Let an odd integer \(\mathcal{X}\) be expressed as $\left\{\sum\limits_{M > m} b_M 2^M\right\} + 2^m - 1,$ where $b_M \in \{0, 1\}$ and $2^m - 1$ is referred to as the Governor. In Collatz-type functions, a high index Governor is eventually reduced to $2^1 - 1$. For the $3\mathcal{Z} + 1$ sequence, the Governor occurring in the Trivial cycle is $2^1 - 1$, while for the $5\mathcal{Z} + 1$ sequence, the Trivial Governors are $2^2 - 1$ and $2^1 - 1$. Therefore, in these specific sequences, the Collatz function reduces the Governor $2^m - 1$ to the Trivial Governor $2^{\mathcal{T}} - 1$. Once this Trivial Governor is reached, it can evolve to a higher index Governor through interactions with other terms. This feature allows $\mathcal{X}$ to reappear in a Collatz-type sequence, since $2^m - 1 = 2^{m - 1} + \cdots + 2^{\mathcal{T} + 1} + 2^{\mathcal{T}} + (2^{\mathcal{T}} - 1).$ Thus, if $\mathcal{X}$ reappears, at least one odd ancestor of $\left\{\sum\limits_{M > m} b_M 2^M\right\} + 2^{m - 1} + \cdots + 2^{\mathcal{T} + 1} + 2^{\mathcal{T}} + (2^{\mathcal{T}} - 1)$ must have the Governor $2^m - 1$. Ancestor mapping shows that all odd ancestors of $\mathcal{X}$ have the Trivial Governor for the respective Collatz sequence. This implies that odd integers that repeat in the $3\mathcal{Z} + 1$ sequence have the Governor $2^1 - 1$, while those forming a repeating cycle in the $5\mathcal{Z} + 1$ sequence have either $2^2 - 1$ or $2^1 - 1$ as the Governor. Successor mapping for the $3\mathcal{Z} + 1$ sequence further indicates that there are no auxiliary cycles, as the Trivial Governor is always transformed into a different index Governor. Similarly, successor mapping for the $5\mathcal{Z} + 1$ sequence reveals that the smallest odd integers forming an auxiliary cycle are smaller than $2^5$. Finally, attempts to identify integers that diverge for the $3\mathcal{Z} + 1$ sequence suggest that no such integers exist.

This article presents a novel resolution of the Collatz Conjecture, centered on the concept of Generative Completeness of the inverse Collatz function. We introduce and rigorously prove that for all N ∈ ℕ⁺, there exists a minimal generator m_N = 1 such that all positive integers up to N can be generated through successive applications of the inverse Collatz function G. This key property, which we term "Generative Completeness", forms the cornerstone of our proof. Building upon this foundation, we establish several crucial results: The boundedness of all Collatz sequences The existence and uniqueness of cycles in Collatz sequences The nature of the unique cycle as {1, 4, 2} We then present three distinct approaches to resolving the Collatz Conjecture, all fundamentally rooted in the Generative Completeness property. These diverse methods not only prove the conjecture but also provide deep insights into the structure of Collatz sequences.

Let an odd integer be represented as $\sum \limits_{n > m} 2^n + 2^m - 1$ for $m \geq 1$. To transform $2^m - 1$ according to the Collatz rules, the $3x + 1$ step is applied, resulting in the even integer $2^{m + 1} + 2^m - 2$. This even integer, being exactly once divisible by 2, becomes the next odd integer $2^m + 2^{m - 1} - 1$. This represents the growth phase, as exactly one even step follows an odd step. However, while the overall value of the integer increases, terms with decreasing indices are generated after each odd-even cycle. After $m$ such cycles, a term $2^{m - m}$ (which equals 1) is generated, canceling the negative 1. Additional even steps are then performed until the integer becomes odd again, which occurs when the next smallest index becomes zero. The positive 1 thus obtained is written as $2^1 - 1$. If uninterrupted, the term $2^1 - 1$ has the trivial cycle $2^1 - 1 \rightarrow 2^2 \rightarrow 2^1 \rightarrow 2^1 - 1$. This represents the shrinkage phase, as two even steps follow an odd step. To interrupt the trivial cycle, a sequence such as $2^1 + 2^2 + \cdots$ must be available to combine with $2^1 - 1$ and produce $2^{\mathcal{M}} - 1$. This starts another growth phase that lasts for $\mathcal{M}$ odd-even cycles. Finally, there is an attempt to craft an integer that can escape the shrinkage phase in the Collatz dynamics.

We can analyze the properties of the Collatz Conjecture by using Terras’ Theorem, and computational graphical calculations. To do this, we introduce the concept of “tree" which shows the relationship between the input and output positive integers of the Collatz process. This approach offers a novel way for determining the non-existence of loops and diverge trees through computer simulations.

