Bidirectional Dynamics in the Collatz Problem: A Complete Resolution

1. Introduction

1.1. The Collatz Problem: Definition and Historical Context

The Collatz conjecture inhabits that rare mathematical territory where elementary formulation meets extraordinary resistance to resolution. At its core lies a deceptively simple iterative process defined on the positive integers.

Definition 1

(Collatz Function). For any positive integer n, the Collatz function $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ is defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}},\hfill & ifniseven\hfill \\ 3n+1,\hfill & ifn\mathit{is}\mathit{odd}\hfill \end{array}\right.$$

(1)

Conjecture 1

(Collatz Conjecture). For any positive integer n, the sequence ${\left(C^k\left(n\right)\right)}_{k\ge 0}$ obtained by repeatedly applying the Collatz function eventually reaches 1.

To illustrate, consider the sequence starting from $n=7$:

$$7\stackrel{3n+1}{\to }22\stackrel{n/2}{\to }11\stackrel{3n+1}{\to }34\stackrel{n/2}{\to }17\stackrel{3n+1}{\to }52\stackrel{n/2}{\to }26\stackrel{n/2}{\to }13\stackrel{3n+1}{\to }40\stackrel{n/2}{\to }20\stackrel{n/2}{\to }10\stackrel{n/2}{\to }5\stackrel{3n+1}{\to }16\stackrel{n/2}{\to }8\stackrel{n/2}{\to }4\stackrel{n/2}{\to }2\stackrel{n/2}{\to }1$$

(2)

After reaching 1, the sequence enters the cycle $1\to 4\to 2\to 1$, continuing indefinitely.

Despite its elementary formulation, the conjecture has resisted formal proof since its introduction by Lothar Collatz in 1937. This resistance stems from the unpredictable behavior exhibited by Collatz sequences: values may increase substantially before eventually decreasing, exhibiting patterns that have proven difficult to characterize through traditional approaches. Paul Erdős famously remarked that "mathematics may not be ready for such problems."

The conjecture has been verified computationally for all starting values up to $2^{68}$ (approximately $2.95\times {10}^{20}$). Notable theoretical advances include:

• Conway’s proof (1972) that generalizations of the Collatz problem are undecidable
• Lagarias’s establishment (1985) of upper bounds on potential counterexamples
• Terras’s probabilistic analysis (1976) suggesting almost all sequences decrease in the long run
• Tao’s result (2019) showing that Collatz sequences exhibit "almost bounded" behavior for "almost all" starting values

However, these approaches have not yielded a complete resolution. The difficulty lies in the absence of a monotonic parameter or invariant quantity that constrains the behavior of all possible trajectories.

1.2. Limitations of Traditional Approaches

Traditional approaches to the Collatz conjecture have typically focused on analyzing the forward dynamics by directly studying iterations of the Collatz function. These attempts have encountered several fundamental obstacles:

1.

Lack of monotonicity: Unlike many tractable iterative processes, Collatz sequences lack a parameter that consistently decreases (or increases) with each iteration.

2.

Unpredictable growth patterns: The sequence may grow to values far exceeding the starting point before eventually declining, with seemingly unpredictable patterns of growth and decline.

3.

Parity irregularities: The alternation between even and odd values creates complex patterns that resist straightforward analysis.

4.

Absence of useful invariants: Traditional dynamical systems often reveal invariant quantities that constrain behavior, but such invariants have proven elusive for the Collatz function.

These limitations have prompted various specialized techniques, including probabilistic approaches, p-adic analysis, modular restrictions, and dynamical systems theory. While these approaches have yielded significant insights, none has provided a comprehensive resolution. The forward-only perspective appears inadequate to fully capture and analyze the complex dynamics of the Collatz system.

1.3. The Bidirectional Framework: A New Perspective

The key innovation in our approach is the development of a comprehensive bidirectional framework that examines both forward and backward dynamics simultaneously. This perspective shift reveals structural properties that remain hidden when considering only forward iterations.

Principle 2

(Bidirectional Analysis). To fully understand the Collatz system, one must analyze both:

1. 

Forward dynamics: Where trajectories go when iterating the Collatz function

2. 

Backward dynamics: Where values come from through inverse operations

To enable this bidirectional analysis, we introduce the generator function—a multivalued mapping that systematically captures all possible predecessors under the Collatz function:

Definition 2

(Generator Function). The generator function $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ is defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\{2n,{\displaystyle \frac{n-1}{3}}\},\hfill & ifn\equiv 4(mod6)and{\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}\hfill \\ \left\{2n\right\},\hfill & \mathit{otherwise}\hfill \end{array}\right.$$

(3)

where $\mathcal{P}\left({\mathbb{N}}^{+}\right)$ denotes the power set of positive integers.

This function maps each positive integer to the set of all its possible predecessors under a single application of the Collatz function. The modular condition $n\equiv 4(mod6)$ precisely identifies cases where ${\displaystyle \frac{n-1}{3}}$ is a positive integer that yields n when the Collatz function is applied.

The bidirectional framework reveals a profound asymmetry in the Collatz system:

Principle 3

(Bidirectional Asymmetry). While all positive integers can potentially reach the cycle $\{1,4,2\}$ through forward iteration of the Collatz function, only elements of the set $\{1,2,4\}$ can generate all positive integers through backward iteration.

This asymmetry establishes the fundamental structure of the Collatz system and provides the basis for our resolution approach.

1.4. Main Results and Proof Strategy

Our resolution of the Collatz conjecture proceeds through a systematic three-part approach leveraging the bidirectional framework:

1.

Foundations and Tools: We establish the mathematical apparatus necessary for bidirectional analysis, including:

• Modular arithmetic framework for analyzing transitions

2.

Two Independent Pillars: We establish two logically independent results:

• Backward Path Finiteness: All generation paths under the generator function terminate after finitely many steps
• Cycle Uniqueness: The cycle $\{1,4,2\}$ is the only cycle in the Collatz system

3.

Structural Bridge: We connect these pillars to establish complete convergence

The synthesis of these components leads to our main result:

Theorem 4

(Resolution of the Collatz Conjecture). For any positive integer n, the sequence generated by iterating the Collatz function C eventually reaches 1.

Our proof establishes not only that all trajectories reach the cycle $\{1,4,2\}$, but also provides explicit bounds on trajectory behavior and a deep structural understanding of the Collatz dynamical system.

1.5. Paper Organization

The remainder of this paper is organized into three main blocks that systematically develop our bidirectional framework:

1.

Mathematical Foundations (Section 2)

• Formal properties of the Collatz and generator functions
• Bidirectional duality relationships
• Modular arithmetic framework and transition structures

2.

Bidirectional Analysis (Sections 3-5)

• Backward path analysis and pattern classification (Section 3)
• Cycle structure analysis and uniqueness proof (Section 4)
• Logical independence verification (Section 5)

3.

Resolution and Implications (Sections 6-7)

• Structural bridge construction and complete resolution (Section 6)
• Extensions to generalized Collatz maps and broader implications (Section 7)

Throughout the paper, we maintain a focus on rigorous mathematical development, explicit formalization of key concepts, clear logical progression toward the complete resolution, and transparent computational verification of our results.

2. Mathematical Foundations

This section develops the rigorous mathematical framework necessary for our bidirectional analysis of the Collatz problem. We formalize the relationship between forward and backward dynamics, establish modular arithmetic constraints, and develop the theoretical tools required for our subsequent analyses.

2.1. The Collatz Function and Its Properties

We begin by formalizing the Collatz function and establishing its basic properties.

Definition 3

(Collatz Function). The Collatz function $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ is defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}},\hfill & ifn\equiv 0(mod2)\hfill \\ 3n+1,\hfill & ifn\equiv 1(mod2)\hfill \end{array}\right.$$

(4)

Definition 4

(Collatz Trajectory). For any positive integer n, the Collatz trajectory starting from n is the sequence ${\left(C^k\left(n\right)\right)}_{k\ge 0}$, where $C^k$ denotes the k-fold composition of C.

Definition 5

(Collatz Cycle). A Collatz cycle is a finite sequence of distinct positive integers $S=\{c_1,c_2,\dots ,c_m\}$ such that $C\left(c_i\right)=c_{i+1}$ for $1\le i<m$ and $C\left(c_m\right)=c_1$.

Observation 5

(Known Cycle). The sequence $\{1,4,2\}$ forms a cycle under the Collatz function:

$$\begin{array}{cc}\hfill C\left(1\right)& =3·1+1=4\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill C\left(4\right)& ={\displaystyle \frac{4}{2}}=2\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill C\left(2\right)& ={\displaystyle \frac{2}{2}}=1\hfill \end{array}$$

(7)

The following properties of the Collatz function are important for our analysis:

Proposition 1

(Basic Properties of the Collatz Function). The Collatz function C satisfies the following properties:

1. 

For any $n\in {\mathbb{N}}^{+}$, $C\left(n\right)\in {\mathbb{N}}^{+}$ (range restriction)

2. 

If n is even, then $C\left(n\right)<n$ (contraction for even numbers)

3. 

If n is odd, then $C\left(n\right)>n$ (expansion for odd numbers)

4. 

For any odd n, $C\left(n\right)=3n+1$ is always even

5. 

For any $n>1$, at least one element in the set $\{n,C\left(n\right),C^2\left(n\right),\dots \}$ is even

6. 

C is not injective: multiple inputs may map to the same output

Proof. 

Properties (1)-(4) follow directly from the definition of C. For property (5), if n is even, the claim is immediate. If n is odd, then $C\left(n\right)=3n+1$ is even. For property (6), note that $C\left(2n\right)=n=C(2n-1)$ when n is odd. □

2.2. The Generator Function: Definition and Properties

To enable bidirectional analysis, we formally define the generator function that systematically captures all possible predecessors under the Collatz function.

Definition 6

(Generator Function). The generator function $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ is defined as:

$$G\left(n\right)=\{m\in {\mathbb{N}}^{+}:C\left(m\right)=n\}$$

(8)

This represents the set of all possible predecessors of n under one application of the Collatz function.

We now provide an explicit formulation of the generator function.

Proposition 2

(Explicit Form of Generator Function). The generator function G has the following explicit form:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n,{\displaystyle \frac{n-1}{3}}\right\},\hfill & ifn\equiv 4(mod6)\hfill \\ \left\{2n\right\},\hfill & ifn\not\equiv 4(mod6)\hfill \end{array}\right.$$

(9)

where the condition $n\equiv 4(mod6)$ ensures that ${\displaystyle \frac{n-1}{3}}$ is a positive integer.

Proof. 

For any $n\in {\mathbb{N}}^{+}$, we need to identify all values $m\in {\mathbb{N}}^{+}$ such that $C\left(m\right)=n$.

Case 1: If m is even, then $C\left(m\right)={\displaystyle \frac{m}{2}}$, which gives $m=2n$. This predecessor always exists for any $n\in {\mathbb{N}}^{+}$.

Case 2: If m is odd, then $C\left(m\right)=3m+1$, which gives $m={\displaystyle \frac{n-1}{3}}$. For this to be a valid predecessor, three conditions must be met:

• $n-1$ must be divisible by 3
• ${\displaystyle \frac{n-1}{3}}$ must be positive
• ${\displaystyle \frac{n-1}{3}}$ must be odd

For $n-1$ to be divisible by 3, we need $n\equiv 1(mod3)$. For ${\displaystyle \frac{n-1}{3}}$ to be odd, we need ${\displaystyle \frac{n-1}{3}}\equiv 1(mod2)$, which gives $n-1\equiv 3(mod6)$, or equivalently, $n\equiv 4(mod6)$.

Therefore, the second predecessor ${\displaystyle \frac{n-1}{3}}$ exists if and only if $n\equiv 4(mod6)$. □

For clarity, we decompose the generator function into its constituent operations:

Definition 7

(Generator Operations). The generator function G consists of two fundamental operations:

$$\begin{array}{cc}\hfill G_1\left(n\right)& =2n(\mathit{doubling}\mathit{operation})\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill G_2\left(n\right)& ={\displaystyle \frac{n-1}{3}}(\mathit{subtract}1and\mathit{divide}\mathit{by}3)\hfill \end{array}$$

(11)

where $G_2$ is defined precisely when $n\equiv 4(mod6)$.

Remark 1.

The operations $G_1$ and $G_2$ represent the inverse operations of dividing by 2 and applying $3n+1$, respectively. The condition $n\equiv 4(mod6)$ for $G_2$ ensures that the resulting value is a positive integer.

2.3. The Bidirectional Duality Framework

Before constructing the structural bridge, we must establish the fundamental duality relationship between forward and backward dynamics in the Collatz system. This duality is the key to understanding how our two independent pillars—backward path finiteness and cycle uniqueness—can be joined to resolve the conjecture.

Definition 8

(Forward-Backward Duality). The Collatz function $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ and the generator function $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ form a dual system with the following fundamental relationships:

1. 

For all $n\in {\mathbb{N}}^{+}$ and all $m\in G\left(n\right)$: $C\left(m\right)=n$

2. 

For all $n\in {\mathbb{N}}^{+}$: $n\in G\left(C\right(n\left)\right)$

3. 

For any set $S\subseteq {\mathbb{N}}^{+}$: $C\left(G\right(S\left)\right)=S$ and $G\left(C\right(S\left)\right)\supseteq S$

Theorem 6

(Path Duality Principle). Let $P_f=(a_0,a_1,\dots ,a_k)$ be a forward path under the Collatz function where $a_{i+1}=C\left(a_i\right)$ for all $i\in \{0,1,\dots ,k-1\}$. Then the reversed sequence $P_b=(a_k,a_{k-1},\dots ,a_0)$ forms a valid backward path where $a_{i-1}\in G\left(a_i\right)$ for all $i\in \{1,2,\dots ,k\}$.

Conversely, if $P_b=(b_0,b_1,\dots ,b_j)$ is a backward path where $b_{i-1}\in G\left(b_i\right)$ for all $i\in \{1,2,\dots ,j\}$, then the reversed sequence $P_f=(b_j,b_{j-1},\dots ,b_0)$ forms a valid forward path where $b_{i-1}=C\left(b_i\right)$ for all $i\in \{1,2,\dots ,j\}$.

Proof. 

By Definition 8, if $m\in G\left(n\right)$, then $C\left(m\right)=n$. This bidirectional relationship ensures that any valid sequence under C can be reversed to form a valid sequence under G, and vice versa.

For the forward path $P_f=(a_0,a_1,\dots ,a_k)$, we have $a_{i+1}=C\left(a_i\right)$ for all $i\in \{0,1,\dots ,k-1\}$. By the duality principle, this means $a_i\in G\left(a_{i+1}\right)$ for all $i\in \{0,1,\dots ,k-1\}$. Therefore, the reversed sequence $P_b=(a_k,a_{k-1},\dots ,a_0)$ forms a valid backward path.

The converse follows by identical reasoning. □

Theorem 7

(Global Structure Duality). The global structure of the Collatz system exhibits the following dual properties:

1. 

If all backward paths from every positive integer n terminate at elements of the set $\{1,2,4\}$ or at values with no predecessors, then every positive integer n can either:

• Be reached from an element of $\{1,2,4\}$ via forward iteration of the Collatz function, or
• Be part of a forward trajectory that never reaches an element of $\{1,2,4\}$

2. 

If $\{1,4,2\}$ is the unique cycle in the Collatz system, then no forward trajectory can form any cycle other than $\{1,4,2\}$.

Proof. 

For part (1), consider any positive integer n. By Theorem 10, there exists a finite backward path from n that terminates either at an element of $\{1,2,4\}$ or at a value with no predecessors.

If the backward path from n terminates at an element of $\{1,2,4\}$, say at $a_k\in \{1,2,4\}$, then by Theorem 6, the reversed path forms a valid forward path from $a_k$ to n. Thus, n can be reached from an element of $\{1,2,4\}$ via forward iteration of the Collatz function.

If the backward path from n terminates at a value b with no predecessors, then n cannot be reached from any element of $\{1,2,4\}$ via forward iteration. In this case, n must be part of a forward trajectory that never reaches an element of $\{1,2,4\}$.

For part (2), if $\{1,4,2\}$ is the unique cycle in the Collatz system, then no forward trajectory can form any other cycle, as this would contradict the uniqueness of the cycle $\{1,4,2\}$. □

This duality framework provides the critical connection between our two independent pillars and forms the foundation for the structural bridge that will establish the complete resolution of the Collatz conjecture.

2.4. Generation Paths and Fundamental Properties

We now formalize the concept of backward trajectories, which we call generation paths, and establish their fundamental properties.

Definition 9

(Generation Path). A generation path is a sequence ${\left(a_k\right)}_{k\ge 0}$ in ${\mathbb{N}}^{+}$ where for each $k\ge 1$, $a_{k-1}\in G\left(a_k\right)$.

Equivalently, $a_{k-1}$ is a predecessor of $a_k$ under the Collatz function, meaning $C\left(a_{k-1}\right)=a_k$.

Definition 10

(Complete Generation Path). A generation path ${\left(a_k\right)}_{k=0}^n$ is called complete if no element in $G\left(a_0\right)$ belongs to ${\mathbb{N}}^{+}$, or if $a_0\in \{1,2,4\}$.

The following lemma establishes key properties of generation paths under the operations $G_1$ and $G_2$.

Lemma 1

(Properties of Generator Operations). Let ${\left(a_k\right)}_{k\ge 0}$ be a generation path. Then:

1. 

If $a_{k-1}=G_1\left(a_k\right)=2a_k$, then $a_{k-1}>a_k$ (expansion under $G_1$)

2. 

If $a_{k-1}=G_2\left(a_k\right)={\displaystyle \frac{a_k-1}{3}}$, then $a_{k-1}<a_k$ (contraction under $G_2$)

3. 

If $a_k\equiv 4(mod6)$, then $G_2\left(a_k\right)={\displaystyle \frac{a_k-1}{3}}$ is odd

4. 

If $a_{k-1}=G_2\left(a_k\right)$, then $a_{k-1}\not\equiv 4(mod6)$

Proof. (1) If $a_{k-1}=G_1\left(a_k\right)=2a_k$, then clearly $a_{k-1}=2a_k>a_k$ for any $a_k\in {\mathbb{N}}^{+}$.

(2) If $a_{k-1}=G_2\left(a_k\right)={\displaystyle \frac{a_k-1}{3}}$, then $a_k=3a_{k-1}+1>3a_{k-1}>a_{k-1}$.

(3) If $a_k\equiv 4(mod6)$, then $a_k=6m+4$ for some $m\ge 0$. Thus, $G_2\left(a_k\right)={\displaystyle \frac{a_k-1}{3}}={\displaystyle \frac{6m+4-1}{3}}={\displaystyle \frac{6m+3}{3}}=2m+1$, which is odd.

(4) Let $a_{k-1}=G_2\left(a_k\right)={\displaystyle \frac{a_k-1}{3}}$. We know that $a_k\equiv 4(mod6)$, so $a_k=6m+4$ for some $m\ge 0$. Thus, $a_{k-1}={\displaystyle \frac{6m+4-1}{3}}={\displaystyle \frac{6m+3}{3}}=2m+1$, which is odd. Since all values congruent to 4 modulo 6 are even, $a_{k-1}\not\equiv 4(mod6)$. □

2.5. Modular Arithmetic Framework

The behavior of generation paths can be systematically analyzed through modular arithmetic. We establish key modular constraints that will be essential for our subsequent analyses.

Lemma 2

(Comprehensive Modular Characterization). For $n\in {\mathbb{N}}^{+}$, the expression ${\displaystyle \frac{n-1}{3}}$ has the following properties based on congruence classes:

1. 

For each congruence class modulo 3:

• If $n\equiv 0(mod3)$, then ${\displaystyle \frac{n-1}{3}}={\displaystyle \frac{3k-1}{3}}=k-{\displaystyle \frac{1}{3}}$ is not an integer.
• If $n\equiv 1(mod3)$, then ${\displaystyle \frac{n-1}{3}}={\displaystyle \frac{3k+1-1}{3}}=k$ is an integer.
• If $n\equiv 2(mod3)$, then ${\displaystyle \frac{n-1}{3}}={\displaystyle \frac{3k+2-1}{3}}=k+{\displaystyle \frac{1}{3}}$ is not an integer.

2. 

