Collatz Conjecture

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Collatz Conjecture

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Asset Durmagambetov^*

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Advanced Array Signal Processing for B5G/6G: Models, Algorithms, and Applications

Submitted:

30 December 2023

Posted:

04 January 2024

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Abstract

This paper presents an analysis of the number of zeros in the binary representation of natural numbers. The primary method of analysis involves the use of the concept of the fractional part of a number, which naturally emerges in the determination of binary representation. This idea is grounded in the fundamental property of the Riemann zeta function, constructed using the fractional part of a number. Understanding that the ratio between the fractional and integer parts of a number, analogous to the Riemann zeta function, reflects the profound laws of numbers becomes the key insight of this work. The findings suggest a new perspective on the interrelation between elementary properties of numbers and more complex mathematical concepts, potentially opening new directions in number theory and analysis."

Keywords:

Subject: Computer Science and Mathematics - Analysis

MSC: [2010] 11M26

1. Introduction

The paper analyzes the number of zeros in the binary representation of a natural number. The analysis is carried out using the concept of the fractional part of a number, which naturally arises when finding a binary representation. This idea relies on the fundamental property of the Riemann zeta function, which is constructed using the fractional part of a number. Understanding that the ratio of the fractional and integer parts, by analogy with the Riemann zeta function, expresses the deep laws of numbers, will explain the essence of this work.

2. Materials and Methods

This work is based on the following methods

Analysis of simple cases of natural numbers
A process of expansion of a natural number in powers of two is created.
The proximity to the completion of decomposition is analyzed at each stage
The number of zeros in the binary expansion of a natural number is calculated

3. Results

This document reveals a comprehensive solution to the Collatz Conjecture, as first proposed in [1]. The Collatz Conjecture, a well-known unsolved problem in mathematics, questions whether iterative application of two basic arithmetic operations can invariably convert any positive integer into 1. It deals with integer sequences generated by the following rule: if a term is even, the subsequent term is half of it; if odd, the next term is the previous term tripled plus one. The conjecture posits that all such sequences culminate in 1, regardless of the initial positive integer.

Named after mathematician Lothar Collatz, who introduced the concept in 1937, this conjecture is also known as the 3n + 1 problem, the Ulam conjecture, Kakutani’s problem, the Thwaites conjecture, Hasse’s algorithm, or the Syracuse problem. The sequence is often termed the hailstone sequence due to its fluctuating nature, resembling the movement of hailstones.

Paul Erdős and Jeffrey Lagarias have commented on the complexity and mathematical depth of the Collatz Conjecture, highlighting its challenging nature.

Consider an operation applied to any positive integer:

Divide it by two if it’s even.
Triple it and add one if it’s odd.

This operation is mathematically defined as:

f (n) = \{\begin{matrix} \frac{n}{2}, & if n \equiv 0 mod 2, \\ 3 n + 1, & if n \equiv 1 mod 2 . \end{matrix}

A sequence is formed by continuously applying this operation, starting with any positive integer, where each step’s result becomes the next input. The Collatz Conjecture asserts that this sequence will always reach 1 Recent substantial advancements in addressing the Collatz problem have been documented in works [2].

Now let’s move on to our research, which we will conduct according to the announced plan. For this, we will start with the following

Theorem 1.

Let

x \in N, [α_{j}] - [α_{j + 1}] = δ_{j} > 0, ϵ_{1} < 1 / 2,

α_{j} = [α_{j}] + ϵ_{j}, ϵ_{j} < 1,

x = \sum_{i = 1}^{j - 1} 2^{[α_{i}]} + 2^{α_{j}}, x = \sum_{i = 1}^{j} 2^{[α_{i}]} + 2^{α_{j + 1}}, σ_{j} = 1 - ϵ_{j}

Then

δ_{j} = 1

σ_{j + 1} l n 2 = \frac{2 σ_{j} l n 2}{1 - σ_{j + 1} l n 2} + o (σ_{j + 1}^{2} / 4)

δ_{j} > 1

σ_{j + 1} l n 2 = - 2^{δ_{j} - 1} \frac{l n 2 - 2^{- δ_{j} - 1}}{1 - σ_{j + 1} l n 2 / 2} + 2^{δ_{j} - 1} σ_{j} l n 2 \frac{1}{1 - σ_{j + 1} l n 2 / 2} + o (σ_{j + 1}^{2})

Proof.

