The Collatz Conjecture "The Nonexistence of Divergent Orbits, through a Dynamic Diophantine Linear Analysis of Codification

Giovanny Fuentes

doi:10.20944/preprints202408.0985.v1

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The Collatz Conjecture "The Nonexistence of Divergent Orbits, through a Dynamic Diophantine Linear Analysis of Codification

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Giovanny Fuentes^*

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Submitted:

12 August 2024

Posted:

14 August 2024

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Abstract

We define the function $Col: \mathbb{N} \to \mathbb{N}$ as the Collatz function, given by $3n + 1$ if $n$ is odd and $\displaystyle\frac{n}{2}$ if $n$ is even \cite{Lagarias1990}. The conjecture postulates that for any positive integer, at some point, its iteration will reach 1, or equivalently, every orbit will fall into the periodic cycle $\{4, 2, 1\}$. Two conditions would invalidate the conjecture: the existence of a divergent orbit or the presence of another cycle. In 2019 Terence Tao \cite{tao} proved that almost all orbits converge to the trivial cycle of $\{1,4,2\}$, in this work we will prove the non-existence of divergent orbits. The central idea is to decompose the dynamic system $Col$ into a binary dynamic system, generated by different compositions of two functions. This differs from a typical dynamic system, which is generated by the iteration of a single function. We consider the system generated by the functions $\theta, \psi: \mathbb{R} \to \mathbb{R}$, defined as $\displaystyle\theta(x) = \frac{x}{2}$ and $\displaystyle\psi(x) = \frac{3x + 1}{2}$, denoted as $\langle \theta, \psi \rangle$. We examine sequences of functions in $\langle \theta, \psi \rangle$ of the form $S_k(x) = s_k \circ S_{k-1}(x)$ with $s_k \in \{\theta, \psi\}$. We define a function that assigns to each element of these sequences an integer, called the minimum value positive integer or simply the minimum value. This corresponds to the smallest positive integer $n$ such that $S_k(n)$ is an integer. As we will prove later, the minimum value is monotonically increasing, meaning that increasing the terms of the sequence will never result in a value lower than the previous one. Based on this behavior, we distinguish two types of sequences: stable ones, where the minimum value is constant from a certain point, and unstable ones, where the minimum value is divergent. The main result for unstable sequences is that when the slope of the sequence is divergent, the sequence is unstable. From this result, we can prove that there are no divergent orbits for the function $Col$.

Keywords:

Collatz Conjecture

;

Dynamical System

Subject:

Computer Science and Mathematics - Mathematics

1. Notations and Conventions

In this work, we are going to denote the set of positive integers as

N

, the set of non-negative integers as

N_{0}

and the greatest common divisor of a and b as

(a, b)

and the least common multiple of a and b as

[a, b]

We use the following symbology to refer to an arbitrary composition of functions:

\begin{matrix} ⨀_{i = 1}^{n} f_{i} (x) = f_{n} \circ f_{n - 1} \circ \dots \circ f_{2} \circ f_{1} . \end{matrix}

2. Introduction

2.1. Collatz’s Conjecture

The Collatz Conjecture, also known as the

3 n + 1

Conjecture or the Ulam Problem. Formulated in simple terms, the conjecture proposes that by applying a specific rule to any positive natural number, one will inevitably reach the number 1.

The conjecture is named after Lothar Collatz, a German mathematician who first presented it in 1937. Despite its apparent simplicity, the Collatz conjecture has resisted formal proof for over 80 years. However, its appeal lies in the fascinating tension between its elementary statement and the complexity involved in its demonstration. The Collatz’s Conjecture is formally enunciated as:

An equivalent formulation of the conjecture argues that starting from any positive integer, one will eventually reach cycles

{4, 2, 1}

, and then this cycle will repeat indefinitely. Despite its apparent simplicity, the Collatz conjecture has perplexed mathematicians for decades. The validity of the conjecture has been confirmed for extremely large numbers; however, to date, no one has managed to demonstrate in a general sense that all positive integers adhere to the conjecture. Additionally, no counterexample has been found to invalidate the assertion.

Example 1.

\begin{matrix} 27 & \to 82 \to 41 \to 124 \to 62 \to 31 \to 94 \to 47 \to 142 \to 71 \\ \to 214 \to 107 \to 322 \to 161 \to 484 \to 242 \to 121 \to 364 \to 182 \\ \to 91 \to 274 \to 137 \to 412 \to 206 \to 103 \to 310 \to 155 \to 466 \\ \to 233 \to 700 \to 350 \to 175 \to 526 \to 263 \to 790 \to 395 \to 1186 \\ \to 593 \to 1780 \to 890 \to 445 \to 1336 \to 668 \to 334 \to 167 \to 502 \\ \to 251 \to 754 \to 377 \to 1132 \to 566 \to 283 \to 850 \to 425 \to 1276 \\ \to 638 \to 319 \to 958 \to 479 \to 1438 \to 719 \to 2158 \to 1079 \to 3238 \\ \to 1619 \to 4858 \to 2429 \to 7288 \to 3644 \to 1822 \to 911 \to 2734 \\ \to 1367 \to 4102 \to 2051 \to 6154 \to 3077 \to 9232 \to 4616 \to 2308 \\ \to 1154 \to 577 \to 1732 \to 866 \to 433 \to 1300 \to 650 \to 325 \to 976 \\ \to 488 \to 244 \to 122 \to 61 \to 184 \to 92 \to 46 \to 23 \to 70 \\ \to 35 \to 106 \to 53 \to 160 \to 80 \to 40 \to 20 \to 10 \to 5 \\ \to 16 \to 8 \to 4 \to 2 \to 1 \end{matrix}

Example 2.

\begin{matrix} 871 & \to 2614 \to 1307 \to 3922 \to 1961 \to 5884 \to 2942 \to 1471 \to 4414 \to 2207 \\ \to 6622 \to 3311 \to 9934 \to 4967 \to 14902 \to 7451 \to 22354 \to 11177 \to 33532 \\ \to 16766 \to 8383 \to 25150 \to 12575 \to 37726 \to 18863 \to 56590 \to 28295 \to 84886 \\ \to 42443 \to 127330 \to 63665 \to 190996 \to 95498 \to 47749 \to 143248 \to 71624 \to 35812 \\ \to 17906 \to 8953 \to 26860 \to 13430 \to 6715 \to 20146 \to 10073 \to 30220 \to 15110 \\ \to 7555 \to 22666 \to 11333 \to 34000 \to 17000 \to 8500 \to 4250 \to 2125 \to 6376 \\ \to 3188 \to 1594 \to 797 \to 2392 \to 1196 \to 598 \to 299 \to 898 \to 449 \to 1348 \\ \to 674 \to 337 \to 1012 \to 506 \to 253 \to 760 \to 380 \to 190 \to 95 \to 286 \\ \to 143 \to 430 \to 215 \to 646 \to 323 \to 970 \to 485 \to 1456 \to 728 \to 364 \\ \to 182 \to 91 \to 274 \to 137 \to 412 \to 206 \to 103 \to 310 \to 155 \to 466 \\ \to 233 \to 700 \to 350 \to 175 \to 526 \to 263 \to 790 \to 395 \to 1186 \to 593 \\ \to 1780 \to 890 \to 445 \to 1336 \to 668 \to 334 \to 167 \to 502 \to 251 \to 754 \\ \to 377 \to 1132 \to 566 \to 283 \to 850 \to 425 \to 1276 \to 638 \to 319 \to 958 \\ \to 479 \to 1438 \to 719 \to 2158 \to 1079 \to 3238 \to 1619 \to 4858 \to 2429 \to 7288 \\ \to 3644 \to 1822 \to 911 \to 2734 \to 1367 \to 4102 \to 2051 \to 6154 \to 3077 \to 9232 \\ \to 4616 \to 2308 \to 1154 \to 577 \to 1732 \to 866 \to 433 \to 1300 \to 650 \to 325 \\ \to 976 \to 488 \to 244 \to 122 \to 61 \to 184 \to 92 \to 46 \to 23 \to 70 \\ \to 35 \to 106 \to 53 \to 160 \to 80 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \\ \to 4 \to 2 \to 1 \end{matrix}

2.2. Extension of the Collatz Function

The Collatz function can be naturally extended to the set of integers, as the concept of parity is a concept defined for integers. This concept is not trivially extended to the set of rational numbers, as there is no unique representation, We are going to consider a modification of extension on the rationals proposed by Lagaria in [5]. We will distinguish two subsets of

Q

. The set of rationals with odd denominators, denoted by

Q_{o d d}

, and the set of rationals with even denominators such that the numerator and denominator are co-prime, denoted by

Q_{e v e n}

. We will say that a rational number in

Q_{odd}

is odd if the numerator is odd, and it is even if its numerator is even. In the case of

Q_{even}

, since the denominator is already even and due to coprimality, all elements are odd. We are going to consider the following extension of the Collatz function.

We are going to show that the extension of the Collatz function that we defined is well-defined on

Q_{o d d} .

Proof: We are going to show that

C o l

is well-defined over

Q_{o d d}

. Let

p, q \in Z

with

q \neq 0

and

(q, 2) = (p, q) = 1

. Let

λ

an odd number, then

\begin{matrix} C o l (\frac{λ p}{λ q}) = \{\begin{matrix} \frac{3 λ p + λ q}{λ q} & if λ p odd \\ \frac{λ p}{2 λ q} & if λ p even, \end{matrix} = \{\begin{matrix} \frac{3 p + q}{q} & if p odd \\ \frac{p}{2 q} & if p even, \end{matrix} = C o l (\frac{p}{q}) \end{matrix}

QED.

Let’s observe that when we apply the Collatz function to

\frac{p}{q} \in Q_{odd}

with odd number, we always obtain a fraction with an even numerator, and when applied to

Q_{even}

, we always obtain an odd number. This will be very important since in Section 5, we are going to define how to coding the orbits, assigning 1 if it is odd and 0 if it is even, in the case of

Q_{even}

, we will have that all its elements have the same encoding which is

1111 \dots

unlike

Q_{odd}

, where the codings will be generated by 10 and 0. For this reason we are going to work mainly on

Q_{odd}

, let’s simplify the Collatz function a bit, as

C o l : Q_{odd} \to Q_{odd} \cup Q_{even}

given by

\begin{matrix} C o l (\frac{p}{q}) = \{\begin{matrix} \frac{3 p + q}{2 q} & if p odd \\ \frac{p}{2 q} & if p even, \end{matrix} \end{matrix}

Proof: We will show that

C o l

does not change the parity of the numerator.

if p is odd, we have $3 p + q = 2 r$ with $r \in Z$ , then

$\begin{matrix} \frac{3 p + q}{2 q} = \frac{2 r}{2 q} = \frac{r}{q} \end{matrix}$

Since q is odd, we have independent of the simplification $\frac{r}{q} \in Q_{o d d}$ .
if $p = 2 r$ with $r \in Z$ , we have

$\begin{matrix} \frac{p}{2 q} = \frac{2 r}{2 q} = \frac{r}{q} \end{matrix}$

Since q is odd, we have independent of the simplification, we have $\frac{r}{q} \in Q_{o d d}$ .

QED.

Considering the proposition above, we are going to define the Collatz function on

Q_{o d d}

Example 3.

Let

\frac{5}{7} \in Q_{o d d}

and

\frac{5}{2} \in Q_{e v e n}

we have:

\begin{matrix} O (\frac{5}{7}) : \frac{5}{7} \to \frac{11}{7} \to \frac{20}{7} \to \frac{10}{7} \to \frac{5}{7} \end{matrix}

and

\begin{matrix} O (\frac{5}{2}) : \frac{5}{2} \to \frac{17}{2} \to \frac{53}{2} \to \frac{161}{2} \to \frac{485}{2} \to \dots \to \frac{3^{k} 5 + (3^{k} - 1)}{2} \to \dots \to \infty \end{matrix}

The first objective of the work is to show that there are no divergent orbits in

Q_{o d d}

3. Set Generate by $θ$ and $ψ^{q}$

In this section, we delve into functions generated by the composition of two real linear functions,

θ

and

ψ^{q}

, focusing on their properties over integers. We define the set

〈 θ, ψ^{q} 〉

, representing compositions of these functions, and examine their orbits and associated sets of integers. Before delving into their properties, we introduce the crucial concept of the integer set of a function. Denoted as

E (S)

, this set represents the integers generated by the orbit of the function S. We emphasize the one-to-one correspondence between functions of the same length and the partition of integers into sets based on this length. These results provide a solid foundation for a detailed understanding of the properties of these functions and their application in the study of iterative functions over rational numbers.

3.1. Summary of Propositions in the Section

1: Definition 3: Introduces the set $〈 θ, ψ^{q} 〉$ , generated by two real linear functions $θ$ and $ψ^{q}$ .
2: Definition 4 : Defines the integer set of a function $S \in 〈 θ, ψ^{q} 〉$ as $O_{S} (x) = {x, s_{1} (x), \dots, S (x)}$ , where $s_{i}$ are functions in the composition of S.
3: Definition 5 : Defines the entire set of a function.
4: Lemma 1 Monotony of Integer Set Lemma.
5: Proposition 3 : Establishes a relation of monotony in the entire sets concerning the composition of functions.
6: Lemma 2 Establishes a characterization of the integer sets.
7: Proposition 4 : Establishes a one-to-one correspondence between functions of the same length and integer sets of the same length, and Affirms that the integer sets of functions of the same length are disjoint.
8: Theorem 1: Ensures that the integer sets are the disjoint union of the integer sets of functions in $〈 θ, ψ^{q} 〉$ with the same length.
9: Proposition 5: Guarantees the existence of a unique sequence of elements for a function $S \in 〈 θ, ψ^{q} 〉$ .

3.2. Set Generate by $θ$ and $ψ^{q}$

The Generated Spaces, denoted as

〈 θ, ψ^{q} 〉

. These spaces arise from the iterative composition of functions, where the individual contributions of

θ

and

ψ^{q}

combine to form an enriched dynamic structure.

