or 6.

* Group C: Numbers ending in 2 or 7.

* Group D: Numbers ending in 3 or 8.

* Group E: Numbers ending in 4 or 9.

**Base Case:**

Consider the numbers from 0 to 9. By definition:

* 0 and 5 belong to Group A.

* 1 and 6 belong to Group B.

* 2 and 7 belong to Group C.

* 3 and 8 belong to Group D.

* 4 and 9 belong to Group E.

Adding 1 to any of these numbers moves them to the next group as follows:

* 0 (Group A) becomes 1 (Group B).
• 1 (Group B) becomes 2 (Group C).

• 2 (Group C) becomes 3 (Group D).

• 3 (Group D) becomes 4 (Group E).

• 4 (Group E) becomes 5 (Group A).

• 5 (Group A) becomes 6 (Group B).

• 6 (Group B) becomes 7 (Group C).

• 7 (Group C) becomes 8 (Group D).

• 8 (Group D) becomes 9 (Group E).

• 9 (Group E) becomes 0 (Group A).

Thus, the base case holds true for all single-digit numbers.

**Inductive Step:**

Assume that for a number $n$, where $n \ge 0$, the number $n$ belongs to a specific group based on its final digit, and adding 1 to $n$ transitions it to the next appropriate group.

Now, let's consider the number $n+1$. The final digit of $n+1$ is $(n \bmod 10) + 1$ (modulo 10):

• If $n$ ends in 0, then $n+1$ ends in 1, moving from Group A to Group B.

• If $n$ ends in 1, then $n+1$ ends in 2, moving from Group B to Group C.

• If $n$ ends in 2, then $n+1$ ends in 3, moving from Group C to Group D.

• If $n$ ends in 3, then $n+1$ ends in 4, moving from Group D to Group E.

• If $n$ ends in 4, then $n+1$ ends in 5, moving from Group E to Group A.

• If $n$ ends in 5, then $n+1$ ends in 6, moving from Group A to Group B.

• If $n$ ends in 6, then $n+1$ ends in 7, moving from Group B to Group C.

• If $n$ ends in 7, then $n+1$ ends in 8, moving from Group C to Group D.

• If $n$ ends in 8, then $n+1$ ends in 9, moving from Group D to Group E.

• If $n$ ends in 9, then $n+1$ ends in 0, moving from Group E to Group A.

Thus, by induction, if $n$ belongs to a specific group and adding 1 moves it to the next group, then $n+1$ will indeed belong to the next group.

**Conclusion:**

By induction, every natural number $n$ belongs to a specific group based on its final digit, and adding 1 to the number will move it to the next appropriate group. $\square$

**Lemma 10.2.** Each and every number belongs to one and only one color group, Red or Green, based on its properties.

*Proof.* Induction will be used to prove that every number $n$ belongs to one and only one color group, Red or Green, based on its properties.

**Red Assignment:** If $n \bmod 4 \in \{0, 1\}$, then $n$ is Red.

**Green Assignment:** If $n \bmod 4 \in \{2, 3\}$, then $n$ is Green.

**Base Case:**

Consider the numbers from 0 to 3:

• 0 $\bmod 4 = 0$, so 0 is Red.

• 1 $\bmod 4 = 1$, so 1 is Red.

• 2 $\bmod 4 = 2$, so 2 is Green.

• 3 $\bmod 4 = 3$, so 3 is Green.

Thus, the base case holds true for numbers 0 to 3.

**Inductive Step:**
Assume that for a number $n$, where $n \ge 0$, the number $n$ belongs to one and only one color group (Red or Green) based on its properties. That is, $n \bmod 4$ determines whether $n$ is Red or Green.

Now, consider the number $n+1$. We must show that $n+1$ belongs to the correct color group based on $(n+1) \bmod 4$:

* If $n \bmod 4 = 0$, then $(n+1) \bmod 4 = 1$, so $n+1$ is Red.

* If $n \bmod 4 = 1$, then $(n+1) \bmod 4 = 2$, so $n+1$ is Green.