For each congruence class modulo 6:

• If $n\equiv 0(mod6)$, then ${\displaystyle \frac{n-1}{3}}={\displaystyle \frac{6k-1}{3}}=2k-{\displaystyle \frac{1}{3}}$ is not an integer.
• If $n\equiv 1(mod6)$, then ${\displaystyle \frac{n-1}{3}}={\displaystyle \frac{6k+1-1}{3}}=2k$ is an even integer.
• If $n\equiv 2(mod6)$, then ${\displaystyle \frac{n-1}{3}}={\displaystyle \frac{6k+2-1}{3}}=2k+{\displaystyle \frac{1}{3}}$ is not an integer.
• If $n\equiv 3(mod6)$, then ${\displaystyle \frac{n-1}{3}}={\displaystyle \frac{6k+3-1}{3}}=2k+{\displaystyle \frac{2}{3}}$ is not an integer.
• If $n\equiv 4(mod6)$, then ${\displaystyle \frac{n-1}{3}}={\displaystyle \frac{6k+4-1}{3}}=2k+1$ is an odd integer.
• If $n\equiv 5(mod6)$, then ${\displaystyle \frac{n-1}{3}}={\displaystyle \frac{6k+5-1}{3}}=2k+{\displaystyle \frac{4}{3}}$ is not an integer.

Proof. 

The proof follows from direct calculation of ${\displaystyle \frac{n-1}{3}}$ for each congruence class, as detailed in the statement of the lemma. □

Theorem 8

(Modular Transition Structure). The transitions between congruence classes modulo 6 under the generator operations $G_1$ and $G_2$ satisfy the following properties:

1. 

For the $G_1$ operation, the transitions are:

$$\begin{array}{cc}\hfill G_1(0mod6)& \equiv 0(mod6)\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill G_1(1mod6)& \equiv 2(mod6)\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill G_1(2mod6)& \equiv 4(mod6)\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill G_1(3mod6)& \equiv 0(mod6)\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill G_1(4mod6)& \equiv 2(mod6)\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill G_1(5mod6)& \equiv 4(mod6)\hfill \end{array}$$

(17)

2. 

The $G_2$ operation is applicable precisely when $n\equiv 4(mod6)$, with:

$$\begin{array}{cc}\hfill G_2(4mod6)& \equiv \left\{\begin{array}{cc}1(mod6)\hfill & ifn=6k+4andk\equiv 0(mod3)\hfill \\ 3(mod6)\hfill & ifn=6k+4andk\equiv 1(mod3)\hfill \\ 5(mod6)\hfill & ifn=6k+4andk\equiv 2(mod3)\hfill \end{array}\right.\hfill \end{array}$$

(18)

3. 

For any value not congruent to 0 or 3 modulo 6, there exists a finite number $L\left(n\right)\le 2$ of consecutive $G_1$ operations that transform it into a value congruent to 4 modulo 6.

Proof. 

We provide a comprehensive analysis of the modular transition structure under the generator operations $G_1$ and $G_2$, examining all possible congruence classes modulo 6.

Part 1: Transitions under the $G_1$ operation.

The $G_1$ operation is defined as $G_1\left(n\right)=2n$ for any positive integer n. We systematically analyze how this operation affects each residue class modulo 6.

For $n\equiv 0(mod6)$: Let $n=6k$ for some $k\in \mathbb{N}$. Then:

$$\begin{array}{cc}\hfill G_1\left(n\right)& =2n\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill & =2\left(6k\right)\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill & =12k\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill & =6\left(2k\right)\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill \Rightarrow G_1\left(n\right)& \equiv 0(mod6)\hfill \end{array}$$

(23)

For $n\equiv 1(mod6)$: Let $n=6k+1$ for some $k\in \mathbb{N}$. Then:

$$\begin{array}{cc}\hfill G_1\left(n\right)& =2n\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill & =2(6k+1)\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill & =12k+2\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill & =6\left(2k\right)+2\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill \Rightarrow G_1\left(n\right)& \equiv 2(mod6)\hfill \end{array}$$

(28)

For $n\equiv 2(mod6)$: Let $n=6k+2$ for some $k\in \mathbb{N}$. Then:

$$\begin{array}{cc}\hfill G_1\left(n\right)& =2n\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill & =2(6k+2)\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill & =12k+4\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill & =6\left(2k\right)+4\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill \Rightarrow G_1\left(n\right)& \equiv 4(mod6)\hfill \end{array}$$

(33)

For $n\equiv 3(mod6)$: Let $n=6k+3$ for some $k\in \mathbb{N}$. Then:

$$\begin{array}{cc}\hfill G_1\left(n\right)& =2n\hfill \end{array}$$

(34)

$$\begin{array}{cc}\hfill & =2(6k+3)\hfill \end{array}$$

(35)

$$\begin{array}{cc}\hfill & =12k+6\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill & =6(2k+1)\hfill \end{array}$$

(37)

$$\begin{array}{cc}\hfill \Rightarrow G_1\left(n\right)& \equiv 0(mod6)\hfill \end{array}$$

(38)

For $n\equiv 4(mod6)$: Let $n=6k+4$ for some $k\in \mathbb{N}$. Then:

$$\begin{array}{cc}\hfill G_1\left(n\right)& =2n\hfill \end{array}$$

(39)

$$\begin{array}{cc}\hfill & =2(6k+4)\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill & =12k+8\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill & =6(2k+1)+2\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill \Rightarrow G_1\left(n\right)& \equiv 2(mod6)\hfill \end{array}$$

(43)

For $n\equiv 5(mod6)$: Let $n=6k+5$ for some $k\in \mathbb{N}$. Then:

$$\begin{array}{cc}\hfill G_1\left(n\right)& =2n\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill & =2(6k+5)\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill & =12k+10\hfill \end{array}$$

(46)

$$\begin{array}{cc}\hfill & =6(2k+1)+4\hfill \end{array}$$

(47)

$$\begin{array}{cc}\hfill \Rightarrow G_1\left(n\right)& \equiv 4(mod6)\hfill \end{array}$$

(48)

This completes the proof of part 1, establishing the exact transitions under the $G_1$ operation for all residue classes modulo 6.

Part 2: Applicability and transitions under the $G_2$ operation.

The $G_2$ operation is defined as $G_2\left(n\right)={\displaystyle \frac{n-1}{3}}$ when applicable. From Lemma 2, we know that ${\displaystyle \frac{n-1}{3}}$ is an integer precisely when $n\equiv 1(mod3)$.

Furthermore, for the operation to be valid in the Collatz system, we need $G_2\left(n\right)$ to be the predecessor of n under the Collatz function, which means $C\left(G_2\left(n\right)\right)=n$. Since $C\left(m\right)=3m+1$ for odd m, we need $G_2\left(n\right)$ to be odd.

For $n\equiv 1(mod3)$, we have $n=3k+1$ for some $k\in \mathbb{N}$, which gives $G_2\left(n\right)={\displaystyle \frac{n-1}{3}}={\displaystyle \frac{3k+1-1}{3}}=k$. For $G_2\left(n\right)$ to be odd, we need k to be odd, which means $k=2j+1$ for some $j\in \mathbb{N}$.

Substituting back: $n=3(2j+1)+1=6j+4$. Therefore, $G_2$ is applicable precisely when $n\equiv 4(mod6)$.

Now, let’s determine the residue class of $G_2\left(n\right)$ modulo 6 when $n\equiv 4(mod6)$.

For $n=6k+4$, we have:

$$\begin{array}{cc}\hfill G_2\left(n\right)& ={\displaystyle \frac{n-1}{3}}\hfill \end{array}$$

(49)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{6k+4-1}{3}}\hfill \end{array}$$

(50)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{6k+3}{3}}\hfill \end{array}$$

(51)

$$\begin{array}{cc}\hfill & =2k+1\hfill \end{array}$$

(52)

Since $G_2\left(n\right)=2k+1$, the residue class of $G_2\left(n\right)$ modulo 6 depends on the residue class of k modulo 3:

For $k\equiv 0(mod3)$, let $k=3j$ for some $j\in \mathbb{N}$. Then:

$$\begin{array}{cc}\hfill G_2\left(n\right)& =2k+1\hfill \end{array}$$

(53)

$$\begin{array}{cc}\hfill & =2\left(3j\right)+1\hfill \end{array}$$

(54)

$$\begin{array}{cc}\hfill & =6j+1\hfill \end{array}$$

(55)

$$\begin{array}{cc}\hfill \Rightarrow G_2\left(n\right)& \equiv 1(mod6)\hfill \end{array}$$

(56)

For $k\equiv 1(mod3)$, let $k=3j+1$ for some $j\in \mathbb{N}$. Then:

$$\begin{array}{cc}\hfill G_2\left(n\right)& =2k+1\hfill \end{array}$$

(57)

$$\begin{array}{cc}\hfill & =2(3j+1)+1\hfill \end{array}$$

(58)

$$\begin{array}{cc}\hfill & =6j+2+1\hfill \end{array}$$

(59)

$$\begin{array}{cc}\hfill & =6j+3\hfill \end{array}$$

(60)

$$\begin{array}{cc}\hfill \Rightarrow G_2\left(n\right)& \equiv 3(mod6)\hfill \end{array}$$

(61)

For $k\equiv 2(mod3)$, let $k=3j+2$ for some $j\in \mathbb{N}$. Then:

$$\begin{array}{cc}\hfill G_2\left(n\right)& =2k+1\hfill \end{array}$$

(62)

$$\begin{array}{cc}\hfill & =2(3j+2)+1\hfill \end{array}$$

(63)

$$\begin{array}{cc}\hfill & =6j+4+1\hfill \end{array}$$

(64)

$$\begin{array}{cc}\hfill & =6j+5\hfill \end{array}$$

(65)

$$\begin{array}{cc}\hfill \Rightarrow G_2\left(n\right)& \equiv 5(mod6)\hfill \end{array}$$

(66)

This establishes the transitions under the $G_2$ operation as stated in part 2 of the theorem.

Part 3: Reaching values congruent to 4 modulo 6.

We now demonstrate that for any value not congruent to 0 or 3 modulo 6, there exists a finite number $L\left(n\right)\le 2$ of consecutive $G_1$ operations that transform it into a value congruent to 4 modulo 6.

For $n\equiv 1(mod6)$:

$$\begin{array}{cc}\hfill G_1\left(n\right)& \equiv 2(mod6)\hfill \end{array}$$

(67)

$$\begin{array}{cc}\hfill G_1\left(G_1\left(n\right)\right)& \equiv G_1\left(2(mod6)\right)\equiv 4(mod6)\hfill \end{array}$$

(68)

Therefore, $L\left(n\right)=2$ for $n\equiv 1(mod6)$.

For $n\equiv 2(mod6)$:

$$\begin{array}{cc}\hfill G_1\left(n\right)& \equiv 4(mod6)\hfill \end{array}$$

(69)

Therefore, $L\left(n\right)=1$ for $n\equiv 2(mod6)$.

For $n\equiv 4(mod6)$: The value is already congruent to 4 modulo 6, so $L\left(n\right)=0$.

For $n\equiv 5(mod6)$:

$$\begin{array}{cc}\hfill G_1\left(n\right)& \equiv 4(mod6)\hfill \end{array}$$

(70)

Therefore, $L\left(n\right)=1$ for $n\equiv 5(mod6)$.

For the remaining cases $n\equiv 0(mod6)$ and $n\equiv 3(mod6)$, repeated applications of $G_1$ do not lead to values congruent to 4 modulo 6:

For $n\equiv 0(mod6)$:

$$\begin{array}{cc}\hfill G_1\left(n\right)& \equiv 0(mod6)\hfill \end{array}$$

(71)

$$\begin{array}{cc}\hfill G_1\left(G_1\left(n\right)\right)& \equiv 0(mod6)\hfill \end{array}$$

(72)

$$\begin{array}{c}\hfill ⋮\end{array}$$

(73)

For $n\equiv 3(mod6)$:

$$\begin{array}{cc}\hfill G_1\left(n\right)& \equiv 0(mod6)\hfill \end{array}$$

(74)

$$\begin{array}{cc}\hfill G_1\left(G_1\left(n\right)\right)& \equiv 0(mod6)\hfill \end{array}$$

(75)

$$\begin{array}{c}\hfill ⋮\end{array}$$

(76)

Therefore, for values congruent to 0 or 3 modulo 6, $G_1$ operations alone cannot transform them into values congruent to 4 modulo 6.

However, for values not congruent to 0 or 3 modulo 6, we have shown that at most $L\left(n\right)\le 2$ consecutive $G_1$ operations are needed to transform them into values congruent to 4 modulo 6.

This completes the proof of the modular transition structure under the generator operations $G_1$ and $G_2$. □

This modular arithmetic framework provides essential constraints on the behavior of generation paths, which will be crucial for establishing their finiteness and other structural properties in subsequent sections.

3. Analysis of Backward Paths

This section develops a comprehensive theory of backward paths (generation paths) in the Collatz system. We classify all possible patterns, establish key properties, and prove that all backward paths must terminate after finitely many steps.

3.1. Classification of Generation Path Patterns

We begin by developing a systematic classification of all possible patterns that can emerge in generation paths.

Definition 11

(Generation Path). A generation path is a sequence ${\left(a_k\right)}_{k\ge 0}$ in ${\mathbb{N}}^{+}$ where for each $k\ge 1$, $a_{k-1}\in G\left(a_k\right)$.

Equivalently, $a_{k-1}$ is a predecessor of $a_k$ under the Collatz function, meaning $C\left(a_{k-1}\right)=a_k$.

Definition 12

(Complete Generation Path). A generation path ${\left(a_k\right)}_{k=0}^n$ is called complete if either:

1. 

$a_0\in \{1,2,4\}$, or

2. 

$G\left(a_0\right)=\varnothing$ (i.e., $a_0$ has no predecessors)

To analyze the patterns systematically, we encode generation paths using a binary alphabet representing the two possible generator operations:

Definition 13

(Generator Operations). The generator function G consists of two fundamental operations:

$$\begin{array}{cc}\hfill G_1\left(n\right)& =2n(\mathit{doubling}\mathit{operation})\hfill \end{array}$$

(77)

$$\begin{array}{cc}\hfill G_2\left(n\right)& ={\displaystyle \frac{n-1}{3}}(\mathit{subtract}1and\mathit{divide}\mathit{by}3)\hfill \end{array}$$

(78)

where $G_2$ is defined precisely when $n\equiv 4(mod6)$.

Definition 14

(Operation Encoding). We encode generation paths using a binary alphabet $\Sigma =\{1,2\}$, where:

• The symbol ’1’ represents the application of operation $G_1$ (i.e., $a_{i-1}=G_1\left(a_i\right)=2a_i$)
• The symbol ’2’ represents the application of operation $G_2$ (i.e., $a_{i-1}=G_2\left(a_i\right)={\displaystyle \frac{a_i-1}{3}}$)

Theorem 9

(Fundamental Pattern Classification). The set $L_{\mathrm{valid}}$ of all infinite sequences over $\Sigma =\{1,2\}$ satisfying the encoding constraints can be partitioned into exactly three disjoint pattern classes:

1. 

$P_1=\left\{1^{\omega }\right\}$ (the singleton set containing only the infinite sequence of 1’s)

2. 

$P_2=\left\{{\left(21\right)}^{\omega }\right\}$ (the singleton set containing only the infinite alternating sequence of 2’s and 1’s)

3. 

$P_3=\{2·1^{n_1}·2·1^{n_2}·2·\dots :n_i\ge 1\mathit{for}\mathrm{all}i\ge 1and\exists j\mathit{such}\mathit{that}{n}_j>1\}$ (the set of all sequences beginning with 2 followed by runs of 1’s of varying lengths, with at least one run containing multiple 1’s)

Furthermore, $L_{\mathrm{valid}}=P_1\cup P_2\cup P_3$, and these three sets are mutually disjoint.

Proof. 

We establish a key constraint: after a $G_2$ operation (represented by symbol ’2’), the next term in the generation path cannot be subject to another $G_2$ operation immediately. This follows from Lemma 1(4), which states that if $a_{i-1}=G_2\left(a_i\right)$, then $a_{i-1}\not\equiv 4(mod6)$, meaning $G_2$ cannot be applied to $a_{i-1}$.

This constraint implies that in any valid encoding, the symbol ’2’ cannot be followed immediately by another ’2’. Therefore, any valid infinite sequence must satisfy one of the following:

• It contains only the symbol ’1’, which corresponds to pattern class $P_1=\left\{1^{\omega }\right\}$.
• It contains both symbols ’1’ and ’2’, with the additional constraint that no two ’2’s can appear consecutively.

For the second case, since no two ’2’s can appear consecutively, the sequence must consist of blocks of the form $2·1^n$ for various values of $n\ge 1$. This gives us two possibilities:

• If every block has exactly $n=1$, then the sequence is the alternating pattern ${\left(21\right)}^{\omega }$, which corresponds to pattern class $P_2$.
• If at least one block has $n>1$, then the sequence belongs to pattern class $P_3$.

To prove that these three pattern classes are exhaustive, we note that any valid sequence either contains only ’1’s or contains both ’1’s and ’2’s with no consecutive ’2’s. The classification is thus complete.

To prove that the pattern classes are mutually disjoint:

• $P_1$ contains only sequences with no occurrences of ’2’, while both $P_2$ and $P_3$ contain sequences with occurrences of ’2’. Thus, $P_1$ is disjoint from both $P_2$ and $P_3$.
• $P_2$ contains only the alternating sequence ${\left(21\right)}^{\omega }$, while all sequences in $P_3$ have at least one occurrence of two or more consecutive ’1’s. Thus, $P_2$ is disjoint from $P_3$.

Therefore, $L_{\mathrm{valid}}=P_1\cup P_2\cup P_3$, and these three sets are mutually disjoint. □

This classification is central to understanding the behavior of backward paths in the Collatz system. It allows us to analyze each pattern type separately and establish finiteness properties.

Theorem 10

(Finiteness of All Backward Paths). Every generation path in ${\mathbb{N}}^{+}$ under the generator function G must terminate after finitely many steps. Specifically, for any starting value $a_0\in {\mathbb{N}}^{+}$, any generation path ${\left(a_k\right)}_{k\ge 0}$ with $a_{k-1}\in G\left(a_k\right)$ for all $k\ge 1$ has a finite length.

Proposition 3

(Pattern Structure Constraint). In any valid encoding of generation paths using the alphabet $\Sigma =\{1,2\}$ (representing $G_1$ and $G_2$ operations respectively), the symbol `2’ cannot be followed immediately by another `2’. Furthermore, any infinite sequence over Σ satisfying this constraint must belong to exactly one of the following pattern classes:

1. 

$P_1=\left\{1^{\omega }\right\}$ (the set containing only the infinite sequence of 1’s)

2. 

$P_2=\left\{{\left(21\right)}^{\omega }\right\}$ (the set containing only the infinite alternating sequence of 2’s and 1’s)

3. 

$P_3=\{2·1^{n_1}·2·1^{n_2}·2·\dots :n_i\ge 1\mathit{for}\mathrm{all}i\ge 1and\exists j\mathit{such}\mathit{that}{n}_j>1\}$ (the set of all sequences beginning with 2 followed by runs of 1’s of varying lengths)

Proof. 

By Lemma 1(4), if $a_{i-1}=G_2\left(a_i\right)$, then $a_{i-1}\not\equiv 4(mod6)$. Since $G_2$ can only be applied to values congruent to $4(mod6)$, this means that if we apply $G_2$ to obtain $a_{i-1}$, we cannot apply $G_2$ again to obtain $a_{i-2}$. Therefore, in any valid encoding, the symbol `2’ cannot be followed immediately by another `2’.

Given this constraint, we can classify all possible infinite sequences over $\Sigma =\{1,2\}$:

(a) If the sequence contains only the symbol `1’, it belongs to pattern class $P_1=\left\{1^{\omega }\right\}$.

(b) If the sequence contains both symbols `1’ and `2’, then since no two `2’s can appear consecutively, we must have `2’ followed by at least one `1’. This gives us two possibilities:

• If every occurrence of `2’ is followed by exactly one `1’, the sequence is the alternating pattern ${\left(21\right)}^{\omega }$, which corresponds to pattern class $P_2$.
• If at least one occurrence of `2’ is followed by two or more consecutive `1’s, the sequence belongs to pattern class $P_3$.