Now let’s define the S-Orbits. These orbits are ordered sequences that reveal how each element evolves under the iterative action of the functions

θ

and

ψ^{q}

Now let’s define The integer sets of a function, denoted by

E (S)

, represent the integer values that a specific function takes on its domain. Examining

E (S)

allows for the identification of patterns and regularities in the interaction of the function with integers, which is essential for understanding the structure of spaces generated by such functions.

In the following Proposition we are going to see that integer sets have a monotonic behavior concerning the composition of linear functions, this property will be fundamental to studying integer S-orbits.

Proof: In fact, we have

H \circ S (x) = \frac{A B x + B a + 2^{α} b}{2^{α + β}}

Since

(A B, 2) = 1

there are solutions.

Let

x_{0} \in E (H \circ S)

then by definition

\frac{A B x_{0} + B a + 2^{α} b}{2^{α + β}} \in Z

, then we have

\begin{matrix} \frac{A B x_{0} + B a + 2^{α} b}{2^{α + β}} = B (\frac{A x_{0} + a}{2^{α + β}}) + b \in Z \Rightarrow B (\frac{A x_{0} + a}{2^{α + β}}) \in Z \end{matrix}

(B, 2) = 1

then we have

\begin{matrix} \frac{A x_{0} + a}{2^{α + β}} \in Z \Rightarrow \frac{A x_{0} + a}{2^{α}} \in Z \end{matrix}

QED.

Proof: As the functions generated by

〈 θ, ψ^{q} 〉

are linear of the form

\frac{3^{b} x + N}{2^{a}}

with

a, b . N \in N

. The result follows inductively from Lemma 1.

QED.

Example 4.

Let

S_{n} (x) = θ^{2 n} ψ (x) \in 〈 θ, ψ 〉

. We will calculate the integer set of

S_{n}

\begin{matrix} 2^{2 n} y - 3 x = 1 \end{matrix}

we have that

y = 1

and

\frac{2^{n} - 1}{3} \in N

are solutions of the Diophantine Equation, then the integer set is:

\begin{matrix} E (θ^{2 n} ψ) = \frac{2^{n} - 1}{3} + 2^{2 n} Z \end{matrix}

Let

m \geq n

then

\begin{matrix} E (θ^{2 m} ψ) \subset E (θ^{2 n} ψ) \end{matrix}

indeed

\begin{matrix} \frac{3 (\frac{2^{m} - 1}{3}) + 1}{2^{n}} = 2^{m - n} \in N \end{matrix}

The following lemma states that an

S -

orbit is integer if and only if its last value is an integer.

Proof: Let

u \in E (⨀_{i = 1}^{j} s_{i} (x))

then by proposition 3 we have

u \in E (⨀_{i = 1}^{r} s_{i} (x))

with

r < j

, then

⨀_{i = 1}^{r} s_{i} (u) \in Z

. On the other hand, we have

⨀_{i = r + 1}^{j} s_{i} (⨀_{i = 1}^{r} s_{i} (u)) \in Z

then

⨀_{i = 1}^{r} s_{i} (u) \in E (⨀_{i = r + 1}^{j} s_{i} (x))

⨀_{i = 1}^{r} s_{i} (x) \in E (⨀_{i = r + 1}^{j} s_{i} (x))

with

r < j

then

⨀_{i = 1}^{r} s_{i} (u) \in Z

on the other hand

\begin{matrix} ⨀_{i = r + 1}^{j} s_{i} (⨀_{i = 1}^{r} s_{i} (u)) = ⨀_{i = 1}^{j} s_{i} (u) \in Z \end{matrix}

then

u \in E (⨀_{i = 1}^{j} s_{i} (x))

QED.

In the following proposition, We will demonstrate that the integer sets associated with functions of the same length are disjoint. That is, if two integer sets share at least one element, then the functions must be the same. This result plays a crucial role in understanding the structure of integer sets within the context of linear functions and their compositions.

Proof: Let

S (x) = ⨀_{i = 1}^{j} s_{i} (x)

and

H (x) = ⨀_{i = 1}^{j} h_{i} (x)

with

s_{i}, h_{i} \in {θ (x), ψ (x)}

and let

k_{0}

be the largest index such that

s_{j} (x) = h_{j} (x)

for all

j < k_{0}

k_{0} = 1

This means that they have different first terms. Then

E (S) \subset E (θ) = 2 Z

and

E (H) \subset E (ψ^{q}) = 2 Z + 1

or,

E (H) \subset E (θ) = 2 Z

and

E (S) \subset E (ψ^{q}) = 2 Z + 1

in either case we have

E (H) \cap E (S) = \emptyset

Suppose that exist

u \in E (H) \cap E (S)

by proposition 3 we have

\begin{matrix} u \in E (⨀_{i = 1}^{r} s_{i} (x)) \cap E (⨀_{i = 1}^{r} h_{i} (x)) with r \leq j . \end{matrix}

Taken

r = k_{0}

\begin{matrix} u \in E (⨀_{i = 1}^{k_{0}} s_{i} (x)) \cap E (⨀_{i = 1}^{k_{0}} h_{i} (x)) . \end{matrix}

and by lemma 2

\begin{matrix} ⨀_{i = 1}^{k_{0} - 1} s_{i} (u) = ⨀_{i = 1}^{k_{0} - 1} h_{i} (u) \in E (s_{k_{0}} (x)) \cap E (s_{k_{0}} (x)) = 2 Z \cap (2 Z + 1) = \emptyset \end{matrix}

which is a contradiction, On the other hand if

E (H) \cap E (S) = \emptyset

then

S \neq H

otherwise we would have

\begin{matrix} E (S) = E (H) = \emptyset \end{matrix}

however, neither set can be empty

QED.

As a consequence of the above proposition we have

Proof: it is evident that

\begin{matrix} ⨆_{S with length k} E (S) \subset Z \end{matrix}

To prove the other contention we consider the

q -

Collatz function defined by

C o l_{q} : Z \to Z

given by

\begin{matrix} C o l_{q} (u) = \{\begin{matrix} \frac{3 u + q}{2} & if u odd \\ \frac{u}{2} & if u even, \end{matrix} \end{matrix}

let’s take an integer u and calculate its k-th orbit, this orbit can be written as compositions of functions in

〈 θ, ψ^{q} 〉

, let’s call the resulting function S since all the values of the orbit are integers, we have by the lemma 2 we can conclude that u is in the entire set of the function S.

\begin{matrix} u \in E (S) \subset ⨆_{S with length k} E (S) . \end{matrix}

QED.

So far we know that if two functions have the same length, then their integer sets are disjoint, this means that if we take elements of each integer set, the S-orbits of these integers are disjoint sets. Now, if we remove the condition that the lengths are the same, could it be that there is some number such that, given two different functions without being a part of another, it generates the same S-orbit? The answer is in the following proposition.

Proof: Since

S \in 〈 θ, ψ^{q} 〉

, then there exists a sequence of

k_{0}

elements

{s_{j}}_{j = 1}^{k_{0}}

with

s_{j} \in {θ, ψ^{q}}

such that

S (x) = ⨀_{j = 1}^{k_{0}} s_{j} (x)

. Suppose for absurdity, that there is another sequence but of

l_{0}

elements

{h_{j}}_{j = 1}^{l_{0}}

such that

S (x) = ⨀_{j = 1}^{l_{0}} h_{j} (x)

. Let

u \in E (S)

, we have the following cases

If $k_{0} > l_{0}$ . We have

$\begin{matrix} S (x) = ⨀_{j = 1}^{k_{0}} s_{j} (x) = ⨀_{j = 1}^{l_{0}} h_{j} (x) \end{matrix}$

$\begin{matrix} u \in E (⨀_{j = 1}^{k_{0}} s_{j} (x)) = E (⨀_{j = 1}^{l_{0}} h_{j} (x)) \end{matrix}$

by Proposition 3 we have $u \in E (s_{1}) \cap E (h_{1})$ , since $s_{1}, h_{1} \in {θ, ψ^{q}}$ and $E (θ) \cap E (ψ^{q})$ then $s_{1} = h_{1}$ . Since $s_{1}$ and $h_{1}$ are invertible functions, we have

$\begin{matrix} ⨀_{j = 2}^{k_{0}} s_{j} (x) = ⨀_{j = 2}^{l_{0}} h_{j} (x) . \end{matrix}$

Following the same idea up to $k_{0}$ , we have

$\begin{matrix} ⨀_{j = k_{0}}^{l_{0}} s_{j} (x) = x \end{matrix}$

The latter is impossible since the slope of the resulting line is of the form $\frac{3^{b}}{2^{a}}$ with $a, b \in N$ . The case $k_{0} > l_{0}$ is completely analogous, therefore the case where $k_{0}$ and $l_{0}$ are different is not possible.
If $k_{0} = l_{0}$ . Since the sequences are different, there must exist some $k < k_{0}$ such that

$\begin{matrix} ⨀_{j = 1}^{k} s_{j} (x) \neq ⨀_{j = 1}^{k} h_{j} (x) \end{matrix}$

then by Proposition 4 we have

$\begin{matrix} E (⨀_{j = 1}^{k} s_{j} (x)) \cap E (⨀_{j = 1}^{k} h_{j} (x)) = \emptyset \end{matrix}$

However, this is a contradiction to the Proposition 3, because $u \in E (⨀_{j = 1}^{k} s_{j} (x)) \cap E (⨀_{j = 1}^{k} h_{j} (x))$ for all $k \leq k_{0}$ . Then both sequences must be identical.

QED.

4. Stability and Instability of Integer Set

In this section, we delve into the stability and instability of sequences associated with integer sets. We begin by defining functions

ρ_{0}

and

ρ_{1}

that map real functions to integers. We introduce the concepts of positive and negative stability for sequences

{S_{j} (x)}_{j = 1}^{\infty}

. The monotonicity of

ρ_{0}

and

ρ_{1}

is established through Proposition 3, demonstrating the non-decreasing of

ρ_{0}

and the non-increasing of

ρ_{1}

for a given sequence

S_{j} (x)

. Further, the Proposition formally defines positive and negative stability, incorporating limits and intersections of sets. The ensuing Stability Limit Theorem (2) establishes the asymptotic behavior of the integer set of an iterative sequence.

4.1. Summary of Propositions in the Section

1: Definition 6 : Definition of functions $ρ_{0}$ and $ρ_{1}$ .
2: Proposition 6: Monotonicity of the functions $ρ_{0}$ and $ρ_{1}$ .
3: Definition 7: Definition of positively (negatively) stable (unstable) sequences.
4: Theorem 2: Establishes the asymptotic behavior of the integer set when we have a positively (negatively) stable (unstable) sequence.

4.2. Stability and Instability of Integer Set

We initiate this section by introducing functions that associate each integer set with its minimum positive integer value and maximum negative integer value. These values are determined by the solutions closest to zero for the variable x in the Diophantine equation

2^{a} y - 3^{b} x = N

. This equation is representative of the Diophantine equation linked to an element within the space generated by

ψ

and

θ

As a consequence of the Proposition 3. We have that the functions

ρ_{0}

and

ρ_{1}

are monotone.

Proof: By the proposition 3 we have that

\begin{matrix} ρ_{0} (S_{j + 1}), ρ_{1} (S_{j + 1}) \in E (S_{j + 1}) \subset E (S_{j}) \end{matrix}

then

ρ_{0} (S_{j}) \leq ρ_{0} (S_{j + 1})

and

ρ_{1} (S_{j}) \geq ρ_{1} (S_{j + 1})

QED.

From the result of the proposition above, we are going to make a classification of the sequences

S_{j} = ⨀_{i = 1}^{j} s_{i} (x)

according to the behavior of the functions

ρ_{0}

and

ρ_{1}

Now we will give the central theorem of this section, which establishes the asymptotic behavior of the integer sets of

S_{j}

from the stability or stability of this.

Proof: Let

ρ_{0}^{k} = ρ_{0} (⨀_{j = 1}^{k} S_{j} (x))

and

⨀_{j = 1}^{k} S_{j} (x) = \frac{3^{B_{k}} x + N}{2^{A_{k}}}

with

A_{k} =

numbers from

θ

⨀_{j = 1}^{k} S_{j} (x)

and

B_{k} =

numbers from

ψ^{q}

⨀_{j = 1}^{k} S_{j} (x)

. We have

\begin{matrix} E (⨀_{j = 1}^{k} S_{j} (x)) = ρ_{0}^{k} + 2^{A_{k}} N . \end{matrix}

supposed that

{S_{j} (x)}_{j = 1}^{\infty}

is stable, we will first prove that

lim_{k \to \infty} E^{+} (⨀_{j = 1}^{k} S_{j} (x))

is non-empty. By Proposition 3 the sequence of sets

E^{+} (⨀_{j = 1}^{k} S_{j} (x))

is a decreasing sequence of sets i.e. that the next set is a subset of the previous one, then the limit set corresponds to the intersection of all the sets of the sequence.

\begin{matrix} lim_{k \to \infty} E^{+} (⨀_{j = 1}^{k} S_{j} (x)) = ⋂_{k \in N} E^{+} (⨀_{j = 1}^{k} S_{j} (x)) \end{matrix}

Now since

ρ_{0}^{k}

is a function of the natural ones in the natural ones and is convergent, it implies that this function reaches its limit in a finite amount of steps

\begin{matrix} ρ_{0}^{k} = c t s for all k > K \end{matrix}

this implies

\begin{matrix} lim_{k \to \infty} ρ_{0}^{k} \in E^{+} (⨀_{j = 1}^{k} S_{j} (x)) for all k \in N \end{matrix}

then the limit set is non-empty.

