Collatz Conjecture

Asset Durmagambetov

doi:10.20944/preprints202401.0227.v19

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Asset Durmagambetov^*

,Aniyar Durmagambetova

Asset Durmagambetov^*

,Aniyar Durmagambetova

This version is not peer-reviewed.

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Submitted:

04 March 2024

Posted:

05 March 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} & < 0.45, \\ | F_{j} (x) | & < | x |, \\ α_{j} & = [α_{j}] + ϵ_{j}, \\ ϵ_{j} & < 1, \\ σ_{j} & = 1 - ϵ_{j} . \end{matrix}

Then for

δ_{j} = 1

σ_{j} = 2^{- 1} σ_{j + 1} (1 - \frac{σ_{j + 1} ln 2}{2}) + F_{j} (\frac{σ_{j + 1}^{3}}{12}),

(1)

and for

δ_{j} > 1

σ_{j} = 2^{- δ_{j}} σ_{j + 1} + 1 - \frac{2^{- δ_{j}} - 2^{- 2 δ_{j} + 1}}{ln 2} - 2^{- 2 δ_{j}} \frac{σ_{j + 1}^{2} ln 2}{4} + 2^{- 2 δ_{j}} R_{j} (\frac{{ln}^{2} 2 σ_{j + 1}^{3}}{8}) .

(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}

Then, we proceed to 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}

Evaluating 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 2 + ln (1 - \frac{σ_{j + 1} ln 2}{2} + \frac{σ_{j + 1}^{2} {ln}^{2} 2}{4} + F_{j} (\frac{σ_{j + 1}^{3}}{12})) . \end{matrix}

Continuing the computations 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} + 2^{- δ_{j} + 1} F_{j} (σ_{j + 1}^{2} + 2^{- δ_{j} + 1})) \\ = 2^{- δ_{j}} - 2^{- 2 δ_{j} + 1} - 2^{- δ_{j}} \frac{σ_{j + 1} ln 2}{2} + 2^{- 2 δ_{j}} F_{j} (σ_{j + 1}^{2}) . \end{matrix}

Thus, we obtain the final formulas. □

Theorem 2.

Let

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

1 - {α} > 0.55, 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 construct the 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 the binary decomposition process, according to formula (1), stops at the j-th step. It immediately follows that the remaining terms of the decomposition are zeros, and we immediately achieve the truth of the Theorem’s statement. Therefore, we consider the case when the generation of the decomposition according to formula (1) does not stop, and j reaches n. This means that all

σ_{j} > 0, j < n

Let’s conduct a more detailed analysis of the number of zeros and ones in our binary representation. Introduce the following notation:

l- the number of zeros in the binary representation.

m- the number of ones in the binary representation.

n- the binary decomposition bit size, then

n=l+m.

δ_{j} = 1, α_{j} = 0, β_{j} = {((1 - \frac{ln 2 σ_{j + 1}}{2}) / 2 + F_{j} (\frac{σ_{j + 1}^{2}}{12})))}^{- 1}

δ_{j} > 1, α_{j} = - 2^{δ_{j}} (1 - \frac{2^{- δ_{j}} - 2^{- 2 δ_{j} + 1}}{ln 2} + 2^{- δ_{j}} R_{j} (\frac{{ln}^{2} 2 σ_{j + 1}^{3}}{8} + \frac{2^{- 2 δ_{j} + 1}}{ln 2})), β_{j} = 2^{δ_{j}}

To solve the following equations

σ_{j + 1} = α_{j} + β_{j} σ_{j}

we introduce the notation

λ_{m}

- the number of ones after the appearance of

α_{m} > 0

and before the next appearance of zero in the binary decomposition and

γ_{m} = \prod_{k = m}^{m + λ_{m} - 1} β_{k}, α_{m + λ_{m} + 1} > 0

Consider the set

λ_{1}, λ_{2}, . . . . λ_{n}, .

by definition

λ_{1} \geq 1

Define:

m (λ_{*}, i) = inf_{k} {k | λ_{k} > 1 + i}

m (λ^{*}, i) = inf_{k} {k | λ_{k} > 1 + i}

if the set k satisfying the condition is not empty. Let’s perform a series of transformations to understand the next steps.