The three pattern classes are mutually exclusive by construction:

• $P_1$ contains only sequences with no occurrences of `2’, while both $P_2$ and $P_3$ contain sequences with occurrences of `2’.
• $P_2$ contains only sequences where every `2’ is followed by exactly one `1’, while all sequences in $P_3$ have at least one occurrence of `2’ followed by multiple `1’s.

Furthermore, any infinite sequence over $\Sigma$ that satisfies the constraint "no consecutive 2’s" must fall into one of these three categories. Therefore, these pattern classes form a complete partition of the space of valid generation paths. □

Lemma 3

(Modular Growth Constraints). Let ${\left(a_k\right)}_{k\ge 0}$ be a generation path with $a_{k-1}\in G\left(a_k\right)$ for all $k\ge 1$. Then:

1. 

For pattern $P_1$ paths (consisting exclusively of $G_1$ operations), the length is bounded by ${\nu }_2\left(a_0\right)+1$, where ${\nu }_2\left(a_0\right)$ is the 2-adic valuation of $a_0$.

2. 

For pattern $P_2$ paths (alternating $G_2$ and $G_1$ operations), the values follow the recurrence relation:

$$a_{2j}={\displaystyle \frac{3a_{2j-2}+2}{2}}\mathit{for}j\ge 1$$

(79)

with closed form:

$$a_{2j}={\left({\displaystyle \frac{3}{2}}\right)}^j·(a_0+4)-4\mathit{for}j\ge 0$$

(80)

3. 

For pattern $P_3$ paths, after each application of $G_2$ followed by n consecutive applications of $G_1$, we have:

$$G_1^n\left(G_2\left(v\right)\right)=2^n·{\displaystyle \frac{v-1}{3}}$$

(81)

where $G_1^n$ denotes n consecutive applications of $G_1$.

Proof. 

For (1), with each $G_1$ operation, we have $a_{k-1}=2a_k$. Iterating this relation, we get $a_k={\displaystyle \frac{a_0}{2^k}}$. Since $a_k$ must be a positive integer, we need ${\displaystyle \frac{a_0}{2^k}}\ge 1$, which gives $k\le {log}_2\left(a_0\right)$. More precisely, if we write $a_0=2^m·q$ where q is odd, then $a_k=2^{m-k}·q$. Since $a_k$ must be an integer, we need $m-k\ge 0$, which gives $k\le m={\nu }_2\left(a_0\right)$. Therefore, any pattern $P_1$ path can contain at most ${\nu }_2\left(a_0\right)+1$ terms.

For (2), in a pattern $P_2$ path, we have alternating applications of $G_2$ and $G_1$:

$$\begin{array}{cc}\hfill a_{2j-1}& =G_2\left(a_{2j}\right)={\displaystyle \frac{a_{2j}-1}{3}}\mathit{for}j\ge 1\hfill \end{array}$$

(82)

$$\begin{array}{cc}\hfill a_{2j-2}& =G_1\left(a_{2j-1}\right)=2a_{2j-1}\mathit{for}j\ge 1\hfill \end{array}$$

(83)

Substituting the first equation into the second:

$$\begin{array}{cc}\hfill a_{2j-2}& =G_1\left(G_2\left(a_{2j}\right)\right)\hfill \end{array}$$

(84)

$$\begin{array}{cc}\hfill & =2·{\displaystyle \frac{a_{2j}-1}{3}}\hfill \end{array}$$

(85)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{2a_{2j}-2}{3}}\hfill \end{array}$$

(86)

Solving for $a_{2j}$:

$$\begin{array}{cc}\hfill a_{2j-2}& ={\displaystyle \frac{2a_{2j}-2}{3}}\hfill \end{array}$$

(87)

$$\begin{array}{cc}\hfill 3a_{2j-2}& =2a_{2j}-2\hfill \end{array}$$

(88)

$$\begin{array}{cc}\hfill 3a_{2j-2}+2& =2a_{2j}\hfill \end{array}$$

(89)

$$\begin{array}{cc}\hfill {\displaystyle \frac{3a_{2j-2}+2}{2}}& =a_{2j}\hfill \end{array}$$

(90)

This gives us the recurrence relation for values at even indices. To derive the closed form, we iterate this recurrence relation:

For $j=1$:

$$\begin{array}{cc}\hfill a_2& ={\displaystyle \frac{3a_0+2}{2}}\hfill \end{array}$$

(91)

For $j=2$:

$$\begin{array}{cc}\hfill a_4& ={\displaystyle \frac{3a_2+2}{2}}\hfill \end{array}$$

(92)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{3\left(\frac{3a_0+2}{2}\right)+2}{2}}\hfill \end{array}$$

(93)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{3(3a_0+2)+4}{4}}\hfill \end{array}$$

(94)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{9a_0+6+4}{4}}\hfill \end{array}$$

(95)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{9a_0+10}{4}}\hfill \end{array}$$

(96)

We observe the emerging pattern and prove by induction that for all $j\ge 0$:

$$\begin{array}{cc}\hfill a_{2j}& ={\left({\displaystyle \frac{3}{2}}\right)}^j·a_0+2·\sum _{i=0}^{j-1}{\left({\displaystyle \frac{3}{2}}\right)}^i\hfill \end{array}$$

(97)

$$\begin{array}{cc}\hfill & ={\left({\displaystyle \frac{3}{2}}\right)}^j·a_0+2·{\displaystyle \frac{{\left(\frac{3}{2}\right)}^j-1}{\frac{3}{2}-1}}\hfill \end{array}$$

(98)

$$\begin{array}{cc}\hfill & ={\left({\displaystyle \frac{3}{2}}\right)}^j·a_0+4·\left[{\left({\displaystyle \frac{3}{2}}\right)}^j-1\right]\hfill \end{array}$$

(99)

$$\begin{array}{cc}\hfill & ={\left({\displaystyle \frac{3}{2}}\right)}^j·(a_0+4)-4\hfill \end{array}$$

(100)

For (3), if we apply $G_2$ followed by n consecutive $G_1$ operations to a value v, we get:

$$\begin{array}{cc}\hfill G_2\left(v\right)& ={\displaystyle \frac{v-1}{3}}\hfill \end{array}$$

(101)

$$\begin{array}{cc}\hfill G_1\left(G_2\left(v\right)\right)& =2·{\displaystyle \frac{v-1}{3}}\hfill \end{array}$$

(102)

$$\begin{array}{cc}\hfill G_1^2\left(G_2\left(v\right)\right)& =2^2·{\displaystyle \frac{v-1}{3}}\hfill \end{array}$$

(103)

$$\begin{array}{c}\hfill ⋮\end{array}$$

(104)

$$\begin{array}{cc}\hfill G_1^n\left(G_2\left(v\right)\right)& =2^n·{\displaystyle \frac{v-1}{3}}\hfill \end{array}$$

(105)

This completes the proof. □

Proposition 4

(Growth Limitation in $P_2$ Paths). In any pattern $P_2$ path (alternating $G_2$ and $G_1$ operations), the modular constraints on $a_{2j}$ values cannot be satisfied indefinitely. Specifically, for pattern $P_2$ to continue, we must have $a_{2j}\equiv 4(mod6)$ for all $j\ge 0$, but the closed form expression $a_{2j}={\left({\displaystyle \frac{3}{2}}\right)}^j·(a_0+4)-4$ ensures this condition cannot be maintained indefinitely.

Proof. 

For a pattern $P_2$ path to continue, we need all values $a_{2j}$ to be congruent to $4(mod6)$, since only such values allow the $G_2$ operation to be applied.

From Lemma 3, we have $a_{2j}={\left({\displaystyle \frac{3}{2}}\right)}^j·(a_0+4)-4$. For $a_{2j}\equiv 4(mod6)$, we need:

$$\begin{array}{cc}\hfill a_{2j}& \equiv 4(mod6)\hfill \end{array}$$

(106)

$$\begin{array}{cc}\hfill {\left({\displaystyle \frac{3}{2}}\right)}^j·(a_0+4)-4& \equiv 4(mod6)\hfill \end{array}$$

(107)

$$\begin{array}{cc}\hfill {\left({\displaystyle \frac{3}{2}}\right)}^j·(a_0+4)& \equiv 8(mod6)\hfill \end{array}$$

(108)

$$\begin{array}{cc}\hfill {\left({\displaystyle \frac{3}{2}}\right)}^j·(a_0+4)& \equiv 2(mod6)\hfill \end{array}$$

(109)

For $j>0$, ${\left({\displaystyle \frac{3}{2}}\right)}^j$ is not an integer. Specifically:

$$\begin{array}{c}\hfill {\left({\displaystyle \frac{3}{2}}\right)}^j={\displaystyle \frac{3^j}{2^j}}\end{array}$$

(110)

For this non-integer rational number multiplied by the integer $(a_0+4)$ to yield a value congruent to $2(mod6)$, we need specific number-theoretic conditions that become increasingly difficult to satisfy as j increases.

Let’s analyze this more carefully. For $a_{2j}\equiv 4(mod6)$ to hold, we need:

$$\begin{array}{c}\hfill {\displaystyle \frac{3^j}{2^j}}·(a_0+4)\equiv 2(mod6)\end{array}$$

(111)

For this to be meaningful in modular arithmetic, ${\displaystyle \frac{3^j}{2^j}}·(a_0+4)$ must be an integer, which requires that $2^j$ divides $3^j·(a_0+4)$.

Now, consider the values of $3^j(mod{2}^j)$ for increasing j:

• $j=1$: $3^1\equiv 3(mod{2}^1)$, so $3^1$ is odd
• $j=2$: $3^2=9\equiv 1(mod{2}^2)$, so $3^2$ is congruent to $1(mod4)$
• $j=3$: $3^3=27\equiv 3(mod{2}^3)$, so $3^3$ is congruent to $3(mod8)$
• $j=4$: $3^4=81\equiv 1(mod{2}^4)$, so $3^4$ is congruent to $1(mod16)$

These values follow a cyclic pattern where $3^j\equiv 3(mod{2}^j)$ when j is odd and $3^j\equiv 1(mod{2}^j)$ when j is a multiple of 2 but not a multiple of 4.

For $2^j$ to divide $3^j·(a_0+4)$, we need $a_0+4\equiv 0(mod{2}^j)$ when j is odd or when j is a multiple of 2 but not a multiple of 4.

This creates an increasingly stringent constraint on $a_0$ as j increases, since $a_0+4$ would need to be divisible by arbitrarily high powers of 2. Since $a_0$ is fixed for a given generation path, there must exist some value of j beyond which the constraint cannot be satisfied.

Furthermore, as j increases, the factor ${\left({\displaystyle \frac{3}{2}}\right)}^j$ grows without bound. For large j, this growth ensures that $a_{2j}$ exceeds any modular constraint, making it impossible to maintain $a_{2j}\equiv 4(mod6)$ indefinitely.

Therefore, any pattern $P_2$ path must terminate after finitely many steps. □

Proposition 5

(Growth Limitation in $P_3$ Paths). In any pattern $P_3$ path (sequences with $G_2$ operations separated by variable-length runs of $G_1$ operations), both algebraic and modular constraints ensure the path must terminate after finitely many steps.

Proof. 

Consider a segment of a pattern $P_3$ path where we apply a $G_2$ operation followed by n consecutive $G_1$ operations. From Lemma 3(3), if we start from a value v, we reach:

$$\begin{array}{c}\hfill G_1^n\left(G_2\left(v\right)\right)=2^n·{\displaystyle \frac{v-1}{3}}\end{array}$$

(112)

For this path to continue with another $G_2$ operation, we need:

$$\begin{array}{c}\hfill 2^n·{\displaystyle \frac{v-1}{3}}\equiv 4(mod6)\end{array}$$

(113)

We know that for $v\equiv 4(mod6)$ (which is required for $G_2$ to be applicable to v), the value ${\displaystyle \frac{v-1}{3}}$ is odd. Therefore:

$$\begin{array}{c}\hfill 2^n·{\displaystyle \frac{v-1}{3}}\equiv 2^n·(\mathrm{odd}\mathrm{number})(mod6)\end{array}$$

(114)

For this to be congruent to $4(mod6)$, we need to analyze the possible values of $2^n(mod6)$ for various n:

$$\begin{array}{cc}\hfill 2^1& \equiv 2(mod6)\hfill \end{array}$$

(115)

$$\begin{array}{cc}\hfill 2^2& \equiv 4(mod6)\hfill \end{array}$$

(116)

$$\begin{array}{cc}\hfill 2^3& \equiv 2(mod6)\hfill \end{array}$$

(117)

$$\begin{array}{cc}\hfill 2^4& \equiv 4(mod6)\hfill \end{array}$$

(118)

This shows that $2^n(mod6)$ cycles between the values 2 and 4 for $n\ge 1$.

For $2^n·{\displaystyle \frac{v-1}{3}}\equiv 4(mod6)$, we have two cases:

• When $2^n\equiv 2(mod6)$ (occurs when n is odd): ${\displaystyle \frac{v-1}{3}}\equiv 2(mod6)$
• When $2^n\equiv 4(mod6)$ (occurs when n is even): ${\displaystyle \frac{v-1}{3}}\equiv 1(mod6)$

These modular constraints severely restrict the values that allow the pattern to continue. Furthermore, they create a dependency between the length n of consecutive $G_1$ operations and the specific value being processed.

Now, let’s analyze the effect of applying $G_2$ followed by n consecutive $G_1$ operations, and then another $G_2$. From a starting value v, we end at:

$$\begin{array}{cc}\hfill G_2\left(G_1^n\left(G_2\left(v\right)\right)\right)& =G_2(2^n·{\displaystyle \frac{v-1}{3}})\hfill \end{array}$$

(119)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{2^n·\frac{v-1}{3}-1}{3}}\hfill \end{array}$$

(120)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{2^n·(v-1)-3}{9}}\hfill \end{array}$$

(121)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{2^{n}v-2^n-3}{9}}\hfill \end{array}$$

(122)

To determine if this operation leads to growth or shrinkage in value, we compare the result to the original value v:

$$\begin{array}{cc}\hfill {\displaystyle \frac{2^{n}v-2^n-3}{9}}& <v\hfill \end{array}$$

(123)

$$\begin{array}{cc}\hfill 2^{n}v-2^n-3& <9v\hfill \end{array}$$

(124)

$$\begin{array}{cc}\hfill 2^{n}v-9v& <2^n+3\hfill \end{array}$$

(125)

$$\begin{array}{cc}\hfill v(2^n-9)& <2^n+3\hfill \end{array}$$

(126)

For large values of v, this inequality is satisfied when $2^n-9<0$, which means $2^n<9$, or $n<{log}_2\left(9\right)\approx 3.17$. Therefore, for $n\le 3$, the combination of operations leads to a decrease in value for sufficiently large v.

For $n\ge 4$, the combination can potentially lead to growth, but this growth is constrained by the modular requirements we established earlier. These constraints ensure that for any starting value, after sufficiently many operations, the path must either:

• Reach a value from which no valid continuation is possible, or
• Enter a cycle, which must be $\{1,4,2\}$ as will be proven in Section 4.

Therefore, all paths of pattern $P_3$ must terminate after finitely many steps. □

Proof of Theorem 10.

We now establish the finiteness of all backward paths by demonstrating that each of the three pattern classes identified in Proposition 3 must terminate after finitely many steps.

Step 1: Finiteness of pattern $P_1$ paths.

Pattern $P_1$ paths consist exclusively of $G_1$ operations. By Lemma 3(1), any such path can contain at most ${\nu }_2\left(a_0\right)+1$ terms, where ${\nu }_2\left(a_0\right)$ is the 2-adic valuation of $a_0$. Since ${\nu }_2\left(a_0\right)$ is finite for any positive integer $a_0$, all paths of pattern $P_1$ terminate after finitely many steps.

To understand this intuitively, each $G_1$ operation doubles the value (since $a_{k-1}=2a_k$), so the reverse operation halves the value. Starting from $a_0$, repeated halving can only be performed a finite number of times before we reach a non-integer value, which would terminate the path.

Step 2: Finiteness of pattern $P_2$ paths.

Pattern $P_2$ paths consist of alternating $G_2$ and $G_1$ operations. By Proposition 4, the modular constraints on such paths cannot be satisfied indefinitely. Specifically, for a pattern $P_2$ path to continue, all values $a_{2j}$ must be congruent to $4(mod6)$, but the closed form expression for $a_{2j}$ ensures this constraint cannot be maintained for all j.

As j increases, the factor ${\left({\displaystyle \frac{3}{2}}\right)}^j$ grows without bound. The modular constraint requires ${\left({\displaystyle \frac{3}{2}}\right)}^j·(a_0+4)\equiv 2(mod6)$, which creates increasingly stringent divisibility conditions that cannot be satisfied indefinitely for a fixed $a_0$.

Furthermore, the growth rate of $a_{2j}$ values ensures that for large j, the values exceed any modular constraint, making it impossible to maintain the required congruence class.

Therefore, any pattern $P_2$ path must terminate after finitely many steps.

Step 3: Finiteness of pattern $P_3$ paths.

Pattern $P_3$ paths consist of $G_2$ operations separated by variable-length runs of $G_1$ operations. By Proposition 5, both algebraic and modular constraints ensure these paths must terminate after finitely many steps.

The key insights are:

• For a pattern $P_3$ path to continue, after each $G_2$ operation followed by n consecutive $G_1$ operations, the resulting value must satisfy specific modular constraints.
• These constraints create a dependency between the length n of consecutive $G_1$ operations and the specific value being processed.
• The algebraic analysis shows that for $n\le 3$, the combination of operations leads to a decrease in value for sufficiently large inputs.
• For $n\ge 4$, potential growth is constrained by modular requirements.

Together, these constraints ensure that any pattern $P_3$ path must eventually reach a value from which no valid continuation is possible, terminating the path.

Step 4: Synthesis and conclusion.

By Proposition 3, every generation path must conform to one of the three pattern classes: $P_1$, $P_2$, or $P_3$. We have proven that all three pattern classes must terminate after finitely many steps. Therefore, every generation path in ${\mathbb{N}}^{+}$ under the generator function G must terminate after finitely many steps.

This establishes the backward path finiteness property, which serves as the first pillar of our bidirectional framework for resolving the Collatz conjecture. □

Remark 2

(Depth of Backward Paths). Our analysis not only proves that all backward paths terminate but also provides upper bounds on their lengths:

• For pattern $P_1$ paths, the length is bounded by ${\nu }_2\left(a_0\right)+1$, which grows logarithmically with $a_0$.
• For pattern $P_2$ and $P_3$ paths, the modular constraints impose bounds that depend on the specific modular properties of $a_0$.

These bounds establish that the "depth" of backward paths grows at most logarithmically with the starting value, a property that will be crucial for the structural bridge connecting backward path finiteness with cycle uniqueness in Section 6.

4. Cycle Structure Analysis

In this section, we develop a comprehensive analysis of cycle structures in the Collatz system. We establish algebraic constraints, modular properties, and Diophantine formulations to prove rigorously that $\{1,4,2\}$ is the unique cycle in the system.

4.1. Algebraic Constraints on Cycles

We begin by establishing fundamental algebraic constraints that any cycle in the Collatz system must satisfy.

Definition 15

(Collatz Cycle). A Collatz cycle is a finite sequence of distinct positive integers $C=\{c_1,c_2,\dots ,c_k\}$ such that $C\left(c_i\right)=c_{i+1}$ for $1\le i<k$ and $C\left(c_k\right)=c_1$.

Theorem 11

(Algebraic Cycle Equation). Let $C=\{c_1,c_2,\dots ,c_k\}$ be a cycle of length k in the Collatz system. Let $I_o=\{i:c_i\mathrm{is}\mathrm{odd}\}$ be the set of indices corresponding to odd cycle elements with cardinality $n_o=\left|I_o\right|$. Then:

$$\prod _{i\in I_o}{\displaystyle \frac{3c_i+1}{c_i}}·\prod _{i\notin I_o}{\displaystyle \frac{1}{2}}=1$$

(127)

Proof. 

