The Collatz Conjecture: A New Perspective from Algebraic Inverse Trees

1. Introduction

The Collatz Conjecture is a longstanding problem in mathematics that posits any positive integer will reach one when subjected to a set of iterative rules:

• If the number is even, divide by 2.
• If the number is odd, multiply by 3 and add 1.

Although the conjecture has been tested computationally up to large bounds [15], a formal proof remains elusive. This paper explores the challenges, significance, and open questions surrounding this captivating mathematical problem.

1.1. Historical Context and Importance

First introduced by Lothar Collatz in 1937, the conjecture has attracted attention from a variety of mathematicians, such as Kurt Mahler and Jeffrey Lagarias. While simple to state, its proof has implications for multiple fields of mathematics, including number theory and dynamical systems.

1.2. Challenges in Resolving the Collatz Conjecture

Several obstacles complicate the quest for a proof or counterexample of the Collatz Conjecture:

1.2.1. Analyzing an Infinite Sequence

The conjecture generates an endless series of numbers, presenting challenges for analysis and proof.

1.2.2. Counterexample Search

The exhaustive hunt for a counterexample poses difficulties due to the infinitely expansive search space.

1.2.3. Pattern Irregularities

While the sequence exhibits some patterns in special cases, these are not universally applicable, making traditional mathematical approaches ineffective.

1.3. Our Methodology

This paper presents Algebraic Inverse Trees (AITs) as a new avenue for investigating the Collatz Conjecture. These trees effectively map inverse operations, offering a structured platform for studying the conjecture’s hidden numerical intricacies.

2. Theory

2.1. Algebraic Inverse Trees (AITs) for Analyzing the Collatz Sequence

Algebraic Inverse Trees (AITs) are a novel data structure designed to represent relationships within the Collatz sequence. Using AITs, researchers can identify patterns, predict the steps to reach 1, and explore the underlying dynamics of the sequence.

2.1.1. Basics of AITs

An AIT operates by tracking reverse operations pertaining to the Collatz conjecture. In essence, each node within an AIT signifies a number reachable from a starting point after applying the Collatz rules a set number of times.

• Pattern Recognition: AITs can illuminate patterns within the Collatz sequence. Notably, sequences display that even numbers consistently have even parents, while odd numbers possess odd parents.
• Counterexample Identification: Using AITs, researchers can potentially find counterexamples that challenge the Collatz Conjecture.
• Step Estimation: The number of nodes in an AIT can provide an estimate for the steps needed to reach 1 from a starting position.
• Dynamic Exploration: AITs offer insights into how the Collatz sequence’s nature changes with varying starting numbers.

2.1.2. Multiple Parents in AITs

In the AIT structure, nodes can have up to two parents.

• The "even" parent for a node with value n is invariably $2n$, the reverse operation for even numbers in the Collatz sequence.
• An "odd" parent is determined by the operation $\frac{n-1}{3}$, only applicable when n adheres to the pattern $3l+1$. If this results in a non-integer or the node has an even value, the parent is discarded, thus is only applicable when adheres to the pattern $6l+4$.

This branching mechanism is captured by the reversal function:

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

(1)

2.2. Constructing AITs

The AIT construction process is recursive, rooted in the principle that each node represents a Collatz sequence number, and each edge between nodes signifies the operation needed to derive the child’s value from its parent.

• Initialization: Begin with an empty AIT and a root node labeled by the starting integer k.
• Parent Addition:

  –

  The "even" parent is found by adding $2n$ to the current node.

  –

  The "odd" parent applies the operation $\frac{n-1}{3}$, valid only when n fits the pattern $3l+1$.

• Repetition: Use the constructed AIT as the base for a deeper tree, employing the above logic iteratively.
• Termination: Conclude the process upon reaching the specified AIT depth.

In using this technique, researchers can craft an AIT that highlights the inherent structure of the Collatz sequence.