Now we will prove the limit set contains a single element. Suppose there exists another element

u_{0}

that is contained in all positive integer sets, then there exists a non-negative integer t such that

\begin{matrix} u_{0} = ρ_{0}^{k} + 2^{A_{k}} t \end{matrix}

without loss of generality, we can assume that

ρ_{0}^{k}

is constant. Solving the equation in terms of t, we have

\begin{matrix} t = \frac{u_{0} - ρ_{0}^{k}}{2^{A_{k}}} \end{matrix}

This solution is a fraction less than 1 for k large enough., which contradicts the fact that t is an integer.

Now let us take the unstable case. Suppose there exists an element

u_{0}

in the limiting set i.e. an element that is contained in all non-negative integer sets, then there exists t a non-negative integer such that

\begin{matrix} u_{0} = ρ_{0}^{k} + 2^{A_{k}} t \geq ρ_{0}^{k} \end{matrix}

ρ_{0}^{k}

diverges and

u_{0}

constant, then there exists a K such that

ρ_{0}^{k}

is greater than

u_{0}

, then

u_{0}

cannot belong to any integer set with

k > K

, which is a contradiction. Analogously for the other case.

QED.

Example 5.

S_{n} (x) = θ^{2 n} ψ (x)

is positively and negatively unstable. Indeed, by example 4, we have

E (θ^{2 n} ψ (x)) = \frac{2^{n} - 1}{3} + 2^{2 n} Z

where

ρ_{0} (θ^{2 n} ψ (x)) = \frac{2^{n} - 1}{3} \to \infty

n \to \infty

then

lim_{n \to \infty} E^{+} (θ^{2 n} ψ (x)) = \emptyset

. On the other hand

ρ_{1} (θ^{2 n} ψ (x)) = \frac{2^{n} - 1}{3} - 2^{2 n} \to - \infty

Example 6.

S_{n} (x) = (θ ψ(x {))}^{n}

is positively unstable and negatively stable. Let’s calculate the integer set

E ((θ ψ(x))^{n})

, let’s observe that

\begin{matrix} θ ψ (- 1) = \frac{3 (- 1) + 1}{2} = - 1 \end{matrix}

Then

E (θ ψ(x {))}^{n} = - 1 + 2^{n} Z

then we have

ρ_{1} ((θ ψ(x))^{n}) = - 1

and

ρ_{0} ((θ ψ(x))^{n}) = 2^{n} - 1 \to \infty

n \to \infty

5. Coding of the Orbits

In this section, we will delve into the study of the coding of the orbits of the Collatz function. The main results of this section are the invariance of the coding between

C o d_{q}

and

C o d

on the fractions with denominator q and the one-to-one identification of each element of

Q_{o d d}

with its coding.

5.1. Summary of Propositions in the Section

Definition 8: Coding maps and $Σ_{2}^{*}$ the space of sequences 0 and 10.
Proposition 7: General form of the elements generated by $ψ$ and $θ$ .
Proposition 8: Fist Cod invariance: $C o d (S) = C o d_{q} (S_{q})$ .
Definition 9: Definition of $C o l_{q} : Z \to Z$ .
Proposition 9: $C o l - C o l_{q}$ equivalence : if $\frac{p}{q} \in Q_{o d d}$ then $C o l^{k} (\frac{p}{q}) = \frac{1}{q} C o l_{q}^{k} (p)$ .
Proposition 10: Second Cod invariance: $C o d^{k} (\frac{p}{q}) = C o d_{q}^{k} (p)$ .
Proposition 11: $p \in E (S)$ if and only if $C o d_{q}^{k} (p) = C o d_{q} (S) .$
Proposition 12 $p \in E (S)$ if and only if $C o d^{k} (\frac{p}{q}) = C o d_{q} (S) .$
Proposition 13: $C o d^{k} (\frac{p}{q}) = C o d^{k} (\frac{p}{q} + \frac{2^{a_{k}} T}{q})$ .
Definition 10: The Coding set $C o d^{k} (ξ)$ .
Proposition 14: Monotony of the coding set $C o d^{k + 1} (ξ) \subset C o d^{k} (ξ)$ .
Proposition 15: Generating property:if $\frac{p}{q}, \frac{t}{r} \in C o d^{k} (ξ)$ then $\frac{p}{q} = \frac{t}{r} + \frac{2^{a_{k}} T}{q r}$ .
Theorem 3: Uniqueness of the full coding $Q_{o d d}$ .

5.2. Coding of the Orbits

It is a common practice in dynamical systems to encode orbits based on specific criteria. In our case, we will encode the orbits of the Collatz function according to the parity of its elements, assigning the value 1 when they are odd and 0 when they are even. Since our primary focus is on the Collatz function over

Q_{o d d}

, we will modify the initial coding by assigning 10 when it is odd, as opposed to just 1. We will denote the space where these encodings reside as

Σ_{2}^{*}

since it is a subset of the sequence space consisting of 0s and 1s, denoted in dynamics as

Σ_{2}

. Formally, we express this as

To rigorously examine the properties of the coding, it is essential to establish a precise form for the elements generated by

ψ

and

θ

Proof: We will prove by induction on k (length of S) for

k = 1

we have

$C o d_{q} (ψ^{q} (x)) = 10$ , then, $b = 1, a_{1} = 0, a = 1$ and $N = 3^{0} 2^{0} = 1$ then $S (x) = \frac{3 x + q}{2} = ψ^{q} (x)$ .
$C o d_{q} (θ (x)) = 0$ , then, $b = 0, a = 1$ and $N = 0$ then $S (x) = \frac{3^{0} x + 0 q}{2} = \frac{x}{2} = θ (x)$ .

Suppose the statement is true up to k, let

S \circ H \in 〈 θ, ψ^{q} 〉

of length

k + 1

with H of length k and let b be the quantity of 1 and a be the quantity of 0 of

C o d_{q} (H)

Claim 1:

C o d_{q} (ψ^{q} \circ H) = C o d_{q} (H) 10

. We have:

the quantity of 1 of $C o d_{q} (H) 10$ is $b + 1$
$a_{b + 1} (C o d_{q} (H) 10) = A$ and $a_{k} (C o d_{q} (H) 10) = a_{k} (C o d_{q} (H))$ for $k \leq b$ and the quantity of 0 of $C o d_{q} (H) 10$ is $A + 1$
$N (C o d_{q} (H) 10) = 3^{b} 2^{a_{1}} + 3^{b - 1} 2^{a_{2}} + \dots + 3^{1} 2^{a_{b - 1}} + 2^{A}$

On the other hand, we have:

\begin{matrix} ψ^{q} (\frac{3^{b} x + q N (C o d_{q} (H))}{2^{A}}) & = \frac{3^{b + 1} x + q 3 N (C o d_{q} (H)) + q 2^{A}}{2^{A + 1}} \\ = \frac{3^{b + 1} x + q 3 (3^{b - 1} 2^{a_{1}} + 3^{b - 2} 2^{a_{2}} + \dots + 2^{a_{b - 1}}) + q 2^{A}}{2^{A + 1}} \\ = \frac{3^{b + 1} x + q (3^{b} 2^{a_{1}} + 3^{b_{k} - 1} 2^{a_{2}} + \dots + 3^{1} 2^{a_{b - 1}} + 2^{A})}{2^{A + 1}} \\ = \frac{3^{b + 1} x + q N (C o d_{q} (H) 10)}{2^{A + 1}} \end{matrix}

where we observe that the values coincide with those calculated.

Claim 2:

C o d_{q} (θ \circ H) = C o d_{q} (H) 0

. We have

the quantity of 1 of $C o d_{q} (H) 0$ is b.
$a_{k} (C o d_{q} (H) 0) = a_{k} (C o d_{q} (H))$ for $k \leq b$
$N (C o d_{q} (H) 0) = 3^{b - 1} 2^{a_{1}} + 3^{b - 1} 2^{a_{2}} + \dots + 2^{a_{b - 1}} = N (C o d_{q} (H))$ .

On the other hand, we have

\begin{matrix} θ (\frac{3^{b} x + q N (C o d_{q} (H))}{2^{A}}) & = \frac{3^{b} x + q N (C o d_{q} (H) 0)}{2^{A + 1}} \end{matrix}

where we observe that the values coincide with those calculated, then the statement is true.

QED.

Now we will see the first property of the coding

Proof: Let

q \in Z

. To prove that they have the same coding, we have to prove that they have the same decomposition in principle, except that where there is

ψ

we have a

ψ^{q}

. let us observe that q has commutative properties with

θ

and

ψ

$q θ (x) = \frac{q x}{2} = θ (q x)$ .
$q ψ (x) = q (\frac{3 x + 1}{2}) = \frac{3 (q x) + q}{2} = ψ^{q} (q x)$

S \in 〈 θ, ψ 〉

then there exists

s_{j} \in {θ, ψ}

such that

\begin{matrix} S (x) = ⨀_{j = 1}^{b} s_{j} (x) . \end{matrix}

For convenience we will denote

s_{j}^{q} \in {θ, ψ^{q}}

. Then we have

\begin{matrix} S_{q} (x) = \frac{3^{b} x + q N}{2^{a}} = q (\frac{3^{b} (\frac{x}{q}) + N}{2^{a}}) = q ⨀_{j = 1}^{b} s_{j} (\frac{x}{q}) = ⨀_{j = 1}^{b} s_{j}^{q} (q \frac{x}{q}) = ⨀_{j = 1}^{b} s_{j}^{q} (x) \end{matrix}

since q does not permute any element

s_{j}

, we have that if

s_{j}

θ

then

s_{j}^{q}

is still

θ

and if

s_{j}

ψ

then

s_{j}^{q}

corresponds to

ψ^{q}

. By Proposition 5 we have that the coding of

S_{q}

has to be the same as that of S.

QED.

Let us contemplate a generalization of the Collatz function applied to integers. In this variant, rather than adding 1, the function adds

q \in Z

, where q is an odd integer. Subsequently, we will establish the compatibility of this generalization with the extension of the Collatz function to

Q_{o d d}

Now, we will demonstrate the compatibility of this generalization

Proof: We let’s observe that

\begin{matrix} C o l (\frac{p}{q}) = \{\begin{matrix} \frac{\frac{p}{q} + 1}{2} & if p = 1 mod 2 \\ \frac{p}{2 q} & if p = 0 mod 2 \end{matrix} = \frac{1}{q} \{\begin{matrix} \frac{3 p + q}{2} & if p = 1 mod 2 \\ \frac{p}{2} & if p = 0 mod 2 \end{matrix} = \frac{1}{q} C o l_{q} (p) \end{matrix}

Suppose first that

(q, 3) = 1

. This fraction is irreducible. Indeed we have that

(3 p + q, q) = (3 p, q) = 1

. Then the parity of the fraction depends only on the numerator since there is no possibility of simplification that changes the parity of the numerator and we can continue with the iteration for all k since the irreducibility of the iterations only depends on the initial fraction is irreducible. Then we have

\begin{matrix} C o l^{k} (\frac{p}{q}) = \frac{1}{q} C o l_{q}^{k} (p) \end{matrix}

Now suppose that

(q, 3) \neq 1

, for this case, the resulting fraction is not irreducible. However, as we are going to prove below, this does not change the parity of the orbits, so the formula would continue to be valid for this case. Suppose that,

q = 3^{l} v

with

(v, 2) = (v, p) = 1

and let

C o l_{q}^{k} (p) = \frac{3^{b} p + 3^{l} v N}{2^{a}}

. We will divide this proof into two parts.

Case one $b \leq l$ : We are going to prove the statement by induction. To

k = 1

\begin{matrix} \frac{1}{3^{l} v} C o l_{3^{l} v} (p) = \frac{1}{3^{l} v} (\frac{3 p + 3^{l} v}{2}) = \frac{3 (\frac{p}{3^{l} v}) + 1}{2} = C o l (\frac{p}{3^{l} v}) \end{matrix}

Now suppose that the statement is true for k, observe before continuing that the expressions

(\frac{3^{b} p + 3^{l} v N}{2^{a}})

and

(\frac{p + 3^{l - b} v N}{2^{a}})

have the same parity. Indeed,

\begin{matrix} \frac{3^{b} p + 3^{l} v N}{2^{a}} = 3^{b} (\frac{p + 3^{l - b} v N}{2^{a}}) \end{matrix}

if the expression on the left-hand side is even, if and only if

(\frac{p + 3^{l - b} v N}{2^{a}})

it is even. On the other hand, if the left side is odd,

(\frac{p + 3^{l - b} v N}{2^{a}})

must be odd and if

(\frac{p + 3^{l - b} v N}{2^{a}})

is odd, since the product of odd is odd, the left side is odd, so the expressions have the same parity.

if $(\frac{3^{b} p + 3^{l} v N}{2^{a}})$ it is odd. Expanding the left-hand side of the proposition,

$\begin{matrix} \frac{1}{3^{l} v} C o l^{k + 1} (p) = \frac{1}{3^{l} v} ψ (\frac{3^{b} p + 3^{l} v N}{2^{a}}) & = \frac{1}{3^{l} v} (\frac{3^{b + 1} p + 3^{l + 1} v N + 2^{a} 3^{l} v}{2^{a + 1}}) \\ = \frac{1}{3^{l - (b + 1)} v} (\frac{p + 3^{l + 1 - (b + 1)} v N + 2^{a} 3^{l - (b + 1)} v}{2^{a + 1}}) \\ = \frac{1}{3^{l - (b + 1)} v} (\frac{p + 3^{l - b} v N + 2^{a} 3^{l - (b + 1)} v}{2^{a + 1}}) \end{matrix}$

developing the right-hand side of the proposition,

$\begin{matrix} C o l^{k + 1} (\frac{p}{3^{l} v}) = C o l (C o l^{k} (\frac{p}{3^{l} v})) & = C o l (\frac{1}{3^{l} v} (\frac{3^{b} p + 3^{l} v N}{2^{a}})) \\ = C o l (\frac{1}{3^{l - b} v} (\frac{p + 3^{l - b} v N}{2^{a}})) \\ = \frac{1}{2} (\frac{3}{3^{l - b} v} (\frac{p + 3^{l - b} v N}{2^{a}}) + 1) \\ = \frac{1}{2} (\frac{1}{3^{l - (b + 1)} v} (\frac{p + 3^{l - b} v N + 2^{a} 3^{l - (b + 1)} v}{2^{a}})) \\ = \frac{1}{3^{l - (b + 1)} v} (\frac{p + 3^{l - b} v N + 2^{a} 3^{l - (b + 1)} v}{2^{a + 1}}) \end{matrix}$

We conclude in this case that both parts are equal
if $(\frac{3^{b} p + 3^{l} v N}{2^{a}})$ it is even. Expanding the left-hand side of the proposition,

$\begin{matrix} \frac{1}{3^{l} v} C o l^{k + 1} (p) = \frac{1}{3^{l} v} θ (\frac{3^{b} p + 3^{l} v N}{2^{a}}) & = \frac{1}{3^{l} v} (\frac{3^{b} p + 3^{l} v N}{2^{a + 1}}) \\ = \frac{1}{3^{l - p} v} (\frac{p + 3^{l - b} v N}{2^{a + 1}}) \end{matrix}$

developing the right-hand side of the proposition,

$\begin{matrix} C o l^{k + 1} (\frac{p}{3^{l} v}) = C o l (C o l^{k} (\frac{p}{3^{l} v})) & = C o l (\frac{1}{3^{l} v} (\frac{3^{b} p + 3^{l} v N}{2^{a}})) \\ = C o l (\frac{1}{3^{l - b} v} (\frac{p + 3^{l - b} v N}{2^{a}})) \\ = \frac{1}{3^{l - b} v} (\frac{p + 3^{l - b} v N}{2^{a + 1}}) \end{matrix}$

We conclude in this case that both parts are equal. Since in both cases it gave equality, we conclude that the proposition is true.

