Collatz Conjecture

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Collatz Conjecture

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Asset Durmagambetov^*

,Aniyar Durmagambetova

Asset Durmagambetov^*

,Aniyar Durmagambetova

This version is not peer-reviewed

This preprints belongs to the Topic

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

Submitted:

21 January 2024

Posted:

22 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

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

\begin{matrix} M & \in N, \\ α_{j}] - [α_{j + 1}] & = δ_{j} > 0, \\ ϵ_{1} & < 0.65, \\ | F (x) | & < | x |, \\ 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} . \end{matrix}

Then, for

δ_{j} = 1

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

(1)

and for

δ_{j} > 1

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

(2)

Proof.

Consider

\begin{matrix} M - M & = 0 = \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}]} + 2^{[α_{j + 1}] - [α_{j}] + [α_{j}] + ϵ_{j + 1}} . \end{matrix}

Next, we move to the functional relations between

σ_{j}

and

σ_{j + 1}

\begin{matrix} 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) . \end{matrix}

Calculating for

δ_{j} = 1

, we get:

\begin{matrix} ln (2^{- δ_{j} + 1 - σ_{j + 1}} + 1) |_{δ_{j} = 1} & = ln (2^{- σ_{j + 1}} + 1) \\ = ln (1 - \frac{σ_{j + 1} ln 2}{2} + \frac{σ_{j + 1}^{2} {ln}^{2} 2}{4} + F (\frac{σ_{j + 1}^{2}}{4})) . \end{matrix}

Continuing the calculations for

δ_{j} > 1

, we obtain:

\begin{matrix} ln (2^{- δ_{j} + 1 - σ_{j + 1}} + 1) & = ln (1 + 2^{- δ_{j} + 1} - 2^{- δ_{j} + 1} \frac{σ_{j + 1} ln 2}{2} + F (σ_{j + 1}^{2} + 2^{- δ_{j} + 1})) \\ = 2^{- δ_{j} + 1} - 2^{- δ_{j} + 1} \frac{σ_{j + 1} ln 2}{2} + F (σ_{j + 1}^{2} + 2^{- δ_{j} + 1}) . \end{matrix}

Thus, we arrive at the final formulas. □

Theorem 2.

Let

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

1 - {α} > 0.67, n^{*} = [n \frac{ln (3)}{ln (2)}],

(3)

then

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

Proof.

Let

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 for such cases we have

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

Repeating j times, we obtain

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 F (ln 2 \frac{σ_{k + 1}^{2}}{4}) .

By Theorem 1 and the conditions of this theorem, we get

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

From the smallness of

α

and the last estimate,

\Rightarrow \exists δ_{j} > 1 & σ_{j} > 1 / 2 .

Now, let’s consider the number of terms where

δ_{j} > 1

. Assume our zero-generation process stopped within d steps, i.e.,

δ_{j + 1} = δ_{j + 2} = \dots = δ_{j + d} = 1, δ_{j + d + 1} > 1 .

Then applying formula (2) we get

σ_{j} = 2^{- d} σ_{j + d} + \sum_{k = 0}^{d} 2^{- 1 - k} \cdot F (ln 2 \frac{σ_{j + 1 + k}^{2}}{4}),

1 / 2 \leq σ_{j} \leq 2^{- d} σ_{j + d} + \sum_{k = 0}^{d} 2^{- 1 - k} \cdot F (ln 2 \frac{σ_{j + 1 + k}^{2}}{4}) .

This implies that

d \leq 3

. According to formula (1) and Theorem (1), for each value of d, we have

δ_{j + d} (d) = d

. Assuming the binary decomposition process has stopped and

2 \leq δ_{j + k} = 1, 0 < k < n - j,

then we can apply reasoning about the appearance of a zero and get another zero. Continuing this process, we find that each group of ones is accompanied by at least one zero. We show that in fact, the group of zeros will be no less than the corresponding group of ones. Analyzing the situation of the group of zeros using formulas (1-2), assume we have a group consisting of three ones and

\Rightarrow σ_{j + d} \geq 0.9 \Rightarrow 0.9 \leq σ_{j + d} = 2^{- δ_{j + d} + 1} σ_{j + 1 + d} +

(1 - \frac{2^{- δ_{j + d} + 1}}{ln 2}) + 2^{- δ_{j + d} + 1} F (ln 2 \frac{σ_{j + d}^{2}}{4}) \Rightarrow 4 \leq δ_{j + d} \Rightarrow i > n / 2 .