2^{ϵ_{j}} = 2^{- δ_{j} + ϵ_{j + 1}} + 1 \Rightarrow 2^{1 - σ_{j}} = 2^{- δ_{j} + 1 - σ_{j + 1}} + 1 \Rightarrow

l n (2^{1 - σ_{j}}) = l n 2 - σ_{j} l n 2 = l n (2^{- δ_{j} + 1 - σ_{j + 1}} + 1)

Computing as

δ_{j} = 1

l n (2^{- δ_{j} + 1 - σ_{j + 1}} + 1) |_{δ_{j} = 1} = l n (2^{- σ_{j + 1}} + 1) = l n ((1 - σ_{j + 1} l n 2 + σ_{j + 1}^{2} l n^{2} / 2) + 1 + o (σ_{j + 1}^{2} / 4))

l n (2 - σ_{j + 1} l n 2 + σ_{j + 1}^{2} l n 2^{2} / 2) = l n 2 + l n (1 - σ_{j + 1} l n 2 / 2 + σ_{j + 1}^{2} l n^{2} / 4 + o (σ_{j + 1}^{2} / 4))

l n (2^{- σ_{j + 1}} + 1) = l n 2 - l n 2 σ_{j + 1} / 2 + l n^{2} 2 σ_{j + 1}^{2} / 2 + o (σ_{j + 1}^{2} / 4)

l n 2 - σ_{j} l n 2 = l n 2 - l n 2 σ_{j + 1} / 2 + l n^{2} 2 σ_{j + 1}^{2} / 4 + o (σ_{j + 1}^{2} / 4)

σ_{j + 1} l n 2 = \frac{2 σ_{j} l n 2}{1 - σ_{j + 1} l n 2 / 2} + o (σ_{j + 1}^{2} / 4)

Repeating computing as

δ_{j} > 1

we get

l n (2^{- δ_{j} + 1 - σ_{j + 1}} + 1) = l n (2^{- δ_{j} + 1} 2^{- σ_{j + 1}} + 1) =

l n (1 + 2^{- δ_{j} + 1} + 2^{- δ_{j} + 1} [- σ_{j + 1} l n 2 + σ_{j + 1}^{2} l n 2^{2} / 2] + o (σ_{j + 1}^{2} / 4 + 2^{- δ_{j} + 1}) =

2^{- δ_{j} + 1} + 2^{- δ_{j} + 1} [- σ_{j + 1} l n 2 + σ_{j + 1}^{2} l n 2^{2} / 2] + o (σ_{j + 1}^{2} + 2^{- δ_{j} + 1}) \Rightarrow

l n 2 - σ_{j} l n 2 = 2^{- δ_{j} + 1} + 2^{- δ_{j} + 1} [- σ_{j + 1} l n 2 + σ_{j + 1}^{2} l n 2^{2} / 2] + o (σ_{j + 1}^{2} + 2^{- δ_{j} + 1})

σ_{j + 1} l n 2 = - 2^{δ_{j} - 1} \frac{l n 2 - 2^{- δ_{j} + 1}}{1 - σ_{j + 1} l n 2 / 2} + 2^{δ_{j} - 1} σ_{j} l n 2 \frac{1}{1 - σ_{j + 1} l n 2 / 2} + o (σ_{j + 1}^{2} + 2^{- δ_{j} + 1})

□

Theorem 2.

Let

x \in N, α_{j} = [α_{j}] x = \sum_{i = 1}^{j - 1} 2^{[α_{i}]} + 2^{α_{j}}

Then the number of zeros in the binary representation

C_{z}

is calculated by the following formula

C_{z} = \sum_{i = 1}^{j - 1} [δ_{i} - 1] + α_{j} - 1

Proof.

C_{z} = \sum_{i = 1}^{j - 1} [α_{i} - α_{i + 1} - 1] + α_{j} - 1

By definition

δ_{i}

C_{z} = \sum_{i = 1}^{j - 1} [δ_{i} - 1] + α_{j} - 1

□

Let’s introduce

μ_{k}, ν_{k}

for

x = \sum_{i = 0}^{n} γ_{i} 2^{i}

by following rules

γ_{k} + γ_{k + 1} = 1, γ_{k + μ_{k}} + γ_{k + μ_{k} + 1} = 1, \prod_{i = k + 1}^{i = μ_{k}} γ_{i} = 1;

γ_{j} + γ_{j + 1} = 1, γ_{j + μ_{j}} + γ_{j + ν_{j} + 1} = 1, ν_{j} = \sum_{i = j + 1}^{i = ν_{j}} (1 - γ_{i})

another words

$μ_{k},$ is count of ones starting at point k with no zeros in between until the first zero or until the end of the sequence
$ν_{j}$ is count of zeros starting at point j with no ones in between until the first zero or until the end of the sequence

Theorem 3.