σ_{n + 1} = α_{n} + β_{1} γ_{1} \frac{α_{1}}{β_{1} γ_{1}} \prod_{k = 0}^{n - 2} γ_{n - k} β_{n - k} + \sum_{m = 1}^{n - 2} γ_{n - m} β_{n - m} \frac{α_{n - m}}{β_{n - m} γ_{n - m}} \prod_{k = 0}^{m - 1} β_{n - k} γ_{n - k} + σ_{1} \prod_{k = 0}^{n - 1} β_{n - k} γ_{n - k}

σ_{n + 1} = α_{n} + \frac{α_{1}}{β_{1} γ_{1}} \prod_{k = 0}^{n - 1} β_{n - k} γ_{n - k} + \sum_{m = 1}^{n - 2} \frac{α_{n - m}}{β_{n - m} γ_{n - m}} \prod_{k = 0}^{m} β_{n - k} γ_{n - k} + σ_{1} \prod_{k = 0}^{n - 1} β_{n - k} γ_{n - k}

(4)

With the consideration of the definition of

m (λ_{*}, i)

, in case of existence

λ_{i} > 1

σ_{n + 1} = α_{n} + \frac{α_{1}}{β_{1} γ_{1}} \prod_{k = 0}^{n - 1} β_{n - k} γ_{n - k} + \sum_{m = 1}^{m (λ_{*}, i) - 1} \frac{α_{n - m}}{β_{n - m} γ_{n - m}} \prod_{k = 0}^{m} β_{n - k} γ_{n - k} +

\sum_{m = m (λ_{*}, i)}^{n - 2} \frac{α_{n - m}}{β_{n - m} γ_{n - m}} \prod_{k = 0}^{m} β_{n - k} γ_{n - k} + σ_{1} \prod_{k = 0}^{n - 1} β_{n - k} γ_{n - k}

σ_{n + 1} = α_{n} + \sum_{m = 1}^{m (λ_{*}, i) - 1} \frac{α_{n - m}}{β_{n - m} γ_{n - m}} \prod_{k = 0}^{m} β_{n - k} γ_{n - k} +

\sum_{m = m (λ_{*}, i)}^{n - 1} \frac{α_{n - m}}{β_{n - m} γ_{n - m}} \prod_{k = 0}^{m} β_{n - k} γ_{n - k} + σ_{1} \prod_{k = 0}^{n - 1} β_{n - k} γ_{n - k}

\sum_{m = m (λ_{*}, i)}^{n - 1} \frac{γ_{n - m (λ_{*}, i)} α_{n - m}}{β_{n - m} γ_{n - m}} \prod_{k = 0}^{m} β_{n - k} γ_{n - k} + σ_{1} γ_{n - m (λ_{*}, i)} \prod_{k = 0}^{m (λ_{*}, i) - 1} β_{n - k} γ_{n - k}

Introduce the notation

α_{*} = inf_{δ_{i}} \frac{| α_{i} |}{β_{i}}, α^{*} = sup_{0 \leq i \leq n} \frac{| α_{i} |}{β_{i}}

β_{*} = inf_{0 \leq i \leq n, δ_{i} = 1} β_{i}, β^{*} = sup_{0 \leq i \leq n, δ_{i} = 1} β_{i}

A (m) = \sum_{k = 1, δ_{j} = 1}^{m} l n_{2} (β_{j}) + \sum_{k = 1, δ_{j} > 1}^{m} l n_{2} (β_{j}) = A_{1} (m) + A_{2} (m)

by definition

α_{i}, γ_{i}

1 < α_{*} < α^{*} < 1.3

2 < β * < β^{*} < 2 / (1 - ln 2 / 2)

Rewrite equation (6) using

i, m (λ_{*}, i)

and assuming that we have only one zero

σ_{1} \leq (\frac{σ_{n}}{{β *}^{i} 2^{A (n - 1)}} - \frac{α^{*}}{{β *}^{i} 2^{A (n - 1)}} \sum_{m = 1}^{m = m (λ_{*}, i)} 2^{A (m)}) +

(\frac{σ_{n}}{{β *}^{i} 2^{A (n - 1)}} - \frac{α^{*}}{({β^{*}}^{i} 2^{A (n - 1)}} \sum_{m = m (λ_{*}, i)}^{m = n - 1} 2^{A (m)})

σ_{1} < 2 \frac{1.3}{β^{i}}

It follows that after zero there cannot be more than three ones. Suppose that between two zeros there are two ones, using Theorem 1 and denoting

x (k) = 2^{- σ_{i + k}}, k \in {1, 2, 3, 4, 5}

we get the system of equations

A x = b

(5)

where

A, A^{- 1}, b

are defined below.

