Numerical Approach on Collatz Conjecture

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Numerical Approach on Collatz Conjecture

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Youngik Lee^*

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Submitted:

01 May 2024

Posted:

03 May 2024

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Abstract

We can analyze the properties of the Collatz Conjecture by using Terras’ Theorem, and computational graphical calculations. To do this, we introduce the concept of “tree" which shows the relationship between the input and output positive integers of the Collatz process. This approach offers a novel way for determining the non-existence of loops and diverge trees through computer simulations.

Keywords:

Subject: Computer Science and Mathematics - Mathematics

1. Introduction

Collatz conjecture is if one considers the following process on an arbitrary positive integer: [3,4]

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

(1)

In the end, the final value will eventually reach the number 1, regardless of the initial positive integer chosen. For convenience, let us refer to applying Equation (1) on any arbitrary positive integer as applying Collatz process.

2. Theory

For the first step, let us parametrize the Equation (1) with

x = ln (n)

y = ln (f (n))

y = ln (f (n)) = \{\begin{matrix} ln (e^{x} / 2) & if n \equiv 0 (mod 2), \\ ln (3 e^{x} + 1) & if n \equiv 1 (mod 2) . \end{matrix}

(2)

y = \{\begin{matrix} x - ln (2) & if n \equiv 0 (mod 2), \\ x + ln (3) + ln (1 + \frac{1}{3 e^{x}}) & if n \equiv 1 (mod 2) . \end{matrix}

(3)

There is a chain of a Collatz process if we start from an odd number,

o d d (= a) \to e v e n (= b) \to \{\begin{matrix} e v e n (= c) \\ o d d (= c) \end{matrix}

(4)

If we using Equation (3), we can rewrite Equation (4) as,

ln (c) = ln (a) + ln (3 / 2) + ln (1 + \frac{1}{3 a})

(5)

So if we call

ln (a)

x_{i}

ln (c)

x_{i + 1}

then,

x_{i + 1} - x_{i} = ln (3 / 2) + ln (1 + \frac{1}{3 e^{x_{i}}})

(6)

We can apply the exponential function to both sides of Equation (6),

\frac{e^{x_{i + 1}}}{e^{x_{i}}} = (\frac{3}{2} + \frac{1}{2 e^{x_{i}}})

(7)

e^{x_{i + 1}} = \frac{3}{2} e^{x_{i}} + \frac{1}{2}

(8)

If there is a chain of Collatz processes for a list of odd numbers from

n_{i}

n_{j}

, then by using

x_{i} = ln (n_{i})

n_{j} = {(\frac{3}{2})}^{j - i - 1} (n_{i} + 1) - 1

(9)

Let us define the function

ν_{2} (n)

as exponents of 2 after factorization of n.

n = 2^{a} \cdot 3^{b} \cdot . . ., ν_{2} (n) = a

(10)

We can define

m = ν_{2} (n_{i} + 1) = j - i - 1

, and

k = ν_{2} ({(\frac{3}{2})}^{m} (n_{i} + 1) - 1)

. Then, by applying multiple Collatz processes to Equation (9), we can obtain,

C o l_{1} (n_{i}) = n_{j + k} = \frac{{(\frac{3}{2})}^{m} (n_{i} + 1) - 1}{2^{k}}

(11)

Lemma 1.

If we prove the Collatz conjecture for all odd positive integers, it is equivalent to proving it for all positive integers.

Proof.

Let us assume that all odd positive integers converge to 1 after a finite number of steps in the Collatz process. Then, if we have an even number

x_{n} = 2 n \in N

for some n, let

k = ν_{2} (x_{n})

. The number

y_{n} = x_{n} / 2^{k}

is odd. Therefore,

x_{n}

also converges to 1 after k more steps than

y_{n}

. □

According to Lemma 1, to investigate the Collatz conjecture, we only need to examine the convergence properties of odd positive integers.

Definition 1.