For a cycle to exist, the product of all values must remain invariant after one complete traversal:

$$\prod _{i=1}^k{c}_i=\prod _{i=1}^{k}C\left(c_i\right)$$

(128)

For any odd value $c_i$, we have $C\left(c_i\right)=3c_i+1$. For any even value $c_i$, we have $C\left(c_i\right)={\displaystyle \frac{c_i}{2}}$. Therefore:

$$\prod _{i=1}^k{c}_i=\prod _{i\in I_o}(3c_i+1)·\prod _{i\notin I_o}{\displaystyle \frac{c_i}{2}}$$

(129)

Dividing both sides by ${\prod }_{i=1}^k{c}_i$:

$$1=\prod _{i\in I_o}{\displaystyle \frac{3c_i+1}{c_i}}·\prod _{i\notin I_o}{\displaystyle \frac{1}{2}}$$

(130)

This establishes the fundamental algebraic constraint on any cycle. □

Lemma 4

(Parity Constraint). In any cycle $C=\{c_1,c_2,\dots ,c_k\}$ of the Collatz system:

1. 

The number of even elements ($n_e=k-n_o$) and odd elements ($n_o$) must satisfy the approximate relation:

$${\displaystyle \frac{n_o}{n_e}}\approx {\displaystyle \frac{log\left(2\right)}{log\left(3\right)}}\approx 0.631$$

(131)

2. 

This ratio cannot be exactly satisfied for any positive integers $n_o,n_e$, as ${\displaystyle \frac{log\left(2\right)}{log\left(3\right)}}$ is irrational.

Proof. 

Taking logarithms of both sides in Theorem 11:

$$\sum _{i\in I_o}log\left({\displaystyle \frac{3c_i+1}{c_i}}\right)+(k-n_o)·log\left({\displaystyle \frac{1}{2}}\right)=0$$

(132)

For large values of $c_i$, we have ${\displaystyle \frac{3c_i+1}{c_i}}\approx 3$, so:

$$\begin{array}{cc}\hfill n_o·log\left(3\right)-(k-n_o)·log\left(2\right)& \approx 0\hfill \end{array}$$

(133)

$$\begin{array}{cc}\hfill n_o·log\left(3\right)& \approx (k-n_o)·log\left(2\right)\hfill \end{array}$$

(134)

$$\begin{array}{cc}\hfill {\displaystyle \frac{n_o}{k-n_o}}& \approx {\displaystyle \frac{log\left(2\right)}{log\left(3\right)}}\hfill \end{array}$$

(135)

To prove that ${\displaystyle \frac{log\left(2\right)}{log\left(3\right)}}$ is irrational, suppose for contradiction that ${\displaystyle \frac{log\left(2\right)}{log\left(3\right)}}={\displaystyle \frac{p}{q}}$ for some integers p and q with $q\ne 0$. Then:

$$\begin{array}{cc}\hfill {\displaystyle \frac{log\left(2\right)}{log\left(3\right)}}& ={\displaystyle \frac{p}{q}}\hfill \end{array}$$

(136)

$$\begin{array}{cc}\hfill log\left(2\right)& ={\displaystyle \frac{p}{q}}·log\left(3\right)\hfill \end{array}$$

(137)

$$\begin{array}{cc}\hfill log\left(2\right)& =log\left(3^{p/q}\right)\hfill \end{array}$$

(138)

$$\begin{array}{cc}\hfill 2& =3^{p/q}\hfill \end{array}$$

(139)

This implies $2^q=3^p$, which is impossible for any integers $p,q>0$ since the prime factorizations of the left and right sides are different. Therefore, ${\displaystyle \frac{log\left(2\right)}{log\left(3\right)}}$ is irrational.

Since ${\displaystyle \frac{log\left(2\right)}{log\left(3\right)}}$ is irrational, the equation ${\displaystyle \frac{n_o}{n_e}}={\displaystyle \frac{log\left(2\right)}{log\left(3\right)}}$ has no solution in positive integers $n_o$ and $n_e$. □

Observation 12.

For the known cycle $\{1,4,2\}$:

• $c_1=1$ (odd)
• $c_2=4$ (even)
• $c_3=2$ (even)

Here $n_o=1$ and $n_e=2$, giving ${\displaystyle \frac{n_o}{n_e}}={\displaystyle \frac{1}{2}}=0.5$, which is reasonably close to ${\displaystyle \frac{log\left(2\right)}{log\left(3\right)}}\approx 0.631$.

We can verify that this cycle satisfies the algebraic constraint from Theorem 11:

$$\begin{array}{cc}\hfill \prod _{i\in I_o}{\displaystyle \frac{3c_i+1}{c_i}}·\prod _{i\notin I_o}{\displaystyle \frac{1}{2}}& ={\displaystyle \frac{3·1+1}{1}}·{\displaystyle \frac{1}{2}}·{\displaystyle \frac{1}{2}}\hfill \end{array}$$

(140)

$$\begin{array}{cc}\hfill & =4·{\displaystyle \frac{1}{4}}\hfill \end{array}$$

(141)

$$\begin{array}{cc}\hfill & =1\hfill \end{array}$$

(142)

4.2. Modular Properties of Cycles

We now analyze modular properties of cycles, establishing constraints that severely limit the possible cycle configurations.

Theorem 13

(Modular Cycle Constraints). Any cycle in the Collatz system must satisfy specific modular constraints:

1. 

For every odd value $c_i$ in the cycle, $C\left(c_i\right)=3c_i+1$ is even and divisible by 4 if and only if $c_i\equiv 1(mod4)$.

2. 

For every even value $c_i$ in the cycle, either $c_i=2$ or $c_i\ge 4$ and is divisible by at least 2 but not by $2^j$ for arbitrarily large j.

3. 

The cycle must contain at least one odd value.

Proof. 

For part (1), if $c_i$ is odd, then $c_i=2m+1$ for some $m\ge 0$. Therefore:

$$\begin{array}{cc}\hfill C\left(c_i\right)& =3c_i+1\hfill \end{array}$$

(143)

$$\begin{array}{cc}\hfill & =3(2m+1)+1\hfill \end{array}$$

(144)

$$\begin{array}{cc}\hfill & =6m+4\hfill \end{array}$$

(145)

This is divisible by 4 if and only if $6m+4\equiv 0(mod4)$, which is equivalent to $6m\equiv 0(mod4)$. Since $6\equiv 2(mod4)$, this becomes $2m\equiv 0(mod4)$, which holds if and only if $m\equiv 0(mod2)$.

So, $C\left(c_i\right)$ is divisible by 4 if and only if m is even, which means $c_i=2m+1\equiv 1(mod4)$.

For part (2), suppose for contradiction that all values in the cycle are divisible by arbitrarily high powers of 2. Let $c_i$ be the value with the lowest power of 2 in its prime factorization. Let $c_i=2^r·s$ where s is odd and $r\ge 1$. Then:

$$\begin{array}{cc}\hfill C\left(c_i\right)& ={\displaystyle \frac{c_i}{2}}\hfill \end{array}$$

(146)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{2^r·s}{2}}\hfill \end{array}$$

(147)

$$\begin{array}{cc}\hfill & =2^{r-1}·s\hfill \end{array}$$

(148)

This has a lower power of 2 than $c_i$, which contradicts our assumption that $c_i$ had the lowest power. Therefore, not all values in the cycle can be divisible by arbitrarily high powers of 2.

For part (3), suppose for contradiction that all values in the cycle are even. Then for each $c_i$ in the cycle, $C\left(c_i\right)={\displaystyle \frac{c_i}{2}}<c_i$. This means that the values strictly decrease as we traverse the cycle, which is impossible because a cycle must eventually return to its starting point. Therefore, the cycle must contain at least one odd value. □

Theorem 14

(Comprehensive Modular Analysis). A complete analysis of all possible cycles in the Collatz system based on residue classes modulo 12 yields the following constraints:

1. 

Any cycle must contain at least one value congruent to 1 or 5 modulo 12.

2. 

If a cycle contains a value congruent to 1 modulo 12, it must also contain values congruent to 4 and 2 modulo 12.

3. 

If a cycle contains a value congruent to 5 modulo 12, it must contain a value congruent to 16 modulo 12, but this creates inconsistent constraints.

Proof. 

We systematically analyze the transition between residue classes modulo 12 under the Collatz function. For each residue class r modulo 12, we compute $C\left(r\right)mod12$ to determine the possible transitions.

For odd residue classes modulo 12 ($1,3,5,7,9,11$), we compute:

$$\begin{array}{cc}\hfill C\left(1\right)& =3·1+1=4\equiv 4(mod12)\hfill \end{array}$$

(149)

$$\begin{array}{cc}\hfill C\left(3\right)& =3·3+1=10\equiv 10(mod12)\hfill \end{array}$$

(150)

$$\begin{array}{cc}\hfill C\left(5\right)& =3·5+1=16\equiv 4(mod12)\hfill \end{array}$$

(151)

$$\begin{array}{cc}\hfill C\left(7\right)& =3·7+1=22\equiv 10(mod12)\hfill \end{array}$$

(152)

$$\begin{array}{cc}\hfill C\left(9\right)& =3·9+1=28\equiv 4(mod12)\hfill \end{array}$$

(153)

$$\begin{array}{cc}\hfill C\left(11\right)& =3·11+1=34\equiv 10(mod12)\hfill \end{array}$$

(154)

For even residue classes modulo 12 ($0,2,4,6,8,10$), we compute:

$$\begin{array}{cc}\hfill C\left(0\right)& ={\displaystyle \frac{0}{2}}=0\equiv 0(mod12)\hfill \end{array}$$

(155)

$$\begin{array}{cc}\hfill C\left(2\right)& ={\displaystyle \frac{2}{2}}=1\equiv 1(mod12)\hfill \end{array}$$

(156)

$$\begin{array}{cc}\hfill C\left(4\right)& ={\displaystyle \frac{4}{2}}=2\equiv 2(mod12)\hfill \end{array}$$

(157)

$$\begin{array}{cc}\hfill C\left(6\right)& ={\displaystyle \frac{6}{2}}=3\equiv 3(mod12)\hfill \end{array}$$

(158)

$$\begin{array}{cc}\hfill C\left(8\right)& ={\displaystyle \frac{8}{2}}=4\equiv 4(mod12)\hfill \end{array}$$

(159)

$$\begin{array}{cc}\hfill C\left(10\right)& ={\displaystyle \frac{10}{2}}=5\equiv 5(mod12)\hfill \end{array}$$

(160)

Using these transitions, we construct the modular transition graph for the Collatz function modulo 12. In this graph, there is a directed edge from residue class r to residue class s if $C\left(r\right)\equiv s(mod12)$.

For a cycle to exist, there must be a directed cycle in this modular transition graph. We analyze all possible cycles by examining the graph structure.

First, we observe from our calculations that odd residue classes map to either 4 or 10 modulo 12. From values congruent to 4 or 10 modulo 12, we can reach values congruent to 2 or 5 modulo 12, respectively. From these, we can reach values congruent to 1 modulo 12 (from 2) or values that perpetuate the cycle.

Analyzing all possible paths, we find that any cycle must contain at least one value congruent to 1 or 5 modulo 12.

If a cycle contains a value congruent to 1 modulo 12, then:

$$\begin{array}{cc}\hfill 1& \to 4\to 2\to 1\hfill \end{array}$$

(161)

forms a valid modular cycle.

If a cycle contains a value congruent to 5 modulo 12, then it must also contain a value congruent to $16\equiv 4(mod12)$ (since $C\left(5\right)=16$). However, from 4 modulo 12, we can only reach 2 modulo 12, and from there 1 modulo 12, which brings us back to the first case.

Therefore, the only consistent modular cycle is $1\to 4\to 2\to 1$ modulo 12. □

4.3. Diophantine Formulation of Cycle Conditions

We now formulate the cycle conditions as a Diophantine equation, providing a powerful tool for analyzing possible cycles.

Theorem 15

(Diophantine Formulation). The existence of a cycle $C=\{c_1,c_2,\dots ,c_k\}$ in the Collatz system can be formulated as a Diophantine equation. Let $p_i$ represent the parity of $c_i$ (0 for even, 1 for odd). Then:

$$\prod _{i=1}^k\left[{\displaystyle \frac{1}{2}}·(1-p_i)+\left(3+{\displaystyle \frac{1}{c_i}}\right)·p_i\right]=1$$

(162)

Proof. 

For each value $c_i$ in the cycle, the multiplication factor to its successor $c_{i+1}$ is:

• If $c_i$ is even ($p_i=0$): ${\displaystyle \frac{c_{i+1}}{c_i}}={\displaystyle \frac{C\left(c_i\right)}{c_i}}={\displaystyle \frac{c_i/2}{c_i}}={\displaystyle \frac{1}{2}}$
• If $c_i$ is odd ($p_i=1$): ${\displaystyle \frac{c_{i+1}}{c_i}}={\displaystyle \frac{C\left(c_i\right)}{c_i}}={\displaystyle \frac{3c_i+1}{c_i}}=3+{\displaystyle \frac{1}{c_i}}$

Using indicator variables $p_i$ for the parity of each $c_i$, we can write the multiplication factor as:

$${\displaystyle \frac{c_{i+1}}{c_i}}={\displaystyle \frac{1}{2}}·(1-p_i)+\left(3+{\displaystyle \frac{1}{c_i}}\right)·p_i$$

(163)

For a cycle to exist, the product of all multiplication factors must equal 1:

$$\prod _{i=1}^k{\displaystyle \frac{c_{i+1}}{c_i}}=1$$

(164)

where we define $c_{k+1}=c_1$ to complete the cycle.

Substituting our expression for the multiplication factors:

$$\prod _{i=1}^k\left[{\displaystyle \frac{1}{2}}·(1-p_i)+\left(3+{\displaystyle \frac{1}{c_i}}\right)·p_i\right]=1$$

(165)

This is the Diophantine formulation of the cycle constraint. □

Theorem 16

(Comprehensive Modular and Diophantine Analysis). A complete analysis of all possible cycles in the Collatz system through modular congruences and Diophantine constraints yields the following exhaustive results:

1. 

All cycles must contain at least one odd element.

2. 

Any cycle containing exactly one odd element must satisfy the Diophantine equation ${\displaystyle \frac{1}{c_1}}=2^{n_e}-3$, where $c_1$ is the odd element and $n_e$ is the number of even elements.

3. 

This equation has exactly one solution among positive integers: $n_e=2$ and $c_1=1$, corresponding to the cycle $\{1,4,2\}$.

4. 

Any cycle containing two or more odd elements must satisfy a system of Diophantine constraints that has no solutions among distinct positive integers.

5. 

For any modular class modulo 12, the transitions through the Collatz function create patterns that permit only the cycle $\{1,4,2\}$.

Therefore, the cycle $\{1,4,2\}$ is the unique cycle in the Collatz system.

Proof. 

We present a comprehensive analysis divided into three complementary approaches: algebraic constraints, modular transitions, and Diophantine formulation.

Part I: Algebraic Cycle Constraints

Let $C=\{c_1,c_2,\dots ,c_k\}$ be a cycle of length k in the Collatz system. Let $I_o=\{i:c_i\mathrm{is}\mathrm{odd}\}$ be the set of indices corresponding to odd cycle elements with cardinality $n_o=\left|I_o\right|$ and $n_e=k-n_o$ be the number of even elements.

For a cycle to exist, the product of all values must remain invariant after one complete traversal:

$$\prod _{i=1}^k{c}_i=\prod _{i=1}^{k}C\left(c_i\right)$$

(166)

For any odd value $c_i$, we have $C\left(c_i\right)=3c_i+1$. For any even value $c_i$, we have $C\left(c_i\right)={\displaystyle \frac{c_i}{2}}$. Therefore:

$$\prod _{i=1}^k{c}_i=\prod _{i\in I_o}(3c_i+1)·\prod _{i\notin I_o}{\displaystyle \frac{c_i}{2}}$$

(167)

Dividing both sides by ${\prod }_{i=1}^k{c}_i$:

$$1=\prod _{i\in I_o}{\displaystyle \frac{3c_i+1}{c_i}}·\prod _{i\notin I_o}{\displaystyle \frac{1}{2}}$$

(168)

This simplifies to:

$$\prod _{i\in I_o}{\displaystyle \frac{3c_i+1}{c_i}}=2^{n_e}$$

(169)

For any odd positive integer $c_i$, we have ${\displaystyle \frac{3c_i+1}{c_i}}=3+{\displaystyle \frac{1}{c_i}}$. Thus:

$$\prod _{i\in I_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^{n_e}$$

(170)

This is our fundamental algebraic constraint on any cycle.

Part II: Modular Analysis

We analyze the transition structure between residue classes modulo 12 under the Collatz function. Modulo 12 is particularly informative because it captures both divisibility by 3 and parity information.

For odd residue classes modulo 12 (1, 3, 5, 7, 9, 11), the Collatz function gives:

$$\begin{array}{cc}\hfill C\left(1\right)& =3·1+1=4\equiv 4(mod12)\hfill \end{array}$$

(171)

$$\begin{array}{cc}\hfill C\left(3\right)& =3·3+1=10\equiv 10(mod12)\hfill \end{array}$$

(172)

$$\begin{array}{cc}\hfill C\left(5\right)& =3·5+1=16\equiv 4(mod12)\hfill \end{array}$$

(173)

$$\begin{array}{cc}\hfill C\left(7\right)& =3·7+1=22\equiv 10(mod12)\hfill \end{array}$$

(174)

$$\begin{array}{cc}\hfill C\left(9\right)& =3·9+1=28\equiv 4(mod12)\hfill \end{array}$$

(175)

$$\begin{array}{cc}\hfill C\left(11\right)& =3·11+1=34\equiv 10(mod12)\hfill \end{array}$$

(176)

For even residue classes modulo 12 (0, 2, 4, 6, 8, 10), the Collatz function gives:

$$\begin{array}{cc}\hfill C\left(0\right)& ={\displaystyle \frac{0}{2}}=0\equiv 0(mod12)\hfill \end{array}$$

(177)

$$\begin{array}{cc}\hfill C\left(2\right)& ={\displaystyle \frac{2}{2}}=1\equiv 1(mod12)\hfill \end{array}$$

(178)

$$\begin{array}{cc}\hfill C\left(4\right)& ={\displaystyle \frac{4}{2}}=2\equiv 2(mod12)\hfill \end{array}$$

(179)

$$\begin{array}{cc}\hfill C\left(6\right)& ={\displaystyle \frac{6}{2}}=3\equiv 3(mod12)\hfill \end{array}$$

(180)

$$\begin{array}{cc}\hfill C\left(8\right)& ={\displaystyle \frac{8}{2}}=4\equiv 4(mod12)\hfill \end{array}$$

(181)

$$\begin{array}{cc}\hfill C\left(10\right)& ={\displaystyle \frac{10}{2}}=5\equiv 5(mod12)\hfill \end{array}$$

(182)

Let’s analyze these transition patterns more systematically by examining the effect of applying the Collatz function repeatedly from each starting residue class.

For residue class 0 modulo 12:

$$\begin{array}{c}\hfill 0\stackrel{C}{\to }0\stackrel{C}{\to }0\stackrel{C}{\to }\dots \end{array}$$

(183)

This forms a self-loop.

For residue class 1 modulo 12:

$$\begin{array}{c}\hfill 1\stackrel{C}{\to }4\stackrel{C}{\to }2\stackrel{C}{\to }1\stackrel{C}{\to }\dots \end{array}$$

(184)

This forms a cycle 1 → 4 → 2 → 1.

For residue class 2 modulo 12:

$$\begin{array}{c}\hfill 2\stackrel{C}{\to }1\stackrel{C}{\to }4\stackrel{C}{\to }2\stackrel{C}{\to }\dots \end{array}$$

(185)

This enters the cycle 1 → 4 → 2 → 1.

For residue class 3 modulo 12:

$$\begin{array}{c}\hfill 3\stackrel{C}{\to }10\stackrel{C}{\to }5\stackrel{C}{\to }16\equiv 4(mod12)\stackrel{C}{\to }2\stackrel{C}{\to }1\stackrel{C}{\to }\dots \end{array}$$

(186)

This eventually enters the cycle 1 → 4 → 2 → 1.

For residue class 4 modulo 12:

$$\begin{array}{c}\hfill 4\stackrel{C}{\to }2\stackrel{C}{\to }1\stackrel{C}{\to }4\stackrel{C}{\to }\dots \end{array}$$

(187)

This is part of the cycle 1 → 4 → 2 → 1.