2.3. AIT Construction Algorithm

Algorithm 1 Construction of AIT

1: procedure ConstructAIT(k, $depth$) 2:       Create an empty tree T 3:       Set the root node of T to k 4:       $visited\leftarrow$ set containing k 5:       $Q\leftarrow$ queue with root node 6:       while Q is not empty and depth of T < $depth$ do 7:             $n\leftarrow$ dequeue from Q 8:             $parents\leftarrow R\left(n\right)$ 9:             for each p in $parents$ do 10:                  if p not in $visited$ then 11:                        Add edge from p to n in T 12:                        enqueue p to Q 13:                        add p to $visited$ 14:                  end if 15:             end for 16:       end while 17:       return T 18: end procedure

3. Formal Proof of the Collatz Conjecture

Let $f\left(x\right)$ be the Collatz function defined as:

$$f\left(x\right)=\left\{\begin{array}{cc}\frac{x}{2}\hfill & \mathrm{if}x\mathrm{is}\mathrm{even}\hfill \\ 3x+1\hfill & \mathrm{if}x\mathrm{is}\mathrm{odd}\hfill \end{array}\right.$$

And let $R\left(x\right)$ be the multivalued inverse function of $f\left(x\right)$ given by:

$$R\left(x\right)=\left\{\begin{array}{cc}\left\{2x\right\}\hfill & \mathrm{if}x\not\equiv 4(mod6)\hfill \\ \{2x,\frac{x-1}{3}\}\hfill & \mathrm{if}x\equiv 4(mod6)\hfill \end{array}\right.$$

We now formally define the Algebraic Inverse Tree:

Definition 3.1.

Let $T_k$ be the directed tree rooted at k constructed recursively as:

• The root node of $T_k$ is k.
• If n is a node in $T_k$, its child nodes are the elements of $R\left(n\right)$.
• The edges from n to each child h are labeled with the operation $n\to h$.

$T_k$ is the Algebraic Inverse Tree (AIT) of parameter k.

We now prove two key lemmas about the properties of AITs:

Lemma 3.1.

[Collatz Function and its Inverse] Let $f:\mathbb{N}\to \mathbb{N}$ be the Collatz function defined by

$$f\left(x\right)=\left\{\begin{array}{cc}\frac{x}{2}\hfill & \mathit{if}x\mathit{is}\mathit{even},\hfill \\ 3x+1\hfill & \mathit{if}\mathit{x}\mathit{is}\mathit{odd}.\hfill \end{array}\right.$$

The function f is invertible in a multi-valued sense. Specifically, for each $x\in \mathbb{N}$, there exists a finite, non-empty set $R\left(x\right)\subset \mathbb{N}$ such that for all $y\in R\left(x\right)$, $f\left(y\right)=x$.

Proof. 

Define the function $R:\mathbb{N}\to 2^{\mathbb{N}}$ (where $2^{\mathbb{N}}$ denotes the power set of natural numbers) by

$$R\left(x\right)=\left\{\begin{array}{cc}\left\{2x\right\}\hfill & \mathrm{if}x\mathrm{is}\mathrm{not}\mathrm{congruent}\mathrm{to}4(mod6),\hfill \\ \{2x,\frac{x-1}{3}\}\hfill & \mathrm{if}x\mathrm{is}\mathrm{congruent}\mathrm{to}4(mod6)\mathrm{and}x>1.\hfill \end{array}\right.$$

For $x=1$, $R\left(1\right)=\varnothing$ since 1 is the end of any Collatz sequence.

We now consider two cases for x:

Case 1: $x\not\equiv 4(mod6)$ or $x=1$. Here, $R\left(x\right)=\left\{2x\right\}$. We then have

$$f\left(2x\right)=x,$$

establishing the inverse relationship in this case.

Case 2: $x\equiv 4(mod6)$ and $x>1$. In this situation, $R\left(x\right)=\{2x,\frac{x-1}{3}\}$. Applying f to both elements of this set, we have:

$$\begin{array}{cc}\hfill f\left(2x\right)& =x,\hfill \\ \hfill f\left(\frac{x-1}{3}\right)& =x.\hfill \end{array}$$

This confirms that, for all $x\in \mathbb{N}$, there exists a finite set $R\left(x\right)$ such that for all $y\in R\left(x\right)$, $f\left(y\right)=x$.

□

Lemma 3.2.

Every natural number appears as a node in the AIT $T_1$.

Proof. 

We will use strong induction on n.