* If $n \bmod 4 = 2$, then $(n+1) \bmod 4 = 3$, so $n+1$ is Green.

* If $n \bmod 4 = 3$, then $(n+1) \bmod 4 = 0$, so $n+1$ is Red.

Thus, by the inductive hypothesis, if $n$ belongs to one and only one color group, then $n+1$ also belongs to one and only one correct color group.

**Conclusion:**

By induction, every natural number $n$ belongs to one and only one color group, Red or Green, based on its properties. $\square$

**Theorem 10.3.** Each and every number has a unique direction (Up, Down, Left, Right) based on its properties.

*Proof.* Induction will be used to prove that each and every number has a unique direction (Up, Down, Left, Right) based on its properties.

**Definitions:**

* An Up number is defined as $n \bmod 8 \in \{0, 2\}$.

* An Down number is defined as $n \bmod 8 \in \{1, 3\}$.

* An Left number is defined as $n \bmod 8 \in \{4, 6\}$.

* An Right number is defined as $n \bmod 8 \in \{5, 7\}$.

**Base Case:**

Consider the numbers from 0 to 7:

* 0 $\bmod 8 = 0$, so 0 is Up.

* 1 $\bmod 8 = 1$, so 1 is Down.

* 2 $\bmod 8 = 2$, so 2 is Up.

* 3 $\bmod 8 = 3$, so 3 is Down.

* 4 $\bmod 8 = 4$, so 4 is Left.

* 5 $\bmod 8 = 5$, so 5 is Right.

* 6 $\bmod 8 = 6$, so 6 is Left.

* 7 $\bmod 8 = 7$, so 7 is Right.

Thus, the base case holds true for numbers 0 to 7.

**Inductive Step:**

Assume that for some number $n$, where $n \ge 0$, the number $n$ has a unique direction based on its properties as defined above. That is, $n \bmod 8$ determines whether $n$ is Up, Down, Left, or Right.

Now, let's consider the number $n+1$. We must show that $n+1$ belongs to the correct direction based on $(n+1) \bmod 8$:

* If $n \bmod 8 = 0$, then $(n+1) \bmod 8 = 1$, so $n+1$ is Down.

* If $n \bmod 8 = 1$, then $(n+1) \bmod 8 = 2$, so $n+1$ is Up.

* If $n \bmod 8 = 2$, then $(n+1) \bmod 8 = 3$, so $n+1$ is Down.

* If $n \bmod 8 = 3$, then $(n+1) \bmod 8 = 4$, so $n+1$ is Left.

* If $n \bmod 8 = 4$, then $(n+1) \bmod 8 = 5$, so $n+1$ is Right.

* If $n \bmod 8 = 5$, then $(n+1) \bmod 8 = 6$, so $n+1$ is Left.
• If $n \bmod 8 = 6$, then $(n+1) \bmod 8 = 7$, so $n+1$ is Right.

• If $n \bmod 8 = 7$, then $(n+1) \bmod 8 = 0$, so $n+1$ is Up.

Thus, by induction, if $n$ has a unique direction, then $n+1$ also has a unique direction based on its properties.

The proof involves calculating $n \bmod 8$ and assigning the direction based on the result, ensuring that each number is uniquely assigned to one direction.

**Conclusion:**

By induction, each and every number has a unique direction (Up, Down, Left, Right) based on its properties. □

**Lemma 10.4.** Whenever using the universal index, $n_{i+1} = n_i + 1$, the number, when not in universal index notation calculated by the Universal sequence formula, will change by 5.

*Proof.* We induction to prove that whenever using the universal index, $n_{i+1} = n_i + 1$, the number, when not in universal index notation calculated by the Universal sequence formula, will change by 5.

Let $n_i$ be a number in the universal index.

**Base Case:**

Consider the initial number $n_0$. By the Universal sequence formula, this number corresponds to some value $n_0$ that is consistent with the group to which it belongs. When we apply the sequence formula, $n_1 = n_0 + 1$, the change is determined by the group's rule, and specifically, $n_1$ will be 5 more than the previous number if it is not calculated in universal index notation within the same applicable group.