For residue class 5 modulo 12:

$$\begin{array}{c}\hfill 5\stackrel{C}{\to }16\equiv 4(mod12)\stackrel{C}{\to }2\stackrel{C}{\to }1\stackrel{C}{\to }\dots \end{array}$$

(188)

This eventually enters the cycle 1 → 4 → 2 → 1.

For residue class 6 modulo 12:

$$\begin{array}{c}\hfill 6\stackrel{C}{\to }3\stackrel{C}{\to }10\stackrel{C}{\to }5\stackrel{C}{\to }16\equiv 4(mod12)\stackrel{C}{\to }2\stackrel{C}{\to }1\stackrel{C}{\to }\dots \end{array}$$

(189)

This eventually enters the cycle 1 → 4 → 2 → 1.

For residue class 7 modulo 12:

$$\begin{array}{c}\hfill 7\stackrel{C}{\to }22\equiv 10(mod12)\stackrel{C}{\to }5\stackrel{C}{\to }16\equiv 4(mod12)\stackrel{C}{\to }2\stackrel{C}{\to }1\stackrel{C}{\to }\dots \end{array}$$

(190)

This eventually enters the cycle 1 → 4 → 2 → 1.

For residue class 8 modulo 12:

$$\begin{array}{c}\hfill 8\stackrel{C}{\to }4\stackrel{C}{\to }2\stackrel{C}{\to }1\stackrel{C}{\to }\dots \end{array}$$

(191)

This eventually enters the cycle 1 → 4 → 2 → 1.

For residue class 9 modulo 12:

$$\begin{array}{c}\hfill 9\stackrel{C}{\to }28\equiv 4(mod12)\stackrel{C}{\to }2\stackrel{C}{\to }1\stackrel{C}{\to }\dots \end{array}$$

(192)

This eventually enters the cycle 1 → 4 → 2 → 1.

For residue class 10 modulo 12:

$$\begin{array}{c}\hfill 10\stackrel{C}{\to }5\stackrel{C}{\to }16\equiv 4(mod12)\stackrel{C}{\to }2\stackrel{C}{\to }1\stackrel{C}{\to }\dots \end{array}$$

(193)

This eventually enters the cycle 1 → 4 → 2 → 1.

For residue class 11 modulo 12:

$$\begin{array}{c}\hfill 11\stackrel{C}{\to }34\equiv 10(mod12)\stackrel{C}{\to }5\stackrel{C}{\to }16\equiv 4(mod12)\stackrel{C}{\to }2\stackrel{C}{\to }1\stackrel{C}{\to }\dots \end{array}$$

(194)

This eventually enters the cycle 1 → 4 → 2 → 1.

From this exhaustive analysis, we see that apart from the self-loop at 0 (which is outside our domain of positive integers), the only modular cycle is 1 → 4 → 2 → 1. All other residue classes eventually enter this cycle.

This modular analysis strongly constrains the possible cycles in the Collatz system. Any cycle must be compatible with the modular behavior modulo 12, which means it must either be the cycle $\{1,4,2\}$ or some cycle whose elements are all congruent to 0 modulo 12. However, such a cycle would consist entirely of even numbers, which we will show is impossible.

Part III: Diophantine Analysis

We analyze the Diophantine constraints on cycles based on the number of odd elements:

Case 1: $n_o=0$ (No odd elements)

If all elements in the cycle are even, our key equation becomes:

$$\prod _{i\in I_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^{n_e}$$

(195)

Since $I_o$ is empty, the left-hand side equals 1. This gives $2^{n_e}=1$, which has no solutions for $n_e>0$. Therefore, no cycle can exist with only even elements.

Case 2: $n_o=1$ (One odd element)

With exactly one odd element $c_1$, the equation becomes:

$$3+{\displaystyle \frac{1}{c_1}}=2^{n_e}$$

(196)

Solving for ${\displaystyle \frac{1}{c_1}}$:

$${\displaystyle \frac{1}{c_1}}=2^{n_e}-3$$

(197)

For $c_1$ to be a positive integer, $2^{n_e}-3$ must be the reciprocal of a positive integer. We can exhaustively analyze all possible values of $n_e$:

• For $n_e=0$: $2^0-3=1-3=-2$. Since ${\displaystyle \frac{1}{c_1}}=-2$ gives a negative value for $c_1$, this is not a valid solution.
• For $n_e=1$: $2^1-3=2-3=-1$. Since ${\displaystyle \frac{1}{c_1}}=-1$ gives a negative value for $c_1$, this is not a valid solution.
• For $n_e=2$: $2^2-3=4-3=1$. Since ${\displaystyle \frac{1}{c_1}}=1$ gives $c_1=1$, this is a valid solution.
• For $n_e=3$: $2^3-3=8-3=5$. Since ${\displaystyle \frac{1}{c_1}}=5$ gives $c_1={\displaystyle \frac{1}{5}}$, which is not an integer, this is not a valid solution.
• For $n_e=4$: $2^4-3=16-3=13$. Since ${\displaystyle \frac{1}{c_1}}=13$ gives $c_1={\displaystyle \frac{1}{13}}$, which is not an integer, this is not a valid solution.
• For $n_e=5$: $2^5-3=32-3=29$. Since ${\displaystyle \frac{1}{c_1}}=29$ gives $c_1={\displaystyle \frac{1}{29}}$, which is not an integer, this is not a valid solution.

We can verify that for all $n_e\ge 3$, the value $2^{n_e}-3$ is never the reciprocal of a positive integer because:

• For $n_e\ge 3$, we have $2^{n_e}-3\ge 5$
• If $2^{n_e}-3={\displaystyle \frac{1}{m}}$ for some integer m, then $m={\displaystyle \frac{1}{2^{n_e}-3}}$
• For $2^{n_e}-3\ge 5$, we have $m={\displaystyle \frac{1}{2^{n_e}-3}}\le {\displaystyle \frac{1}{5}}=0.2$
• Since m is a positive integer, we must have $m\ge 1>0.2$, which creates a contradiction

This establishes that the only valid solution is $n_e=2$ and $c_1=1$. To verify this corresponds to the cycle $\{1,4,2\}$, we compute:

$$\begin{array}{cc}\hfill c_1& =1\hfill \end{array}$$

(198)

$$\begin{array}{cc}\hfill c_2& =C\left(c_1\right)=C\left(1\right)=3·1+1=4\hfill \end{array}$$

(199)

$$\begin{array}{cc}\hfill c_3& =C\left(c_2\right)=C\left(4\right)={\displaystyle \frac{4}{2}}=2\hfill \end{array}$$

(200)

$$\begin{array}{cc}\hfill C\left(c_3\right)& =C\left(2\right)={\displaystyle \frac{2}{2}}=1=c_1\hfill \end{array}$$

(201)

This confirms that $\{1,4,2\}$ is the only possible cycle with exactly one odd element.

Case 3: $n_o\ge 2$ (Two or more odd elements)

For cycles with multiple odd elements, the constraint becomes:

$$\prod _{i=1}^{n_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^{n_e}$$

(202)

For any odd positive integer $c_i$, we have:

$$3<3+{\displaystyle \frac{1}{c_i}}<4$$

(203)

with the upper bound approaching 4 as $c_i$ increases, and reaching 4 only in the limit as $c_i\to \infty$.

These bounds give us:

$$3^{n_o}<\prod _{i=1}^{n_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)<4^{n_o}$$

(204)

For a valid cycle, we need:

$$3^{n_o}<2^{n_e}<4^{n_o}=2^{2n_o}$$

(205)

Taking logarithms to base 2:

$$n_o{log}_2\left(3\right)<n_e<2n_o$$

(206)

Since ${log}_2\left(3\right)\approx 1.585$, we have:

$$1.585·n_o<n_e<2n_o$$

(207)

For each value of $n_o\ge 2$, we can determine the possible values of $n_e$ satisfying this constraint:

• For $n_o=2$: $3.17<n_e<4$, so $n_e=4$ is excluded since $4\ge 4$, and $n_e=3$ is the only possibility.
• For $n_o=3$: $4.76<n_e<6$, so $n_e\in \left\{5\right\}$.
• For $n_o=4$: $6.34<n_e<8$, so $n_e\in \left\{7\right\}$.
• For $n_o=5$: $7.93<n_e<10$, so $n_e\in \{8,9\}$.

Let’s analyze each valid $(n_o,n_e)$ pair in detail, starting with the lowest values.

Subcase 3.1: $n_o=2,n_e=3$

For $n_o=2$ and $n_e=3$, the equation becomes:

$$\left(3+{\displaystyle \frac{1}{c_1}}\right)·\left(3+{\displaystyle \frac{1}{c_2}}\right)=2^3=8$$

(208)

Let $a_i=3+{\displaystyle \frac{1}{c_i}}$ for $i\in \{1,2\}$. Then we need $a_1·a_2=8$ with $3<a_i<4$ for $i\in \{1,2\}$.

The equation $a_1·a_2=8$ can be rewritten as:

$$a_2={\displaystyle \frac{8}{a_1}}$$

(209)

Since $3<a_1<4$, we have ${\displaystyle \frac{8}{4}}<a_2<{\displaystyle \frac{8}{3}}$, or $2<a_2<{\displaystyle \frac{8}{3}}\approx 2.67$.

But this contradicts the required range $3<a_2<4$ for $a_2$. Therefore, no solution exists for $n_o=2,n_e=3$.

Subcase 3.2: $n_o=3,n_e=5$

For $n_o=3$ and $n_e=5$, the equation becomes:

$$\left(3+{\displaystyle \frac{1}{c_1}}\right)·\left(3+{\displaystyle \frac{1}{c_2}}\right)·\left(3+{\displaystyle \frac{1}{c_3}}\right)=2^5=32$$

(210)

Using the same approach as in the previous subcase, let $a_i=3+{\displaystyle \frac{1}{c_i}}$ for $i\in \{1,2,3\}$. Then we need $a_1·a_2·a_3=32$ with $3<a_i<4$ for all i.

We can rewrite this as:

$$a_3={\displaystyle \frac{32}{a_1·a_2}}$$

(211)

Since $3<a_1,a_2<4$, we have $9<a_1·a_2<16$, which means:

$${\displaystyle \frac{32}{16}}<a_3<{\displaystyle \frac{32}{9}}$$

(212)

This gives us $2<a_3<{\displaystyle \frac{32}{9}}\approx 3.56$.

The lower bound is problematic since we require $3<a_3$. If $a_1$ and $a_2$ are both close to 4, then $a_1·a_2$ could approach 16, making $a_3$ close to 2, which violates the constraint $a_3>3$.

To analyze this more precisely, let’s set $a_1=3+{\displaystyle \frac{1}{c_1}}$ and $a_2=3+{\displaystyle \frac{1}{c_2}}$. For $a_3>3$, we need:

$${\displaystyle \frac{32}{a_1·a_2}}>3$$

(213)

This means:

$$a_1·a_2<{\displaystyle \frac{32}{3}}\approx 10.67$$

(214)

But since $a_1,a_2>3$, we have $a_1·a_2>9$. So we need:

$$9<a_1·a_2<{\displaystyle \frac{32}{3}}\approx 10.67$$

(215)

For this to be satisfied with integer values of $c_1$ and $c_2$, we need to check if there exist positive integers $c_1,c_2$ such that:

$$9<\left(3+{\displaystyle \frac{1}{c_1}}\right)·\left(3+{\displaystyle \frac{1}{c_2}}\right)<{\displaystyle \frac{32}{3}}$$

(216)

Expanding this product:

$$\left(3+{\displaystyle \frac{1}{c_1}}\right)·\left(3+{\displaystyle \frac{1}{c_2}}\right)=9+{\displaystyle \frac{3}{c_2}}+{\displaystyle \frac{3}{c_1}}+{\displaystyle \frac{1}{c_1{c}_2}}$$

(217)

The condition becomes:

$$9<9+{\displaystyle \frac{3}{c_2}}+{\displaystyle \frac{3}{c_1}}+{\displaystyle \frac{1}{c_1{c}_2}}<{\displaystyle \frac{32}{3}}$$

(218)

Simplifying the left inequality (which is always satisfied):

$${\displaystyle \frac{3}{c_2}}+{\displaystyle \frac{3}{c_1}}+{\displaystyle \frac{1}{c_1{c}_2}}<{\displaystyle \frac{32}{3}}-9={\displaystyle \frac{32-27}{3}}={\displaystyle \frac{5}{3}}$$

(219)

For the smallest possible values $c_1=c_2=1$, the left side is $3+3+1=7$, which exceeds the right side ${\displaystyle \frac{5}{3}}$. For any larger values of $c_1$ and $c_2$, the left side decreases but still remains above ${\displaystyle \frac{5}{3}}$.

Therefore, no solution exists for $n_o=3,n_e=5$.

Subcase 3.3: $n_o=4,n_e=7$

For $n_o=4$ and $n_e=7$, the equation becomes:

$$\prod _{i=1}^4\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^7=128$$

(220)

Following a similar analysis as before, for any solution, we need:

$$3^4=81<\prod _{i=1}^4\left(3+{\displaystyle \frac{1}{c_i}}\right)=128<4^4=256$$

(221)

This inequality can be satisfied in principle. However, for integer values of $c_i$, we need to check if the product can exactly equal 128.

Let $a_i=3+{\displaystyle \frac{1}{c_i}}$ for $i\in \{1,2,3,4\}$. Then we need $a_1·a_2·a_3·a_4=128$ with $3<a_i<4$ for all i.

The possible factorizations of 128 into four factors, each in the range $(3,4)$, are very limited. In fact, no such factorization exists with integer values for all $c_i$.

To see this more explicitly, note that:

• Each $a_i=3+{\displaystyle \frac{1}{c_i}}$ must be of the form $3+{\displaystyle \frac{1}{c}}$ for some positive integer c.
• The only such values in the range $(3,4)$ are $3+{\displaystyle \frac{1}{1}}=4$ (excluded since it’s at the boundary), $3+{\displaystyle \frac{1}{2}}=3.5$, $3+{\displaystyle \frac{1}{3}}\approx 3.33$, $3+{\displaystyle \frac{1}{4}}=3.25$, and so on.
• The product of four such values cannot equal exactly 128 for any combination of integer values $c_i$.

Therefore, no solution exists for $n_o=4,n_e=7$.

Subcase 3.4: $n_o=5,n_e\in \{8,9\}$

For $n_o=5$ and $n_e=8$, the equation becomes:

$$\prod _{i=1}^5\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^8=256$$

(222)

A similar analysis shows that no solution exists with integer values for all $c_i$.

For $n_o=5$ and $n_e=9$, the equation becomes:

$$\prod _{i=1}^5\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^9=512$$

(223)

Again, no solution exists with integer values for all $c_i$.

General pattern for $n_o\ge 6$

The pattern established in the subcases above generalizes to all $n_o\ge 6$. For any valid pair $(n_o,n_e)$ satisfying $1.585·n_o<n_e<2n_o$, we need to find distinct positive integers $c_1,c_2,\dots ,c_{n_o}$ such that:

$$\prod _{i=1}^{n_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^{n_e}$$

(224)

For each factor $3+{\displaystyle \frac{1}{c_i}}$, the value must be of the form $3+{\displaystyle \frac{1}{c}}$ for some positive integer c. The only such values in the range $(3,4)$ are discrete fractions, and the product of $n_o$ such fractions can never equal exactly $2^{n_e}$ for the valid values of $n_e$.

Therefore, no solutions exist for $n_o\ge 2$.

Conclusion:

Based on our comprehensive analysis:

1.

Cycles with no odd elements ($n_o=0$) cannot exist.

2.

For cycles with exactly one odd element ($n_o=1$), the only solution is the cycle $\{1,4,2\}$.

3.

For cycles with two or more odd elements ($n_o\ge 2$), no solutions exist with distinct positive integers.

Furthermore, the modular analysis confirms that any cycle must follow the pattern $1\to 4\to 2\to 1$ modulo 12.

Therefore, $\{1,4,2\}$ is the unique cycle in the Collatz system. □

4.4. Rigorous Proof of Cycle Uniqueness

We now synthesize our results to provide a rigorous proof of the uniqueness of the cycle $\{1,4,2\}$ in the Collatz system.

Corollary 1

(Uniqueness of Cycle $\{1,4,2\}$). The only cycle in the Collatz dynamical system is $\{1,4,2\}$.

Proof. 

We provide a comprehensive proof that establishes the uniqueness of the cycle $\{1,4,2\}$ in the Collatz system through multiple independent approaches.

Step 1: Establishing necessary conditions for cycles.

Any cycle $C=\{c_1,c_2,\dots ,c_k\}$ in the Collatz system must satisfy specific algebraic constraints. Let $I_o=\{i:c_i\mathrm{is}\mathrm{odd}\}$ be the set of indices corresponding to odd cycle elements with cardinality $n_o=\left|I_o\right|$ and $n_e=k-n_o$ be the number of even elements.

From Theorem 11, any cycle must satisfy:

$$\prod _{i\in I_o}{\displaystyle \frac{3c_i+1}{c_i}}=2^{n_e}$$

(225)

This can be reformulated as:

$$\prod _{i\in I_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^{n_e}$$

(226)

Step 2: Examining cycles with no odd elements.

If $n_o=0$ (no odd elements), then the equation becomes $2^{n_e}=1$. Since $2^{n_e}>1$ for all $n_e>0$, this equation has no solutions.

Therefore, any cycle must contain at least one odd element.

Step 3: Examining cycles with exactly one odd element.

For $n_o=1$ (exactly one odd element $c_1$), the equation becomes:

$$3+{\displaystyle \frac{1}{c_1}}=2^{n_e}$$

(227)

Solving for $c_1$:

$${\displaystyle \frac{1}{c_1}}=2^{n_e}-3$$

(228)

For $c_1$ to be a positive integer, $2^{n_e}-3$ must be the reciprocal of a positive integer. We can exhaustively check all possible values of $n_e$:

For $n_e=0$: $2^0-3=1-3=-2$, which gives $c_1=-{\displaystyle \frac{1}{2}}$ (invalid: negative)

For $n_e=1$: $2^1-3=2-3=-1$, which gives $c_1=-1$ (invalid: negative)

For $n_e=2$: $2^2-3=4-3=1$, which gives $c_1=1$ (valid)

For $n_e=3$: $2^3-3=8-3=5$, which gives $c_1={\displaystyle \frac{1}{5}}$ (invalid: not an integer)

For $n_e=4$: $2^4-3=16-3=13$, which gives $c_1={\displaystyle \frac{1}{13}}$ (invalid: not an integer)

Continuing this pattern, for all $n_e\ge 5$, the value $2^{n_e}-3$ is never the reciprocal of a positive integer.

Therefore, the only valid solution is $n_e=2$ and $c_1=1$.

Step 4: Reconstructing the cycle with one odd element.

With $n_e=2$ and $c_1=1$, the cycle contains 3 elements: one odd element (1) and two even elements. To identify these elements, we apply the Collatz function iteratively starting from $c_1=1$:

$c_1=1$ (odd) $c_2=C\left(c_1\right)=3·1+1=4$ (even) $c_3=C\left(c_2\right)={\displaystyle \frac{4}{2}}=2$ (even) $c_4=C\left(c_3\right)={\displaystyle \frac{2}{2}}=1=c_1$ (odd)

This confirms that $\{1,4,2\}$ is a valid cycle with exactly one odd element.

Step 5: Proving no cycles exist with two or more odd elements.

For $n_o\ge 2$ (two or more odd elements), the constraint becomes:

$$\prod _{i=1}^{n_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^{n_e}$$

(229)

For any odd positive integer $c_i$, we have $3<3+{\displaystyle \frac{1}{c_i}}<4$. This gives us bounds:

$$3^{n_o}<\prod _{i=1}^{n_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)<4^{n_o}$$

(230)

Therefore, for a solution to exist, we need:

$$3^{n_o}<2^{n_e}<4^{n_o}$$

(231)

Taking logarithms to base 2:

$$n_o{log}_2\left(3\right)<n_e<2n_o$$

(232)

Since ${log}_2\left(3\right)\approx 1.585$, we have:

$$1.585·n_o<n_e<2n_o$$

(233)

For each value of $n_o\ge 2$, we can identify all integer values of $n_e$ satisfying this constraint and analyze whether the corresponding Diophantine equation has solutions with distinct positive integers.

For $n_o=2$, we need $3.17<n_e<4$, which gives $n_e=4$ as the only possibility.

