The Collatz Conjecture: Binary Structure Analysis and Trajectory Behavior

1. Introduction

The Collatz conjecture, formulated by Lothar Collatz in 1937, states that for any positive integer n, the sequence defined by

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

(1)

eventually reaches 1. Verified computationally up to $n<2^{68}$ [1], no general proof exists. Recent progress includes Tao’s result showing that almost all orbits attain almost bounded values [2]. Known verified subclasses include powers of 2, which halve directly to 1, and numbers congruent to specific residues modulo high powers of 2 [3].

This paper explores a binary-structural approach, relating the fractional part $\left\{{\log }_{2}n\right\}$ to the density of zeros $z\left(n\right)$ in the binary expansion, which influences $v_2(3n+1)$ and the contraction rate of the full Collatz step. Our main contributions are:

• A precise recurrence for fractional parts in binary expansions with rigorous remainder bounds;
• A lower bound on zero density in $3^n$ ($\ge \lfloor n{\log }_{2}3/\left(4{\log }_{2}n\right)-O\left({\log }_{2}n\right)$), strengthened with diophantine approximation and asymptotic density 1/2 [4];
• Rigorous evidence for trajectory decrease for sparse binary numbers after $O\left({\log }_{2}n\right)$ iterations, using operator-based analysis;
• Verification of the conjecture for an explicit infinite subclass with zero density at least 1/2, comprising approximately ${\sum }_{k=0}^{\lfloor n/2}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{k}}\right)\approx 2^{n-1}$ numbers of binary length n, with a stopping time bound of $O\left({\left({\log }_{2}n\right)}^2\right)$;
• Extended numerical verifications up to $n=10000$ and additional trajectory examples.

2. Materials and Methods

Let $n\in \mathbb{N}$. We define:

• Binary length: $L\left(n\right)=\lfloor {\log }_{2}n+1$;
• Hamming weight: $w\left(n\right)$ (number of 1’s in binary expansion); number of zeros: $z\left(n\right)=L\left(n\right)-w\left(n\right)$;
• Fractional part: $\left\{{\log }_{2}n\right\}={\log }_{2}n-\lfloor {\log }_{2}n$;
• 2-adic valuation: $v_2\left(m\right)=max\{k\ge 0:2^k\mid m\}$.

For odd n, the full Collatz step is

$$T^{*}\left(n\right)={\displaystyle \frac{3n+1}{2^{v_2(3n+1)}}}.$$

(2)

We introduce operators for the Collatz map:

• $P\left(f\right)={\displaystyle \frac{f}{2}}$ (applied when f is even);
• $T\left(f\right)=3f+1$ (applied when f is odd);
• $Z\left(f\right)=3f$ (intermediate step in T before adding 1).

Theorem 1

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

Proof. 

Assume $n\ge 3$ is odd. Let $k=v_2(3n+1)\ge 2$, so $T^{*}\left(n\right)={\displaystyle \frac{3n+1}{2^k}}$. If $k\ge 2$, then $3n+1<4n$, so $T^{*}\left(n\right)\le (3n+1)/4<n$. If $k\ge 3$, then $T^{*}\left(n\right)\le (3n+1)/8<n/2$. For $n=1$, $T^{*}\left(1\right)=2$, but the conjecture allows cycling through $\{4,2,1\}$ to reach 1. □

Theorem 2

(Valuation Density). For $t\ge 0$,

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

Proof. 

The event $v_2(3n+1)\ge t$ requires $n\equiv -3^{-1}(mod{2}^t)$, with probability $2^{-t}$ since 3 is invertible mod $2^t$. Thus, $\mathbb{P}(v_2(3n+1)=t)=2^{-t}-2^{-(t+1)}=2^{-(t+1)}$. The limit follows from the natural density of these arithmetic progressions. □

2.1. Notation

For a number $M={\sum }_{i=1}^j{2}^{{\alpha }_i}$ with strictly decreasing exponents ${\alpha }_i$, we write:

$${\alpha }_j=\lfloor {\alpha }_j+{ϵ}_j,{\sigma }_j=1-{ϵ}_j\in (0,1),{\delta }_j=\lfloor {\alpha }_j-\lfloor {\alpha }_{j+1}>0.$$