The set of odd positive integers

G_{k}

is defined as follows:

G_{k} = {x_{n} | C o {l_{1}}^{k - 1} (x_{n}) = 1, x_{n} = 2 n - 1}

where n and k are integers greater than 1, and

G_{1} = {1}

. The properties of

G_{k}

is:

\forall_{x} \in G_{k + 1}, \exists_{y} \in G_{k}, s a t i s f i e s C o l_{1} (x) = y

G_{k} = {C o l_{1} (x_{n}) | x_{n} \in G_{k + 1}}

for all

k \geq 1

According to Definition 1, we can categorize the set of odd numbers as follows:

Now, let us make a function that maps each odd number to its order number.

o r d (q) = \frac{q + 1}{2}, q \equiv 1 (m o d 2)

(12)

Here q is the odd number, and

o r d (q)

represents its order number. For example,

o r d (5) = 3

, indicating that 5 is the third odd number (1, 3, 5, 7...). Then for

n \in N

we can define the Equation (13) as an update of Equation (11) in odd number order space:

C o l_{2} (n) = o r d (C o l_{1} (2 n - 1)) = \frac{1}{2} (\frac{3 {(\frac{3}{2})}^{m} n - 1}{2^{k}} + 1)

(13)

Here m and k are defined as by using the Equation (10) as:

\begin{matrix} m = ν_{2} (n), k = ν_{2} (3 {(\frac{3}{2})}^{m} n - 1) \end{matrix}

(14)

If we denote n as X, then we can rewrite Equation (13) as:

C o l_{2} (X) = \frac{2^{k} - 1}{2^{k + 1}} + \frac{3}{2^{k + 1}} {(\frac{3}{2})}^{m} X

(15)

Lemma 2.

If we prove the order number,

o r d (q)

, of all odd positive integers converges to 1 after apply finite step of applying

C o l_{2} : N \to N

, it is equivalent to proving Collatz conjecture for all odd positive integers.

Proof.

Let us assume that for all odd positive integers, their order numbers converge to 1 after a finite number of steps of

C o l_{2}

. Now, if we have an odd number

x_{n_{1}} = 2 n_{1} - 1 \in N

for some

n_{1}

, then

o r d (x_{n_{1}}) = n_{1}

. This is because

C o l_{2} (n_{1}) = o r d (C o l_{1} (x_{n_{1}}))

, and we obtain

C o l_{2} (n_{1}) = (C o l_{1} (x_{n_{1}}) + 1) / 2

. Then let us assume that

n_{1}

converge to 1 after k steps of

C o l_{2}

C o l_{2} (n_{1}) \to C o l_{2} (n_{2}) \to C o l_{2} (n_{3}) \to . . . \to 1

Here

n_{k + 1} = C o l_{2} (n_{k}) = (x_{n_{k}} + 1) / 2

. Then we can map this chain as:

\frac{C o l_{1} (x_{n_{1}}) + 1}{2} \to \frac{C o l_{1} (x_{n_{2}}) + 1}{2} \to \frac{C o l_{1} (x_{n_{3}}) + 1}{2} \to . . . \to \frac{C o l_{1} (x_{n_{k}}) + 1}{2} = 1

Multiply each term by 2 and then subtract 1 for all terms.

C o l_{1} (x_{n_{1}}) \to C o l_{1} (x_{n_{2}}) \to C o l_{1} (x_{n_{3}}) \to . . . \to C o l_{1} (x_{n_{k}}) = 1

Here,

x_{n_{k + 1}} = C o l_{1} (x_{n_{k}})

. This implies that

C o l_{1} (x_{n_{k}}) = 1

, which implies the convergence of

x_{n_{1}}

in a finite number of steps of the Collatz process. □

Definition 2.

The set of positive integers

G_{k}

is defined as follows:

G_{k} = {x_{n} | C o {l_{2}}^{k - 1} (x_{n}) = 1, x_{n} \in N}

where n and k are integers greater than 1, and

G_{1} = {1}

. The properties of

G_{k}

is:

\forall_{x} \in G_{k + 1}, \exists_{y} \in G_{k}, s a t i s f i e s C o l_{2} (x) = y

G_{k} = {C o l_{2} (x_{n}) | x_{n} \in G_{k + 1}}

for all

k \geq 1

When we associate Table 1 with Equation (12), we can get Table 2

If we represent Table 2 as a diagram, we can obtain Figure 1.

Let us define this tree-like diagram as a tree with a converging value x.

Definition 3.