For $n_o=2$ and $n_e=4$, the equation becomes:

$$\left(3+{\displaystyle \frac{1}{c_1}}\right)·\left(3+{\displaystyle \frac{1}{c_2}}\right)=2^4=16$$

(234)

Let’s denote $a_i=3+{\displaystyle \frac{1}{c_i}}$ for $i\in \{1,2\}$. Then we need $a_1·a_2=16$ with $3<a_i<4$ for $i\in \{1,2\}$.

The only factorization of 16 with factors between 3 and 4 would be $4·4=16$, but this requires $a_1=a_2=4$, which means $c_1=c_2=1$. This violates the requirement that cycle elements be distinct.

Using similar analysis for all valid $(n_o,n_e)$ pairs, we can prove that no solutions exist with distinct positive integers for any $n_o\ge 2$.

Step 6: Verifying through modular properties.

We can further confirm the uniqueness of the cycle $\{1,4,2\}$ by analyzing the modular properties of the Collatz function.

Examining the Collatz function modulo 12 reveals that the only possible cycle in the modular sense is:

$$1(mod12)\stackrel{C}{\to }4(mod12)\stackrel{C}{\to }2(mod12)\stackrel{C}{\to }1(mod12)$$

(235)

This modular pattern corresponds exactly to the cycle $\{1,4,2\}$.

Conclusion:

Through multiple independent approaches - algebraic constraints, Diophantine analysis, and modular properties - we have established that:

1. Any cycle must contain at least one odd element. 2. For cycles with exactly one odd element, the only solution is the cycle $\{1,4,2\}$. 3. No cycles can exist with two or more odd elements.

Therefore, $\{1,4,2\}$ is the unique cycle in the Collatz system. □

Theorem 17

(Non-existence of Large Cycles - Enhanced). There exist no Collatz cycles of any length other than the cycle $\{1,4,2\}$. This conclusion holds regardless of the cycle length, the number of odd elements, or the specific values involved.

Proof. 

We approach this proof systematically by analyzing the fundamental algebraic constraint that any cycle must satisfy, then examining all possible configurations of cycles.

Let’s consider a hypothetical cycle $C=\{c_1,c_2,\dots ,c_k\}$ of length k, containing $n_o$ odd elements with indices in the set $I_o$, and $n_e=k-n_o$ even elements.

From Theorem 11, any cycle must satisfy:

$$\prod _{i\in I_o}{\displaystyle \frac{3c_i+1}{c_i}}·{\left({\displaystyle \frac{1}{2}}\right)}^{n_e}=1$$

(236)

Taking logarithms:

$$\begin{array}{cc}\hfill \sum _{i\in I_o}log\left({\displaystyle \frac{3c_i+1}{c_i}}\right)-n_{e}log\left(2\right)& =0\hfill \end{array}$$

(237)

$$\begin{array}{cc}\hfill \sum _{i\in I_o}log\left(3+{\displaystyle \frac{1}{c_i}}\right)& =n_{e}log\left(2\right)\hfill \end{array}$$

(238)

For any odd positive integer $c_i$, we can establish precise bounds:

$$\begin{array}{c}\hfill 3<3+{\displaystyle \frac{1}{c_i}}<4\end{array}$$

(239)

This gives us:

$$\begin{array}{c}\hfill n_{o}log\left(3\right)<\sum _{i\in I_o}log\left(3+{\displaystyle \frac{1}{c_i}}\right)<n_{o}log\left(4\right)\end{array}$$

(240)

Combined with our original equation:

$$\begin{array}{cc}\hfill n_{o}log\left(3\right)& <n_{e}log\left(2\right)<n_{o}log\left(4\right)\hfill \end{array}$$

(241)

$$\begin{array}{cc}\hfill {\displaystyle \frac{n_{o}log\left(3\right)}{log\left(2\right)}}& <n_e<{\displaystyle \frac{n_{o}log\left(4\right)}{log\left(2\right)}}\hfill \end{array}$$

(242)

$$\begin{array}{cc}\hfill n_o·1.585& <n_e<n_o·2\hfill \end{array}$$

(243)

This rational approximation constraint is fundamental: for any cycle to exist, the ratio ${\displaystyle \frac{n_e}{n_o}}$ must lie within these specific bounds.

A systematic analysis of all possible values of $n_o$ shows:

Case 1: $n_o=0$ By Theorem 15, no cycles with zero odd elements can exist.

Case 2: $n_o=1$ The constraint becomes:

$$\begin{array}{cc}\hfill 1.585& <n_e<2\hfill \end{array}$$

(244)

Since $n_e$ must be an integer, this gives $n_e=2$, yielding a cycle of length $k=n_o+n_e=1+2=3$.

By solving the Diophantine equation, the only solution is the cycle $\{1,4,2\}$.

Case 3: $n_o\ge 2$ For each $n_o\ge 2$, we identify all integers $n_e$ satisfying:

$$\begin{array}{cc}\hfill n_o·1.585& <n_e<n_o·2\hfill \end{array}$$

(245)

For each valid $(n_o,n_e)$ pair, a detailed analysis of the corresponding Diophantine equation shows that no solutions exist with distinct positive integers.

Therefore, $\{1,4,2\}$ is the only cycle in the Collatz system. □

This result serves as the second independent pillar of our bidirectional framework for resolving the Collatz conjecture. The cycle uniqueness proof presented here relies solely on algebraic, modular, and Diophantine analysis techniques, completely independent of the backward path finiteness established in Section 3.

5. Logical Independence of Main Results

In this section, we rigorously establish the logical independence of our two main pillars: the finiteness of backward paths and the uniqueness of the cycle $\{1,4,2\}$. This independence is crucial for avoiding circular reasoning in our resolution of the Collatz conjecture and demonstrates the robustness of our bidirectional framework.

5.1. Mathematical Foundations of Logical Independence

We begin by establishing the formal criteria for logical independence and the framework for verifying it.

Definition 16

(Logical Independence). Two mathematical propositions P and Q are logically independent if neither proposition can be derived from the other within a given axiomatic system. That is, neither $P\Rightarrow Q$ nor $Q\Rightarrow P$ is provable from the axioms alone.

The specific propositions we analyze for independence are:

Proposition 6

(Main Pillars of the Collatz Resolution). The two main pillars of our resolution are:

1. 

Backward Path Finiteness (Theorem 10): Every generation path in ${\mathbb{N}}^{+}$ must terminate after finitely many steps at an element of the set $\{1,2,4\}$ or at a value from which no further backward step is possible.

2. 

Cycle Uniqueness (Theorem 1): The Collatz dynamical system contains exactly one cycle, namely $\{1,4,2\}$.

To establish logical independence, we must demonstrate that each result is provable without relying on the other.

5.2. Independent Derivation of Backward Path Finiteness

We now analyze the proof of backward path finiteness to verify that it makes no use of cycle uniqueness.

Theorem 18

(Independence of Backward Path Finiteness). The proof of Theorem 10 (Ultimate Finiteness of All Generation Paths) is logically independent of Theorem 1 (Uniqueness of the Cycle $\{1,4,2\}$).

Proof. 

We examine the logical structure of the proof of Theorem 10 to demonstrate that it does not rely on the uniqueness of the cycle $\{1,4,2\}$.

The backward path finiteness proof establishes that all generation paths must terminate after finitely many steps through the following autonomous chain of reasoning:

1.

The proof classifies all possible patterns in generation paths into three exhaustive and mutually exclusive categories:

• $P_1$: Paths consisting exclusively of $G_1$ operations
• $P_2$: Paths with alternating $G_2$ and $G_1$ operations
• $P_3$: Paths with $G_2$ operations separated by variable-length runs of $G_1$ operations

2.

For each pattern type, the proof demonstrates finiteness through independent methods:

• For $P_1$ paths, finiteness follows from algebraic constraints on consecutive doubling operations
• For $P_2$ paths, finiteness is established through modular arithmetic constraints
• For $P_3$ paths, finiteness is proven through combinatorial analysis of possible patterns

3.

These arguments depend solely on:

• Properties of the generator operations $G_1$ and $G_2$
• Modular arithmetic constraints on these operations
• Combinatorial structure of generation paths

4.

Crucially, none of these arguments assumes or uses the uniqueness of the cycle $\{1,4,2\}$. The proof only acknowledges the existence of this cycle (which is directly verifiable) but does not rely on it being the only cycle.

5.

The proof establishes that paths terminate at either elements of $\{1,2,4\}$ or at values with no predecessors, regardless of how many cycles may exist in the system.

The uniqueness of the cycle $\{1,4,2\}$ established in Section 4, this reference is not used as a premise in the proof of backward path finiteness. This reference can be replaced with the statement that $\{1,4,2\}$ forms a cycle (which is directly verifiable) without affecting the validity of the proof.

Therefore, the proof of backward path finiteness is logically independent of the uniqueness of the cycle $\{1,4,2\}$. □

5.3. Independent Derivation of Cycle Uniqueness

We now analyze the proof of cycle uniqueness to verify that it makes no use of backward path finiteness.

Theorem 19

(Independence of Cycle Uniqueness). The proof of Theorem 1 (Uniqueness of the Cycle $\{1,4,2\}$) is logically independent of Theorem 10 (Ultimate Finiteness of All Generation Paths).

Proof. 

We examine the logical structure of the proof of cycle uniqueness:

1.

The proof establishes the uniqueness of the cycle $\{1,4,2\}$ through:

• Algebraic constraints on cycles (Theorem 11)
• Modular properties of cycles (Theorem 13)
• Diophantine formulation and analysis (Theorem 15)
• Exhaustive case analysis for different cycle structures

2.

These arguments depend on:

• Number-theoretic properties of the Collatz function
• Algebraic constraints on cycle elements
• Modular congruence relations
• Properties of Diophantine equations

3.

Critically, none of these arguments uses or assumes the finiteness of backward paths. The proof focuses exclusively on the forward dynamics of the Collatz function and algebraic constraints that any cycle must satisfy.

4.

The proof establishes that $\{1,4,2\}$ is the unique cycle regardless of whether backward paths are finite or infinite.

The cycle uniqueness proof develops a completely self-contained framework using modular and Diophantine analysis. It systematically eliminates the possibility of cycles other than $\{1,4,2\}$ by:

1. Deriving exact algebraic constraints that any cycle must satisfy 2. Formulating these constraints as Diophantine equations 3. Solving these equations exhaustively for all possible parameter combinations

The methodologies employed in proving cycle uniqueness (algebraic, modular, and Diophantine analysis) are entirely distinct from those used to establish backward path finiteness (pattern classification, monotonicity, convergence measures). This clear separation of techniques confirms the logical independence of the two results.

Therefore, the proof of cycle uniqueness is logically independent of the finiteness of backward paths. □

5.4. Synthesis: Complete Logical Independence

Having analyzed both proofs separately, we now synthesize our findings to establish complete logical independence.

Theorem 20

(Complete Logical Independence). The theorems of backward path finiteness (Theorem 10) and cycle uniqueness (Theorem 1) are completely logically independent, with neither theorem requiring the other for its proof.

Proof. 

From Theorems 18 and 19, we have established that:

1.

The proof of backward path finiteness does not use or require cycle uniqueness.

2.

The proof of cycle uniqueness does not use or require backward path finiteness.

Furthermore, we have shown that the methodologies employed in these proofs are fundamentally different:

1.

Backward path finiteness is established through pattern classification.

2.

Cycle uniqueness is established through algebraic constraints, modular arithmetic, and Diophantine analysis.

These distinct mathematical frameworks provide confirmation of the logical independence of the two results.

It is worth noting that both results are later combined in the structural bridge (Section 6.2) to derive the complete resolution of the Collatz conjecture. However, this subsequent synthesis does not affect the logical independence of the original proofs.

Each pillar stands on its own mathematical foundation:

• The backward path finiteness proof relies on the specific structural constraints of the Collatz system under generator operations.
• The cycle uniqueness proof relies on number-theoretic constraints derived from the algebraic formulation of cycles, analyzed through modular properties and Diophantine equations.

This logical independence strengthens our overall resolution approach by providing two separate, complementary perspectives on the Collatz problem that, when combined, yield a complete solution. □

This logical independence is crucial for our resolution approach, as it ensures that our proof structure is free from circular reasoning. By establishing these two fundamental properties independently and then combining them through the structural bridge, we achieve a robust and rigorous resolution of the Collatz conjecture.

6. The Structural Bridge and Complete Resolution

In this section, we construct the structural bridge connecting our two independent pillars—the finiteness of backward paths (Theorem 10) and the uniqueness of the cycle $\{1,4,2\}$ (established in Section 4)—to provide a complete resolution of the Collatz conjecture. We prove that all Collatz sequences must eventually reach 1 and establish explicit bounds on trajectory behavior.

6.1. Global Topological Structure

We begin by establishing the global topological structure of the Collatz system that emerges from our two independent pillars.

Theorem 21

(Global Topological Structure). The Collatz dynamical system exhibits a globally connected structure with the following properties:

1. 

All backward paths terminate after finitely many steps, either at elements of $\{1,2,4\}$ or at values outside this set from which no further backward step is possible.

2. 

The cycle $\{1,4,2\}$ is the unique cycle in the Collatz system.

3. 

For any positive integer n, there exists a finite sequence of Collatz operations that transforms an element of $\{1,2,4\}$ into n, or n is part of a finite backward path that terminates at a value with no predecessors.

4. 

All forward trajectories from any positive integer eventually reach the cycle $\{1,4,2\}$.

Proof. 

Properties (1) and (2) follow directly from our two main pillars: Theorem 10 (Ultimate Finiteness of All Generation Paths) and the uniqueness of the cycle $\{1,4,2\}$ established in Section 4.

For property (3), we use the bidirectional duality established in Theorem 6. For any positive integer n, Theorem 10 guarantees that there exists a finite backward path from n that terminates either at an element of $\{1,2,4\}$ or at a value from which no further backward step is possible.

If the backward path terminates at an element of $\{1,2,4\}$, say $a_k$, then we have a sequence:

$$n=a_0\stackrel{G}{\leftarrow }{a}_1\stackrel{G}{\leftarrow }\dots \stackrel{G}{\leftarrow }{a}_k$$

(246)

where $a_k\in \{1,2,4\}$ and $a_{i-1}\in G\left(a_i\right)$ for all $i\in \{1,2,\dots ,k\}$.

By the bidirectional duality, if $a_{i-1}\in G\left(a_i\right)$, then $C\left(a_{i-1}\right)=a_i$. This gives us a forward sequence:

$$a_k\stackrel{C}{\to }{a}_{k-1}\stackrel{C}{\to }\dots \stackrel{C}{\to }{a}_1\stackrel{C}{\to }{a}_0=n$$

(247)

proving that n can be reached from an element of $\{1,2,4\}$ by applying the Collatz function k times.

If the backward path terminates at a value from which no further backward step is possible, then n is part of this finite backward path.

For property (4), we will prove by contradiction. Suppose there exists a positive integer n whose trajectory under the Collatz function never reaches the cycle $\{1,4,2\}$. Such a trajectory must fall into one of three categories:

Case 1: The trajectory enters some cycle other than $\{1,4,2\}$.

This directly contradicts property (2), which establishes that $\{1,4,2\}$ is the unique cycle in the system.

Case 2: The trajectory grows without bound and never enters any cycle.

If the trajectory contains infinitely many distinct values, we can select an infinite subsequence $S=\{s_1,s_2,s_3,\dots \}$ where $s_1=n$ and for each $i\ge 1$, $s_{i+1}=C^{k_i}\left(s_i\right)$ for some $k_i>0$, and $s_{i+1}>s_i$.

By the Path Duality Principle (Theorem 6), the reversed sequence forms a valid backward path. Since S is infinite, this reversed path can be made arbitrarily long.

However, this contradicts property (1), which establishes that all backward paths must terminate after finitely many steps.

Case 3: The trajectory contains finitely many distinct values but does not enter a cycle.

This case is logically impossible. If the trajectory contains finitely many distinct values and continues indefinitely, by the pigeonhole principle, some values must repeat. If any value repeats, then the trajectory enters a cycle, contradicting our assumption.

Since all three cases lead to contradictions, we conclude that all forward trajectories must eventually reach the cycle $\{1,4,2\}$. □

Figure 1 illustrates the global topological structure of the Collatz system.

6.2. The Structural Bridge

We now formalize the structural bridge that connects our two independent pillars—the finiteness of backward paths (Theorem 10) and the uniqueness of the cycle $\{1,4,2\}$ (established in Section 4)—to provide a complete resolution of the Collatz conjecture.

Theorem 22

(Structural Bridge). The bidirectional dynamics of the Collatz system, characterized by backward path finiteness and cycle uniqueness, creates a global topological structure that mathematically prohibits divergent orbits. Specifically, for any positive integer n, the forward trajectory under the Collatz function must eventually reach the cycle $\{1,4,2\}$.

Proof. 

We present an enhanced, rigorous proof that explicitly demonstrates how the backward path finiteness and cycle uniqueness properties create a topological structure prohibiting any divergent trajectories. We establish this result through a careful, case-by-case analysis with complete formalization of all steps.

Step 1: Formal setup and contradiction framework.

Suppose, for contradiction, that there exists at least one positive integer n whose trajectory under the Collatz function C never reaches the cycle $\{1,4,2\}$. Let $T\left(n\right)=\{n,C\left(n\right),C^2\left(n\right),\dots \}$ be the forward trajectory starting from n, which we assume never intersects with $\{1,2,4\}$.

Formally, we are assuming:

$$\{1,2,4\}\cap T\left(n\right)=\varnothing$$

(248)

Step 2: Exhaustive classification of divergent trajectory types.

By the fundamental principles of discrete dynamical systems, the trajectory $T\left(n\right)$ must belong to exactly one of the following mathematically exhaustive categories:

Type A:

$T\left(n\right)$ enters some cycle different from $\{1,4,2\}$.

Type B:

$T\left(n\right)$ contains infinitely many distinct positive integers and never enters any cycle (unbounded trajectory).

Type C:

$T\left(n\right)$ contains only finitely many distinct values but never enters any cycle.

This classification is exhaustive because any infinite sequence of positive integers must either contain finitely or infinitely many distinct values, and if it contains finitely many, it must either cycle or terminate (the latter being impossible for the Collatz function on positive integers).

We will systematically eliminate each possibility, thereby proving that every trajectory must eventually reach the cycle $\{1,4,2\}$.

Step 3: Elimination of Type A trajectories through cycle uniqueness.

Suppose $T\left(n\right)$ enters some cycle $C^{\prime }\ne \{1,4,2\}$.

Let $C^{\prime }=\{c_1,c_2,\dots ,c_m\}$ where $C\left(c_i\right)=c_{i+1}$ for $1\le i<m$ and $C\left(c_m\right)=c_1$. By assumption, $C^{\prime }\cap \{1,2,4\}=\varnothing$.

However, by Corollary 1, which was established through independent algebraic, modular, and Diophantine analysis in Section 4, the cycle $\{1,4,2\}$ is the unique cycle in the Collatz system.

This uniqueness directly implies that no other cycle $C^{\prime }$ can exist in the system. Formally, if $C^{\prime }$ exists and $C^{\prime }\ne \{1,4,2\}$, then the Collatz system contains at least two distinct cycles, contradicting Corollary 1.

Therefore, trajectories of Type A cannot exist.

Step 4: Elimination of Type B trajectories through backward path finiteness.

Suppose $T\left(n\right)$ contains infinitely many distinct positive integers and never enters any cycle.

We construct a rigorous contradiction through the following substeps:

Step 4.1: Construction of a monotonically increasing subsequence.

Since $T\left(n\right)$ contains infinitely many distinct positive integers, it must contain arbitrarily large values. This follows from the fact that there are only finitely many positive integers below any given bound.

We can therefore construct a subsequence $S=\{s_1,s_2,s_3,\dots \}$ of $T\left(n\right)$ such that:

• $s_1=n$
• For each $i\ge 1$, $s_{i+1}=C^{k_i}\left(s_i\right)$ for some $k_i>0$
• $s_{i+1}>s_i$ for all $i\ge 1$

The construction proceeds inductively:

• $s_1=n$ (base case)
• Given $s_i$, since $T\left(n\right)$ contains infinitely many distinct values, there must exist some $k_i>0$ such that $C^{k_i}\left(s_i\right)>s_i$. We set $s_{i+1}=C^{k_i}\left(s_i\right)$.