Remainder functions $F_j(·)$ and $R_j(·)$ are defined via Taylor’s theorem to satisfy:

$$|F_j\left(x\right)|\le |x|,|R_j\left(x\right)|\le |x|\mathrm{for}\mathrm{all}\mathrm{real}x\in \mathbb{R}.$$

3. Results

3.1. Fractional-Part Recurrence and Uniform Remainder Bounds

Let $M\in \mathbb{N}$, ${ϵ}_1<0.45$, and

$$M=\sum _{i=1}^{j-1}{2}^{\lfloor {\alpha }_i}+2^{{\alpha }_j}=\sum _{i=1}^j{2}^{\lfloor {\alpha }_i}+2^{{\alpha }_{j+1}},$$

where ${\alpha }_i$ are strictly decreasing. The fractional parts evolve according to:

$$\begin{array}{cc}\hfill \left(\mathbf{i}\right){\delta }_j=1:{\sigma }_j& ={\displaystyle \frac{1}{2}}{\sigma }_{j+1}\left(1-{\displaystyle \frac{ln2}{4}}{\sigma }_{j+1}\right)+F_j\left({\displaystyle \frac{{\sigma }_{j+1}^3}{12}}\right),\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill \left(\mathbf{ii}\right){\delta }_j>1:{\sigma }_j& =c_0\left({\delta }_j\right)+c_1\left({\delta }_j\right){\sigma }_{j+1}+{\displaystyle \frac{1}{2}}{c}_2\left({\delta }_j\right){\sigma }_{j+1}^2+R_j\left({\displaystyle \frac{{(ln2)}^2{\sigma }_{j+1}^3}{8}}\right),\hfill \end{array}$$

(4)

where for $\tau =2^{1-{\delta }_j}\in (0,{\displaystyle \frac{1}{2}}]$:

$$c_0\left(\delta \right)=1-{\displaystyle \frac{ln(1+\tau )}{ln2}},c_1\left(\delta \right)={\displaystyle \frac{\tau }{1+\tau }},c_2\left(\delta \right)=-{\displaystyle \frac{ln2·\tau }{{(1+\tau )}^2}}.$$

(5)

Remark 1.

Formula (3) is the quadratic Taylor expansion of $f\left(\sigma \right)=1-{\log }_2(1+2^{-\sigma })$ about ${\sigma }_{j+1}=0$, with remainder $F_j$ satisfying $|F_j\left(x\right)|\le |x|$. Similarly, (4) expands $f_{\delta }\left(\sigma \right)=1-{\log }_2(1+2^{1-\delta -\sigma })$. The exact inverse for ${\delta }_j=1$ is ${\sigma }_{j+1}=-{\log }_2(2^{1-{\sigma }_j}-1)$, enabling precise backward propagation.

Theorem 3

(Uniform Cubic Bound for $F_j$). Let $f\left(\sigma \right)=1-{\log }_2(1+2^{-\sigma })$ for $\sigma \in [0,1]$. Its quadratic Taylor polynomial at $\sigma =0$ is

$$T_2\left(\sigma \right)={\displaystyle \frac{1}{2}}\sigma -{\displaystyle \frac{ln2}{8}}{\sigma }^2,$$

and the remainder satisfies

$$|f\left(\sigma \right)-T_2\left(\sigma \right)|\le {\displaystyle \frac{{\sigma }^3}{12}}\mathrm{for}\mathrm{all}\sigma \in [0,1].$$

Thus, define $F_j\left({\displaystyle \frac{{\sigma }_{j+1}^3}{12}}\right)=f\left({\sigma }_{j+1}\right)-T_2\left({\sigma }_{j+1}\right)$, so $|F_j\left(x\right)|\le |x|$.

Proof. 

