Collatz Conjecture

Preprint

Article

Collatz Conjecture

Altmetrics

Downloads

899

Views

275

Comments

Asset Durmagambetov^*

,Aniyar Durmagambetova

Asset Durmagambetov^*

,Aniyar Durmagambetova

This version is not peer-reviewed

Submitted:

10 January 2024

Posted:

11 January 2024

Read the latest preprint version here

Alerts

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

1. Introduction

We will use the following well-known fact: if, for the members of the Collatz sequence, zeros predominate in their binary representation, then these members will lead to a decrease in the subsequent members according to the Collatz rule. A striking example is when the initial number in the Collatz sequence is equal to

2^{n}

. Let’s write the solution of the equation

n = 2^{x}

in the form

x = x + [x]

and note that the smaller x, the more zeros in the corresponding binary representation for n. Developing this idea, we come to the following steps.

Analysis of the binary representation of simple cases of natural numbers.
Creation of a process for decomposing an arbitrary natural number into powers of two.
Analysis of the proximity of the process to binary decomposition at the completion of decomposition at each stage.
Calculation of the number of zeros in the binary decomposition of a natural number.
Estimation of the Collatz sequence members depending on the number of ones in the binary decomposition.

2. 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

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

α_{j} = [α_{j}] + ϵ_{j}, ϵ_{j} < 1,

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

Then, for

δ_{j} = 1

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

and for

δ_{j} > 1

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

Proof.

0 = M - M = \sum_{i = 1}^{j} 2^{[α_{i}]} + 2^{α_{j + 1}} - [\sum_{i = 1}^{j - 1} 2^{[α_{i}]} + 2^{α_{j}}] = 2^{[α_{j}]} + 2^{α_{j + 1}} - 2^{α_{j}}

2^{α_{j}} = 2^{[α_{j}]} + 2^{α_{j + 1}}, 2^{[α_{j}] + ϵ_{j}} = 2^{[α_{j}]} + 2^{α_{j + 1}} = 2^{[α_{j}]} + 2^{[α_{j + 1}] - [α_{j}] + [α_{j}] + ϵ_{j + 1}}

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

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

Calculating for

δ_{j} = 1

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

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

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

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

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

Continuing the calculation for

δ_{j} > 1

, we get

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

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

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

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

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

□

Theorem 2.

Let

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

1 - {α} > \frac{1}{2}, n^{*} = n \times [\frac{ln (3)}{ln (2)}],

(1)

then

\sum_{γ_{i} = 0} 1 \geq \frac{n^{*}}{2} - 5 .

Proof.

Suppose

3^{n} = 2^{α} \Rightarrow α = \frac{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}}

Assume

δ_{j} = 1

, and in such cases, we have

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

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

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

Repeating j times, we get

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

From Theorem 1 and the conditions of this theorem, we obtain

(1 - α) ln 2 = σ_{1} ln 2 < o (ln 2 \frac{σ_{k + 1}^{2}}{4})

From the smallness of

α

and the last estimate,

\Rightarrow \exists δ_{j} > 1

Now let’s consider the question of the number of

δ_{j} > 1

elements. For this, we consider the general case where there are

δ_{j} > 1

and

δ_{k} = 1

. Let’s introduce the following notations:

for δ_{j} = 1 : α_{j} = 0, β_{j} = \frac{1}{1 - σ_{j + 1} ln 2 / 2}

for δ_{j} > 1 : α_{j} = - 2^{δ_{j} - 1} \frac{ln 2 - 2^{- δ_{j} + 1}}{1 - σ_{j + 1} ln 2 / 2}, β_{j} = \frac{2^{δ_{j} - 1}}{1 - σ_{j + 1} ln 2 / 2}

ln 2 σ_{j + 1} = α_{j} + ln 2 β_{j} σ_{j}

Solving the last equation for

σ_{j}

, we get

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

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

According to the condition

α_{j} \leq 0

, we have

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

Denoting by

T = inf_{k < j} {β_{k}},

we obtain

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

We come to evaluating the most important term:

I_{j} = - \sum_{m = 1}^{j - 1} \frac{α_{j - m}}{\prod_{l = m}^{j - 1} β_{j - l}}

The sum is defined by non-zero terms

α_{j - m}

ξ_{*} \leq ξ_{j - m} = \frac{ln 2 - 2^{- δ_{j - m} + 1}}{1 - σ_{j + 1 - m} ln 2 / 2} \leq ξ^{*}, α_{j - m} = - 2^{δ_{j - m} - 1} ξ_{j - m}

μ_{*} \leq μ_{j} = \frac{1}{1 - σ_{j + 1} ln 2 / 2} \geq μ^{*}, β_{j} = 2^{δ_{j} - 1} μ_{j}

I_{j} = \sum_{m = 1}^{j - 1} \frac{2^{δ_{j - m} - 1} ξ_{j - m}}{\prod_{l = m}^{j - 1} 2^{δ_{j - l} - 1} μ_{j}}, I_{j} = \sum_{m = 1}^{j - 1} \frac{ξ_{j - m}}{\prod_{l = m + 1}^{j - 1} 2^{δ_{j - l} - 1} μ_{j}}

I_{*} (n) = \frac{ξ_{*}}{μ^{*}} \sum_{m = 1}^{n - 1} 2^{A_{m}}, A_{m} = \sum_{l = m + 1}^{n - 1} [- δ_{j - l} + 1], S^{*} (n) = \frac{ξ^{*}}{μ_{*}} \sum_{m = 1}^{n - 1} 2^{A_{m}},

S (n) = 2^{A_{n}} \sum_{m = 1}^{n - 1} 2^{A_{m} - A_{n}}, B_{n} = A_{m} - A_{n}

Consider the cases

B_{m} = i \leq \frac{n}{2}

according to the definition

B_{n}

By definition, each Collatz sequence corresponds to a set

A_{m}

; therefore, by iterating over all the sets, we iterate over all corresponding variants of the Collatz sequence. Consider now all possible variants of reaching the maximum level

B_{m_{*}} = i

, due to the monotonicity of the sequence

B_{m_{*} + k} = i, k > 0

and for each variant, calculate

S (n) = 2^{A_{n}} \sum_{m = 1}^{m_{*}} 2^{B_{m}} + 2^{A_{n}} \sum_{m = m_{*}}^{n} 2^{B_{m}},

S (n) = 2^{- i} \sum_{m = 1}^{m_{*}} 2^{m \frac{i}{m_{*}}} + 2^{i} \sum_{m = m_{*}}^{n} 2^{i},

J (n) = 2^{- i} \sum_{m = 1}^{n} 2^{m \frac{i}{m_{*}}} \leq S (n)

J (n) = n \frac{2^{- i} (2^{i} - 1 + O (1))}{i ln 2}

\frac{n}{i ln 2} \leq σ_{1}, \Rightarrow the theorem statement

□

Theorem 3.

Let

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

Then

\exists j^{*} < 0.1 n

and

a_{4 n - j^{*}} < a_{n} .

Proof.

Let’s introduce operators defined by the formulas

P f = \frac{f}{2}, T f = 3 f + 1, Z f = 3 f .

T_{i} \in {P, T}, R_{i} \in {Z, P} .

Consider all possible scenarios of the behavior of the Collatz sequence, which can be written in the following form:

\begin{matrix} a_{n + n} = T_{1} T_{2} \dots T_{n} a_{n} \end{matrix}

We need to estimate each 2n-th term of the Collatz sequence based on the number of P, T, Z operators applied during n steps.

\begin{matrix} a_{n + n} = T_{n} T_{n - 1} \dots T_{1} a_{n} \end{matrix}

Let

a_{n}

have m ones in its binary representation, then we count the number of Z operator applications by the formula:

m = \sum_{R_{i} = Z, i \leq n} 1

and count the number of P operator applications by the formula:

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

Since each application of Z is accompanied by operator P, and the number of P applications corresponds to the number of zeros in

a_{n}

, which equals n - m. According to the rules of Collatz, after n steps we have:

a_{n + n} = \frac{3^{m}}{2^{n}} a_{n} + T_{n} T_{n - 1} \dots T_{1} 1 = \frac{3^{m}}{2^{n}} a_{n} + B_{n} .