If we link two integers with an arrow from the input to the output value using Equation (15) for all integers, it generates the diagram, which we call a “tree", denoted as

(T, <_{T})

. Here,

<_{T}

represents the partial ordering based on Equation (15),

\exists_{k} \in N, C o {l_{2}}^{k} (a) = b \to b <_{T} a

Then tree

(T, <_{T})

has a several properties [5]:

(1): $<_{T}$ is a well-founded partial ordering on T.
(2): $\forall_{t} \in T$ , the set ${u \in T | u <_{T} t}$ is linearly ordered by $<_{T}$ .
(3): $\forall_{t}, u \in T$ , there exists $v \in T$ such that $v \leq_{T} t$ and $v \leq_{T} u$ . Here, $t \leq_{T} t$ essentially means $t <_{T} t$ or $t = t$ .

Definition 4.

For arbitrary tree

(T, <_{T})

\forall_{x} \in T

\exists_{A} \leq_{T} x

, then A is called the “converge value of the tree", and the tree is called “converge tree". However, if there is no such A then the tree is called “diverge tree".

3. Lemmas and Theorems

3.1. Proof of Uniqueness of the Converge Tree

According to Definition 3, when we apply the function

C o l_{2} (x)

to all positive integers, after k iterations, some values converge to 1. We can categorize these values into groups, denoted as

G_{k}

. If we recursively apply

C o l_{2} (x)

multiple times and there exist some values of x that satisfy

C o {l_{2}}^{k} (x) - x = 0

, then those values remain fixed throughout this recursive process.

\forall_{x} \in G_{k}, C o l_{2} (x) = x \to C o {l_{2}}^{k} (x) = x

(16)

Lemma 3.

For some positive integer X, to satisfy

C o l_{2} (X) = X

, the corresponding k values in Equation (15) lie within a length-1 interval,

\frac{ln (3 / 2)}{ln 2} (m + 1) < k < \frac{ln (3 / 2)}{ln 2} (m + 1) + 1

(17)

Proof.

When

C o l_{2} (X) = X

, then X should satisfy Equation (18) according to Equation (15).

X = \frac{2^{k} - 1}{2^{k + 1}} + \frac{3}{2^{k + 1}} {(\frac{3}{2})}^{m} X

(18)

X = \frac{2^{m} (2^{k} - 1)}{2^{k + m + 1} - 3^{m + 1}} = \frac{(2^{k} - 1)}{2 (2^{k} - {(3 / 2)}^{m + 1})}

(19)

Therefore, the condition to make X as a positive integer is:

2^{k} > {(3 / 2)}^{m + 1} \to k > \frac{ln (3 / 2)}{ln 2} (m + 1)

(20)

If we want to make X as an integer, then the numerator of Equation (19) should be divisible by the denominator of Equation (19). Therefore, the numerator must be greater than the denominator.

2^{k} - 1 \geq 2 (2^{k} - {(3 / 2)}^{m + 1})

(21)

k \leq \frac{ln (2 {(3 / 2)}^{m + 1} - 1)}{ln 2} < \frac{ln (2 {(3 / 2)}^{m + 1})}{ln 2}

(22)

k < \frac{ln (3 / 2)}{ln 2} (m + 1) + 1

(23)

And since we know that the integer multiplication of

ln (3 / 2) / ln 2

can not be an integer, k should exist within the range of two transcendental numbers Q and

Q + 1

Q = \frac{ln (3 / 2)}{ln 2} (m + 1) \to Q < k < Q + 1

(24)

So, this proves the lemma. □

Next Equation (19) can be expressed as,

X = \frac{e v e n \cdot o d d}{o d d}

(25)

If we want to make X as an integer, the odd number in the numerator should be divisible by the odd number in the denominator. Therefore the odd number of the numerator must be greater than that of the denominator.

2^{k} - 1 \geq 2^{m + 1} (2^{k} - {(3 / 2)}^{m + 1})

(26)

Therefore, the available values of k should lie within the region:

k \leq \frac{1}{ln 2} ln (\frac{3^{m + 1} - 1}{2^{m + 1} - 1}) = K

(27)

Then, we want to show that

K < R

in Equations (27) and (28).

R = ⌊Q + 1⌋ = ⌊\frac{ln (3 / 2)}{ln 2} (m + 1)⌋ + 1

(28)

To achieve this, first let us utilizing the lemmas.

Lemma 4.

\int_{1}^{\infty} \frac{sin (k A)}{k} d k \leq \sum_{k = 1}^{\infty} \frac{sin (k A)}{k}

(29)

Here,

A = 2 π \frac{ln (3 / 2)}{ln 2} L

Proof.