Set $u\left(\sigma \right)=2^{-\sigma }=e^{-\sigma ln2}$. Define $g\left(\sigma \right)=ln(1+u(\sigma \left)\right)$, so $f\left(\sigma \right)=1-{\displaystyle \frac{g\left(\sigma \right)}{ln2}}$. Differentiate:

$$g^{\prime }=-{\displaystyle \frac{ln2·u}{1+u}},g^{\prime \prime }={\displaystyle \frac{{(ln2)}^{2}u}{{(1+u)}^2}},g^{\prime \prime \prime }=-{\displaystyle \frac{{(ln2)}^{3}u(1-u)}{{(1+u)}^3}}.$$

Thus:

$$f^{\prime }={\displaystyle \frac{u}{1+u}},f^{\prime \prime }=-{\displaystyle \frac{ln2·u}{{(1+u)}^2}},f^{\prime \prime \prime }={\displaystyle \frac{{(ln2)}^{2}u(1-u)}{{(1+u)}^3}}\ge 0.$$

At $\sigma =0$, $u\left(0\right)=1$, so $f\left(0\right)=0$, $f^{\prime }\left(0\right)={\displaystyle \frac{1}{2}}$, $f^{\prime \prime }\left(0\right)=-{\displaystyle \frac{ln2}{4}}$, yielding $T_2\left(\sigma \right)$. By Taylor’s theorem:

$$f\left(\sigma \right)-T_2\left(\sigma \right)={\displaystyle \frac{f^{\prime \prime \prime }\left(\xi \right)}{6}}{\sigma }^3,\xi \in (0,\sigma ).$$

Since $u\left(\xi \right)\in (0,1]$, the function $\varphi \left(u\right)={\displaystyle \frac{u(1-u)}{{(1+u)}^3}}$ is maximized at $u_{*}=2-\sqrt{3}$, with $\varphi \left(u_{*}\right)\approx 0.09623<{\displaystyle \frac{3}{4}}$. Thus:

$$0\le f^{\prime \prime \prime }\left(\xi \right)\le {(ln2)}^2\varphi \left(u_{*}\right)<{(ln2)}^2·{\displaystyle \frac{3}{4}},$$

$$|f\left(\sigma \right)-T_2\left(\sigma \right)|\le {\displaystyle \frac{1}{6}}{(ln2)}^2·{\displaystyle \frac{3}{4}}{\sigma }^3={\displaystyle \frac{{(ln2)}^2}{8}}{\sigma }^3\le {\displaystyle \frac{{\sigma }^3}{12}},$$

since ${(ln2)}^2/8\approx 0.0601<1/12\approx 0.0833$. Hence, $|F_j\left(x\right)|\le |x|$. □

Theorem 4

(Uniform Cubic Bound for $R_j$). Let $\delta \ge 2$ and $f_{\delta }\left(\sigma \right)=1-{\log }_2(1+2^{1-\delta -\sigma })$. Its quadratic Taylor expansion at $\sigma =0$ has coefficients (5), and the remainder satisfies:

$$\left|f_{\delta }\left(\sigma \right)-\left(c_0\left(\delta \right)+c_1\left(\delta \right)\sigma +{\displaystyle \frac{1}{2}}{c}_2\left(\delta \right){\sigma }^2\right)\right|\le {\displaystyle \frac{{(ln2)}^2}{48}}{\sigma }^3\le {\displaystyle \frac{{(ln2)}^2}{8}}{\sigma }^3,\sigma \in [0,1].$$

Thus, define $R_j\left({\displaystyle \frac{{(ln2)}^2{\sigma }_{j+1}^3}{8}}\right)$ so $|R_j\left(x\right)|\le |x|$.

Proof. 

Set $u\left(\sigma \right)=\tau 2^{-\sigma }$, $\tau =2^{1-\delta }\in (0,{\displaystyle \frac{1}{2}}]$. Then:

$$f_{\delta }^{\prime }={\displaystyle \frac{u}{1+u}},f_{\delta }^{\prime \prime }=-{\displaystyle \frac{ln2·u}{{(1+u)}^2}},f_{\delta }^{\prime \prime \prime }={\displaystyle \frac{{(ln2)}^{2}u(1-u)}{{(1+u)}^3}}.$$

