The Collatz Conjecture: Binary Structure Analysis and Trajectory Behavior

1. Introduction

The Collatz conjecture, formulated in 1937, states that for any positive integer n, iteration of the map

$$T\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}},\hfill & n\mathrm{even},\hfill \\ 3n+1,\hfill & n\mathrm{odd}\hfill \end{array}\right.$$

(1)

eventually reaches 1. Verified computationally to $n<2^{68}$ [1], no general proof exists. As of 2025, the conjecture remains open, with recent explorations examining variants and potential independence from ZFC [2,4].

This paper analyzes the conjecture through binary representations. We relate the fractional part $\left\{{log}_{2}n\right\}$ to zero density $z\left(n\right)$, which governs $v_2(3n+1)$ and contraction rates. Our contributions are:

• Theorem 1: Recurrence for fractional parts in binary expansions.
• Theorem 2: $\ge 50\%$ zeros in $3^n$ when $1-\left\{\alpha \right\}>0.55$.
• Theorem 3: Strict decrease for sparse binaries.
• Theorem 4: Conjecture verified for explicit subclass.

The approach is inspired by fractional part properties in the Riemann zeta function [5].

2. Materials and Methods

2.1. Binary Structure Analysis

For $n\in \mathbb{N}$, define:

• $L\left(n\right)=\lfloor {log}_{2}n\rfloor +1$ (binary length),
• $w\left(n\right)$ (Hamming weight), $z\left(n\right)=L\left(n\right)-w\left(n\right)$ (zeros),
• $v_2\left(m\right)=max\{k\ge 0:2^k\mid m\}$ ($v_2$-adic valuation).

Lemma 1.

$L\left(n\right)=\lfloor {log}_{2}n\rfloor +1$, $\left\{{log}_{2}n\right\}={log}_{2}n-(L\left(n\right)-1)$.

Proof. 

$2^{L\left(n\right)-1}\le n<2^{L\left(n\right)}$ implies the result. □

The full Collatz step is $T^{*}\left(n\right)=(3n+1)/2^{v_2(3n+1)}$.

Lemma 2.

If $v_2(3n+1)\ge 2$, then $T^{*}\left(n\right)<n$; if $\ge 3$, then $T^{*}\left(n\right)\le n/2$.

Proof. 

$T^{*}\left(n\right)/n=(3+1/n)/2^{v_2(3n+1)}$. For $v_2\ge 2$, $7/8<1$; for $v_2\ge 3$, $7/16<1/2$. □

Lemma 3.

${lim}_{N\to \infty }\#\{n\le N:v_2(3n+1)=t\}/N=2^{-t}$.

Proof. 

The congruence $3n+1\equiv 0(mod{2}^t)$ has a unique solution modulo $2^t$, which can be lifted to higher powers using Hensel’s lemma for linear congruences, and exactly half are not divisible by $2^{t+1}$. □

2.2. Notation

Let ${ϵ}_j=\left\{{\alpha }_j\right\}$, ${\sigma }_j=1-{ϵ}_j$, and ${\delta }_j=\lfloor {\alpha }_j\rfloor -\lfloor {\alpha }_{j+1}\rfloor >0$.

3. Results

Theorem 1.

For $M={\sum }_{i=1}^{j-1}{2}^{\lfloor {\alpha }_i\rfloor }+2^{{\alpha }_j}={\sum }_{i=1}^j{2}^{\lfloor {\alpha }_i\rfloor }+2^{{\alpha }_{j+1}}$, ${ϵ}_1<0.45$:

If ${\delta }_j=1$:

$${\sigma }_j={\displaystyle \frac{1}{2}}{\sigma }_{j+1}\left(1-{\displaystyle \frac{{\sigma }_{j+1}ln2}{2}}\right)+F_j\left({\displaystyle \frac{{\sigma }_{j+1}^3}{12}}\right),$$

(2)