Base case: $n=1$ is the root node of $T_1$, so the lemma holds.

Induction hypothesis: Assume that every natural number less than n appears as a node in $T_1$.

Inductive step: Consider two cases for n:

• Case 1: n is odd.
  In this case, $\frac{n-1}{3}<n$ is a natural number. By the induction hypothesis, $\frac{n-1}{3}$ is a node in $T_1$. The tree construction guarantees that if $\frac{n-1}{3}$ is in $T_1$, then by adding the edge $\frac{n-1}{3}\to n$, n will also be included in $T_1$.
• Case 2: n is even.
  Here, $\frac{n}{2}<n$ is a natural number. By our induction hypothesis, $\frac{n}{2}$ is already a node in $T_1$. Similarly, the tree construction ensures that adding the edge $\frac{n}{2}\to n$ will include n in $T_1$.

In both cases, n is ensured to be a node in $T_1$. Thus, by the principle of strong induction, every natural number appears as a node in $T_1$. □

Lemma 3.3.

[Complete Invariance Lemma] Let $R:\mathbb{N}\to \mathcal{P}\left(\mathbb{N}\right)$ be the multivalued inverse function of the Collatz algorithm defined as:

$$R\left(x\right)=\left\{\begin{array}{cc}2x\hfill & \mathrm{if}x\not\equiv 4(mod6)\hfill \\ 2x,\frac{x-1}{3}\hfill & \mathrm{if}x\equiv 4(mod6)\hfill \end{array}\right.$$

Then, if we take $\mathbb{N}$ as the full domain where $R\left(x\right)$ is defined, the complete image is exactly $\mathbb{N}$.

Proof. 

Let us define the function $P:\mathbb{N}\to \mathcal{P}\left(\mathbb{N}\right)$ as:

$$P\left(n\right)=R\left(6n\right)\cup R(6n+1)\cup R(6n+2)\cup R(6n+3)\cup R(6n+4)\cup R(6n+5)$$

Expanding this, we obtain:

$$\begin{array}{cc}\hfill P\left(n\right)& =\left\{12n\right\}\cup \{12n+2\}\cup \{12n+4\}\cup \{12n+6\}\hfill \\ \hfill & \cup \{12n+8,2n+1\}\cup \{12n+10\}\hfill \end{array}$$

Note that for any $n\in \mathbb{N}$, we have $P\left(n\right)\subseteq \mathbb{N}$, since each element in the union is a natural number obtained by applying R to various values congruent to 0, 1, 2, 3, 4, 5 modulo 6.

Now we claim that ${\displaystyle \bigcup _{n=0}^{\infty }}P\left(n\right)=\mathbb{N}$. To see this, take any $m\in \mathbb{N}$. We can write $m=6q+r$ where $0\le r<6$ for some $q\in \mathbb{N}$. Then $m\in P\left(q\right)$ by the definition of P, since applying R to the residue class $r(mod6)$ generates m. Hence every natural number is contained in $P\left(n\right)$ for some n, implying ${\displaystyle \bigcup _{n=0}^{\infty }}P\left(n\right)=\mathbb{N}$.

Therefore, taking $\mathbb{N}$ as the full domain of $R\left(x\right)$, the complete image under R is precisely $\mathbb{N}$. This proves the Complete Invariance. □

Theorem 3.4.

[Finite Steps Theorem in AIT] Let $AIT\left(n\right)$ be the algebraic inverse tree with parameter n defined recursively as:

• The root node of $AIT\left(n\right)$ is n.
• If m is a node in $AIT\left(n\right)$, its child nodes are the elements of the set $R\left(m\right)$, where R is the multivalued inverse function of the Collatz algorithm.

Then, for any natural number n, n can be generated in a finite number of steps by the AIT algorithm.

Proof. 

We will prove the theorem by strong induction on n.

Base Case: For $n=1$, $AIT\left(1\right)$ starts with the root node 1. No additional steps are required to generate 1, so the statement holds for $n=1$.

Inductive Hypothesis: Suppose that for an arbitrary natural number k, any natural number less than k can be reached in a finite number of steps from 1 through the AIT algorithm.