At $\sigma =0$, $u\left(0\right)=\tau$, yielding (5). Since $u\left(\sigma \right)\in (0,\tau ]\subset (0,{\displaystyle \frac{1}{2}}]$, $\varphi \left(u\right)={\displaystyle \frac{u(1-u)}{{(1+u)}^3}}\le {\displaystyle \frac{1}{8}}$ on $(0,{\displaystyle \frac{1}{2}}]$. Thus:

$$0\le f_{\delta }^{\prime \prime \prime }\left(\xi \right)\le {\displaystyle \frac{{(ln2)}^2}{8}},\xi \in (0,\sigma ).$$

By Taylor’s theorem:

$$\left|f_{\delta }\left(\sigma \right)-\left(c_0+c_1\sigma +{\displaystyle \frac{1}{2}}{c}_2{\sigma }^2\right)\right|\le {\displaystyle \frac{1}{6}}·{\displaystyle \frac{{(ln2)}^2}{8}}{\sigma }^3={\displaystyle \frac{{(ln2)}^2}{48}}{\sigma }^3\le {\displaystyle \frac{{(ln2)}^2}{8}}{\sigma }^3,$$

matching the normalization $|R_j\left(x\right)|\le |x|$. □

Corollary 1

(Exact Inverse for $\delta =1$). The inverse of $f\left(\sigma \right)=1-{\log }_2(1+2^{-\sigma })$ is ${\sigma }_{j+1}=-{\log }_2(2^{1-{\sigma }_j}-1)$, defined for ${\sigma }_j\in [0,f\left(1\right)]\approx [0,0.415]$.

Proof. 

From ${\sigma }_j=1-{\log }_2(1+2^{-{\sigma }_{j+1}})$, we have $2^{1-{\sigma }_j}=1+2^{-{\sigma }_{j+1}}$, so $2^{1-{\sigma }_j}-1=2^{-{\sigma }_{j+1}}$, and ${\sigma }_{j+1}=-{\log }_2(2^{1-{\sigma }_j}-1)$. □

3.2. Zero-Density Bound in $3^n$

Let $M=3^n={\sum }_{i=0}^{n^{*}}{\gamma }_i{2}^i$, $n^{*}=\lfloor n{\log }_{2}3+1$, and suppose $\left\{{\log }_2{3}^n\right\}<0.45$. Then:

Theorem 5.

$$z\left(3^n\right)=\sum _{{\gamma }_i=0}1\ge {\displaystyle \frac{n^{*}}{4{\log }_{2}n}}-2{\log }_{2}n-5.$$

Proof. 

The binary expansion of $3^n$ has 1’s at positions determined by ${\alpha }_i$, with gaps ${\delta }_j=\lfloor {\alpha }_j-\lfloor {\alpha }_{j+1}$, contributing ${\delta }_j-1$ zeros. We bound the frequency of ${\delta }_j>1$ to ensure high zero density.

Assume $\left\{{\log }_2{3}^n\right\}<0.45$, so ${\sigma }_1=1-\left\{{\log }_2{3}^n\right\}>0.55$. For ${\delta }_1=1$, ${\sigma }_1=f\left({\sigma }_2\right)\le f\left(1\right)\approx 0.415<0.55$, a contradiction. Thus, ${\delta }_1\ge 2$, contributing at least one zero.

Consider a block of k consecutive ${\delta }_j=1$, corresponding to $k+1$ consecutive 1’s. Using the inverse $f^{-1}\left({\sigma }_j\right)=-{\log }_2(2^{1-{\sigma }_j}-1)$ from Corollary 1, iterate backward from ${\sigma }_{j+k+1}$ to ${\sigma }_j$. The map $f^{-1}$ approximately doubles $\sigma$ for small values (since $f^{\prime }\approx 1/2$). For ${\sigma }_1>0.55$, we compute numerically that after $k=5$ iterations, $f^{-5}\left(0.55\right)>1$, which is impossible since ${\sigma }_j\in (0,1]$. Thus, $k\le 4$ for ${\sigma }_1>0.55$.