Let

x = 3^{n} = 2^{[α] + {α}} = \sum_{i = 1}^{n *} γ_{i} 2^{i},

{α} > l n 2, n^{*} = n * [ln (3) / ln (2)]

(1)

then

\sum_{γ_{i} = 0} 1 \geq n^{*} / 2 - 5

Proof.

3^{n} = 2^{α} \Rightarrow α = n / ln (3) / ln (2) \Rightarrow 3^{n} = 2^{[α] + {α}}

Using Theorem 1, we create a sequence

ϵ_{i}, m_{i}, ϵ_{1} = {α}

2^{ϵ_{1}} = \sum_{k = 0}^{i - 1} 2^{[α_{k}] - α_{1}} + 2^{α_{i} - α_{1}}

Suppose

\sum_{γ_{i} = 0} 1 = 0

then by Theorem 1

\Rightarrow σ_{j + 1} l n 2 = \frac{2 σ_{j} l n 2}{1 - σ_{j + 1} l n 2} + o (ln 2 σ_{j + 1}^{2} / 4) \Rightarrow

2^{- 1} σ_{j + 1} l n 2 = \frac{σ_{j} l n 2}{1 - σ_{j + 1} l n 2} + 2^{- 1} * o (ln 2 σ_{j + 1}^{2} / 4)

After repeating j times we get

2^{- j} σ_{j + 1} l n 2 = \frac{σ_{1} l n 2}{\prod_{1}^{j} (1 - σ_{k + 1} l n 2 / 2)} + \sum_{1}^{j} 2^{- k} * o (ln 2 σ_{k + 1}^{2} / 4)

By Theorems (1-2) and condition of the current Theorem proceed

l n 2 / 2 < σ_{1} l n 2 < o (ln 2 σ_{k + 1}^{2} / 4)

immediately

\Rightarrow \sum_{γ_{i} = 0} 1 > 0

Let’s introduce

a s δ_{k} = 1 : α_{k} = 0, β_{k} = \frac{1}{1 - σ_{j + 1} l n 2}

a s δ_{k} > 1 : α_{k} = - 2^{δ_{j} - 1} \frac{l n 2 - 2^{- δ_{j} - 1}}{1 - σ_{j + 1} l n 2 / 2}, β_{k} = \frac{2^{δ_{j} - 1}}{1 - σ_{j + 1} l n 2 / 2}

l n 2 σ_{k + 1} = α_{k} + β_{k} l n 2 σ_{k}

σ_{j + 1} l n 2 = α_{j} + \sum_{m = 1}^{m = j - 1} α_{j - m} \prod_{l = 1}^{l = m} β_{j - l + 1} + \prod_{l = 0}^{l = j - 1} β_{j - l} l n 2 σ_{1} \Rightarrow

\frac{σ_{j + 1} l n 2}{\prod_{l = 0}^{l = j - 1} β_{j - l}} = \sum_{m = 0}^{m = j - 1} \frac{α_{j - m}}{\prod_{l = m + 1}^{l = j - 1} β_{j - l}} \geq σ_{1} l n 2 \Rightarrow

By condition the theorem

\frac{σ_{j + 1} l n 2}{\prod_{l = 0}^{l = n - 1} β_{n - l}} - \sum_{m = 0}^{m = n - 1} \frac{α_{j - m}}{\prod_{l = m}^{l = n - 1} β_{j - l}} \geq σ_{1} l n 2 \Rightarrow

\frac{σ_{n} l n 2}{\prod_{l = 0}^{l = n - 1} β_{n - l}} - \sum_{m = 0}^{m = n - 1} \frac{l n 2 - 2^{- δ_{j} + 1}}{\prod_{l = m + 1}^{l = n - 1} β_{j - l}} \geq σ_{1} l n 2 \Rightarrow

Suppose

\forall i \in (1, n) δ_{i} = 2 \Rightarrow

\frac{σ_{n} l n 2}{\prod_{l = 0}^{l = n - 1} β_{n - l}} + \sum_{m = 0}^{m = n - 1} \frac{l n 2 - 1 / 2}{2^{m}} > σ_{1} l n 2 \Rightarrow

2 (l n 2 - 1 / 2) > σ_{1} \Rightarrow

\exists δ_{j} > 2

In common case

\frac{σ_{n} l n 2}{\prod_{l = 0}^{l = n - 1} β_{n - l}} + \sum_{m = 0}^{m = n - 1} \frac{l n 2 - \frac{2}{2^{δ_{m}}}}{2^{\sum_{m = 0}^{m = n - 1} β_{m}}} > σ_{1} l n 2 \Rightarrow

and we see in case

δ_{m} > 2

With the growth of

δ_{m} > 2

, we see an exponential growth of the denominator, so the influence of

δ_{m} > 2

terms with large values does not have a significant effect on the estimate of the left side of the inequality. The reason is the accumulation of the