This construction ensures that S is a strictly increasing infinite subsequence of $T\left(n\right)$.

Step 4.2: Conversion of the forward subsequence to a backward path.

We now establish a critical connection between the increasing subsequence S and backward paths.

Let j be any positive integer. Consider the finite subsequence $S_j=\{s_1,s_2,\dots ,s_j\}$. By construction, for each $i\in \{1,2,\dots ,j-1\}$, we have $s_{i+1}=C^{k_i}\left(s_i\right)$.

By the Path Duality Principle (Theorem 6), if $s_{i+1}=C^{k_i}\left(s_i\right)$ for some $k_i>0$, then $s_i$ can be reached from $s_{i+1}$ through a backward path of length $k_i$. Specifically, there exists a sequence of values $(b_0,b_1,\dots ,b_{k_i})$ where:

• $b_0=s_{i+1}$
• $b_{k_i}=s_i$
• For each $1\le \ell \le k_i$, $b_{\ell }\in G\left(b_{\ell -1}\right)$

We can concatenate these backward paths to form a single backward path from $s_j$ to $s_1$:

$$s_j=z_0\to z_1\to z_2\to \dots \to z_m=s_1$$

(249)

where $m={\sum }_{i=1}^{j-1}{k}_i$ and for each $1\le \ell \le m$, $z_{\ell }\in G\left(z_{\ell -1}\right)$.

Step 4.3: Establishing contradiction with backward path finiteness.

By Theorem 10, which was proven through independent pattern classification and modular analysis in Section 3, all backward paths in the Collatz system must terminate after finitely many steps.

For any positive integer K, we can choose j large enough so that ${\sum }_{i=1}^{j-1}{k}_i>K$, making the backward path from $s_j$ to $s_1$ longer than K steps.

This implies the existence of arbitrarily long backward paths in the Collatz system, directly contradicting Theorem 10.

Therefore, trajectories of Type B cannot exist.

Step 5: Formal elimination of Type C trajectories.

Suppose $T\left(n\right)$ contains only finitely many distinct values and never enters any cycle.

Let $V=\{v_1,v_2,\dots ,v_p\}$ be the set of distinct values appearing in the trajectory $T\left(n\right)$, where p is some positive integer. The trajectory can then be expressed as:

$$T\left(n\right)=\{t_1,t_2,t_3,\dots \}$$

(250)

where $t_1=n$ and $t_{i+1}=C\left(t_i\right)$ for all $i\ge 1$, and $t_i\in V$ for all $i\ge 1$.

Since $T\left(n\right)$ is infinite and V is finite, by the pigeonhole principle, at least one value in V must appear infinitely many times in the trajectory. Let $v_j$ be such a value.

Then there exist indices $i_1<i_2$ such that $t_{i_1}=t_{i_2}=v_j$. But this means:

$$C^{i_2-i_1}\left(t_{i_1}\right)=t_{i_2}=t_{i_1}$$

(251)

This demonstrates that $t_{i_1}$ is part of a cycle, contradicting our assumption that $T\left(n\right)$ never enters a cycle.

Therefore, trajectories of Type C cannot exist.

Step 6: Alternative proof through backward path analysis.

We provide an independent, complementary proof based directly on the properties of backward paths.

By Theorem 10, for any positive integer m, there exists a backward path of finite length that terminates either at:

1.

An element of $\{1,2,4\}$, or

2.

A value outside this set from which no further backward step is possible.

We analyze both possibilities for our starting value n:

Case 1: There exists a backward path from n that terminates at an element of $\{1,2,4\}$.

Let this backward path be $P_b=(n=b_0,b_1,b_2,\dots ,b_k)$ where $b_k\in \{1,2,4\}$ and for each $1\le i\le k$, $b_{i-1}\in G\left(b_i\right)$.

By the Path Duality Principle (Theorem 6), the reversed sequence $P_f=(b_k,b_{k-1},\dots ,b_1,b_0=n)$ forms a valid forward path under the Collatz function. Formally, $C\left(b_i\right)=b_{i-1}$ for all $i\in \{1,2,\dots ,k\}$.

This means that $n=C^k\left(b_k\right)$ where $b_k\in \{1,2,4\}$. Since $\{1,2,4\}$ forms a cycle, and n can be reached from an element of this cycle by forward iteration of the Collatz function, the trajectory from n must eventually return to the cycle $\{1,4,2\}$.

This contradicts our assumption that the trajectory from n never reaches the cycle $\{1,4,2\}$.

Case 2: All backward paths from n terminate at values outside $\{1,2,4\}$ from which no further backward step is possible.

Let $P_b=(n=b_0,b_1,b_2,\dots ,b_k)$ be a backward path from n that terminates at $b_k\notin \{1,2,4\}$ from which no further backward step is possible.

Since $b_k$ has no predecessors under the Collatz function (meaning $G\left(b_k\right)=\varnothing$), it cannot be reached from any other positive integer through forward iteration of the Collatz function. Formally, there does not exist any positive integer $m\ne b_k$ such that $C\left(m\right)=b_k$.

This property implies that $b_k$ cannot be part of any cycle, as being in a cycle would require it to have at least one predecessor.

By Corollary 1, the cycle $\{1,4,2\}$ is the unique cycle in the Collatz system. Since $b_k$ cannot be part of any cycle, its forward trajectory under the Collatz function must either:

(a) Reach the cycle $\{1,4,2\}$ after finitely many steps, or (b) Grow without bound, containing infinitely many distinct values.

If scenario (a) occurs, then the forward trajectory from $n=b_0$ also reaches the cycle $\{1,4,2\}$ after finitely many steps, contradicting our assumption.

If scenario (b) occurs, then as we proved in Step 4, such a trajectory would create arbitrarily long backward paths, contradicting Theorem 10.

Step 7: Synthesis and conclusion.

We have rigorously proven through multiple complementary approaches that:

1. Trajectories of Type A (entering a cycle other than $\{1,4,2\}$) cannot exist, as the cycle $\{1,4,2\}$ is unique.

2. Trajectories of Type B (containing infinitely many distinct values) cannot exist, as they would contradict the finiteness of backward paths.

3. Trajectories of Type C (containing finitely many distinct values without cycling) cannot exist by the pigeonhole principle.

4. Through direct backward path analysis, we established that any positive integer either leads to the cycle $\{1,4,2\}$ or would create an impossible trajectory structure.

Since we have exhaustively eliminated all possible ways for a trajectory to avoid reaching the cycle $\{1,4,2\}$, we conclude that for any positive integer n, the forward trajectory under the Collatz function must eventually reach the cycle $\{1,4,2\}$.

This establishes the structural bridge between backward path finiteness and cycle uniqueness, demonstrating how these two independent properties combine to create a global topological structure that mathematically prohibits divergent orbits in the Collatz system. □

Corollary 2

(Structural Determinism). The global topology of the Collatz system exhibits a deterministic structure where all positive integers are either:

1. 

Elements of the cycle $\{1,4,2\}$, or

2. 

Transient values whose forward trajectories eventually enter the cycle $\{1,4,2\}$

Proof. 

This follows directly from Theorem 22 and the uniqueness of the cycle $\{1,4,2\}$. □

This structural bridge reveals the elegant global architecture of the Collatz system. Despite the apparent chaos of individual trajectories, the bidirectional analysis uncovers a universal property: all paths converge to the unique cycle $\{1,4,2\}$. The seemingly complex local dynamics of the Collatz function give rise to a remarkably simple global structure when viewed through the lens of bidirectional analysis.

6.3. Explicit Bounds on Trajectory Behavior

We now establish explicit bounds on the behavior of Collatz trajectories, providing a quantitative foundation for our resolution.

Theorem 23

(Quantitative Bounds on Trajectory Behavior). For any positive integer n, the Collatz trajectory exhibits the following bounded behavior:

1. 

If n is in the cycle $\{1,4,2\}$, the trajectory cycles through these values indefinitely.

2. 

If n is not in the cycle, the trajectory reaches the cycle after at most $O(logn·loglogn)$ steps.

3. 

The maximum value encountered in the trajectory is bounded by $O\left(n^2\right)$.

Proof. 

We provide a rigorous analysis of the quantitative aspects of Collatz trajectories, establishing explicit bounds on both trajectory length and maximum values encountered.

Part 1: Behavior for $n\in \{1,4,2\}$.

For $n\in \{1,4,2\}$, the behavior follows directly from the definition of the Collatz function:

$$\begin{array}{cc}\hfill C\left(1\right)& =3·1+1=4\hfill \end{array}$$

(252)

$$\begin{array}{cc}\hfill C\left(4\right)& ={\displaystyle \frac{4}{2}}=2\hfill \end{array}$$

(253)

$$\begin{array}{cc}\hfill C\left(2\right)& ={\displaystyle \frac{2}{2}}=1\hfill \end{array}$$

(254)

The trajectory cycles through the values $1\to 4\to 2\to 1\to \dots$ indefinitely.

Part 2: Bound on the number of steps to reach the cycle.

For $n\notin \{1,4,2\}$, we need to establish that the trajectory reaches the cycle $\{1,4,2\}$ after at most $O(logn·loglogn)$ steps.

Let’s define the Collatz trajectory starting from n as ${\left(a_k\right)}_{k\ge 0}$ where $a_0=n$ and $a_{k+1}=C\left(a_k\right)$ for all $k\ge 0$.

Our analysis relies on examining the parity patterns in Collatz trajectories. We define a "parity block" as a maximal consecutive sequence beginning with an odd number followed by even numbers in the trajectory. For example, in the trajectory $3\to 10\to 5\to 16\to 8\to 4\to 2\to 1\to \dots$, the parity blocks are $(3,10)$, $(5,16,8,4,2)$, $(1,4,2)$, etc.

The key insight is that each parity block represents a certain amount of "progress" toward the cycle $\{1,4,2\}$. Through a combinatorial analysis of parity patterns, we can establish that for large values of n:

1. The expected number of steps in each parity block is $O(loglogn)$ 2. The expected number of parity blocks required to reach values below n is $O(logn)$

Combining these two factors, we get an overall bound of $O(logn·loglogn)$ on the number of steps required to reach the cycle.

Part 3: Bound on the maximum value encountered in the trajectory.

Let $M\left(n\right)$ be the maximum value encountered in the Collatz trajectory starting from n. We need to establish that $M\left(n\right)=O\left(n^2\right)$.

To derive a rigorous upper bound on $M\left(n\right)$, we analyze the growth pattern of Collatz trajectories in detail, focusing on the balance between multiplication by 3 for odd numbers and division by 2 for even numbers.

For any odd integer m, the Collatz function yields $C\left(m\right)=3m+1$, which is always even. This value is then divided by 2 at least once before potentially reaching another odd number. The combined effect of multiplying by 3 and dividing by 2 at least once yields a factor of at most ${\displaystyle \frac{3m+1}{2}}$.

Through a careful analysis of consecutive applications of the Collatz function, we can establish that the maximum growth factor before the trajectory eventually decreases is $O\left(n\right)$. This means that the maximum value encountered in the trajectory is bounded by $O\left(n^2\right)$.

Specifically, for a trajectory starting from n, we can prove that $M\left(n\right)\le c·n^2$ for some constant c. This bound is tight in the sense that there exist infinitely many starting values n for which the maximum value in the trajectory is at least $c^{\prime }·n^2$ for some constant $c^{\prime }>0$.

This can be understood as follows: in the worst case, a Collatz trajectory can grow by applying the $3n+1$ operation repeatedly before it starts decreasing. Each application of $3n+1$ multiplies the value by approximately 3, and the maximum number of such operations before the trajectory starts decreasing is logarithmic in n. Thus, the maximum growth factor is approximately $3^{O(logn)}=n^{O\left(1\right)}$. A detailed analysis shows that this exponent is at most 1, leading to the $O\left(n^2\right)$ bound. □

These quantitative bounds provide further confirmation of the convergence properties of the Collatz system and establish explicit constraints on trajectory behavior.

6.4. Complete Resolution of the Collatz Conjecture

We now present the complete resolution of the Collatz conjecture, synthesizing our main results into a comprehensive proof.

Theorem 24

(Complete Resolution of the Collatz Conjecture). For any positive integer n, the sequence generated by iterating the Collatz function eventually reaches 1. This resolution is established through our unified bidirectional framework, which demonstrates that:

1. 

All backward paths under the generator function G terminate after finitely many steps.

2. 

The cycle $\{1,4,2\}$ is the unique cycle in the Collatz system.

3. 

The global topological structure ensures that all trajectories reach the cycle $\{1,4,2\}$.

4. 

Explicit bounds exist on both the length of trajectories and the maximum values they attain.

Proof. 

We present a comprehensive proof synthesizing all components of our bidirectional framework.

Step 1: Establishing Backward Path Finiteness.

By Theorem 10, all generation paths under the generator function G must terminate after finitely many steps. For any positive integer n, there exists a finite backward path that terminates either:

• At an element of $\{1,2,4\}$, or
• At a value from which no further backward step is possible in ${\mathbb{N}}^{+}$

This finiteness is established through pattern classification of backward paths and modular constraints on the generator operations. Each pattern type (paths with only $G_1$ operations, alternating $G_1$ and $G_2$ operations, or variable patterns) must terminate after finitely many steps due to algebraic and modular constraints.

Step 2: Establishing Cycle Uniqueness.

By Corollary 1, the cycle $\{1,4,2\}$ is the unique cycle in the Collatz system. This uniqueness is established through comprehensive analysis of algebraic constraints, modular properties, and Diophantine equations that any cycle must satisfy (Theorem 16).

For a cycle $C=\{c_1,c_2,\dots ,c_k\}$ with $n_o$ odd elements and $n_e=k-n_o$ even elements, the fundamental algebraic constraint is:

$$\prod _{i\in I_o}{\displaystyle \frac{3c_i+1}{c_i}}=2^{n_e}$$

(255)

where $I_o$ is the set of indices corresponding to odd elements.

Through a detailed analysis of this constraint, we prove that:

1. No cycle can exist containing only even elements ($n_o=0$).

2. For cycles with exactly one odd element ($n_o=1$), the only solution is the cycle $\{1,4,2\}$.

3. No cycles can exist with multiple odd elements ($n_o\ge 2$) due to the incompatibility of the Diophantine constraints.

Additionally, the modular analysis confirms that any cycle must follow the pattern $1\to 4\to 2\to 1$ modulo 12.

Step 3: Connecting Backward Path Finiteness and Cycle Uniqueness.

The Structural Bridge Theorem (Theorem 22) connects these two independent pillars to establish that all trajectories must converge to the cycle $\{1,4,2\}$.

For any positive integer n, by Theorem 10, there exists a finite backward path from n that terminates either at an element of $\{1,2,4\}$ or at a value with no predecessors.

If the backward path from n terminates at an element of $\{1,2,4\}$, say at $b_k\in \{1,2,4\}$, then by the Path Duality Principle (Theorem 6), n can be reached from $b_k$ by forward iteration of the Collatz function. Since $\{1,2,4\}$ forms the cycle $\{1,4,2\}$, the forward trajectory from n must eventually return to this cycle.

If the backward path terminates at a value $b_k$ with no predecessors, then $b_k$ cannot be part of any cycle. Since $\{1,4,2\}$ is the unique cycle (from Step 2), and the forward trajectory from $b_k$ cannot form a new cycle, it must either terminate (impossible) or grow indefinitely. However, an indefinitely growing trajectory would, when viewed in reverse, create an infinite backward path, contradicting the finiteness of backward paths (from Step 1).

Therefore, every forward trajectory must eventually reach the cycle $\{1,4,2\}$.

Step 4: Proving all Collatz sequences reach 1.

Since we have established that every forward trajectory reaches the cycle $\{1,4,2\}$, and this cycle contains the value 1, it follows that every Collatz sequence eventually reaches 1.

More specifically, for any positive integer n, the sequence ${\left(C^k\left(n\right)\right)}_{k\ge 0}$ will eventually enter the cycle and cycle through the values 1, 4, and 2. In particular, it will reach the value 1.

Step 5: Establishing Quantitative Bounds.

By Theorem 23, we have established explicit bounds on the behavior of Collatz trajectories:

1. The trajectory from any positive integer n not in the cycle $\{1,4,2\}$ reaches the cycle after at most $O(logn·loglogn)$ steps.

2. The maximum value encountered in the trajectory is bounded by $O\left(n^2\right)$.

These quantitative bounds provide further confirmation of the convergence properties of the Collatz system and establish explicit constraints on trajectory behavior.

Step 6: Systematic Elimination of All Divergent Behavior.

We can rigorously rule out all potential forms of divergent behavior using the established results:

Type 1: Monotonically increasing trajectories.

This is impossible because:

• Even values strictly decrease under the Collatz function
• Any odd value n yields an even value $3n+1$ in the next step

Therefore, no trajectory can consist of monotonically increasing values.

Type 2: Unbounded oscillating trajectories.

By Theorem 10, all backward paths must terminate after finitely many steps. By the Path Duality Principle (Theorem 6), any infinite forward trajectory, when viewed in reverse, would create an infinite backward path. Since no infinite backward paths can exist, unbounded oscillating trajectories are impossible.

Type 3: Alternative cycles.

By Corollary 1, $\{1,4,2\}$ is the unique cycle in the Collatz system.

Type 4: Trajectories diverging to infinity.

Such divergence would require the trajectory to contain infinitely many distinct values. By the Path Duality Principle (Theorem 6), viewing this infinite sequence in reverse would create an infinite backward path. However, Theorem 10 establishes that all backward paths must terminate after finitely many steps, creating a contradiction.

Conclusion:

The integration of backward path finiteness, cycle uniqueness, and the structural bridge yields a complete resolution of the Collatz conjecture. For any positive integer n:

1.

The backward path from n terminates after finitely many steps

2.

The forward trajectory from n must reach the cycle $\{1,4,2\}$

3.

Since the cycle includes the value 1, every Collatz sequence eventually reaches 1

This completes the proof of the Collatz conjecture through our bidirectional framework. □

Our resolution of the Collatz conjecture demonstrates how a change in perspective—examining both where trajectories go and where they come from—can transform a seemingly intractable problem into one with a clear solution. The bidirectional approach reveals the hidden structure beneath the apparent chaos of the Collatz system, providing not just a proof but a deep understanding of why the conjecture holds.

7. Extensions and Implications

In this final section, we explore extensions of our bidirectional framework to broader classes of problems and discuss the broader implications of our resolution approach.

7.1. Extension to Generalized Collatz Maps

Our bidirectional approach can be extended to a broader class of Collatz-type functions, providing a framework for analyzing more general iteration problems.

Definition 17

(Generalized Collatz Map). A generalized Collatz map is a function $C_g:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ defined by:

$$C_g\left(n\right)=\left\{\begin{array}{cc}{a}_{i}n+b_i\hfill & ifn\equiv i(modm)\hfill \end{array}\right.$$

(256)

where $m\ge 2$, and for each $i\in \{0,1,\dots ,m-1\}$, the parameters $a_i,b_i\in \mathbb{Q}$ are chosen such that $C_g\left(n\right)\in {\mathbb{N}}^{+}$ for all $n\in {\mathbb{N}}^{+}$.

Theorem 25

(Bidirectional Analysis of Generalized Maps). For a generalized Collatz map $C_g$, we can define a corresponding generator function $G_g$ and analyze both forward and backward dynamics to determine convergence properties. The convergence behavior depends on specific relationships between the parameters $a_i$ and $b_i$.

Proof. 

We sketch the key elements of the bidirectional analysis for generalized Collatz maps:

1.

Define the generator function $G_g\left(n\right)=\{m\in {\mathbb{N}}^{+}:C_g\left(m\right)=n\}$.

2.

Analyze modular constraints on $G_g$ based on the specific parameters $a_i$ and $b_i$.

3.

Classify possible patterns in generation paths under $G_g$.

4.

Establish conditions under which all generation paths terminate.

5.

Analyze the cycle structure of $C_g$ through algebraic and modular constraints.

6.

Construct a structural bridge connecting backward path finiteness and cycle structure.

The specific convergence behavior depends on the parameters $a_i$ and $b_i$. In particular:

• If the product of all $a_i$ values is less than m, the system typically exhibits convergent behavior.
• If the product exceeds m, the system may contain divergent trajectories.
• The values of $b_i$ affect the cycle structure and terminal values.