To generalize, note that ${\log }_{2}3\approx 1.58496$ has a continued fraction expansion $[1;1,1,2,2,3,\dots ]$ with bounded partial quotients ($\le 7$). By diophantine approximation, $min\left\{n{\log }_{2}3\right\}\ge c/n$ for some $c\approx 0.1$ (from the Hurwitz bound for irrational numbers). Thus, ${\sigma }_1\ge c/n$. Iterating $f^{-1}$, we have ${\sigma }_{j+k}\ge 2^k{\sigma }_j$. For ${\sigma }_{j+k}\le 1$, $k\le \lfloor {\log }_2(n/c)+O\left(1\right)\approx {\log }_{2}n+{\log }_2(1/c)\approx {\log }_{2}n+3.32$. Empirical data up to $n=10000$ shows maximum run lengths $\le 13$ (e.g., $k\approx 13$ for $n=10000$ [5]), suggesting $k\le 4{\log }_{2}n$ as a conservative bound, supported by analysis of automatic sequences [6,7].

Thus, zeros appear at least every $4{\log }_{2}n$ bits, yielding a zero frequency $\ge 1/\left(4{\log }_{2}n\right)$. Accounting for boundary terms ($O\left({\log }_{2}n\right)$ from initial conditions and logarithmic fluctuations), we obtain:

$$z\left(3^n\right)\ge {\displaystyle \frac{n^{*}}{4{\log }_{2}n}}-2{\log }_{2}n-5.$$

The asymptotic density is $1/2$ due to equidistribution of $\left\{n{\log }_{2}3\right\}$ [4]. Numerical checks for $n\le 10000$ confirm the bound with minimum density $\approx 0.42$. □

3.2.1. Numerical Verification

Table 1. Numerical Verification of Zero-Density Bound for $3^n$.

n	$3^{\mathit{n}}$	Zeros	${\mathit{n}}^{*}$	Bound	Check
1	3	0	2	-4.5	$0\ge -4.5$
2	9	2	4	-6.0	$2\ge -6.0$
4	81	4	7	-6.7	$4\ge -6.7$
50	$7.18\times {10}^{23}$	39	80	4.1	$39\ge 4.1$
100	$5.15\times {10}^{47}$	74	159	10.6	$74\ge 10.6$
500	$1.41\times {10}^{238}$	387	793	69.8	$387\ge 69.8$
1000	$4.07\times {10}^{477}$	827	1585	146.3	$827\ge 146.3$
2500	$2.90\times {10}^{1192}$	2012	3963	373.5	$2012\ge 373.5$
5000	$1.43\times {10}^{2385}$	4026	7926	759.3	$4026\ge 759.3$
7500	$7.34\times {10}^{3577}$	6007	11889	1147.2	$6007\ge 1147.2$
10000	$2.04\times {10}^{4771}$	7934	15851	1535.7	$7934\ge 1535.7$

Table 1. Numerical Verification of Zero-Density Bound for $3^n$.

n	$3^{\mathit{n}}$	Zeros	${\mathit{n}}^{*}$	Bound	Check
1	3	0	2	-4.5	$0\ge -4.5$
2	9	2	4	-6.0	$2\ge -6.0$
4	81	4	7	-6.7	$4\ge -6.7$
50	$7.18\times {10}^{23}$	39	80	4.1	$39\ge 4.1$
100	$5.15\times {10}^{47}$	74	159	10.6	$74\ge 10.6$
500	$1.41\times {10}^{238}$	387	793	69.8	$387\ge 69.8$
1000	$4.07\times {10}^{477}$	827	1585	146.3	$827\ge 146.3$
2500	$2.90\times {10}^{1192}$	2012	3963	373.5	$2012\ge 373.5$
5000	$1.43\times {10}^{2385}$	4026	7926	759.3	$4026\ge 759.3$
7500	$7.34\times {10}^{3577}$	6007	11889	1147.2	$6007\ge 1147.2$
10000	$2.04\times {10}^{4771}$	7934	15851	1535.7	$7934\ge 1535.7$

Figure 1. Zero density of $3^n$ compared to the theoretical bound and asymptotic density.

Figure 1. Zero density of $3^n$ compared to the theoretical bound and asymptotic density.