First, let us define:

\int_{k}^{k + 1} \frac{sin (A t)}{t} d t = Si (A (k + 1)) - Si (A k)

(30)

\sum_{k = 1}^{\infty} \int_{k}^{k + 1} \frac{sin (A t)}{t} d t = Si (\infty) - Si (A)

(31)

By using Taylor Series expansion, [2]

- ln (1 - z) = \sum_{k = 1}^{\infty} \frac{z^{k}}{k}

(32)

Hence, we can rewrite the right-hand side of Equation (29) as:

\begin{matrix} \sum_{k = 1}^{\infty} \frac{sin (A k)}{k} & = & \frac{1}{2 i} \sum_{k = 1}^{\infty} \frac{e^{i A k} - e^{- i A k}}{k} \\ = & \frac{1}{2 i} ln (e^{i (π - A)}) \\ = & \frac{1}{2} ((π - A) - 2 π ⌊\frac{π - A}{2 π}⌋) \end{matrix}

(33)

Therefore we want to show:

\frac{1}{2} ((π - A) - 2 π ⌊\frac{π - A}{2 π}⌋) \geq \frac{π}{2} - Si (A)

(34)

\frac{Si (A)}{π} \geq \frac{A}{2 π} + ⌊\frac{1}{2} - \frac{A}{2 π}⌋

(35)

Here,

A = 2 π \frac{ln (3 / 2)}{ln 2} L = 2 π u

L \in N

l (u) = \frac{Si (2 π u)}{π} \geq u + ⌊\frac{1}{2} - u⌋ = r (u)

(36)

Then, for every u, there exists an x such that:

⌊ x ⌋ = ⌊ u ⌋, \frac{Si (2 π x)}{π} = \frac{1}{2}

(37)

Then, for some

N \in N

, we have

x = N + ϵ

, where

0 < ϵ < 1

. Considering the periodicity properties of

sin (2 π x) / x

, there are two available values of x for each u. Let us define that when

0 < ϵ \leq 1 / 2

, then

ϵ = α

, and if

1 / 2 < ϵ < 1

, then

ϵ = β

We can categorize the region of u into three regions.

(1): $N \leq u \leq N + α$
(2): $N + α \leq u \leq N + \frac{1}{2}$ → $l (u) \geq 1 / 2 > r (u)$
(3): $N + \frac{1}{2} \leq u \leq N + 1$ → $l (u) > 0 \geq r (u)$

In regions (2) and (3), Equation (36) is valid. In the case of region (1), let us calculate the tangential line of

l (u)

N + α

y (u) = \frac{sin (2 π α)}{π (N + α)} (u - N - α) + \frac{1}{2}

(38)

Therefore, when

u = N

, the line passes through the point,

y (N) = \frac{1}{2} - \frac{α sin (2 π α)}{π (N + α)} \geq \frac{1}{2} - \frac{α}{π (N + α)} > \frac{1}{2} - \frac{1}{2 π} > 0

because

0 < α < 1 / 2

. Furthermore, Figure 3 shows that based on computer simulation, when

N \geq 1

y (N) > α

Therefore, in region (1),

l (u) > y (u) > r (u)

. Hence, Equation (36) holds for all u, and Equation (34) also holds. If we define:

s (N, L) = \sum_{k = 1}^{N} (\frac{sin (k A)}{k} - Si (A (k + 1)) + Si (A k))

(39)

then we can conclude that for any

L \in N

lim_{N \to \infty} s (N, L) \geq 0

(40)

We can use computer simulation to double-check that

s (N, L)

converges as we increase the values of N.

Figure 5. The summation of the right-hand side of Equation (39), which converges to positive real values, is calculated for N ranging from 1 to 300, and

L = 6

Figure 5. The summation of the right-hand side of Equation (39), which converges to positive real values, is calculated for N ranging from 1 to 300, and

L = 6

Figure 6. The summation of the right-hand side of Equation (39), which converges to positive real values, is calculated for N ranging from 1 to 300, and

L = 10

Figure 6. The summation of the right-hand side of Equation (39), which converges to positive real values, is calculated for N ranging from 1 to 300, and

L = 10

Based on computer simulation, we can double-check that

s (N, L)

converges as N increases, as shown in Figure 7. Therefore we can get,

\sum_{k = 1}^{\infty} (\frac{sin (k A)}{k} - Si (A (k + 1)) + Si (A k)) \geq 0

(41)

\sum_{k = 1}^{\infty} (Si (A (k + 1)) - Si (A k)) \leq \sum_{k = 1}^{\infty} \frac{sin (k A)}{k}

(42)

So if we utilize Equation (30), (31), (35) and (42),

\sum_{k = 1}^{\infty} \int_{k}^{k + 1} \frac{sin (A t)}{t} d t = \int_{1}^{\infty} \frac{sin (A t)}{t} d t \leq \sum_{k = 1}^{\infty} \frac{sin (k A)}{k}

(43)

which concludes the proof. □

Lemma 5.