For example, consider the following group-specific transformations:

• Group A: $G_A(n) = 5n + 0$

• Group B: $G_B(n) = 5n + 1$

• Group C: $G_C(n) = 5n + 2$

• Group D: $G_D(n) = 5n + 3$

• Group E: $G_E(n) = 5n + 4$

If $n_0$ belongs to Group A, then $G_A(n_1) = 5(n_0 + 1) + 0 = 5n_0 + 5$. Similarly, for Group B, $G_B(n_1) = 5n_0 + 6$, and this follows for each group. The difference between $G_A(n_1)$ and $G_A(n_0)$ is 5, and similarly for the other groups. This confirms the base case.

**Inductive Step:**

Assume that for some number $n_i$, the statement holds true; that is, $n_{i+1} = n_i + 1$, and the number, when not in universal index notation calculated by the Universal sequence formula, will change by 5.

Now, consider $n_{i+2} = n_{i+1} + 1$. By the inductive hypothesis, $n_{i+1}$ has already changed by 5 from $n_i$, and applying the sequence formula again will result in $n_{i+2}$ also being 5 more than $n_{i+1}$ when not calculated in universal index notation.

For example, applying the group-specific transformations again:

• Group A: $G_A(n_{i+2}) = 5(n_{i+1} + 1) + 0 = 5n_{i+1} + 5$

• Group B: $G_B(n_{i+2}) = 5n_{i+1} + 6$

• Group C: $G_C(n_{i+2}) = 5n_{i+1} + 7$

• Group D: $G_D(n_{i+2}) = 5n_{i+1} + 8$

• Group E: $G_E(n_{i+2}) = 5n_{i+1} + 9$

Here, the difference between $G_A(n_{i+2})$ and $G_A(n_{i+1})$ is again 5, confirming that $n_{i+2}$ is 5 more than $n_{i+1}$. The same logic applies to the other groups.
Thus, $n_{i+2}$ will be 5 more than $n_{i+1}$, consistent with the sequence formula and the group's rules.

This completes the lemma.

**Lemma 10.5.** Each transformation rule preserves group structure.

*Proof.* Induction will be used to prove that each transformation rule preserves group structure.

Let $n_i$ be some number in Group A, B, C, D, or E, with each group defined by its last digit number:

**Group Definitions:**

* If the last digit is 0 or 5, $n$ is in Group A.

* If the last digit is 1 or 6, $n$ is in Group B.

* If the last digit is 2 or 7, $n$ is in Group C.

* If the last digit is 3 or 8, $n$ is in Group D.

* If the last digit is 4 or 9, $n$ is in Group E.

**Base Case:**

Consider the number $n_0$ in any of the groups A, B, C, D, or E. For example, if $n_0$ belongs to Group A, its last digit must be 0 or 5. When a transformation is applied, it should preserve the group structure by keeping the last digit in the same set.

For example:

* Group A: $n_0$ ends in 0 or 5, and $T_A(n_0)$ results in a number that still ends in 0 or 5.

* Group B: $n_0$ ends in 1 or 6, and $T_B(n_0)$ results in a number that still ends in 1 or 6.

* Group C: $n_0$ ends in 2 or 7, and $T_C(n_0)$ results in a number that still ends in 2 or 7.

* Group D: $n_0$ ends in 3 or 8, and $T_D(n_0)$ results in a number that still ends in 3 or 8.

* Group E: $n_0$ ends in 4 or 9, and $T_E(n_0)$ results in a number that still ends in 4 or 9.

The base case is verified by showing that the transformation preserves the last digit and thus the group structure.

**Inductive Step:**

Assume that for some number $n_i$, the transformation rule preserves the group structure. That is, if $n_i$ belongs to a specific group, then after applying the transformation, $n_{i+1} = T(n_i)$ will still belong to the corresponding group.