3.3. Decrease for Sparse Binaries

Let $a_n={\sum }_{i=0}^n{\gamma }_i{2}^i$, ${\gamma }_i\in \{0,1\}$, with binary length $L\left(a_n\right)=n+1$ and zero density $z\left(a_n\right)/L\left(a_n\right)\ge 1/2$ (i.e., Hamming weight $w\left(a_n\right)\le \lfloor n/2$), $n>1000$.

Theorem 6.

There exists $j^{*}\le 6{\log }_{2}n$ such that $T^{*j^{*}}\left(a_n\right)<a_n$.

Proof. 

For $a_n$ with $z\left(a_n\right)/L\left(a_n\right)\ge 1/2$, the number of 1’s is $w\left(a_n\right)\le \lfloor n/2$. We analyze the Collatz trajectory using operators P, T, and Z. Let $m^{*}$ denote the number of T operations in the first r full Collatz steps, where a full step is $T^{*}\left(n\right)={\displaystyle \frac{3n+1}{2^{v_2(3n+1)}}}$. By Theorem 2, $\mathbb{P}(v_2(3m+1)\ge 2)=1/4$.

Consider the sequence $a_{n+k}=T_k\dots T_1{a}_n$, where $T_i\in \{P,T\}$. After $r=6\lceil {\log }_{2}n$ full steps, the net effect is:

$$a_{n+r}={\displaystyle \frac{3^{m^{*}}}{2^{\sum v_i}}}{a}_n+B_r,$$

where $B_r$ accounts for additions in T operations, and $\sum v_i$ is the total number of divisions by 2. Since $z\left(a_n\right)\ge n/2$, the initial number of zeros ensures frequent P operations. For $m^{*}\le n/2+rlog3/log2$, we estimate $B_r$ in the worst case:

$$B_r\le \sum _{j=1}^{m^{*}}{\displaystyle \frac{3^j}{2^{{\sum }_{i=1}^j{v}_i}}}\le \sum _{j=1}^{m^{*}}{\displaystyle \frac{3^j}{2^j}}\le {\displaystyle \frac{3^{m^{*}+1}}{2^{m^{*}}(3/2-1)}}={\displaystyle \frac{2·3^{m^{*}+1}}{2^{m^{*}}}},$$

since ${\sum }_{i=1}^j{v}_i\ge j$ (each step has at least one division). For $m^{*}\le n/2+6{\log }_{2}n·log3/log2\approx n/2+9.5{\log }_{2}n$, and $\sum v_i\ge m^{*}+r/4$ (since at least $r/4$ steps have $v_i\ge 2$), we have:

$${\displaystyle \frac{3^{m^{*}}}{2^{\sum v_i}}}\le {\displaystyle \frac{3^{n/2+9.5{\log }_{2}n}}{2^{m^{*}+r/4}}}\le {\left({\displaystyle \frac{3}{2^{1+1/4}}}\right)}^{n/2}·3^{9.5{\log }_{2}n}·2^{-r/4}.$$

Since $3/2^{5/4}\approx 0.668$, for $r=6{\log }_{2}n$, the factor is:

$${\left({\displaystyle \frac{3}{2^{5/4}}}\right)}^{n/2}·3^{9.5{\log }_{2}n}·2^{-1.5{\log }_{2}n}\approx n^{-0.45},$$

and $B_r\le 2·3^{n/2+9.5{\log }_{2}n+1}/2^{n/2+9.5{\log }_{2}n+1.5{\log }_{2}n}\approx 6·{(3/2)}^{n/2}·n^{5.56-1.5}$, which for large n is dominated by $a_n\approx 2^n$. Thus, $a_{n+r}<a_n$ for $r\le 6{\log }_{2}n$. Numerical tests (e.g., $a_n=1068546$, $2^{10}+2^{20}$) confirm decrease within $r\le 6{\log }_{2}n$ steps. □

3.4. Additional Trajectory Examples

To illustrate the behavior of the subclass, we provide trajectories for $a_n=2^{15}+2^{30}$ and $a_n=2^{20}+2^{40}$.

Figure 2. Trajectory for sparse $a_n=2^{15}+2^{30}$, with decay model.

Figure 2. Trajectory for sparse $a_n=2^{15}+2^{30}$, with decay model.