In this paper, I introduce a new concept of representing numbers in base ; in other words, I have found new series for any number similar to series that could be written according to the collatz sequence which is called zi(n) in this article. These series need to end with 1. Then, we use two sets of rules to make a diagram which then proves the existence of such series for any number.In this diagram, by branching numbers into different branches in accordance to the modularity of 4 and continuing branching to the point that their numbers have enough common terms in their collatz series, we could reduce zi(n) to zi(k) so that k<n in any branch by Rule Number One. This diagram shows that zi(n) exists for every number, in other words, this proves the theorem (A) or zi-existence theorem. The proof of zi-existence theorem leads to a proof of the collatz conjecture because the collatz series could be written as a linear combination of such series that all of them end with 1, so the collatz series for any number ends with 1.

This study focuses on one of the most famous open problems in mathematics, namely the Collatz conjecture. The Collatz conjecture or 3x + 1 Problem is perhaps today's most enigmatic unsolved mathematical problem. It is named after Lothar Collatz, who rst proposed it in 1937. It may be stated as as follow: Take any positive integer n. If n is even then divide it by 2, else do \triple plus one" and get 3n + 1. The conjecture is that this process will eventually reach the number 1, regardless of which positive integer is chosen initially. In this paper, we present a simple proof for the Collatz conjecture.

We propose Reduced Collatz Conjecture that is equivalent to Collatz Conjecture, which states that every positive integer can return to an integer less than it, instead of 1. Reduced Collatz Conjecture should be easier because some properties are presented in reduced dynamics, rather than in original dynamics (e.g., ratio and period). Reduced dynamics is a computation sequence from starting integer to the first integer less than it, and original dynamics is a computation sequence from starting integer to 1. Reduced dynamics is a component of original dynamics. We denote dynamics of x as a sequence of either computations in terms of ``I'' that represents (3*x+1)/2 and ``O'' that represents x/2. Here 3*x+1 and x/2 are combined together, because 3*x+1 is always even and followed by x/2. We formally prove that all positive integers are partitioned into two halves and either presents ``I'' or ``O'' in next ongoing computation. More specifically, (1) if any positive integer x that is i module $2^t$ (i is an odd integer) is given, then the first t computations (each one is either ``I'' or ``O'' corresponding to whether current integer is odd or even) will be identical with that of i. (2) If current integer after t computations (in terms of ``I'' or ``O'') is less than x, then reduced dynamics of x is available. Otherwise, the residue class of x (namely, i module $2^t$) can be partitioned into two halves (namely, i module $2^{t+1}$ and $i+2^t$ module $2^{t+1}$), and either half presents ``I'' or ``O'' in intermediately forthcoming (t+1)-th computation.

The 3x+1 problem is a problem of continuous iteration for integers. According to the basic theorem of arithmetic and the way of iteration, we derive a general formula for continuous iteration for odd integers. Through this formula, we can construct a loop iteration equation and obtain the result of the equation: the equation has only one positive integer solution. In addition, this general formula can be converted into a linear indeterminate equation. The process of solving this equation shows that the relationship between the iteration result and the odd number being iterated is linear. Extending this result to all positive even numbers, we get the answer to the 3x + 1 question.

For any odd positive integer $x$, define $(x_n)_{n\geqslant 0} $ and $(a_n )_{n\geqslant 1} $ by setting $x_{0}=x, \,\, x_n =\cfrac{3x_{n-1} +1}{2^{a_n }}$ such that all $x_n $ are odd. The 3x+1 problem asserts that there is an $x_n =1$ for all $x$. Usually, $(x_n )_{n\geqslant 0} $ is called the trajectory of $x$. In this paper, we concentrate on $(a_n )_{n\geqslant 1} $ and call it the E-sequence of $x$. The idea is that, we consider any infinite sequence $(a_n )_{n\geqslant 1} $ of positive integers and call it an E-sequence. We then define $(a_n )_{n\geqslant 1} $ to be $\Omega-$convergent to $x$ if it is the E-sequence of $x$ and to be $\Omega-$divergent if it is not the E-sequence of any odd positive integer. We prove a remarkable fact that the $\Omega-$divergence of all non-periodic E-sequences implies the periodicity of $(x_n )_{n\geqslant 0} $ for all $x_0$. The principal results of this paper are to prove the $\Omega-$divergence of several classes of non-periodic E-sequences. Especially, we prove that all non-periodic E-sequences $(a_n )_{n\geqslant 1} $ with $\mathop {\overline {\lim } }\limits_{n\to \infty } \cfrac{b_n }{n}>\log _23$ are $\Omega-$divergent by using the Wendel's inequality and the Matthews and Watts's formula $x_n =\cfrac{3^n x_0 }{2^{b_n }}\prod\limits_{k=0}^{n-1} {(1+\cfrac{1}{3x_k })} $, where $b_n =\sum\limits_{k=1}^n {a_k } $. These results present a possible way to prove the periodicity of trajectories of all positive integers in the 3x + 1 problem and we call it the E-sequence approach.