Now, consider the next number $n_{i+1}$ after applying the transformation. We must show that the group structure is preserved for $n_{i+1}$ and $n_{i+2} = T(n_{i+1})$.

For example:

* If $n_i$ is in Group A (ends in 0 or 5), then $T_A(n_i)$ will result in a number that ends in 0 or 5, meaning $n_{i+1}$ is still in Group A.

* If $n_i$ is in Group B (ends in 1 or 6), then $T_B(n_i)$ will result in a number that ends in 1 or 6, meaning $n_{i+1}$ is still in Group B.

* If $n_i$ is in Group C (ends in 2 or 7), then $T_C(n_i)$ will result in a number that ends in 2 or 7, meaning $n_{i+1}$ is still in Group C.

* If $n_i$ is in Group D (ends in 3 or 8), then $T_D(n_i)$ will result in a number that ends in 3 or 8, meaning $n_{i+1}$ is still in Group D.
• If $n_i$ is in Group E (ends in 4 or 9), then $T_E(n_i)$ will result in a number that ends in 4 or 9, meaning $n_{i+1}$ is still in Group E.

This confirms that after each transformation, that group structure is preserved, and the last digit (and thus the group) remains consistent.

**Conclusion:**

Each transformation rule preserves group structure, ensuring that a number remains in its original group or transitions predictably within the defined group sets based on its last digit. □

**Theorem 10.6.** For any number in the universal index, applying all of the rules of transformation will eventually lead to the 4-2-1 cycle.

*Proof.* For any number $n$ in the universal index, applying the transformation rules of the index will eventually lead to a known cycle, such as the 4-2-1 loop.

**Sequent Calculus by Group and Parity:**

We will use sequent calculus to track how numbers move between groups:

Group A: $G_A(n) \vdash G_A \text{ or } B(T_A(n))$.

• Odd statement: $G_A(n) \vdash T_A(n)$
$n$ is odd, therefore $T_A(n) = 3n$
Sequent: $G_A(n) \vdash G_B(3n)$

• Even statement: $G_A(n) \vdash T_A(n)$
$n$ is even, therefore $T_A(n) = \frac{n}{2}$
Sequent: $G_A(n) \vdash G_A(\frac{n}{2})$

Group B: $G_B(n) \vdash G_D \text{ or } E(T_B(n))$.

• Odd statement: $G_B(n) \vdash T_B(n)$
$n$ is odd, therefore $T_B(n) = \frac{n-1}{2}$
Sequent: $G_B(n) \vdash G_D(\frac{n-1}{2})$

• Even statement: $G_B(n) \vdash T_B(n)$
$n$ is even, therefore $T_B(n) = 3n$
Sequent: $G_B(n) \vdash G_D(3n)$

Group C: $G_C(n) \vdash G_B \text{ or } C(T_C(n))$.

• Odd statement: $G_C(n) \vdash T_C(n)$
$n$ is odd, therefore $T_C(n) = 3n + 1$
Sequent: $G_C(n) \vdash G_C(3n + 1)$

• Even statement: $G_C(n) \vdash T_C(n)$
$n$ is even, therefore $T_C(n) = \frac{n}{2}$
Sequent: $G_C(n) \vdash G_B(\frac{n}{2})$

Group D: $G_D(n) \vdash G_A \text{ or } E(T_D(n))$.

• Odd statement: $G_D(n) \vdash T_D(n)$
$n$ is odd, therefore $T_D(n) = \frac{n-1}{2}$
Sequent: $G_D(n) \vdash G_E(\frac{n-1}{2})$

• Even statement: $G_D(n) \vdash T_D(n)$
$n$ is even, therefore $T_D(n) = 3n + 2$
Sequent: $G_D(n) \vdash G_A(3n + 2)$
Group E: $G_E(n) \vdash G_C \text{ or } D(T_E(n))$.