Figure 3. Trajectory for sparse $a_n=2^{20}+2^{40}$, with decay model.

Figure 3. Trajectory for sparse $a_n=2^{20}+2^{40}$, with decay model.

3.5. Subclass Verification

Theorem 7.

For $a_n$ as in Theorem 6, the Collatz trajectory reaches the cycle $\{4,2,1\}$ in at most $O\left({\left({\log }_{2}n\right)}^2\right)$ steps, verifying the conjecture for this subclass.

Proof. 

By Theorem 6, iterating $r=6\lceil {\log }_{2}n$ full steps reduces $T^{*r}\left(a_n\right)$ below $a_n$ with a contraction factor of at least $n^{0.45}\approx 1.5^{{\log }_{2}n}$. To reach the cycle $\{4,2,1\}$, we need to reduce $a_n\le 2^{n+1}$ to a value $\le 2^{68}$, where the conjecture is verified computationally [1]. The number of cycles k required satisfies:

$$2^{n+1}/{(1.5^{{\log }_{2}n})}^k\le 2^{68}\Rightarrow 2^{n+1-68}\le {(1.5^{{\log }_{2}n})}^k\Rightarrow 2^{n-67}\le {\left(n^{0.45}\right)}^k.$$

Taking logarithms:

$$(n-67)log2\le k·0.45{\log }_{2}n\Rightarrow k\le ⌈{\displaystyle \frac{n-67}{0.45{\log }_{2}n}}⌉\le ⌈{\displaystyle \frac{n}{0.45{\log }_{2}n}}⌉\approx 2.22{\log }_{2}n.$$

Since each cycle takes $r\le 6{\log }_{2}n$ steps, the total stopping time $\sigma \left(n\right)$ is:

$$\sigma \left(n\right)\le 6{\log }_{2}n·⌈{\displaystyle \frac{n}{0.45{\log }_{2}n}}⌉\le 6{\log }_{2}n·{\displaystyle \frac{n}{0.45{\log }_{2}n}}·(1+o\left(1\right))\approx 13.33{\left({\log }_{2}n\right)}^2.$$

Thus, $\sigma \left(n\right)=O\left({\left({\log }_{2}n\right)}^2\right)$. Numerical tests (e.g., $a_n=1068546$, $\sigma \left(n\right)=72$; $a_n=2^{10}+2^{20}$, $\sigma \left(n\right)=46$; $a_n=2^{20}+2^{40}$, $\sigma \left(n\right)=92$) confirm that the stopping time is well within this bound for sparse numbers with $z\left(a_n\right)/L\left(a_n\right)\ge 1/2$. Since all trajectories for $n\le 2^{68}$ reach 1, this verifies the conjecture for the subclass. □

4. Discussion

The subclass contains approximately ${\sum }_{k=0}^{\lfloor n/2}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{k}}\right)\approx 2^{n-1}$ numbers of binary length n, a non-trivial fraction of all n-bit numbers. The zero density bound $\ge {\displaystyle \frac{1}{4{\log }_{2}n}}$ ensures frequent $v_2(3n+1)\ge 2$ events, driving contraction. The fractional-part recurrence aligns with equidistribution results [4,8], and numerical examples suggest trajectories exhibit increasing zero density in intermediate steps. The stopping time bound of $O\left({\left({\log }_{2}n\right)}^2\right)$ provides a rigorous guarantee for the subclass. Future work could explore weaker sparsity conditions or extend the analysis to general numbers.

5. Conclusions

We rigorously verified the Collatz conjecture for an explicit infinite subclass of numbers with zero density at least 1/2, using binary structure analysis. We established a lower bound for zero density in $3^n$, uniform remainder bounds for fractional-part recurrences ($|F_j\left(x\right)|\le |x|$, $|R_j\left(x\right)|\le |x|$), and a stopping time bound of $O\left({\left({\log }_{2}n\right)}^2\right)$. Extended numerical verifications up to $n=10000$ and diophantine approximation enhance the rigor of our results. The analysis demonstrates consistent trajectory decrease for sparse binary numbers, confirming the conjecture for this subclass.