\Rightarrow The statement of the theorem is true .

□

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}

It is necessary to calculate an estimate for each

2 n

-th member 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 units in its binary representation, then calculate the number of applications of the Z operator by the following formula:

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

and calculate the number of applications of the P operator by the following formula:

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

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

a_{n}

, which is equal to

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^{- n + m} \cdot 3^{m} / 2^{m} \cdot a_{n} \leq 2^{- 2 n + 1} \cdot 3^{m} \cdot a_{n} .

According to the last formula, we see that the growth of each member of the sequence depends on the number of units in the binary representation. Next, we will show that a large number of units on the

2 n

-th step leads to an increase in the number of zeros in the

3 n

-th step for the binary representation according to previous theorems, hence the reduction of subsequent members of the sequence:

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

Repeating the reasoning of Theorem 2, consider the equation

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

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

From the last equation, in order to apply the results of theorem 2, we need

σ_{1} = 1 - {x} > 0.5

. To fulfill the last inequality, consider

m_{j} = m - j,

θ = (a_{n} - 2^{n}) \cdot 2^{- n}

{x} = min_{j \in {0, 1}} \{\frac{(m - j) ln (3)}{ln (2)} + \frac{ln (1 + θ)}{ln 2} + F (\frac{1}{2^{n} ln 2})\},

Consider

p = (m - j) \frac{ln 3}{ln 2} = (2 k + l) 1.5849625007 \dots,

ϵ = 1.5849625007 \dots - 1.5,

we get

p = (2 k + l) (1.5 + ϵ + \frac{ln (1 + θ)}{ln 2}) = 3 k + (2 k + l) \cdot ϵ + \frac{ln (1 + θ)}{ln 2},

{p} = {1.5 \cdot l + (2 k + l) \cdot ϵ + \frac{ln (1 + θ)}{ln 2}} = {1.5 \cdot l + {(2 k + l) \cdot ϵ + \frac{ln (1 + θ)}{ln 2}}},

Choosing

l = 0

, if the inequalities

0 \leq {(2 k) \cdot ϵ + \frac{ln (1 + θ)}{ln 2}} \leq 0.5

are true,

{p} = {2 k \cdot ϵ + \frac{ln (1 + θ)}{ln 2}} = {2 k \cdot ϵ + \frac{ln (1 + θ)}{ln 2}},

Choosing

l = 1

, if the inequalities

0.5 < {2 k \cdot ϵ + \frac{ln (1 + θ)}{ln 2}} < 1

are true,

{p} = {2 k \cdot ϵ + \frac{ln (1 + θ)}{ln 2}} = {0.5 + (2 k + l) \cdot ϵ},

Using

ϵ < 0.1,

we also satisfy the condition

σ_{1} = 1 - {x} > 0.51 .

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

According to theorem 2 we get

\begin{matrix} m^{*} \leq n / 2 + (n - j^{*}) \cdot ln 3 / ln 2 / 2, \end{matrix}

According to our application of the Collatz rules, we have the element

a_{4 n - j^{*}}

, and the order of its binary representation is

n_{2} = n + (n - j^{*}) \cdot ln 3 / ln 2 / 2,

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} \cdot 2^{- 3 n - j^{*}} a_{n} + 3^{m^{*}} \cdot 2^{- 2 n - j^{*}} B_{n} + B_{3 n - j^{*}},

a_{4 n - j^{*}} \leq q_{1} \cdot a_{n},

By definition of

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

we get

q_{1} < 1,

Using

n > 1000

, implies

q_{1} < 1 \Rightarrow a_{4 n - j^{*}} < 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}

the Collatz conjecture is true.

Proof.

The proof follows from Theorems 1-3. □

Proof. Proof follows from theorem 1-3

3. 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

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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