Case two $b \geq l$ : We are going to prove the statement by induction. To

k = 1

\begin{matrix} \frac{1}{3 v} C o l_{3 v} (p) = \frac{1}{3 v} (\frac{3 p + 3 v}{2}) = \frac{3 (\frac{p}{3}) + 1}{2} = C o l (\frac{p}{3 v}) \end{matrix}

Now suppose that the statement is true for k, observe before continuing that the expressions

(\frac{3^{b} p + 3^{l} v N}{2^{a}})

and

(\frac{3^{b - l} p + v N}{2^{a}})

have the same parity. Indeed,

\begin{matrix} \frac{3^{b} p + 3^{l} v N}{2^{a}} = 3^{l} (\frac{3^{b - l} p + v N}{2^{a}}) \end{matrix}

if the expression on the left-hand side is even, if and only if

(\frac{3^{b - l} p + v N}{2^{a}})

it is even. On the other hand, if the left side is odd,

(\frac{3^{b - l} p + v N}{2^{a}})

must be odd and if

(\frac{3^{b - l} p + v N}{2^{a}})

is odd since the product of odd is odd, the left side is odd, so the expressions have the same parity.

if $(\frac{3^{b} p + 3^{l} v N}{2^{a}})$ it is odd. Expanding the left-hand side of the proposition,

$\begin{matrix} \frac{1}{3^{l} v} C o l^{k + 1} (p) = \frac{1}{3^{l} v} ψ (\frac{3^{b} p + 3^{l} v N}{2^{a}}) & = \frac{1}{3^{l} v} (\frac{3^{b + 1} p + 3^{l + 1} v N + 2^{a} 3^{l} v}{2^{a + 1}}) \\ = \frac{3^{b + 1 - l} p + 3 v N + 2^{a} v}{2^{a + 1} v} \end{matrix}$

developing the right-hand side of the proposition,

$\begin{matrix} C o l^{k + 1} (\frac{p}{3^{l} v}) = C o l (C o l^{k} (\frac{p}{3^{l} v})) & = C o l (\frac{1}{3^{l} v} (\frac{3^{b} p + 3^{l} v N}{2^{a}})) \\ = C o l (\frac{3^{b - l} p + v N}{2^{a} v}) \\ = \frac{1}{2} (3 (\frac{3^{b - l} p + v N}{2^{a} v}) + 1) \\ = \frac{3^{b + 1 - l} p + 3 v N + 2^{a} v}{2^{a + 1} v} \end{matrix}$

We conclude in this case that both parts are equal.
if $(\frac{3^{b} p + 3^{l} v N}{2^{a}})$ it is even. Expanding the left-hand side of the proposition,

$\begin{matrix} \frac{1}{3^{l} v} C o l^{k + 1} (p) = \frac{1}{3^{l} v} θ (\frac{3^{b} p + 3^{l} v N}{2^{a}}) & = \frac{1}{3^{l} v} (\frac{3^{b} p + 3^{l} v N}{2^{a + 1}}) \\ = \frac{3^{b - l} p + v N}{2^{a + 1} v} \end{matrix}$

developing the right-hand side of the proposition,

$\begin{matrix} C o l^{k + 1} (\frac{p}{3^{l} v}) = C o l (C o l^{k} (\frac{p}{3^{l} v})) & = C o l (\frac{1}{3^{l} v} (\frac{3^{b} p + 3^{l} v N}{2^{a}})) \\ = C o l (\frac{3^{b - l} p + v N}{2^{a} v}) \\ = \frac{3^{b - l} p + v N}{2^{a + 1} v} \end{matrix}$

We conclude in this case that both parts are equal. Since in both cases it gave equality, we conclude that the proposition is true.

QED.

We will define a coding function for the Collatz q-functions and demonstrate that they produce the same coding as the fractions with denominator q.

Proof: By proposition 9 we have

\begin{matrix} C o l^{k} (\frac{p}{q}) = \frac{1}{q} C o l_{q}^{k} (p) \end{matrix}

Since q it is odd, then, we have coding of

C o l^{k} (\frac{p}{q})

and

C o l_{q}^{k} (p)

must be the same.

QED.

We will now establish the initial connection between sets of integers and coding. Specifically, we will demonstrate that all elements within the integer set S share the same coding.

Proof: Let

p \in E (S)

and

S (x) = ⨀_{i = 1}^{k} s_{i} (x) \in 〈 θ, ψ^{q} 〉

then by definition

S (p) \in Z

by Proposition 3 we have

⨀_{i = 1}^{l} s_{i} (p) \in Z

with

l \leq k

, then

C o d (S) = C o d_{q}^{k} (p)

Suppose that

C o d_{q}^{k} (p) = C o d_{q} (S)

then

{p, s_{1} (p), s_{2} \circ s_{1} (p), \dots, ⨀_{i = 1}^{k - 1} s_{i} (p), ⨀_{i = 1}^{k} s_{i} (p) = S (p)} \in Z^{k + 1}

, then

p \in E (S)

QED.

We show below the second connection between the integer sets and the encoding. Specifically, we demonstrate that all values p within the integer set

S_{q}

indeed have the same coding as the corresponding fraction

\frac{p}{q}

Proof: Let

S (x) \in 〈 θ, ψ^{q} 〉

of length k such that

C o d^{k} (p) = C o d_{q} (S)

, for the proposition 9, we have

\begin{matrix} q C o l^{k} (\frac{p}{q}) = C o l_{q}^{k} (p) = S (p) \end{matrix}

then

C o d^{k} (\frac{n}{q}) = C o d_{q} (S)

finally by the proposition 11, we have

p \in E (S)

if and only if

C o d^{k} (\frac{p}{q}) = C o d_{q} (S)

QED.

The following proposition demonstrates that for a given rational number, we can generate a family of rationals that share the same encoding. This suggests that there exist many rationals with the same k-th encoding

Proof: Let

S (x) \in 〈 θ, ψ^{q} 〉

such that

C o d (S) = C o d_{q}^{k} (p)

then

S (p) = C o l_{q}^{k} (p) \in Z

this implies

\begin{matrix} E (S) = p + 2^{A_{k}} T \Rightarrow C o d_{q}^{k} (p + 2^{A_{k}} T) = C o d_{q}^{k} (p) \end{matrix}

then

\begin{matrix} C o l^{k} (\frac{p + 2^{A_{k}} T}{q}) = C o l^{k} (\frac{p}{q} + \frac{2^{A_{k}} T}{q}) = C o d^{k} (\frac{p}{q}) \end{matrix}

QED.

As we have seen so far, we can characterize the entire set S from its encoding. Exploiting this property, we generalize the entire set S to encompass all fractions sharing the same encoding. We will call the

k -

Coding set.

The encoding set also exhibits the property of monotonicity, similar to the integer set of S.

Proof: Let

\frac{p}{q} \in C o d^{k + 1} {ξ_{j}}_{j = 0}^{\infty}

by definition

C o d^{k + 1} (\frac{p}{q}) = {ξ_{j}}_{j = 1}^{k + 1}

then trivially we have

C o d^{k} (\frac{p}{q}) = {ξ_{j}}_{j = 0}^{k}

, then

\frac{p}{q} \in C o d^{k} {ξ_{j}}_{j = 0}^{\infty}

QED.

Similarly, the behavior of the solutions of Diophantine equations, in which knowing a particular solution allows us to determine other solutions, is reflected in the coding set. This connection is illustrated in the following proposition.

Proof: Let

\frac{p}{q}, \frac{t}{r} \in C o d^{k} {ξ_{j}}_{j = 0}^{\infty}

and

S_{k} (x) = \frac{3^{b} x + N}{2^{A_{k}}} \in 〈 θ, ψ 〉

such that

C o d (S_{k}) = {ξ_{j}}_{j = 0}^{k}

, now consider

S_{k}^{q} \in 〈 θ, ψ^{q} 〉

and

S_{k}^{r} \in 〈 θ, ψ^{r} 〉

such that

C o d_{q}^{k} (S_{k}^{q}) = C o d_{r}^{k} (S_{k}^{r}) = {ξ_{j}}_{j = 0}^{k}

by proposition 12 we have

S_{k}^{q} (p), S_{k}^{r} (t) \in Z

the latter is equivalent

\begin{matrix} \frac{3^{b} p + q N}{2^{A_{k}}}, \frac{3^{b} t + r N}{2^{A_{k}}} \in Z \end{matrix}

We are going to prove that

p r

and

q t

are elements of

E (S_{k}^{q r})

with

s^{r} (x) = \frac{3^{b} x + q r N}{2^{A_{k}}}

. Indeed,

\begin{matrix} \frac{3^{b} p r + q r N}{2^{A_{k}}} = r (\frac{3^{b} p + q N}{2^{A_{k}}}) \in Z \end{matrix}

and

\begin{matrix} \frac{3^{b} q t + q r N}{2^{A_{k}}} = q (\frac{3^{b} t + r N}{2^{A_{k}}}) \in Z \end{matrix}

then

\begin{matrix} p r = q t + 2^{A_{k}} T with T \in Z \Rightarrow \frac{p}{q} = \frac{t}{r} + \frac{2^{A_{k}} T}{q r} with T \in Z \end{matrix}

QED.

Now, we will present the main theorem of this section, establishing that the encoding of a rational number is unique.

Proof: Let

A_{k} =

numbers from 0 to

{ξ_{j}}_{j = 0}^{k}

. Suppose there is another element,

\frac{t}{r} \in Q_{o d d}

such than

C o d (\frac{t}{r}) = lim_{k \to \infty} C o d^{k} (\frac{p}{q}) = {ξ_{j}}_{j = 0}^{\infty}

by proposition 15

\begin{matrix} \frac{t}{r} = \frac{p}{q} + \frac{2^{A_{k}} T}{r q} with T \in Z and k \in N \end{matrix}

Since

\frac{p}{q} \neq \frac{t}{r}

then

T \neq 0

. On the other hand, since

2^{A_{k}} \to \infty

and since T is an integer, so

| 2^{A_{k}} T |

tending to infinity, which implies

\frac{t}{r}

should be infinite, which is a contradiction.

QED.

6. The $π^{1}$ and $π^{2}$ Functions

In this section, we introduce two fundamental functions,

π^{1}

and

π^{2}

, that will play a crucial role in the subsequent sections. The function

π^{1}

maps each sequence from

Σ_{2}^{*}

to a real series. We denote by

G_{\infty}

the set of sequences

ξ

for which

π^{1} (ξ) < \infty

Additionally, the function

π^{2}

associates each sequence with a function defined on the natural numbers. The set

G_{0}

comprises sequences for which

π^{2} (ξ) (k)

represents a bounded function.

6.1. Summary of Propositions in the Section

Definition 12: We will give the definition of the functions $π^{1}$ and $π^{2}$ .
Definition 13: Definition of Null Tail: sequences with a finite number of 1s.
Definition 14: Sets $G_{0}$ , $G_{\infty}$ and $G_{1}$ .
Lemma 3 Characterization of $G_{0}$ , $G_{\infty}$ and $G_{1}$ through accumulation points of $\frac{a_{k}}{k}$ .
Proposition 16 Characterization of $G_{0}$ and $G_{\infty}$ through functions $π^{1}$ and $π^{2}$ .

6.2. The $π^{1}$ and $π^{2}$ Functions

Let S be a function of the space generated by the functions

ψ

and

θ

with length k. According to proposition 7, we have that the general form of S is given by

\frac{3^{k} x + N_{k}}{2^{a_{k}}}

. The function

π^{1}

precisely represents the quotient of

N_{k}

3^{k}

, and the function

π^{2}

represents the quotient of

N_{k}

2^{a_{k}}

. As we will show later, these functions preserve the encoding information of S. One detail to note is that, unlike the function

π^{1}

, which is a real number depending on its convergence, the function

π^{2}

is not a number but a function. This is because the function

π^{1}

is related to the number that generates a certain encoding, and the function

π^{2}

is related to the orbit that it generates.