* Odd statement: $G_E(n) \vdash T_E(n)$
$n$ is odd, therefore $T_E(n) = 3n + 2$
Sequent: $G_E(n) \vdash G_D(3n + 2)$

* Even statement: $G_E(n) \vdash T_E(n)$
$n$ is even, therefore $T_E(n) = \frac{n}{2}$
Sequent: $G_E(n) \vdash G_C(\frac{n}{2})$

**Base Case:**

Consider the number $n_0 = 0$ in the universal index in group E. Applying the transformation rules:

* $G_E(T(0)) = G_C(0) = 4$ (Starting from $n=0$ in Group $G_E$, applying the transformation moves the number to $G_C(0)$, which corresponds to 4 in Group $G_C$).

* $G_C(T(0)) = 2$ (Continuing from $n=0$ in Group $G_C$, applying the transformation gives 2 in Group $G_B$).

* $G_B(T(0)) = 1$ (Continuing from $n=0$ in Group $G_B$, applying the transformation gives 1 in Group $G_E$).

* $G_E(T(0)) = G_C(0) = 4$ (Returning to $n=0$ in Group $G_E$, applying the transformation again moves it to $G_C(0)$, which corresponds to 4 in Group $G_C$).

This sequence leads directly to the known 4-2-1 cycle, confirming the base case.

**Inductive Step:**

Assume that for a number $n_i$ in the universal index, the transformation rules eventually lead to a known cycle. That is, after applying the rules for a finite number of steps, $n_i$ will enter a cycle.

Now, consider the number $n_{i+1}$. We will show that applying the transformation rules to $n_{i+1}$ will also eventually lead to the same cycle.

**Cycle Example:**

The sequence for an arbitrary starting number, such as 6, would proceed as follows:

* $G_D(T(0)) = G_A(2)$ (Starting from $n=0$ in Group $G_D$, applying the transformation for even $n=0$ gives 2 in Group $G_A$).

* $G_A(T(2)) = G_A(1)$ (Continuing from $G_A(2)$, since $n=2$ is even, the transformation gives 1 in Group $G_A$).

* $G_A(T(1)) = G_B(3)$ (Continuing from $G_A(1)$, since $n=1$ is odd, the transformation gives 3 in Group $G_B$).

* $G_B(T(3)) = G_D(1)$ (Continuing from $G_B(3)$, since $n=3$ is odd, the transformation gives 1 in Group $G_D$).

* $G_D(T(1)) = G_E(0)$ (Continuing from $G_D(1)$, since $n=1$ is odd, the transformation gives 0 in Group $G_E$).

* $G_E(T(0)) = G_C(0)$ (Continuing from $G_E(0)$, since $n=0$ is even, the transformation stays within Group $G_C$).

* $G_C(T(0)) = G_B(0)$ (Continuing from $G_C(0)$, since $n=0$ is even, the transformation gives 0 in Group $G_B$).

* $G_B(T(0)) = G_D(0)$ (Finally, returning from $G_B(0)$ to $G_D(0)$, the transformation stays in Group $G_D$, and the sequence can loop or continue depending on the input).
Thus, the transformations bring the sequence into the 4-2-1 loop.

**Conclusion:**

Using sequent calculus to track group transitions, for any number in the universal index, and applying all the rules of transformation will eventually lead to the 4-2-1 cycle. □

## 11. CONCLUSION

By analyzing the behavior of natural numbers under group-based transformations derived from the Collatz operation, and using a structured dynamic transformations that treats each number as a unique element within the natural numbers, given the uniqueness of group assignments, the structure of transformations, and the invariance of directional properties. This analysis can provide a potential pathway for sequence creation and further exploration of the Collatz Conjecture.

$$ \forall n \in \mathbb{N}, \exists m \in \mathbb{N} : T_x^m(n) = T_x^{m+1}(n) \implies \exists G_x(n), G_y(T_x^m(n)) \text{ with } G_y(T_x^m(n)) \subseteq KC $$

## REFERENCES

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[3] Terence Tao. *Almost all orbits of the Collatz map attain almost bounded values*. arXiv preprint arXiv:1909.03562, 2019.

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