β_{m}

value corresponding to

δ_{m} = 1

. Therefore, the effect, taking into account

δ_{m} > 2

with large values, is insignificant for estimating the left part of the inequality. ⇒ statement of Theorem □

Theorem 4. Let

a_{n} = \sum_{i = 0}^{n} γ_{i} 2^{i}, n > 1000, γ_{i} \in {0, 1}

then

a_{8 n} < a_{n}

Proof. In more detail, the estimation process consists of replacing

3^{l}

a_{n + l}

by formula 7 which does not contain powers of the triple which allows one to evaluate the resulting terms of the Syracuse sequence. as a result, we get the following estimate. Let’s introduce operators defined formulas

P f = f / 2, T f = 3 f + 1, Z f = 3 f

Let’s consider all possible scenarios of the behavior of the Syracuse sequence, the same possible scenarios can be written in the following form

\begin{matrix} a_{n + n} = T_{1} T_{2} \dots . T_{n} a_{n} \\ T_{i} \in {P, T}, R_{i} \in {Z, P},, a_{n + n} = R_{1} R_{2} \dots . R_{n} a_{n} + A \end{matrix}

Let’s introduce

m = \sum_{R_{i} = Z} 1

and compute

\sum_{R_{i} = P} 1 = n - m + m = n

By rules of Collatz we have after

2 n

steps

a_{n + n} = 3^{m} / 2^{n} a_{n} + B_{n}

where

\begin{matrix} A_{j} = \sum_{R_{i} = Z, i = 1, j} 1, C_{j} = - \sum_{R_{i} = Z, i = 1, j} 1 - \sum_{R_{i} = P, i = 1, j} 1 \\ B_{n} = \sum_{j = 1, n} 3^{A_{j} 2^{C_{j}}} \\ B_{n} \leq \sum_{j = 1, n} 3^{j} / 2^{j} < 23^{n} / 2^{n} \leq 2 {(3 / 4)}^{n} a_{n} \end{matrix}

\begin{matrix} A = a_{2 n} = 3^{m} (a_{n} * 2^{- n} + B_{n}) = (a_{n} * 2^{- n} + B_{n}) 3^{m} \\ A = \sum_{i = 0}^{[α_{1}]} γ_{i} 2^{i}, γ_{i} \in {0, 1}, α_{1} = m * ln 3 / ln 2 + ln (2^{- n} a_{n}) \end{matrix}

Let

\begin{matrix} m^{*} is count of non zeros of γ_{i} \\ l^{*} is count of zeros of γ_{i} \end{matrix}

by theorem 2 we will have

\begin{matrix} m^{*} \leq [α_{1}] / 2 + 5 = [m ln 3 / ln 2] / 2 + 5 \\ l^{*} \geq [α_{1} / 2 - 5 = [m ln 3 / ln 2] / 2 - 5 \end{matrix}

After

[α_{1}]

steps applying rules of Collatz we have

a_{2 n + [α_{1}]} \leq 3^{m} 3^{α_{1} / 2} 2^{- α_{1}} 2^{5} * 3^{5} (a_{n} * 2^{- n} + B_{n}) = 3^{m} q_{1} * a_{n}

where

q_{1} = 3^{m} 3^{α_{1} / 2} 2^{- α_{1}} 2^{5} * 3^{5}

Repeating the process 3 times and using

n > 1000 \Rightarrow q_{3} < 1 \Rightarrow a_{8 n} < a_{n}

Theorem 5. Let

a_{n} = \sum_{i = 0}^{n} γ_{i} 2^{i}, n > 1000, γ_{i} \in {0, 1}

then for

a_{n}

Collatz conjecture is true

Proof. Proof follows from theorem 1-7

6. Conclusions

Our assertion proves that after

2 n

of steps the sequence comes to a number less than the start one, from which follows the solution of the Collatz conjecture.

References

O’Connor, J.J.; Robertson, E.F. (2006). "Lothar Collatz". St Andrews University School of Mathematics and Statistics, Scotland.
Tao, Terence (2022). "Almost all orbits of the Collatz map attain almost bounded values". Forum of Mathematics, Pi. 10: e12. arXiv:1909.03562. ISSN 2050-5086. [CrossRef]

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Collatz Conjecture

Abstract

1. Introduction

2. Materials and Methods

3. Results

6. Conclusions

References

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