The elements of

Σ_{2}^{*}

have two characteristic forms: either they have an infinite sequence of 1s, or they have a finite amount of 1s. Those that have a finite amount of 1s we will refer to as having a null tail (we will formalize this later). We will normalize the representation of the elements of

Σ_{2}^{*}

to precisely define applications. It is important to note that 0 and 1 are symbols, so the expression

0^{0}

denotes the absence of this element. For example,

0^{0} 10^{2} \dots

is equivalent to

10^{2} \dots

We begin by providing a technical definition that we will use for the subsequent results.

Example 7.

We will give some examples of the functions

π^{1}

and

π^{2}

Let $ξ_{0} = 00101001000000 \dots 0000 \dots \in Σ_{2}^{*}$

(a)

$π^{1} (ξ_{0}) = \frac{2^{2}}{3} + \frac{2^{3}}{9} + \frac{2^{5}}{27}$ .

(b)

$π^{2} (ξ_{0}) (k) = \frac{27}{2^{k + 2}} \{\frac{2^{2}}{3} + \frac{2^{3}}{9} + \frac{2^{5}}{27}\}$ for $k \geq 4$ and $lim_{k \to \infty} π^{2} (ξ_{0}) (k) = 0$
Let $ξ_{1} = 100100100 \dots \in Σ_{2}^{*}$ then we have

(a)

$π^{1} (ξ_{1}) = \sum_{j = 1}^{\infty} \frac{2^{2 (j - 1)}}{3^{j}} = \frac{1}{3} \sum_{j = 0}^{\infty} \frac{2^{2 j}}{3^{j}} = \frac{1}{3} \sum_{j = 0}^{\infty} {(\frac{4}{3})}^{j} = \infty .$

(b)

$π^{2} (ξ_{1}) (k) = \frac{3^{k}}{2^{2 k}} \sum_{j = 1}^{k} \frac{2^{2 (j - 1)}}{3^{j}} = \frac{3^{k - 1}}{2^{2 k}} \sum_{j = 0}^{k - 1} \frac{2^{2 j}}{3^{j}}$

$= \frac{3^{k - 1}}{2^{2 k}} \{\frac{\frac{2^{2 k}}{3^{k}} - 1}{\frac{4}{3} - 1}\} = \{1 - {\{\frac{3}{4}\}}^{k}\} .$
Let $ξ_{2} = 10101010 \dots \in Σ_{2}^{*}$ then we have

(a)

$π^{1} (ξ_{2}) = \sum_{j = 1}^{\infty} \frac{2^{j - 1}}{3^{j}} = \frac{1}{3} \sum_{j = 0}^{\infty} \frac{2^{j}}{3^{j}} = \frac{1}{3} \sum_{j = 0}^{\infty} {(\frac{2}{3})}^{j} = 1$ .

(b)

$π^{2} (ξ_{2}) (k) = \frac{3^{k}}{2^{k}} \sum_{j = 1}^{k} \frac{2^{j - 1}}{3^{j}} = \frac{3^{k - 1}}{2^{k}} \sum_{j = 0}^{k - 1} \frac{2^{j}}{3^{j}}$

$= \frac{3^{k - 1}}{2^{k}} \{\frac{\frac{2^{k}}{3^{k}} - 1}{\frac{3}{3} - 1}\} = \{{\{\frac{3}{2}\}}^{k} - 1\} \to \infty .$
Let $ξ_{3} = 10100100010000 \dots \in Σ_{2}^{*}$

(a)

$π^{1} (ξ_{3}) = \sum_{j = 1}^{\infty} \frac{2^{\frac{j (j + 1)}{2} - 1}}{3^{j}} \to \infty$ . Since, $\frac{2^{\frac{j (j + 1)}{2}}}{3^{j}} \to \infty$ .

(b)

$π^{2} (ξ_{3}) (k) = \frac{3^{k}}{2^{\frac{k (k + 1)}{2}}} \sum_{j = 1}^{k} \frac{2^{\frac{j (j + 1)}{2} - 1}}{3^{j}} < 1$
Let $ξ_{4} = 101000101000 \dots \in Σ_{2}^{*}$ then we have

(a)

$π^{1} (ξ_{4}) = \frac{1}{3} + \frac{2}{3^{2}} + \frac{2^{4}}{3^{3}} + \frac{2^{5}}{3^{4}} + \frac{2^{8}}{3^{5}} + \frac{2^{9}}{3^{6}} + \frac{2^{12}}{3^{7}} + \dots = \frac{1}{3} + \{\frac{2}{3^{2}} + \frac{2^{4}}{3^{3}}\} + \frac{2^{4}}{3^{2}} \{\frac{2}{3^{2}} + \frac{2^{4}}{3^{3}}\} + {(\frac{2^{4}}{3^{2}})}^{2} \{\frac{2}{3^{2}} + \frac{2^{4}}{3^{3}}\} + \dots = \frac{1}{3} + \{\frac{2}{3^{2}} + \frac{2^{4}}{3^{3}}\} \sum_{j = 0}^{\infty} {(\frac{2^{4}}{3^{2}})}^{j} \to \infty$ .

(b)

$π^{2} (ξ_{4}) (1) = \frac{1}{2}$ , $π^{2} (ξ_{4}) (2) = \frac{5}{2^{4}}$ , $π^{2} (ξ_{4}) (2 k + 1) = \frac{3^{2 k + 1}}{2^{4 k + 1}} (\frac{1}{3} + \{\frac{2}{3^{2}} + \frac{2^{4}}{3^{3}}\} \sum_{j = 0}^{k - 1} {(\frac{2^{4}}{3^{2}})}^{j})$ and $π^{2} (ξ_{4}) (2 k) = \frac{3^{2 k}}{2^{4 (k + 1)}} (\frac{1}{3} + \{\frac{2}{3^{2}} + \frac{2^{4}}{3^{3}}\} \sum_{j = 0}^{k - 2} {(\frac{2^{4}}{3^{2}})}^{j} + \frac{2^{4 k + 1}}{3^{2 k}})$

Let

ξ \in Σ_{2}^{*}

define the function

\frac{3^{b_{k} (ξ)}}{2^{a_{k} (ξ)}}

. This function corresponds to the slope of the function

S_{k}

such that

C o d^{k} (S_{k}) = {ξ_{j}}_{j = 0}^{k}

We can consider the following set of

Σ_{2}^{*}

The following result establishes a characterization of the elements of

G_{0}

and

G_{\infty}

Proof:

Let $ξ \in G_{0}$ then $\frac{3^{k}}{2^{a_{k - 1}}} \to 0$ as $k \to \infty$ . Suppose that $\underset{k \to \infty}{lim inf} \frac{a_{k}}{k} < \frac{ln (3)}{ln (2)}$ . Then exist $K, ε > 0$ such that $\frac{a_{k_{l}}}{k_{l}} + ε < \frac{ln (3)}{ln (2)}$ for $k > K$ where $lim_{l \to \infty} \frac{a_{k_{l}}}{k_{l}} = \underset{k \to \infty}{lim inf} \frac{a_{k}}{k}$ , then we have,

$\begin{matrix} \frac{3^{k_{l}}}{2^{a_{k_{l}}}} = {(\frac{3}{2^{\frac{a_{k_{l}}}{k_{l}}}})}^{k_{l}} > {(\frac{3}{2^{\frac{ln (3)}{ln (2)} - ε}})}^{k_{l}} = 2^{k_{l} ε} \to \infty as l \to \infty \end{matrix}$

which is a contradiction with the fact that $ξ \in G_{0}$ , then $\underset{k \to \infty}{lim inf} \frac{a_{k}}{k} > \frac{ln (3)}{ln (2)}$ .

Now, let’s suppose that $\underset{k \to \infty}{lim inf} \frac{a_{k}}{k} > \frac{ln (3)}{ln (2)}$ . Then exist $K, ε > 0$ such that $\frac{a_{k}}{k} - ε > \frac{ln (3)}{ln (2)}$ for all $k > K$ . Then

$\begin{matrix} \frac{3^{k}}{2^{a_{k}}} = {(\frac{3}{2^{\frac{a_{k}}{k}}})}^{k} < {(\frac{3}{2^{\frac{ln (3)}{ln (2)} + ε}})}^{k} = 2^{- k ε} \to 0 as k \to \infty \end{matrix}$
Let $ξ \in G_{\infty}$ then $\frac{3^{k}}{2^{a_{k}}} \to \infty$ as $k \to \infty$ . Suppose that $\underset{k \to \infty}{lim sup} \frac{a_{k}}{k} > \frac{ln (3)}{ln (2)}$ . Then exist $K, ε > 0$ such that $\frac{a_{k_{l}}}{k_{l}} - ε > \frac{ln (3)}{ln (2)}$ for $k > K$ where $lim_{l \to \infty} \frac{a_{k_{l}}}{k_{l}} = \underset{k \to \infty}{lim sup} \frac{a_{k}}{k}$ , then we have,

$\begin{matrix} \frac{3^{k_{l}}}{2^{a_{k_{l}}}} = {(\frac{3}{2^{\frac{a_{k_{l}}}{k_{l}}}})}^{k_{l}} < {(\frac{3}{2^{\frac{ln (3)}{ln (2)} + ε}})}^{k_{l}} = 2^{- k_{l} ε} \to 0 as l \to \infty \end{matrix}$

which is a contradiction with the fact that $ξ \in G_{\infty}$ , then $\underset{k \to \infty}{lim sup} \frac{a_{k}}{k} < \frac{ln (3)}{ln (2)}$ .

Now, let’s suppose that $\underset{k \to \infty}{lim sup} \frac{a_{k}}{k} < \frac{ln (3)}{ln (2)}$ , Then exist $K, ε > 0$ such that $\frac{a_{k}}{k} + ε < \frac{ln (3)}{ln (2)}$ for all $k > K$ . Then

$\begin{matrix} \frac{3^{k}}{2^{a_{k}}} = {(\frac{3}{2^{\frac{a_{k}}{k}}})}^{k} > {(\frac{3}{2^{\frac{ln (3)}{ln (2)} - ε}})}^{k} = 2^{k ε} \to \infty as k \to \infty \end{matrix}$

QED.

The following result establishes a characterization of the elements of

G_{0}

and

G_{\infty}

for the behavior of the functions

π

Proof:Proof of the first statement. Is obvious for the case of null tails with index J, since we have

π^{1}

it is automatically finite, and as we see in the examples

π^{2}

would be of the form

\frac{3^{k_{0}}}{2^{k}} A \to 0

when

k \to \infty

which implies that

π^{2}

is finite.

So we are going to assume that

ξ

has no tail null.

Suppose that

ξ \in G_{\infty}

, then by Lemma 3 and

a_{j}

is strictly increasing we have

\underset{j \to \infty}{lim sup} \frac{a_{j - 1}}{j} \leq \underset{j \to \infty}{lim sup} \frac{a_{j}}{j} < \frac{ln (3)}{ln (2)}

, so, there exists

J, ε > 0

such that for all

j > J

we have that

\frac{a_{j - 1}}{j} < \frac{ln (3)}{ln (2)} - ε

\begin{matrix} \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} = \sum_{j = 1}^{J} \frac{2^{a_{j - 1}}}{3^{j}} + \sum_{j = J + 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} = \sum_{j = 1}^{J} \frac{2^{a_{j - 1}}}{3^{j}} + \sum_{j = J + 1}^{\infty} {(\frac{2^{\frac{a_{j - 1}}{j}}}{3})}^{j} \\ < \sum_{j = 1}^{J} \frac{2^{a_{j - 1}}}{3^{j}} + \sum_{j = J + 1}^{\infty} {(\frac{2^{\frac{ln (3)}{ln (2)} - ε}}{3})}^{j} = \sum_{j = 1}^{J} \frac{2^{a_{j - 1}}}{3^{j}} + \sum_{j = J + 1}^{\infty} 2^{- ε j} < \infty \end{matrix}

Let’s suppose

\sum_{j = 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} < \infty

, then

\sum_{j = 0}^{\infty} \frac{2^{a_{j}}}{3^{j}}

is also convergent since

\sum_{j = 0}^{\infty} \frac{2^{a_{j}}}{3^{j}} = \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j - 1}} = 3 \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} < \infty

, so

lim_{j \to \infty} \frac{2^{a_{j}}}{3^{j}} = 0

then

ξ \in G_{\infty} .

QED of the first statement.

Proof of the second statement. Suppose

π^{2} (ξ) (j)

is bounded, we will prove that

\frac{3^{k}}{2^{a_{k} (ξ)}}

converges to 0. Suppose

\frac{3^{k}}{2^{a_{k} (ξ)}} > ε > 0

for

k > K

for any

K \in N

. Then we have

\begin{matrix} π^{2} (ξ) (j) = \frac{N_{j}}{2^{a_{k}}} = \frac{3^{k}}{2^{a_{k}}} \sum_{j = 1}^{k} \frac{2^{a_{j - 1}}}{3^{j}} \end{matrix}

We have that the sum on the right is divergent since

\frac{2^{a_{k}}}{3^{k}}

is divergent, then

\begin{matrix} \frac{3^{k}}{2^{a_{k}}} \sum_{j = 1}^{k} \frac{2^{a_{j - 1}}}{3^{j}} > ε \sum_{j = 1}^{k} \frac{2^{a_{j - 1}}}{3^{j}} = \infty \end{matrix}

which generates a contradiction to the fact that

π^{2} {ξ} (j)

is bounded.

To demonstrate the other implication, let us consider the following lemmas:

Lemma 4.