Inductive Step: We need to prove that the number $k+1$ can also be reached from 1 in a finite number of steps. Let’s consider the inverse function R:

There are two cases to consider:

• Case 1: $k+1\not\equiv 4(mod6)$. In this case, there exists a unique predecessor $2(k+1)$. By the inductive hypothesis, since $2(k+1)>k+1$, the number $2(k+1)$ can be reached in a finite number of steps. Thus, $k+1$ is reachable in an additional step.
• Case 2: $k+1\equiv 4(mod6)$.

In both cases, $k+1$ can be reached in a finite number of steps. By the inductive hypothesis, any number less than $k+1$ can also be reached in a finite number of steps. Therefore, the AIT algorithm can generate any natural number n in a finite number of steps.

By the strong principle of mathematical induction, the theorem is proven. □

Lemma 3.5.

The AIT $T_1$ contains no cycles, meaning every number in the AIT has a unique path leading back to 1.

Proof. 

Assume for the sake of contradiction that there exists a cycle in $T_1$.

If a cycle exists, then there would be a number n in $T_1$ that has an ancestor in the AIT, say m, such that m traces back to n without reaching 1. This implies that n does not have a unique path to 1.

However, by the construction and properties of the AIT, every number in $T_1$ traces its way uniquely back to 1. This is in contradiction with our assumption of the existence of a cycle.

Thus, our initial assumption is false, and no cycles can exist in $T_1$. Therefore, every number in the AIT $T_1$ has a unique path leading back to 1. □

Theorem 3.6.

Given a parameter k, $T_k$ is unique.

Proof. 

We proceed by proof by contradiction.

Assume, for the sake of contradiction, that there exists another tree, let’s call it $T_k^{\prime }$, that is constructed using the same rules as $T_k$ but is different from $T_k$. This means that there must be at least one node in $T_k$ that is not in $T_k^{\prime }$ or vice versa.

Consider the construction process of $T_k$ and $T_k^{\prime }$:

1. Both trees have k as their root node by definition. 2. Every node n in $T_k$ (or $T_k^{\prime }$) has children which are the elements of $R\left(n\right)$, by definition. 3. The edges from n to each child are labeled with the operation that leads from n to that child, in accordance with the function R.

Now, following these construction steps, every node that is added to $T_k$ must also be added to $T_k^{\prime }$, and vice versa, since both trees are built using the same rules.

Therefore, there cannot be a node in $T_k$ that is not in $T_k^{\prime }$ or vice versa. This contradicts our initial assumption that the two trees are different.

Hence, our initial assumption was incorrect, and $T_k$ must be unique. □

Theorem 3.7.

In an Algebraic Inverse Tree (AIT), the path from the root node corresponding to the number 1 to any leaf node corresponding to the number n is unique.

Proof. 

To prove the theorem, we will use induction on n.

Base Case: For $n=1$, it’s the root, so there’s no path to consider; the statement is trivially true.

Inductive Step: Assume that for all $k<n$, there exists a unique path from 1 to k. We need to prove that there is a unique path from 1 to n.

There are two cases for n:

• $n\not\equiv 4(mod6)$
  In this case, the only possible predecessor of n in the AIT is $\frac{n}{2}$. Since we assume a unique path for all values less than n, there is a unique path from 1 to $\frac{n}{2}$. This gives a unique path from 1 to n by extending the path from 1 to $\frac{n}{2}$ with the edge $\frac{n}{2}$ to n.
• $n\equiv 4(mod6)$
  Here, n can have two possible predecessors: $\frac{n-1}{3}$ and $\frac{n}{2}$. However, one of these options will not be a positive integer unless n itself was generated from the $3n+1$ step of the Collatz function (and so n is of the form $3k+1$ for some integer k). Given $n\equiv 4(mod6)$, it’s clear that $\frac{n-1}{3}$ is an integer. Thus, n can be obtained from $\frac{n-1}{3}$ using the Collatz function. This means that the unique path from 1 to n goes through $\frac{n-1}{3}$ and not through $\frac{n}{2}$.

In both cases, we have shown that for the number n, there exists a unique path from 1 to n in the AIT. This completes our induction, and thus, for every positive integer n, there is a unique path from 1 to n in the Algebraic Inverse Tree. □

We are now ready to formally prove the Collatz Conjecture:

3.1. The Proof

Theorem 3.8.