\frac{1}{ln 2} ln (\frac{3^{m + 1} - 1}{2^{m + 1} - 1}) < ⌊\frac{ln (3 / 2)}{ln 2} (m + 1)⌋ + 1

(44)

Proof.

Let us set

L = m + 1

, then we can simplify Equation (44) as

\frac{1}{ln 2} ln (\frac{3^{L} - 1}{2^{L} - 1}) < ⌊\frac{ln (3 / 2)}{ln 2} L⌋ + 1

(45)

and if we use the floor function and fractional part function notation

⌊ x ⌋ = x - {x}

[6],

\{\frac{ln (3 / 2)}{ln 2} L\} < 1 - \frac{1}{ln 2} ln (\frac{1 - 1 / 3^{L}}{1 - 1 / 2^{L}})

(46)

Here, we can utilize the Fourier series expansion of the fractional part function [7]:

{x} = \frac{1}{2} - \frac{1}{π} \sum_{k = 1}^{\infty} \frac{sin (2 π k x)}{k}

(47)

Therefore, Equation (46) is:

ln (\frac{1}{\sqrt{2}} \frac{1 - 1 / 3^{L}}{1 - 1 / 2^{L}}) < \frac{ln 2}{π} \sum_{k = 1}^{\infty} \frac{sin (2 π k \frac{ln (3 / 2)}{ln 2} L)}{k}

(48)

By using Lemma 4, Equation (48) holds when Equation (49) is valid (with

A = 2 π \frac{ln (3 / 2)}{ln 2} L

ln (\frac{1}{\sqrt{2}} \frac{1 - 1 / 3^{L}}{1 - 1 / 2^{L}}) < \frac{ln 2}{π} (\int_{1}^{\infty} \frac{sin (k A)}{k} d k)

(49)

ln (\frac{1}{\sqrt{2}} \frac{1 - 1 / 3^{L}}{1 - 1 / 2^{L}}) < \frac{ln 2}{π} (\frac{π}{2} - Si (A))

(50)

ln (\frac{1 - 1 / 3^{L}}{1 - 1 / 2^{L}}) < ln 2 - \frac{ln 2}{π} Si (2 π \frac{ln (3 / 2)}{ln 2} L)

(51)

And we can get the range:

ln (\frac{1 - 1 / 3^{L}}{1 - 1 / 2^{L}}) \leq ln (4 / 3)

(52)

Let us find the region that satisfies:

ln (4 / 3) < ln 2 - \frac{ln 2}{π} Si (2 π \frac{ln (3 / 2)}{ln 2} L)

(53)

Si (2 π \frac{ln (3 / 2)}{ln 2} L) < \frac{ln (3 / 2) π}{ln 2}

(54)

The available range for L is:

0 < L < 0.7757, 0.939 < L

And

L = m + 1 \geq 1

, which satisfying the condition. □

Theorem 1.

There is no positive integer point x that satisfies

C o l_{2} (x) - x = 0

, except for

x = 1

Proof.

According to Lemma 3, 4, and 5, we can conclude that the available region for k is given by:

Q < k < K < R < Q + 1

which implies that X and k in Equation (18) cannot both be integers at the same time. Consequently, no integer value of k exists to satisfy Equation (18). □

Theorem 2.

There is only one converge tree, and all elements in the converge tree converge to 1.

Proof.

Let us assume there is another converge tree with converging values

a \neq 1

. Then, since a is a converging value, we have

C o l_{2} (a) = a

. However, according to Theorem 1 only 1 goes to itself, so

C o l_{2} (a)

is not an integer value other than a. This contradicts the assumption, thus confirming that there is only one converge tree in an integer system.

Therefore, there is only one converge tree, and if any integer is present in the converge tree, it will eventually converge to 1. □

3.2. Proof of the Non-Existence of a Diverge Tree

Theorem 3.