Let

\underset{k \to \infty}{lim inf} \frac{a_{k}}{k} = λ \in (0, \infty]

. Then exist

T, L > 0

such that if

T < k - j

we have

\begin{matrix} \frac{a_{k} - a_{j - 1}}{k - j} \geq λ . \end{matrix}

Proof: Let

\underset{k \to \infty}{lim inf} \frac{a_{k}}{k} = λ \in (0, \infty]

and

d = k - j

, then we have

\begin{matrix} \underset{(k - j) \to \infty}{lim inf} \frac{a_{k} - a_{j - 1}}{k - j} = \underset{d \to \infty}{lim inf} \frac{a_{j + d} - a_{j - 1}}{d} = \underset{d \to \infty}{lim inf} \frac{a_{d}}{d} = λ \end{matrix}

On the other hand, by definition of lower limit, we have

\begin{matrix} \underset{(k - j) \to \infty}{lim inf} \frac{a_{k} - a_{j - 1}}{k - j} = lim_{t \to \infty} \{inf_{(k - j) > t} \frac{a_{k} - a_{j - 1}}{k - j}\} \end{matrix}

Then exist

T > 0

such that if

T < k - j

we have

\begin{matrix} \frac{a_{k} - a_{j - 1}}{k - j} \geq inf_{(k - j) > T} \frac{a_{k} - a_{j - 1}}{k - j} = λ \end{matrix}

QED of Lemma.

Lemma 5.

Let

k, j \in N

, if

k > j

then, we have

\frac{a_{k} - a_{j - 1}}{k - j} \geq 1

Proof: Let

ξ \in G_{0}

writing explicitly, we have

ξ = 0^{θ_{0}} \prod_{i = 1}^{\infty} 0^{θ_{i}} 1

with

θ_{0} \geq 0

and

θ_{i} > 0

for

i \geq 1

, then we can write:

\begin{matrix} a_{k} = \sum_{i = 0}^{k} θ_{i} \end{matrix}

Suppose

k > j

. Since the minimum value that

θ

can take is 1, we have

\begin{matrix} \frac{a_{k} - a_{j - 1}}{k - j} = \frac{\sum_{i = j}^{k} θ_{i}}{k - j} \geq \frac{k - j}{k - j} = 1 \end{matrix}

QED of Lemma.

By Claim 3 and 4 we have, exist

T, ε \in N

such that if

k - T > j

we have

\begin{matrix} \frac{a_{k} - a_{j - 1}}{k - j} > \frac{ln (3)}{ln (2)} + ε \end{matrix}

Then by claim 5, Let

k > T

\begin{matrix} \frac{3^{k}}{2^{a_{k}}} \sum_{j = 1}^{k} \frac{2^{a_{j - 1}}}{3^{j}} & = \sum_{j = 1}^{k} \frac{2^{a_{j - 1} - a_{k}}}{3^{j - k}} \\ = \sum_{j = 1}^{k} {(\frac{3}{2^{\frac{a_{k} - a_{j - 1}}{k - j}}})}^{k - j} \\ = \sum_{j = 1}^{k - T} {(\frac{3}{2^{\frac{a_{k} - a_{j - 1}}{k - j}}})}^{k - j} + \sum_{j = k - T + 1}^{k} {(\frac{3}{2^{\frac{a_{k} - a_{j - 1}}{k - j}}})}^{k - j} \\ < \sum_{j = 1}^{k - T} {(\frac{3}{2^{\frac{ln (3)}{ln (2)} + ε}})}^{k - j} + \sum_{j = k - T + 1}^{k} {(\frac{3}{2})}^{k - j} \\ < \sum_{j = 1}^{k} {(\frac{3}{2^{\frac{ln (3)}{ln (2)} + ε}})}^{k - j} + \sum_{j = k - T + 1}^{k} {(\frac{3}{2})}^{k - j} \\ < \frac{1}{2^{ε k}} \sum_{j = 1}^{k} 2^{ε j} + {(\frac{3}{2})}^{k} \sum_{j = k - T + 1}^{k} {(\frac{2}{3})}^{j} \\ = \frac{1}{2^{ε k}} (\frac{1 - 2^{ε (k + 1)}}{1 - 2^{ε}}) + {(\frac{3}{2})}^{k} \{\frac{{(\frac{2}{3})}^{k} - {(\frac{2}{3})}^{k - T + 1 - 1}}{\frac{2}{3} - 1}\} \\ < \frac{\frac{1}{2^{ε k}} - 2^{(\frac{k + 1}{k})}}{1 - 2^{ε}} + \frac{1 - {(\frac{3}{2})}^{T}}{\frac{2}{3} - 1} \\ \leq \frac{4}{2^{ε} - 1} + 3 \{{(\frac{3}{2})}^{T} - 1\} \end{matrix}

Let

M = \frac{4}{2^{ε} - 1} + 3 \{{(\frac{3}{2})}^{T} - 1\} > 0

. Then we have

\frac{3^{k}}{2^{a_{k}}} \sum_{j = 1}^{k} \frac{2^{a_{j - 1}}}{3^{j}} < M

. Then we conclude that

π^{2}

is bounded.

QED of the second statement.

QED.

7. The Sigma Function

In this section, we immerse ourselves in the rigorous study of Diophantine equations of the form

2^{k} y - a x = n

, where

a, k, n

are integers. Solving these equations in the domain of integers x and y is a problem in number theory. Usually, these types of Diophantine equations are solved using Euclid’s algorithm or some similar technique, even by trial and error. However, these techniques begin to have a high degree of complexity for very large values. This mainly complicates when we want to study the behavior of the minimum positive values since in this case, we are interested in asymptotic solutions. We introduce the sigma function, symbolized as

σ_{a}^{k} (n)

to address this challenge. This function, whose detailed analysis will constitute the core of our research, plays a fundamental role in the quest for specific solutions to the aforementioned Diophantine equations. Particularly noteworthy is the sigma function’s remarkable property of delivering solutions that are closest to zero in the context of these equations.

7.1. Summary of Propositions in the Section

Definition 15: Definition of the sigma function.
Theorem 4: Establish that $σ_{a}^{k} (n)$ and $\frac{1}{a} (2^{k} σ_{a}^{k} (n) - n)$ are solutions of the Diophantine equation $2^{k} y - a x = n$ . Additionally, $\frac{1}{a} (2^{k} σ_{a}^{k} (n) - n)$ is the minimum non-negative integer value.
Corollary 1: Establishes that the minimum value grows based on the number of times the sigma function takes odd values.
Corollary 2: $σ_{a}^{k} (n) - σ_{- a}^{k} (n) = a$
Proposition 17: Establishes inequalities that estimate the values of the sigma function
Proposition 18: It establishes the periods for the periodic points.
Proposition 19: Establish algebraic properties of additivity, dependent on the parity of the addends
Corollary 3 Establish algebraic properties’ linearity modulo $a .$
Proposition 20 Establish that the sigma function is homogeneous modulo $a .$
Definition 16: Extension of the sigma function on $Q_{o d d}$
Definition 17: Characteristic Function
Lemma 6: Establishes an invariance in the coding of the orbits of the sigma function.
Proposition 21: Establishes homogeneity properties of the extension of the sigma function.
Proposition 22 Algebraic properties of the Extension of the Sigma function.
Definition 18: Definition of dyadic numbers.
Proposition 23: Characterization of the dyadic representation of rational numbers.
Definition 19: Definition of Cod-Sigma function.
Lemma 7: Invariant coding lemma for Cod-Sigma function.
Proposition 24: Change of basis of the Cod-Sigma function.
Proposition 25: Let $r \in N$ and $γ = 3^{b - 1}$ and let $k \in N$ such that $r < 2^{2 γ k}$ and $δ_{j} \in {0, 1}$ such that $r \{\frac{2^{2 γ k} - 1}{3 γ}\} = \sum_{j = 0}^{2 γ k - 1} δ_{j} 2^{j}$ then $C o d σ_{3 γ}^{2 γ k} (r) = δ_{2 γ k - 1}, \dots, δ_{0} .$
Corollary 4: Let $r_{1}, r_{2} \in N$ . Then $C o d σ_{3^{b}}^{a} (r_{1} + r_{2}) = C o d σ_{3^{b}}^{a} (r_{1}) + C o d σ_{3^{b}}^{a} (r_{2}) mod 2^{a}$ .
Proposition 26 $C o d σ_{3^{b}} : N \to Z_{2}$ is linear.
Lemma 8: Rational equivalence of the Cod-Sigma function.
Lemma 27: $- π^{1} (ξ) \in Z_{2}$ and $C o d σ (π^{1} (ξ)) = - π^{1} (ξ) \in Z_{2} .$

7.2. The Sigma function

We are going to define the sigma function. This function is very similar to the Collatz function except that in this function, we do not multiply by 3.

In the following theorem, we explore solutions to the Diophantine equation

2^{k} y - a x = n

, where a, k, and n are integers. This equation arises frequently in number theory, particularly in the study of Diophantine equations. We’ll demonstrate that the sigma function provides particular solutions for y, shedding light on the behavior of solutions in both positive and negative domains. Additionally, we’ll establish formulas for the smallest non-negative solution

ρ_{0}

and the largest non-positive solution

ρ_{1}

for the variable x, offering valuable insights into the structure of solutions to this equation.

Proof: We can write the sigma function as

\begin{matrix} σ_{\pm a} (n) = \frac{n \pm a δ (n)}{2} where δ (n) = \{\begin{matrix} 1 & if n = 1 mod 2 \\ 0 & if n = 0 mod 2 \end{matrix} \end{matrix}

Since the sigma function is defined on the set of integers in the integers, we have that its k-th composition is also an integer value: Let

δ_{j} = δ (σ_{a}^{j} (n))

then

\begin{matrix} σ_{a}^{k} (n) = \frac{\frac{\frac{n + δ_{0}}{2}}{⋱} + a δ_{k - 1}}{2} = \frac{n + a L}{2^{k}} \in Z where L = \sum_{j = 0}^{k - 1} δ_{j} 2^{j} \end{matrix}

and Let

ε_{j} = δ (σ_{- a}^{j} (n))

then

\begin{matrix} σ_{- a}^{k} (n) = \frac{\frac{\frac{n - a ε_{0}}{2}}{⋱} - a ε_{k - 1}}{2} = \frac{n - a U}{2^{k}} \in Z where U = \sum_{j = 0}^{k - 1} ε_{j} 2^{j} \end{matrix}

replacing the k-th iteration sigma function

σ_{a}

in the equation

2^{k} y - a x = n

and solving for

x_{0}

, we have

\begin{matrix} x_{0} = \frac{1}{a} (2^{k} σ_{a}^{k} (n) - n) = L \in Z \end{matrix}

and replacing the k-th iteration sigma function

σ_{- a}

in the equation

2^{k} y - a x = n

and solving for

x_{0}

, we have

\begin{matrix} x_{0} = \frac{1}{a} (2^{k} σ_{- a}^{k} (n) - n) = - U \in Z \end{matrix}

For the positive case, we have that

0 \leq L = \sum_{j = 0}^{k - 1} δ_{j} 2^{j} \leq 2^{k} - 1

, then due to the uniqueness of solutions in

[0, 2^{k}) \cap Z

, L corresponds to the non-negative minimum value and for the negative case we have

0 \geq - G_{\infty} = - \sum_{j = 0}^{k - 1} ε_{j} 2^{j} \geq - (2^{k} - 1)

, again due to uniqueness of solutions in

(2^{k}, 0] \cap Z

, we have that

- G_{\infty}

is the maximum non-positive solution.

QED.

Example 8.

Let us consider the following Diophantine equation

16 y - 7 x = 45

then

\begin{matrix} σ_{7}^{4} (45) = 5 and \frac{1}{7} (16 \cdot 5 - 45) = 5 \end{matrix}

are solutions of the equation.

We will demonstrate that this minimum value increases every time

σ_{a}^{k} (n)

is an odd number. This result is crucial for understanding how the parity of the sigma function influences the structure of non-negative solutions of the associated Diophantine equation.

Proof: Let

S_{a} (x) = \frac{a x + n}{2^{k}}

and

δ_{j} = δ (σ_{a}^{j} (n))

then by Theorem 4 we have

\begin{matrix} σ_{a}^{k} (n) = \frac{1}{2^{k}} \{n + a \sum_{j = 0}^{k - 1} δ_{j} 2^{j}\} \end{matrix}

then

S_{a}^{- 1} (σ_{a}^{k} (n)) = ρ_{0} (S_{a}) = \sum_{j = 0}^{k - 1} δ_{j} 2^{j}

. So, we have that every time

δ_{j} = 1

, the minimum positive integer value increases, and this only happens when

σ_{a}^{j - 1} (n)

is odd.

QED.

In the following corollary, we explore the relationship between the sigma functions

σ_{a}^{k} (n)

and

σ_{- a}^{k} (n)

in the context of the Diophantine equation

2^{k} y - a x = n

Proof: By definition, we have that

ρ_{0}

is the nearest non-negative solution to 0, and

ρ_{1}

is the nearest non-positive solution to 0, which means that

ρ_{0}

and

ρ_{1}

are consecutive solutions. Therefore,

ρ_{0} - 2^{k} = ρ_{1}

. then we have

\begin{matrix} 2^{k} = ρ_{0} (S) - ρ_{1} (S) = \frac{1}{a} \{2^{k} σ_{a}^{k} (n) - n\} - \frac{1}{a} \{2^{k} σ_{- a}^{k} (n) - n\} = \frac{1}{a} 2^{k} σ_{a}^{k} (n) - \frac{1}{a} 2^{k} σ_{- a}^{k} (n) \end{matrix}

Therefore

σ_{a}^{k} (n) - σ_{- a}^{k} (n) = a

QED.

In the following proposition, we examine the inequalities and estimations for the sigma function

σ_{a}^{k} (n)

and

σ_{- a}^{k} (n)

, where n is an integer. We show that the sigma function lies in the interval

[\frac{n}{2^{k}}, \frac{n}{2^{k}} + a)

for

σ_{a}^{k} (n)

, and in the interval

[\frac{n}{2^{k}} - a, n)

for

σ_{- a}^{k} (n)

. These inequalities are fundamental to understand the range of values the sigma function can take in the context of the considered Diophantine equations.