A = [\begin{matrix} 2 & - s & 0 & 0 & 0 \\ 0 & 2 & - 1 & 0 & 0 \\ 0 & 0 & 2 & - 1 & 0 \\ 0 & 0 & 0 & 2 & - t \\ 0 & 0 & 0 & 0 & 2 \end{matrix}]

(6)

b = [\begin{matrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{matrix}]

(7)

s = 2^{- δ_{i}}, t = 2^{- δ_{i + 3}}

A = [\begin{matrix} 1 / 2 & s / 4 & s / 8 & s / 16 & s * t / 32 \\ 0 & 1 / 2 & 1 / 4 & 1 / 8 & t / 16 \\ 0 & 0 & 1 / 2 & 1 / 4 & t / 8 \\ 0 & 0 & 0 & 1 / 2 & t / 4 \\ 0 & 0 & 0 & 0 & 1 / 2 \end{matrix}]

(8)

A^{- 1} = [\begin{matrix} 1 / 2 & s / 4 & s / 8 & s / 16 & s * t / 32 \\ 0 & 1 / 2 & 1 / 4 & 1 / 8 & t / 16 \\ 0 & 0 & 1 / 2 & 1 / 4 & t / 8 \\ 0 & 0 & 0 & 1 / 2 & t / 4 \\ 0 & 0 & 0 & 0 & 1 / 2 \end{matrix}]

(9)

Using Theorem 1 again

2^{1 - σ_{i + 4}} = 2 [\frac{1}{2} + \frac{t (t + 1)}{4}] = 2^{- δ_{i + 4} - σ_{i + 5}} + 1

\frac{t (t + 1)}{2} = 2^{- δ_{i + 4} - σ_{i + 5}}

\frac{2^{- δ_{i + 4}} + 1}{2} = 2^{- σ_{i + 5}}

Continuing the calculations, we obtain

σ_{i + 4} \geq 1 - 2^{- δ_{i + 4} - 1} / ln 2

from Theorem 1 and the last estimate implies

δ_{i + 4} > 2

By considering three units between zeros, we obtain the following matrix:

A = [\begin{matrix} 2 & - 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & - s & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & - 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & - 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & - t \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 \end{matrix}]

The inverse of this matrix is given by:

A^{- 1} = [\begin{matrix} \frac{1}{2} & \frac{1}{4} & \frac{s}{8} & \frac{s}{16} & \frac{s}{32} & \frac{s}{64} & \frac{s t}{128} \\ 0 & \frac{1}{2} & \frac{s}{4} & \frac{s}{8} & \frac{s}{16} & \frac{s}{32} & \frac{s t}{64} \\ 0 & 0 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{t}{32} \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{t}{16} \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{4} & \frac{t}{8} \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{t}{4} \\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} \end{matrix}]

We can also observe that the number of zeros is greater than the number of ones, which implies the statement of the theorem. □

Theorem 3.

Let

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

then

\exists j^{*} < 10, and a_{4 n - j^{*}} < a_{n} .

Proof.

Introduce operators defined as follows:

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 Collatz sequence behaviors that can be written as follows:

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

We need to calculate an estimate for each

2 n

-th term of the Collatz sequence based on the number of applied

P, T, Z

operators within 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 count the number of Z operator applications by the following formula:

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

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

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

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

a_{n}

, which is

n - m

. According to the Collatz rules, 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 sequence member depends on the number of ones in its binary representation. Next, we show that a large number of ones on the

2 n

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

3 n

-th step for the binary representation, according to the previous theorems, which implies a decrease in subsequent sequence members:

\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, to apply the results of theorem 2, we need

σ_{1} > \frac{1}{2 ln 2}

. To satisfy the last inequality, consider

m_{j} = m - j,

θ = (a_{n} - 2^{n}) \cdot 2^{- n}

{x} = min_{j < 10} \{\frac{(m - j) ln (3)}{ln (2)} + \frac{ln (1 + θ)}{ln 2} + F_{j} (\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 from even numbers less than 10, if the inequalities

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

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

Choosing l from odd numbers less than 10, if the inequalities

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

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

Using

ϵ < 0.1,

also satisfies the condition

σ_{1} = 1 - {x} > 0.55

\begin{matrix} m^{*} 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 an 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 the definition of

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

we obtain

q_{1} < 1,

Using

n > 1000

, it follows that

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 holds.

Proof.

The proof follows from Theorems 1-3. □

Conclusion

Our assertion proves that after

3 n - j^{*}

steps, the sequence with an initial binary length of n arrives at a number strictly less than the initial one, thus resolving the Collatz conjecture. Since applying this process n times will inevitably lead us to 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, 10: e12. arXiv:1909.03562. ISSN 2050-5086. [CrossRef]

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Collatz Conjecture

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