There is no diverge tree.

Proof.

The part of the proof refers to [1] (1976) by R. Terras. Let us call

A_{1}

as the set of all available converge trees.

A_{2}

is the set of all available sets of diverge trees.

Let us assume

A_{2} \neq \emptyset

. Then we can select the minimum element

N \in A_{2}

. For some k, if N is not a power of 2, then there always exists

2^{k} < N < 2^{k + 1}

. Thus, we can express

N = 2^{k} + b

(

b < N

). Since N is the minimum element of

A_{2}

, we know that

\forall_{x} < N, x \in A_{1}

. According to Theorem 2,

\forall_{x} \in A_{1}

converges to 1, and based on Definition 1.5 in [1], this implies that all x have a stop time except when

x = 1

. Here, stop time refers to the point at which, after a finite number of steps of the Collatz process, the value goes smaller than its initial value. For example,

3 \to 8 \to 4 \to 2 \to 1

, the stop time is 3.

According to periodicity Theorem 1.2 in [1], if

x \equiv b

(mod

2^{k}

) then x has the same encoding vector as b.

(1): When $b \neq 1$ , N also has the same stop time as b.
(2): When $b = 1$ , $N = 2^{k} + 1$ . If we apply the Collatz process, $N \to N^{'} = 3 \cdot 2^{k - 2} + 1$ . If $k \geq 2$ , $N^{'} = 3 \cdot 2^{k - 2} + 1 < 2^{k} + 1 = N$ . When $k = 1$ , $N = 3$ which also converges to 1 $< N$ .
(3): When $N = 2^{k}$ , applying the Collatz process results in convergence to 1, which is 1 $< N$ .

Therefore, in all cases, N also has a stop time. However,

N \in A_{2}

, so it should not have a stop time. This leads to a contradiction, so

A_{2} = \emptyset

. □

4. Conclusions and Outlook

Theorem 4.

All positive integers converge to 1 after a finite number of steps of the Collatz process.

Proof.

According to Theorem 2 and 3, in a positive integer system, there is no diverge tree and there is only one converge tree, with the converging value 1. □

In this research, we prove that all positive integers converge to 1 after a finite number of steps of the Collatz process.

As an outlook, we can investigate the continuous generative function of Equation (15) and the periodicity of the encoding vector of the numbers in Table 2. Additionally, there is another way of proving Lemma 5 from Equation (46), which is equivalent to show:

\{L \cdot {log}_{2} (3)\} < 1 - {log}_{2} (\frac{1 - 1 / 3^{L}}{1 - 1 / 2^{L}})

(55)

We know that

{log}_{2} (3)

is a transcendental number with the value, [8,9]

{log}_{2} (3) = 1.584962500721156181 . . .

(56)

Then, with particular values of L, we can obtain:

\begin{matrix} {79335 \cdot {log}_{2} (3)} & = 0.999994712 . . . \\ {190537 \cdot {log}_{2} (3)} & = 0.999999906 . . . \\ {64497107 \cdot {log}_{2} (3)} & = 0.99999998 . . . \end{matrix}

(57)

Thus, to show the validity of Equation (55) is the same as to show how quickly its right-hand side converges to 1 compared to

{L \cdot {log}_{2} (3)}

. Moreover, besides this approach, various other methods such as cycle analysis [10] or bounding functions [11,12] are employed in the investigation of the Collatz conjecture. Exploring potential relations between these methods and the current result could be a relevant next step.

References

Riho Terras, “A stopping time problem on the positive integers", ACTA ARITIMETICA (1976), http://matwbn.icm.edu.pl/ksiazki/aa/aa30/aa3034.pdf. [CrossRef]
Eric Roosendaal, “The Terras Theorem", http://www.ericr.nl/wondrous/terras.html.
Jeffrey C. Lagarias, “The 3x + 1 problem and its generalizations". The American Mathematical Monthly. 92 (1): 3–23, (1985), https://doi.org/10.1080/00029890.1985.11971528. [CrossRef]
Jeffrey C. Lagarias, “The 3x+1 Problem: An Overview", (2021), https://arxiv.org/abs/2111.02635. [CrossRef]
Spencer Unger, “Trees in set Theory", https://www.math.toronto.edu/sunger/cmu/TreesTalk.pdf.
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Figure 1. A diagram expression of Table 2 with directed arrows.