Proof: For

σ_{a}^{k} (n)

we have two possible extreme paths, either we always get even or we always get odd, for the first case we would always have division by 2

\begin{matrix} \frac{n}{2^{k}} \leq σ_{a}^{k} (n) \end{matrix}

for the second we would have

\begin{matrix} σ_{a}^{k} (n) \leq \frac{n + a (1 + 2 + \dots + 2^{k - 1})}{2^{k}} = \frac{n}{2^{k}} + a (\frac{1 + 2 + \dots + 2^{k - 1}}{2^{k}}) = \frac{n}{2^{k}} + a (\frac{2^{k} - 1}{2^{k}}) < \frac{n}{2^{k}} + a \end{matrix}

For

σ_{a}^{k} (n)

, regardless of the cases, we always get a less stringent value to the initial value. If it is always even, we will have that it is always divided by 2, now in the case that it is always odd we have

\begin{matrix} σ_{- a}^{k} (n) \geq \frac{n - a (1 + 2 + \dots + 2^{k - 1})}{2^{k}} = \frac{n}{2^{k}} - a (\frac{1 + 2 + \dots + 2^{k - 1}}{2^{k}}) > \frac{n}{2^{k}} - a \end{matrix}

and clearly, we have

\begin{matrix} \frac{n}{2^{k}} > \frac{n}{2^{k}} - a \end{matrix}

QED.

7.3. Periodicity of the Sigma Function

Proof: we have

Let $σ_{a} (u) = u$ , if u is odd, then $\frac{u + a}{2} = u$ which implies $u = a$ . If u is even, we have, $\frac{u}{2} = u$ which implies $u = 0$ .
Let $u \in N$ such that $u < a$ and $σ_{a}^{k} (u) = u$ , so

$\begin{matrix} \frac{u + a L}{2^{k}} = u for any L = \sum_{j = 0}^{k - 1} δ_{j} 2^{j} where δ_{j} \in {0, 1} \end{matrix}$

Then

$\begin{matrix} u + a L = 2^{k} u \Rightarrow a L = (2^{k} - 1) u \Rightarrow (2^{k} - 1) u = 0 mod a \end{matrix}$

suppose that $(u, a) = 1$ , this implies that u is an invertible $mod a$ then, the equation is equivalent

$\begin{matrix} 2^{k} = 1 mod a \end{matrix}$

The minimum value of k is given by the Carmichael function given by

$\begin{matrix} λ (a) = \{\begin{matrix} φ (a) & if a is prime \\ [φ (p_{1}^{α_{1}}), \dots, φ (p_{r}^{α_{r}})] & if a = \prod_{i = 1}^{r} p_{i}^{α_{i}} with p_{i} prime and α_{i} \in N \end{matrix} \end{matrix}$

Let $k_{0} = λ (a)$ , then

$\begin{matrix} (2^{k_{0}} - 1) u = 0 mod a \Rightarrow (2^{k_{0}} - 1) u = L_{0} a for any L_{0} \in N \end{matrix}$

$\begin{matrix} (2^{k_{0}} - 1) = L_{0} \frac{a}{u} \end{matrix}$

as $\frac{a}{u} > 1$ then $(2^{k_{0}} - 1) > L_{0}$ , which is the necessary and sufficient condition for $L_{0}$ to admit decomposition in base 2 up to the power $k_{0} - 1$ which implies that there exist $δ_{j} \in {0, 1}$ such that $L_{0} = \sum_{j = 0}^{k_{0} - 1} δ_{j} 2^{j}$ .

Now suppose that $(u, a) = d > 1$ , then we divide $a L = (2^{k} - 1) u$ by, d

$\begin{matrix} \frac{a}{d} L = (2^{k} - 1) \frac{u}{d} where (\frac{a}{d}, \frac{a}{d}) = 1 \end{matrix}$

then the development is completely analogous to the first case.

In particular, when $a = 3^{r}$ and u are co-prime with 3, then the period of the orbit of u corresponds to the Euler’s totient function, $ϕ$ which in this case is $3^{r - 1} 2$ .

QED.

Let us observe that for the equation

σ_{a}^{k} (u) = u

to have a solution it is necessary and sufficient that

n < a

since the function is monotonically decreasing for

n > a

Example 9.

For,

a = 7

we have

O (5) = {5, 6, 3}

and

O (4) = {4, 2, 1}

For

a = 15

we have

O (1) = {1, 8, 4, 2}

O (3) = {3, 9, 12, 6}

O (5) = {5, 10}

and

O (7) = {7, 11, 13, 14}

7.4. Linearity of the Sigma Function Modulo a

In this section, we address the linearity of the sigma function modulo a. Proposition 19 establishes the addition rules for the sigma function under different parity conditions of the involved numbers. We will see in the corollary 3 that the function sigma modulo a is an automorphism of

Z / a Z

Furthermore, Proposition 20 establishing a relationship between

σ_{a}^{k} (m)

and

σ_{a}^{k} (1)

Proof: Let

m . n \in Z

, then we have

If $m, n$ are even, we have

$\begin{matrix} σ_{a} (m + n) = \frac{m + n}{2} = \frac{m}{2} + \frac{n}{2} = σ_{a} (m) + σ_{a} (n) \end{matrix}$
If m is even and n is odd, we have

$\begin{matrix} σ_{a} (m + n) = \frac{m + n + a}{2} = \frac{m}{2} + \frac{n + a}{2} = σ_{a} (m) + σ_{a} (n) \end{matrix}$
If $m, n$ are odd, we have

$\begin{matrix} σ_{a} (m + n) = \frac{m + n}{2} = \frac{m}{2} + \frac{n}{2} + a - a = \frac{m + a}{2} + \frac{n + a}{2} - a = σ_{a} (m) + σ_{a} (n) - a . \end{matrix}$

QED.

This corollary states that the sigma function, seen as a function on the set

Z / a Z

and taking values in

Z / a Z

, acts as a group additive automorphism. In other words, it preserves the group structure under modular addition in

Z / a Z

Proof: From the previous proposition we have that the sigma function is linearly distributed except for a term

- a

that appears when both addends are odd, this term is congruent to

0 mod a

QED.

This proposition establishes the concept of homogeneity modulo a for the sigma function. It relates the value of

σ_{a}^{k} (m)

m σ_{a}^{k} (1)

under modular arithmetic. This relationship highlights a consistent behavior of the sigma function concerning scaling by m, providing valuable insights into its algebraic properties.

Proof: Let

a, b \in N

such that

(a, 2) = 1

and consider the following Diophantine Equation

2^{k} y - a x = m

. Since

(2^{k}, a) = 1

we have that, this equation is equivalent to

2^{k} Y - a X = 1

. The Theorem 4 we have

σ_{a}^{k} (1)

a particular solution of Y, then

m σ_{a}^{k} (1)

is a solution for y of

2^{k} y - a x = m

, then

\begin{matrix} m σ_{a}^{k} (1) = σ_{a}^{k} (m) mod a \end{matrix}

QED.

7.5. Extension on the $Q_{o d d}$ of the Sigma Function

We can extend the domain of the sigma function to the set of rationals, in the following way,

We are going to provide a numerical interpretation of the extension of the sigma function. Thus far, we understand that the sigma function provides us with the non-negative solution to the Diophantine equation

2^{k} y - a x = n

through the equation

\frac{a x + n}{2^{k}} = σ_{a}^{k} (n)

. We can utilize the latter equation to extend the sigma function to the set of fractions with odd denominators, employing the following equation on

r \in Z

\begin{matrix} \frac{y u r + v x}{2^{k}} = v y σ_{\frac{u}{v}}^{k} (\frac{x}{y}) \end{matrix}

or equivalently

\begin{matrix} \frac{\frac{u}{v} r + \frac{x}{y}}{2^{k}} = σ_{\frac{u}{v}}^{k} (\frac{x}{y}) \end{matrix}

That is to say, the extension of the sigma function gives the fraction that solves the equation

\begin{matrix} \frac{u}{v} r + \frac{x}{y} = 2^{k} t with r \in Z and t, \frac{u}{v}, \frac{x}{y} \in Q_{o d d} \end{matrix}

7.6. Properties of the Extension of the Sigma Function

The introduction of the sigma function extended to odd rationals is crucial for understanding its behavior in a broader domain. This extension, defined on the set

Q_{o d d}

, allows us to explore the algebraic and arithmetic properties of the sigma function in a more general context. In this section, we delve into this extension and explore its implications, focusing on how the sigma function modifies its behavior when applied to fractions with odd denominators. Additionally, we present an important lemma that establishes an invariant relationship between the characteristic function

δ

and the sigma function, providing a deeper understanding of how the sigma function preserves certain properties under different transformations.

The Invariant Coding Lemma, stated in Lemma 6, establishes a fundamental relationship between the characteristic function

δ

and the sigma function under certain conditions. Specifically, it asserts that for co-prime integers u and v, with u being odd, the characteristic function

δ

remains invariant under iterations of the sigma function. This means that the parity of the output of

σ_{u}^{j} (v)

is the same as the parity of

σ_{1}^{j} (\frac{v}{u})

for all non-negative integers j. Furthermore, if v is odd, the lemma demonstrates that the parity of

σ_{u}^{j} (v)

is identical to the parity of

σ_{\frac{u}{v}}^{j} (1)

for all non-negative integers j.

Proof : We have

1: Let $σ_{u}^{k} (v) = \frac{v + u T_{k}^{u} (v)}{2^{a}}$ where $T_{k}^{u} (v) = \sum_{j = 0}^{k - 1} δ (σ_{u}^{j} (v)) 2^{j}$ and $σ_{1}^{j} (\frac{v}{u}) = \frac{\frac{v}{u} + T_{k}^{1} (\frac{v}{u})}{2^{a}}$ where $T_{k}^{1} (\frac{v}{u}) = \sum_{j = 0}^{k - 1} δ (σ_{1}^{j} (\frac{v}{u})) 2^{j}$ . We will prove by induction that

$\begin{matrix} δ (σ_{u}^{j} (v)) = δ (σ_{1}^{j} (\frac{v}{u})) \end{matrix}$

For $j = 0$ , Since if v is odd (or even) then $\frac{v}{u}$ is odd (or even) then

$\begin{matrix} δ (σ_{u}^{0} (v)) = δ (v) = δ (\frac{v}{u}) = δ (σ_{1}^{0} (\frac{v}{u})) . \end{matrix}$

Suppose $δ (σ_{u}^{j} (v)) = δ (σ_{1}^{j} (\frac{v}{u}))$ for $j \leq k$ , then $T_{k}^{u} (v) = T_{k}^{1} (\frac{v}{u})$ , then we have

$\begin{matrix} σ_{u}^{k} (v) = \frac{v + u T_{k}^{u} (v)}{2^{a}} = \frac{v + u T_{k}^{1} (\frac{v}{u})}{2^{a}} = u (\frac{\frac{v}{u} + T_{k}^{1} (\frac{v}{u})}{2^{a}}) = u σ_{1}^{k} (\frac{v}{u}) \end{matrix}$

Since u is odd, we have that $σ_{u}^{k} (v)$ and $σ_{1}^{k} (\frac{v}{u})$ have the same parity, then

$\begin{matrix} δ (σ_{u}^{k + 1} (v)) = δ (σ_{1}^{k + 1} (\frac{v}{u})) \end{matrix}$
2: Let $σ_{u}^{k} (v) = \frac{v + u T_{k}^{u} (v)}{2^{a}}$ where $T_{k}^{u} (v) = \sum_{j = 0}^{k - 1} δ (σ_{u}^{j} (v)) 2^{j}$ and $σ_{\frac{u}{v}}^{j} (1) = \frac{1 + T_{k}^{\frac{u}{v}} (1)}{2^{a}}$ where $T_{k}^{\frac{u}{v}} (1) = \sum_{j = 0}^{k - 1} δ (σ_{\frac{u}{v}}^{j} (1)) 2^{j}$ . We will prove by induction that

$\begin{matrix} δ (σ_{u}^{j} (v)) = δ (σ_{\frac{u}{v}}^{j} (1)) \end{matrix}$

For $j = 0$ , Since v is odd

$\begin{matrix} δ (σ_{u}^{0} (v)) = δ (v) = δ (1) = δ (σ_{\frac{u}{v}}^{0} (1)) . \end{matrix}$

Suppose $δ (σ_{u}^{j} (v)) = δ (σ_{\frac{u}{v}}^{j} (1))$ for $j \leq k$ , then $T_{k}^{u} (v) = T_{k}^{\frac{u}{v}} (1)$ , then we have

$\begin{matrix} σ_{u}^{k} (v) = \frac{v + u T_{k}^{u} (v)}{2^{a}} = \frac{v + u T_{k}^{\frac{u}{v}} (1)}{2^{a}} = v (\frac{1 + \frac{u}{v} T_{k}^{\frac{u}{v}} (1)}{2^{a}}) = v σ_{\frac{u}{v}}^{k} (1) \end{matrix}$

Since v is odd, we have that $σ_{u}^{k} (v)$ and $σ_{\frac{u}{v}}^{k} (1)$ have the same parity, then

$\begin{matrix} δ (σ_{u}^{k + 1} (v)) = δ (σ_{\frac{u}{v}}^{k + 1} (1)) \end{matrix}$

QED.

In the following proposition, we demonstrate homogeneity properties that leave the coding of the orbits of the sigma function invariant.