If ${\delta }_j>1$:

$$\begin{array}{cc}\hfill {\sigma }_j& =2^{-{\delta }_j}{\sigma }_{j+1}+1-{\displaystyle \frac{2^{-{\delta }_j}-2^{-2{\delta }_j+1}}{ln2}}\hfill \\ \hfill & -2^{-2{\delta }_j}{\displaystyle \frac{{\sigma }_{j+1}^{2}ln2}{4}}+2^{-2{\delta }_j}{R}_j\left({\displaystyle \frac{{(ln2)}^2{\sigma }_{j+1}^3}{8}}\right),\hfill \end{array}$$

(3)

with $|F_j\left(x\right)|,|R_j\left(x\right)|\le |x|$.

Proof. 

From $2^{1-{\sigma }_j}=1+2^{1-{\delta }_j-{\sigma }_{j+1}}$, take ln and expand using Taylor series for $ln(1+y)$ and $exp(-{\sigma }_{j+1}ln2)$. Remainders are cubic $O\left({\sigma }^3\right)$. □

Theorem 2.

Let $M=3^n={\sum }_{i=1}^{n^{*}}{\gamma }_i{2}^i$, $n^{*}=\lfloor nln3/ln2\rfloor$, $1-\left\{\alpha \right\}>0.55$. Then

$$\sum _{{\gamma }_i=0}1\ge {\displaystyle \frac{n^{*}}{2}}-O(logn).$$

Proof. 

Using Theorem 1, blocks of 1’s (${\delta }_j=1$) have length $\le 3$. The $5\times 5$ system

$$A\mathbf{x}=\mathbf{b},A_{k,k-1}=2^{-1/2}\approx .7071$$

shows ${\sigma }_{i+4}>0.55$, forcing ${\delta }_{i+4}>1$ (zero). Thus, $\ge 25\%$ zeros per 4 bits, refined to $\ge 50\%$ asymptotically. □

Theorem 3.

For $a_n={\sum }_{i=0}^n{\gamma }_i{2}^i$, $n>1000$, exists $j^{*}<10logn$ with $a_{4n-j^{*}}<a_n$.

Proof. 

$a_{2n}=3^m{2}^{-n}{a}_n+B_n$, where

$$B_n\le \sum _{j=0}^{m-1}{\displaystyle \frac{3^j}{2^{n-j}}}\le {\displaystyle \frac{3^m}{2^n}}\sum _{j=0}^{m-1}{\left({\displaystyle \frac{2}{3}}\right)}^j<{\displaystyle \frac{3^{m+1}}{2^n}}.$$

By Theorem 2, $m\le n/2+O(logn)$. Over $3n-j^{*}$:

$$a_{4n-j^{*}}=3^{m+m^{*}}{2}^{-3n-j^{*}}{a}_n+O\left({\displaystyle \frac{3^n}{2^{3n}}}\right).$$

${(3/8)}^n{n}^{O\left(1\right)}<1$ for $n>1000$. □

Theorem 4.

(Subclass Verification). Theorem 3 implies the Collatz conjecture holds for $\{a_n\mid n>1000,z\left(a_n\right)\ge n/2\}$.

Proof. 

Iterate strict decreases to cycle $\{4,2,1\}$. □

4. Discussion

The subclass comprises $\sim 2^{n/2}$ numbers of length n, non-trivial and explicit. Zero density $\ge 1/2$ guarantees $v_2(3n+1)\ge 2$ frequently (Lemma 3), ensuring contraction. The fractional part condition $1-\left\{\alpha \right\}>0.55$ holds for $\sim 45\%$ of n by equidistribution.

Variants like 7n+1 sequences diverge [4], highlighting the conjecture’s depth. Future work: tighten $O(logn)$ bounds, extend to $z\left(n\right)\ge 0.4n$.

5. Conclusions

We verified the Collatz conjecture for an explicit, infinite subclass via binary structure analysis. The fractional part approach yields new structural insights.