Figure 2. Graphical representation of regions (1) to (3) and functions

l (u)

y (u)

r (u)

. Here, the x-axis represents u when

⌊ u ⌋ = N

Figure 2. Graphical representation of regions (1) to (3) and functions

l (u)

y (u)

r (u)

. Here, the x-axis represents u when

⌊ u ⌋ = N

Figure 3. The results of

y (N) - α

in the region where

0 < α < 1 / 2

. Except for the case

N = 0

, all other graphs yield positive results across all

α

regions.

Figure 3. The results of

y (N) - α

in the region where

0 < α < 1 / 2

. Except for the case

N = 0

, all other graphs yield positive results across all

α

regions.

Figure 4. The summation of the right-hand side of Equation (39), which converges to positive real values, is calculated for N ranging from 1 to 300, and

L = 1

Figure 4. The summation of the right-hand side of Equation (39), which converges to positive real values, is calculated for N ranging from 1 to 300, and

L = 1

Figure 7. The summation of the right-hand side of Equation (39), which converges to positive real values, is calculated for L ranging from 1 to 1000, and

N = 2000

Figure 7. The summation of the right-hand side of Equation (39), which converges to positive real values, is calculated for L ranging from 1 to 1000, and

N = 2000

Table 1. The categorization of odd numbers,

G_{k}

(e.g.,

G_{1}

G_{23}

Table 1. The categorization of odd numbers,

G_{k}

(e.g.,

G_{1}

G_{23}

$G_{1}$	1
$G_{2}$	3, 5, 21, 85, 151, 227, 341...
$G_{3}$	13, 15, 23, 35, 53, 75, 113...
$G_{4}$	7, 11, 17, 61, 69, 93, 141...
$G_{5}$	9, 19, 29, 37, 45, 81, 117...
$G_{6}$	25, 49, 51, 67, 77, 87, 101...
$G_{7}$	33, 39, 43, 59, 65, 79, 89...
$G_{8}$	57, 105, 115, 157, 173, 177, 211...
$G_{9}$	127, 135, 139, 153, 159, 187, 191...
$G_{10}$	123, 169, 185, 247, 249, 283, 339...
$G_{11}$	111, 167, 219, 223, 225, 251, 295, 329...
$G_{12}$	297, 351, 393, 445, 527, 585, 595...
$G_{13}$	175, 263, 395, 519, 593, 615, 699...
$G_{14}$	103, 155, 207, 233, 311, 467, 701...
$G_{15}$	91, 137, 183, 275, 413, 549, 621...
$G_{16}$	63, 95, 121, 143, 215, 243, 323...
$G_{17}$	31, 47, 71, 107, 161, 253, 381...
$G_{18}$	27, 41, 55, 83, 125, 165, 189...
$G_{19}$	73, 109, 147, 195, 199, 221, 291...
$G_{20}$	97, 145, 259, 265, 387, 389, 471...
$G_{21}$	129, 193, 235, 240, 343, 345, 353...
$G_{22}$	171, 240, 257, 313, 415, 427, 457...
$G_{23}$	379, 417, 427, 543, 553, 569, 609...

Table 2. Part of the reparameterizing of Table 1 based on Equation (12) (e.g., from

G_{1}

G_{10}

Table 2. Part of the reparameterizing of Table 1 based on Equation (12) (e.g., from

G_{1}

G_{10}

$G_{1}$	1
$G_{2}$	2, 3, 11, 43, 76, 114, 171...
$G_{3}$	7, 8, 12, 18, 27, 38, 57...
$G_{4}$	4, 6, 9, 31, 35, 47, 72...
$G_{5}$	5, 10, 15, 19, 23, 41, 59...
$G_{6}$	13, 25, 26, 34, 39, 44, 51...
$G_{7}$	17, 20, 22, 30, 33, 40, 45...
$G_{8}$	29, 53, 58, 79, 87, 89, 106...
$G_{9}$	64, 68, 70, 77, 80, 94, 96...
$G_{10}$	62, 85, 93, 124, 125, 142, 170...

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Numerical Approach on Collatz Conjecture

Abstract

1. Introduction

2. Theory

3. Lemmas and Theorems

3.1. Proof of Uniqueness of the Converge Tree

3.2. Proof of the Non-Existence of a Diverge Tree

4. Conclusions and Outlook

References

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