B_{n} \leq 2^{- n + m} \sum_{j = 1}^{m} \frac{3^{j}}{2^{j}} a_{n} < 2^{2 - n + m} \frac{3^{m}}{2^{m}} a_{n} \leq 2^{- 2 n + 1} 3^{m} a_{n} .

According to the last formula, we see that the growth of each term of the sequence depends on the number of ones in the binary representation. Next, we will show that a large number of ones at the 2n-th step leads to an increase in the number of zeros at the 3n-th step in the binary representation according to previous theorems, from which it follows that subsequent terms of the sequence decrease:

\begin{matrix} a_{2 n} = 3^{m} a_{n} \times 2^{- n} + B_{n} = 3^{m} + 3^{m} (a_{n} - 2^{n}) + B_{n} \end{matrix}

Repeating the reasoning of Theorem 2, we consider the equation

2^{x} = a_{2 n} = 3^{m} a_{n} \times 2^{- n} + B_{n} = 3^{m} + 3^{m} (a_{n} - 2^{n}) 2^{- n} + B_{n} .

x ln 2 = m ln (3) + ln (1 + (a_{n} - 2^{n}) 2^{- n} + B_{n} \times 3^{- m}) .

From the last equation, to apply the results of Theorem 2, we need

σ_{1} = 1 - {x} > 0.5

. To meet the last inequality, consider

m_{j} = m - j,

{x} = inf_{j < 0.1 \times n} \{\frac{(m - j) ln (3)}{ln (2)} + \frac{ln (1 + θ)}{ln 2} + O (\frac{1}{2^{n} ln 2})\} .

Depending on the behavior of

γ_{n - 1}, γ_{n - 2}, \dots

we can always choose a variant where the fractional part of x meets the conditions of Theorem 2. Denoting

\begin{matrix} m^{*} is the number of non - zero γ_{i}, \\ l^{*} is the number of zero γ_{i}, \end{matrix}

according to Theorem 2 we get

\begin{matrix} m^{*} \leq (2 n - j^{*}) \frac{ln 3}{ln 2} / 2 + 5, \\ l^{*} \geq (2 n - j^{*}) \frac{ln 3}{ln 2} / 2 - 5 . \end{matrix}

After

3 n - j^{*}

steps of applying the Collatz rules, we have

a_{4 n - j^{*}} = \frac{3^{m^{*}}}{2^{2 n - j^{*}}} a_{2 n} + T_{3 n - j^{*}} T_{3 n - 1 - j^{*}} \dots T_{1} 1 = \frac{3^{m^{*}}}{2^{2 n}} a_{2 n} + B_{3 n} .

a_{4 n - j^{*}} = \frac{3^{m^{*}}}{2^{2 n}} a_{2 n} + T_{3 n - j^{*}} T_{3 n - j^{*} - 1} \dots T_{1} 1 = \frac{3^{m^{*}}}{2^{2 n}} (\frac{3^{m}}{2^{n - j^{*}}} a_{n} + B_{n}) + B_{3 n - j^{*}} .

a_{4 n - j^{*}} = 3^{m^{*} + m} 2^{- 3 n - j^{*}} a_{n} + 3^{m^{*}} 2^{- 2 n - j^{*}} B_{n} + B_{3 n - j^{*}} .

B_{3 n - j^{*}} = 3^{m^{*}} 2^{- 2 n} - j^{*} B_{n} = 3^{m^{*} + m} 2^{- 2 n - n - j^{*}} a_{n} .

a_{4 n - j^{*}} \leq q_{1} \times a_{n} .

By definition of

m^{*}, l^{*}, B_{n}

, we get

q_{1} < 1 .

Using

n > 1000 \Rightarrow q_{1} < 1 \Rightarrow a_{4 n} < a_{n}

. □

Theorem 4.

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 1-3

6. Conclusions

Our assertion proves that after

3 n

steps, a sequence with an initial binary length of n arrives at a number strictly smaller than the initial one, from which the solution to the Collatz conjecture follows. This is because by applying this process n times, we are guaranteed to arrive at 1.

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]

Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

© 2024 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

Copyright: This open access article is published under a Creative Commons CC BY 4.0 license, which permit the free download, distribution, and reuse, provided that the author and preprint are cited in any reuse.