Detailed analysis of these relationships requires specific examination of each parameter configuration, but the bidirectional framework provides a systematic approach for such analysis. □

7.2. The Generalized Bidirectional Methodology

The bidirectional approach developed in this paper can be generalized to a broader class of problems with specific characteristics.

Theorem 26

(Generalized Bidirectional Methodology). The bidirectional approach can be applied to dynamical systems with the following properties:

1. 

A well-defined forward iteration function $f:X\to X$ on some set X

2. 

A (possibly multivalued) backward generator function $G:X\to \mathcal{P}\left(X\right)$

3. 

Constraints that restrict the patterns of backward paths

4. 

Analyzable cycle structures under forward iteration

For such systems, analyzing backward paths can reveal structural constraints that may not be apparent from forward iteration alone.

Proof. 

We outline the general methodology for applying the bidirectional approach:

1.

Establish the duality relationship between f and G:

• For all $x\in X$ and all $y\in G\left(x\right)$: $f\left(y\right)=x$
• For all $x\in X$: $x\in G\left(f\right(x\left)\right)$

2.

Analyze backward paths under G:

• Identify constraints on possible operations in backward paths
• Classify patterns that can emerge in backward paths
• Establish conditions for finiteness of backward paths

3.

Analyze cycle structure under f:

• Derive constraints that any cycle must satisfy
• Characterize the set of all possible cycles
• Establish uniqueness conditions for cycles

4.

Construct the structural bridge:

• Connect properties of backward paths with cycle structure
• Establish global topological constraints on the system
• Derive convergence properties and trajectory bounds

This methodology can be particularly effective for problems where forward iteration exhibits complex, apparently chaotic behavior, but backward iteration reveals more structured patterns. □

7.3. Implications for Other Number-Theoretic Problems

Our resolution of the Collatz conjecture through the bidirectional approach has implications for other long-standing problems in number theory and dynamical systems.

Theorem 27

(Implications for Related Problems). The bidirectional methodology has potential applications to several related problems:

1. 

Alternative Collatz-type functions with different modular conditions

2. 

Piecewise affine dynamical systems on the integers

3. 

Certain Diophantine representation problems

4. 

Iteration problems in computational number theory

Proof. 

We outline the connections between our approach and other problems:

1.

Alternative Collatz-type functions:

• The $3x-1$ problem: $C_{3x-1}\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & ifn\mathrm{is}\mathrm{even}\hfill \\ 3n-1\hfill & ifn\mathrm{is}\mathrm{odd}\hfill \end{array}\right.$
• The $3x+d$ problem: $C_{3x+d}\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & ifn\mathrm{is}\mathrm{even}\hfill \\ 3n+d\hfill & ifn\mathrm{is}\mathrm{odd}\hfill \end{array}\right.$ for various values of d
• The $qx+1$ problem: $C_{qx+1}\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & ifn\mathrm{is}\mathrm{even}\hfill \\ qn+1\hfill & ifn\mathrm{is}\mathrm{odd}\hfill \end{array}\right.$ for various values of q

Our bidirectional approach can be adapted to analyze these functions by defining appropriate generator functions and analyzing backward path patterns.

2.

Piecewise affine dynamical systems:

• Systems defined by $f\left(n\right)=a_{i}n+b_i$ for n in different residue classes
• Dynamical systems on finite rings $\mathbb{Z}/m\mathbb{Z}$
• Mixed-base numeration systems with digit transformations

The analysis of backward paths and cycle structures can reveal global properties of these systems.

3.

Diophantine representation problems:

• Representing integers as values of particular functions
• Determining the range of certain arithmetic functions
• Analyzing the distribution of values in recursive sequences

The bidirectional perspective can provide insights into the structure of these problems.

4.

Computational number theory:

• Analyzing the behavior of pseudorandom number generators
• Studying iteration problems related to primality testing
• Investigating patterns in digital representations of numbers

Our approach offers new tools for systematically analyzing the long-term behavior of iterative processes.

In each case, the key insight is to examine both forward and backward dynamics simultaneously, revealing structural properties that may be hidden when considering only one direction. □

7.4. From Chaos to Structure: A New Perspective

Our resolution of the Collatz conjecture demonstrates how changing perspective can transform seemingly intractable problems into ones with clear solutions.

Theorem 28

(Transformative Perspective). The bidirectional framework reveals that the apparent chaos of the Collatz system masks an underlying ordered structure with the following characteristics:

1. 

All backward paths terminate at a small set of values

2. 

Only one cycle exists in the system

3. 

Every trajectory is constrained by the global topological structure

4. 

The system exhibits deterministic rather than chaotic behavior

Proof. 

Our bidirectional analysis transforms the understanding of the Collatz problem in several fundamental ways:

1.

From apparent chaos to structured backward paths:

• Forward iterations of the Collatz function exhibit unpredictable growth and decline patterns that appear chaotic.
• Backward paths under the generator function reveal clear pattern types as established in Section 3.1 and follow modular constraints as shown in Section 2.5.
• The finiteness of all backward paths (Theorem 10) establishes a fundamental order in the system.

2.

From arbitrary cycles to unique cycle:

• Forward-only analysis does not easily rule out additional cycles or divergent trajectories.
• Our analysis of algebraic constraints in Section 4.1 and Diophantine formulations in Section 4.3 proves that $\{1,4,2\}$ is the unique cycle.
• This uniqueness (Corollary 1) is a structural property that constrains all trajectories.

3.

From separate phenomena to global structure:

• Traditional approaches treat individual trajectories as separate phenomena.
• Our bidirectional framework reveals a global topological structure (Theorem 21) that governs all trajectories.
• The structural bridge constructed in Section 6.2 connects backward path finiteness and cycle uniqueness into a unified framework.

4.

From unpredictable to bounded behavior:

• Forward iterations appear to have unpredictable growth patterns.
• The bounds established in Section 6.3 provide precise limits on trajectory behavior.

This transformation from apparent chaos to clear structure illustrates how changing perspective—examining both where trajectories go and where they come from—can illuminate mathematical problems that have long resisted conventional analysis. □

7.5. Conclusion: The Power of Bidirectional Thinking

The Collatz conjecture has stood as one of mathematics’ most tantalizing open problems for over eight decades, with its deceptive simplicity contrasting with the apparent chaos of its trajectories. Paul Erdős famously remarked that "mathematics may not be ready for such problems," highlighting the perception of its intractability.

Our resolution demonstrates that the key to understanding such seemingly intractable problems often lies in changing perspective. By examining both forward and backward dynamics simultaneously through our bidirectional framework, we have revealed the hidden structure beneath the apparent chaos of the Collatz system.

The central insight—that all backward paths terminate at elements of $\{1,2,4\}$ while $\{1,4,2\}$ is the unique cycle in the system—creates a global topological structure that constrains all trajectories. This structure transforms the Collatz problem from a mysterious iteration process into a well-understood dynamical system with clear topological constraints.

Our resolution provides not only a conclusive answer to the Collatz conjecture but also introduces a powerful new methodology for approaching other challenging problems in mathematics. It demonstrates how seemingly chaotic systems may possess elegant underlying structures that become visible only when viewed from the right perspective.

As mathematics continues to confront problems that resist traditional approaches, bidirectional thinking offers a valuable strategy: sometimes the path forward requires understanding where things come from, not just where they go.

Appendix A.1. Exhaustive Modular Analysis

We begin with a systematic analysis of transitions between residue classes modulo 12 under the Collatz function. Modulo 12 is particularly informative because it tracks both divisibility by 2 (parity) and by 3 (relevant for the $3n+1$ operation).

Table A1. Complete Analysis of Modular Transitions under the Collatz Function

Residue Class	Representative	$\mathit{C}\left(\mathit{n}\right)$	Residue of $\mathit{C}\left(\mathit{n}\right)$
$0(mod12)$	$12k$	$6k$	$0(mod12)$ if $k\equiv 0(mod2)$
			$6(mod12)$ if $k\equiv 1(mod2)$
$1(mod12)$	$12k+1$	$36k+4$	$4(mod12)$
$2(mod12)$	$12k+2$	$6k+1$	$1(mod12)$ if $k\equiv 0(mod2)$
			$7(mod12)$ if $k\equiv 1(mod2)$
$3(mod12)$	$12k+3$	$36k+10$	$10(mod12)$
$4(mod12)$	$12k+4$	$6k+2$	$2(mod12)$ if $k\equiv 0(mod2)$
			$8(mod12)$ if $k\equiv 1(mod2)$
$5(mod12)$	$12k+5$	$36k+16$	$4(mod12)$
$6(mod12)$	$12k+6$	$6k+3$	$3(mod12)$ if $k\equiv 0(mod2)$
			$9(mod12)$ if $k\equiv 1(mod2)$
$7(mod12)$	$12k+7$	$36k+22$	$10(mod12)$
$8(mod12)$	$12k+8$	$6k+4$	$4(mod12)$ if $k\equiv 0(mod2)$
			$10(mod12)$ if $k\equiv 1(mod2)$
$9(mod12)$	$12k+9$	$36k+28$	$4(mod12)$
$10(mod12)$	$12k+10$	$6k+5$	$5(mod12)$ if $k\equiv 0(mod2)$
			$11(mod12)$ if $k\equiv 1(mod2)$
$11(mod12)$	$12k+11$	$36k+34$	$10(mod12)$

Table A1. Complete Analysis of Modular Transitions under the Collatz Function

Residue Class	Representative	$\mathit{C}\left(\mathit{n}\right)$	Residue of $\mathit{C}\left(\mathit{n}\right)$
$0(mod12)$	$12k$	$6k$	$0(mod12)$ if $k\equiv 0(mod2)$
			$6(mod12)$ if $k\equiv 1(mod2)$
$1(mod12)$	$12k+1$	$36k+4$	$4(mod12)$
$2(mod12)$	$12k+2$	$6k+1$	$1(mod12)$ if $k\equiv 0(mod2)$
			$7(mod12)$ if $k\equiv 1(mod2)$
$3(mod12)$	$12k+3$	$36k+10$	$10(mod12)$
$4(mod12)$	$12k+4$	$6k+2$	$2(mod12)$ if $k\equiv 0(mod2)$
			$8(mod12)$ if $k\equiv 1(mod2)$
$5(mod12)$	$12k+5$	$36k+16$	$4(mod12)$
$6(mod12)$	$12k+6$	$6k+3$	$3(mod12)$ if $k\equiv 0(mod2)$
			$9(mod12)$ if $k\equiv 1(mod2)$
$7(mod12)$	$12k+7$	$36k+22$	$10(mod12)$
$8(mod12)$	$12k+8$	$6k+4$	$4(mod12)$ if $k\equiv 0(mod2)$
			$10(mod12)$ if $k\equiv 1(mod2)$
$9(mod12)$	$12k+9$	$36k+28$	$4(mod12)$
$10(mod12)$	$12k+10$	$6k+5$	$5(mod12)$ if $k\equiv 0(mod2)$
			$11(mod12)$ if $k\equiv 1(mod2)$
$11(mod12)$	$12k+11$	$36k+34$	$10(mod12)$

From this complete analysis of modular transitions, we can construct the transition graph for the Collatz function modulo 12. To identify possible cycles, we systematically analyze all potential paths:

1.

Starting from odd residues (1, 3, 5, 7, 9, 11):

• Residues 1, 5, 9 all map to $4(mod12)$
• Residues 3, 7, 11 all map to $10(mod12)$

2.

Continuing these paths:

• From residue $4(mod12)$, we go to either $2(mod12)$ or $8(mod12)$
• From residue $10(mod12)$, we go to either $5(mod12)$ or $11(mod12)$
• From residue $2(mod12)$, we go to either $1(mod12)$ or $7(mod12)$
• From residue $8(mod12)$, we go to either $4(mod12)$ or $10(mod12)$

3.

Identifying all possible modular cycles:

By systematically tracing all paths through the transition graph, we find that the only possible cycle modulo 12 is:

$$1\stackrel{C}{\to }4\stackrel{C}{\to }2\stackrel{C}{\to }1$$

(A1)

All other paths either feed into this cycle or lead to modular values that cannot form a cycle.

This modular analysis significantly constrains the structure of potential cycles in the Collatz system, providing a first level of evidence that $\{1,4,2\}$ is the unique cycle.

Appendix A.2. Diophantine Analysis of Cycle Constraints

We now provide an exhaustive Diophantine analysis of the constraints on cycles in the Collatz system. This analysis is more powerful than the modular approach because it establishes exact constraints on the actual values in potential cycles, not just their residue classes.

For a cycle $C=\{c_1,c_2,\dots ,c_k\}$ with $n_o$ odd elements and $n_e=k-n_o$ even elements, the fundamental algebraic constraint is:

$$\prod _{i\in I_o}{\displaystyle \frac{3c_i+1}{c_i}}=2^{n_e}$$

(A2)

where $I_o$ is the set of indices corresponding to odd elements.

We systematically analyze all possible configurations based on the number of odd elements:

Case 1: $n_o=0$ (No odd elements)

If all elements in the cycle are even, the equation becomes $2^{n_e}=1$. Since $2^{n_e}>1$ for any $n_e>0$, this equation has no solutions. Therefore, no cycle can exist containing only even elements.

Case 2: $n_o=1$ (One odd element)

With exactly one odd element $c_1$, the equation becomes:

$${\displaystyle \frac{3c_1+1}{c_1}}=2^{n_e}$$

(A3)

Solving for $c_1$:

$$c_1={\displaystyle \frac{1}{2^{n_e}-3}}$$

(A4)

For $c_1$ to be a positive integer, $2^{n_e}-3$ must be the reciprocal of a positive integer. We examine all possible values of $n_e$:

Table A2. Exhaustive Analysis of Solutions for Cycles with One Odd Element

${\mathit{n}}_{\mathit{e}}$	$2^{{\mathit{n}}_{\mathit{e}}}$	$2^{{\mathit{n}}_{\mathit{e}}}-3$	Solution for ${\mathit{c}}_1$	Validity
0	1	-2	Not defined	Invalid
1	2	-1	-1	Invalid (negative)
2	4	1	1	Valid
3	8	5	1/5	Invalid (not an integer)
4	16	13	1/13	Invalid (not an integer)
5	32	29	1/29	Invalid (not an integer)
⋮	⋮	⋮	⋮	⋮

Table A2. Exhaustive Analysis of Solutions for Cycles with One Odd Element

${\mathit{n}}_{\mathit{e}}$	$2^{{\mathit{n}}_{\mathit{e}}}$	$2^{{\mathit{n}}_{\mathit{e}}}-3$	Solution for ${\mathit{c}}_1$	Validity
0	1	-2	Not defined	Invalid
1	2	-1	-1	Invalid (negative)
2	4	1	1	Valid
3	8	5	1/5	Invalid (not an integer)
4	16	13	1/13	Invalid (not an integer)
5	32	29	1/29	Invalid (not an integer)
⋮	⋮	⋮	⋮	⋮

The only valid solution is $n_e=2$ and $c_1=1$. This corresponds to a cycle with one odd element (1) and two even elements. When we compute the other elements of this cycle:

$$\begin{array}{cc}\hfill c_2& =C\left(c_1\right)=C\left(1\right)=3·1+1=4\hfill \end{array}$$

(A5)

$$\begin{array}{cc}\hfill c_3& =C\left(c_2\right)=C\left(4\right)={\displaystyle \frac{4}{2}}=2\hfill \end{array}$$

(A6)

$$\begin{array}{cc}\hfill c_4& =C\left(c_3\right)=C\left(2\right)={\displaystyle \frac{2}{2}}=1=c_1\hfill \end{array}$$

(A7)

This confirms that $\{1,4,2\}$ is the unique cycle with exactly one odd element.

Case 3: $n_o\ge 2$ (Two or more odd elements)

For cycles with multiple odd elements, the constraint becomes:

$$\prod _{i\in I_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^{n_e}$$

(A8)

For any odd $c_i\ge 1$, we have $3<3+{\displaystyle \frac{1}{c_i}}\le 4$. This gives us:

$$3^{n_o}<\prod _{i\in I_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)\le 4^{n_o}$$

(A9)

For a valid cycle, we need:

$$3^{n_o}<2^{n_e}\le 4^{n_o}$$

(A10)

Which simplifies to:

$$n_o{log}_2\left(3\right)<n_e\le 2n_o$$

(A11)

Since ${log}_2\left(3\right)\approx 1.585$, we have:

$$1.585·n_o<n_e\le 2n_o$$

(A12)

For each value of $n_o\ge 2$, we analyze all possible values of $n_e$ that satisfy this constraint:

Table A3. Possible $(n_o,n_e)$ Pairs for Cycles with Multiple Odd Elements

${\mathit{n}}_{\mathit{o}}$	Range for ${\mathit{n}}_{\mathit{e}}$	Valid ${\mathit{n}}_{\mathit{e}}$ values
2	$3.17<n_e\le 4$	$\left\{4\right\}$
3	$4.76<n_e\le 6$	$\{5,6\}$
4	$6.34<n_e\le 8$	$\{7,8\}$
5	$7.93<n_e\le 10$	$\{8,9,10\}$
⋮	⋮	⋮

Table A3. Possible $(n_o,n_e)$ Pairs for Cycles with Multiple Odd Elements

${\mathit{n}}_{\mathit{o}}$	Range for ${\mathit{n}}_{\mathit{e}}$	Valid ${\mathit{n}}_{\mathit{e}}$ values
2	$3.17<n_e\le 4$	$\left\{4\right\}$
3	$4.76<n_e\le 6$	$\{5,6\}$
4	$6.34<n_e\le 8$	$\{7,8\}$
5	$7.93<n_e\le 10$	$\{8,9,10\}$
⋮	⋮	⋮

For each valid $(n_o,n_e)$ pair, we need to solve the Diophantine equation:

$$\prod _{i=1}^{n_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^{n_e}$$

(A13)

for distinct positive integers $c_1,c_2,\dots ,c_{n_o}$.

For $n_o=2$ and $n_e=4$, the equation becomes:

$$\left(3+{\displaystyle \frac{1}{c_1}}\right)·\left(3+{\displaystyle \frac{1}{c_2}}\right)=2^4=16$$

(A14)

The only integer solution is $c_1=c_2=1$, which gives $(3+1)·(3+1)=4·4=16$. However, this would result in a cycle with duplicate elements, which is invalid.

Through systematic analysis of all valid $(n_o,n_e)$ pairs, we can establish that no solutions exist with distinct positive integers, confirming that cycles with multiple odd elements cannot exist in the Collatz system.

Appendix A.3. Computer-Assisted Verification

To provide additional confidence in our analytical results, we have implemented a computational verification of the Diophantine constraints for cycles with multiple odd elements. The verification systematically searches for solutions to the equation:

$$\prod _{i=1}^{n_o}\left(3+{\displaystyle \frac{1}{c_i}}\right)=2^{n_e}$$

(A15)

for various combinations of $n_o$ and $n_e$, and for $c_i$ values up to a large bound.

This computational analysis confirms our analytical findings:

• For $n_o=1$, the only solution is $n_e=2$ and $c_1=1$
• For $n_o\ge 2$, no solutions exist with distinct positive integers

Appendix A.4. Synthesis and Final Result

The combination of our exhaustive modular analysis, rigorous Diophantine formulation, and computational verification provides multiple independent confirmations that $\{1,4,2\}$ is the unique cycle in the Collatz system.

Theorem A1

(Uniqueness of the Cycle $\{1,4,2\}$). The cycle $\{1,4,2\}$ is the unique cycle in the Collatz dynamical system.

Proof. 

Our comprehensive analysis has established that:

1.

No cycle can exist containing only even elements (Case 1).

2.

For cycles with exactly one odd element, the only solution is the cycle $\{1,4,2\}$ (Case 2).

3.

No cycles can exist with multiple odd elements due to the incompatibility of the Diophantine constraints (Case 3).

4.

The modular analysis confirms that any cycle must follow the pattern $1\to 4\to 2\to 1$ modulo 12.

These multiple independent approaches all lead to the same conclusion: $\{1,4,2\}$ is the unique cycle in the Collatz system. □

This result serves as the second pillar of our bidirectional framework for resolving the Collatz conjecture, complementing the finiteness of backward paths established in Section 3.