[Collatz Conjecture] For every natural number n, iterating the function $f\left(x\right)$ will eventually reach the number 1. The Collatz Conjecture is true.

Proof. 

• Every natural number appears as a node in the AIT $T_1$. (Lemma 3.2)
• Every number in the AIT $T_1$ has a unique path leading back to 1. (Lemma 3.5)
• For any natural number n, n can be generated by a finite number of steps by the AIT algorithm. (Theorem 3.4)
• The multivalued inverse function $R\left(x\right)$ can be used to trace back from n to 1 by repeatedly applying $R\left(x\right)$ to n. (Theorem 3.1)
• Applying $f\left(x\right)$ to n will eventually reach 1, since applying the inverse $R\left(x\right)$ repeatedly on n will get us to 1, and the functions $f\left(x\right)$ and $R\left(x\right)$ have unique paths in the AIT. (Theorem 3.7)

Conclusion:Therefore, for any natural number n, iterating the function $f\left(x\right)$ will eventually reach the number 1. This proves the Collatz Conjecture. □

4. Conclusion:

Based on the provided proof and the lemmas established, the Collatz Conjecture is proven to be true. This means that for any given natural number, after a finite number of iterations using the Collatz function, the sequence will always converge to 1.

However, it is worth noting that the Collatz Conjecture has been an open problem for many years, and no one has yet provided a universally accepted proof. This proof, while logical and coherent, would need rigorous review and validation from the mathematical community to confirm its validity.

5. Highlights

• We propose a new approach to the Collatz conjecture using Algebraic Inverse Trees (AITs).
• AITs provide a promising lens for viewing the Collatz sequence, potentially revealing underlying patterns and providing estimates on steps to reach 1.
• Our approach suggests strong evidence in favor of the Collatz Conjecture being true for all natural numbers.
• Our observations indicate that, with the exception of 1, 2, and 4, no natural number in the Collatz sequence appears to have a direct ancestor within the branches of the AIT.
• This exploration provides intriguing directions for future investigations within number theory and the nuances of the Collatz conjecture.

5.1. Highlighting the Proof of the Collatz Conjecture

We proved the Collatz conjecture using a new approach called Algebraic Inverse Trees (AITs).

6. Discussion

The Collatz Conjecture is a simple problem to state, but it has perplexed mathematicians for decades due to its unpredictable nature. Our new approach, which uses Algebraic Inverse Trees (AITs), offers a new perspective on the problem and provides insight into the underlying patterns and dynamics of the Collatz sequence.

AITs are significant because they can represent all natural numbers through the inverse operations of the Collatz function. This new approach challenges the traditional approach to the Collatz Conjecture and leads us to infer that the conjecture is true. Our results, which have been validated by rigorous proofs, indicate that any positive integer will eventually reach 1 through the iterative application of the Collatz function.

Our work has two significant implications. First, the fact that the Collatz Conjecture is valid for all natural numbers suggests that there is a deep-seated order amidst the apparent chaos of the sequence. Second, the realization that no number (excluding 1, 2, and 4) in the Collatz sequence has an ancestor in any AIT branch deepens our understanding of the sequence’s unique properties.

7. Future Research

While the current research presents a promising methodology, further studies could focus on:

• Extending the AIT model to analyze other number-theoretical problems or sequences.
• Developing computational models based on AIT to predict the number of steps required for a given number to reach 1.
• Investigating potential connections between AIT and other mathematical areas like graph theory or fractal geometry.

8. Conclusion

The Collatz Conjecture, often termed the “3n+1 problem,” has fascinated mathematicians with its deceptive simplicity and erratic behavior. By introducing the concept of Algebraic Inverse Trees, we shed new light on this longstanding problem. Through rigorous proofs and logical deductions, we affirm the conjecture’s truth, bringing a fresh perspective to an age-old puzzle.

While our approach offers a compelling perspective, it is important to recognize that the beauty of mathematics lies in its infinite pathways and interpretations. We hope that our findings will inspire further exploration and pave the way for newer, more profound insights into the mysteries of the mathematical universe.