Proof: We have

Let $σ_{u}^{k} (v) = \frac{v + u T_{k}^{u} (v)}{2^{a}}$ where $T_{k}^{u} (v) = \sum_{j = 0}^{k - 1} δ (σ_{u}^{j} (v)) 2^{j}$ and $σ_{1}^{j} (\frac{v}{u}) = \frac{\frac{v}{u} + T_{k}^{1} (\frac{v}{u})}{2^{a}}$ where $T_{k}^{1} (\frac{v}{u}) = \sum_{j = 0}^{k - 1} δ (σ_{1}^{j} (\frac{v}{u})) 2^{j}$ . Then we have

$\begin{matrix} σ_{u}^{k} (v) = \frac{v + u T_{k}^{u} (v)}{2^{a}} = \frac{v + u T_{k}^{1} (\frac{v}{u})}{2^{a}} = u (\frac{\frac{v}{u} + T_{k}^{1} (\frac{v}{u})}{2^{a}}) = u σ_{1}^{k} (\frac{v}{u}) \end{matrix}$
Let $σ_{u}^{k} (v) = \frac{v + u T_{k}^{u} (v)}{2^{a}}$ where $T_{k}^{u} (v) = \sum_{j = 0}^{k - 1} δ (σ_{u}^{j} (v)) 2^{j}$ and $σ_{\frac{u}{v}}^{j} (1) = \frac{1 + T_{k}^{\frac{u}{v}} (1)}{2^{a}}$ where $T_{k}^{\frac{u}{v}} (1) = \sum_{j = 0}^{k - 1} δ (σ_{\frac{u}{v}}^{j} (1)) 2^{j}$ . Then we have

$\begin{matrix} σ_{u}^{k} (v) = \frac{v + u T_{k}^{u} (v)}{2^{a}} = \frac{v + u T_{k}^{\frac{u}{v}} (1)}{2^{a}} = v (\frac{1 + T_{k}^{\frac{u}{v}} (1)}{2^{a}}) = v σ_{\frac{u}{v}}^{k} (1) \end{matrix}$

QED.

Proof: Let’s proved first for

a = 1

. Let

\frac{n}{q}, \frac{m}{p} \in Q_{o d d}

and let

β : Z \times Z \to {0, 1}

given by

β (n, m) = 0

if n or m is even fraction and

β (n, m) = 1

if n and m are odd fraction.

\begin{matrix} p q σ (\frac{n}{q} + \frac{m}{p}) & = p q σ (\frac{n}{q} + \frac{m}{p}) \\ = p q σ (\frac{n p + m q}{p q}) \\ = σ_{p q} (p q \frac{n p + m q}{p q}) \\ = σ_{p q} (n p + m q) \\ = σ_{p q} (n p) + σ_{p q} (m q) - p q β (n, m) \\ = p q σ (\frac{n p}{p q}) + p q σ (\frac{m q}{p q}) - p q β (n, m) \\ = p q σ (\frac{n}{q}) + p q σ (\frac{m}{p}) - p q β (n, m) \end{matrix}

Dividing everything by

p q

, we have

\begin{matrix} σ (\frac{n}{q} + \frac{m}{p}) = σ (\frac{n}{q}) + σ (\frac{m}{p}) - β (n, m) \end{matrix}

Now let a an odd fraction with odd numerator, then we have

\frac{1}{a}

is odd fraction, then multiplying by

\frac{1}{a}

does not change the parity of

\frac{n}{q}

\frac{m}{q}

. Then we have

\begin{matrix} σ (\frac{1}{a} \frac{n}{q} + \frac{1}{a} \frac{m}{p}) = σ (\frac{1}{a} \frac{n}{q}) + σ (\frac{1}{a} \frac{m}{p}) - β (n, m) multiplying by a \\ a σ (\frac{1}{a} \frac{n}{q} + \frac{1}{a} \frac{m}{p}) = a σ (\frac{1}{a} \frac{n}{q}) + a σ (\frac{1}{a} \frac{m}{p}) - a β (n, m) \\ σ_{a} (\frac{n}{q} + \frac{m}{p}) = σ_{a} (\frac{n}{q}) + σ_{a} (\frac{m}{p}) - a β (n, m) \end{matrix}

QED.

7.7. Coding of Sigma Function

In this section we are going to use the space of dyadic integers numbers

Z_{2}

as a support for the coding of the sigma function.

7.7.1. $p -$ Adic Numbers

We have the following properties

if $a, b \in N$ then $a = b mod p^{k}$ if and only if ${∥ a - b ∥}_{p} \leq p^{- k}$ .
Let $β \in Q_{p}$ then this is uniquely represented by convergent series ( with norm ${∥ . ∥}_{p}$ ) as

$\begin{matrix} β = \sum_{n = - k}^{\infty} δ_{j} 2^{j} = {(\dots δ_{j} \dots δ_{2} δ_{1} δ_{0} . δ_{- 1} \dots δ_{- k})}_{p} \end{matrix}$
The $p -$ adic expansion allows us to perform arithmetical operations in $Q_{p}$ in way very similar to that in $R$ . Moreover, we will see that the operations in $Q_{p}$ are, in fact, easier to perform than $R$ .

Let $α = \sum_{j = - k}^{\infty} ε_{j} p^{j}$ and $β = \sum_{j = - k}^{\infty} δ_{j} p^{j}$

$\begin{matrix} α \pm β = \sum_{j = - k}^{\infty} (ε_{j} \pm δ_{j}) p^{j} \end{matrix}$

$\begin{matrix} α \cdot β = (\sum_{j = - k}^{\infty} ε_{j} p^{j}) (\sum_{j = - k}^{\infty} δ_{j} p^{j}) = \sum_{r = - k}^{\infty} (\sum_{i + j = r} ε_{j} δ_{j}) p^{r} \end{matrix}$

.
A $p -$ adic number $β \in Q_{p}$ is said to be a $p -$ adic integer if its canonical expansion contains only non-negative power of p. The set of $p -$ adic integers is denoted by $Z_{p}$ , so

$\begin{matrix} Z_{p} = \{\sum_{i = 0}^{\infty} δ_{i} p^{i}, with δ_{j} = 0, \dots, p - 1\} = {β \in Q_{p} {; ∥ β ∥}_{p} \leq 1} \end{matrix}$

This set has the property of being a complete metric subspace.

One of the main characteristics of

p -

adics numbers is

7.7.2. Coding of Sigma Function

Now we are going to define the coding of the sigma function for

Q_{o d d}

The following lemma is a reformulation of lemma 6 for

C o d σ

Proof: Reformulation of the Lemma 6

QED.

Proof: Let’s prove by induction that if

N_{2}

is odd, then

\begin{matrix} C o d σ_{N_{2}}^{a} (x_{0}) = \underset{a}{\underset{︸}{0, \dots, 0, δ_{n_{1}}, \dots, δ_{0}}} . \end{matrix}

Completing with

δ_{j} = 0

if necessary, we have

N_{1} = \sum_{j = 0}^{n_{1}} δ_{j} 2^{j} = \sum_{j = 0}^{a - 1} δ_{j} 2^{j}

, then

\begin{matrix} \frac{x + N_{1} N_{2}}{2^{a}} = \frac{\frac{\frac{\frac{\frac{x + δ_{0} N_{2}}{2} + δ_{1} N_{2}}{2} + δ_{2} N_{2}}{2}}{⋱} + δ_{a - 1} N_{2}}{2} = ⨀_{j = 1}^{a} (\frac{x + δ_{j - 1} N_{2}}{2}) \in 〈 θ, λ^{N_{2}} 〉 \end{matrix}

Since we trivially have that

(1, 2) = 1

and by hypothesis we have that

x_{0} \in E (H)

, by the Lemma 1 we have

\begin{matrix} x_{0} \in E (⨀_{j = 1}^{K} (\frac{x + δ_{j - 1} N_{2}}{2})) for all K \leq a \end{matrix}

In particular, for

K = 0

, we have

x_{0} \in E (\frac{x + δ_{0} N_{2}}{2})

and

N_{2}

is odd, we have

E (θ) \cap E (λ^{N_{2}}) = \emptyset

then

σ_{N_{2}} (x_{0}) = \frac{x_{0} + δ_{0} N_{2}}{2}

. For

K = 1

, we have

x_{0} \in E (\frac{x + δ_{1} N_{2}}{2} \circ \frac{x + δ_{0} N_{2}}{2})

, then

σ_{N_{2}} (x_{0}) \in E (\frac{x_{0} + δ_{1} N_{2}}{2})

, then

σ_{N_{2}}^{2} (x_{0}) = \frac{x_{0} + N_{2} (δ_{0} + δ_{1} 2)}{2^{2}}

. Suppose this continues until for

K = r

i.e.

\begin{matrix} σ_{N_{2}}^{r} (x_{0}) = \frac{x_{0} + N_{2} \sum_{j = 0}^{r - 1} δ_{j} 2^{j}}{2^{r}} \in Z \end{matrix}

We have that

x_{0} \in E (⨀_{j = 1}^{r + 1} (\frac{x + δ_{j - 1} N_{2}}{2}))

then by definitions

⨀_{j = 1}^{r + 1} (\frac{x_{0} + δ_{j - 1} N_{2}}{2}) \in Z

and by inductive hypothesis

σ_{N_{2}}^{r} (x_{0}) = ⨀_{j = 1}^{r} (\frac{x_{0} + δ_{j - 1} N_{2}}{2}) \in Z

then

σ_{N_{2}}^{r} (x_{0}) \in E (\frac{x + δ_{r} N_{2}}{2})

. Therefore,

σ_{N_{2}}^{r + 1} (x_{0}) = \frac{x_{0} + N_{2} \sum_{j = 0}^{r} δ_{j} 2^{j}}{2^{r + 1}}

. Then we have to

\begin{matrix} σ_{N_{2}}^{a} (x_{0}) = \frac{x_{0} + N_{2} \sum_{j = 0}^{a - 1} δ_{j} 2^{j}}{2^{a}} = \frac{x_{0} + N_{2} N_{1}}{2^{a}} . \end{matrix}

Similarly, Completing with

δ_{j} = 0

if necessary we have. Let

N_{2} = \sum_{j = 0}^{n_{2}} ε_{j} 2^{j} = \sum_{j = 0}^{a - 1} ε_{j} 2^{j}

, then

\begin{matrix} σ_{N_{1}}^{a} (x_{0}) = \frac{x_{0} + N_{1} \sum_{j = 0}^{a - 1} ε_{j} 2^{j}}{2^{a}} = \frac{x_{0} + N_{1} N_{2}}{2^{a}} \end{matrix}

Now. If

N_{1}, N_{2}

are odd numbers, we have

\begin{matrix} σ_{N_{1}}^{a} (x_{0}) = σ_{N_{2}}^{a} (x_{0}) \end{matrix}

In particular, we have

\begin{matrix} σ_{N_{2}}^{a} (x_{0}) & = \frac{\frac{\frac{\frac{\frac{x + δ_{0} N_{2}}{2} + δ_{1} N_{2}}{2} + δ_{2} N_{2}}{2}}{⋱} + δ_{a - 1} N_{2}}{2} \\ = \frac{\frac{\frac{\frac{\frac{x + ε_{0} N_{1}}{2} + ε_{1} N_{1}}{2} + ε_{2} N_{1}}{2}}{⋱} + ε_{a - 1} N_{1}}{2} = σ_{N_{1}}^{a} (x_{0}) \end{matrix}

then we have that the coding of

σ_{N_{2}}^{a} (x_{0})

δ_{a - 1} \dots δ_{1} δ_{0}

and of the

σ_{N_{1}}^{a} (x_{0})

ε_{a - 1} \dots ε_{1} ε_{0} .

QED.

Example 10.

Let

H (x) = \frac{x + 15}{2^{5}}

, we have

x_{0} = 2^{5} - 15 = 17 \in E (H)

, then

$σ_{3} (17) = 10, σ_{3} (10) = 5, σ_{3} (5) = 4, σ_{3} (4) = 2$ , $σ_{3} (2) = 1$ then $σ_{3}^{5} (17) = 1$ . Then we have $C o d σ_{3} (17) = 00101$ taking the coding coefficients, to base 2 we have

$\begin{matrix} 1 \cdot 2^{0} + 0 \cdot 2^{1} + 1 \cdot 2^{2} + 0 \cdot 2^{3} + 0 \cdot 2^{4} = 5 . \end{matrix}$
$σ_{5} (17) = 11, σ_{5} (11) = 8, σ_{5} (8) = 4, σ_{5} (4) = 2, σ_{5} (2) = 1$ , then $σ_{5}^{5} (17) = 1$ . Then we have $C o d σ_{5} (17) = 00011$ taking the coding coefficients, to base 2 we have

$\begin{matrix} 1 \cdot 2^{0} + 1 \cdot 2^{1} + 0 \cdot 2^{2} + 0 \cdot 2^{3} + 0 \cdot 2^{4} = 3 . \end{matrix}$

We observe that the obits are equal from the third iteration, which corresponds to the maximum power of two, where all subsequent coefficients are null. We can also observe that from the third term, the values that appear in the orbits are even, they, unfortunately, cannot continue forever, since as we have seen, the orbit of the sigma functions falls into a periodic orbit with the same number of even and odd numbers, so at some point this orbit must fall into an odd one, which means all the initial values must change.

Proof: Let

H : R \to R

given by

H (x) = \frac{3 γ x + r}{2^{2 γ k}}

, then we want to find the minimum positive value of

H

, then solving the following equation.

\begin{matrix} \frac{3 γ x + r}{2^{2 γ k}} = σ_{3 γ}^{2 γ k} (r) \end{matrix}

By Proposition 18 we have

σ_{3 γ}^{2 γ k} (r) = r

, then

\begin{matrix} \frac{3 γ x + r}{2^{2 γ k}} = r then x = \frac{2^{2 γ k} r - r}{3 γ} = r \{\frac{2^{2 γ k} - 1}{3 γ}\} \end{matrix}

Let

δ_{j} \in {0, 1}

such that

r \{\frac{2^{2 γ k} - 1}{3^{3 γ}}\} = \sum_{j = 0}^{2 γ k - 1} δ_{j} 2^{j}

. By Proposition 24 we have

\begin{matrix} C o d σ_{3 γ}^{2 γ k} (r) = δ_{2 γ k - 1}, \dots, δ_{0} \end{matrix}

QED.

Proof: Let

γ = 3^{b - 1}

r = r_{1} + r_{2}

and

H : R \to R

given by

H (x) = \frac{3 γ x + r}{2^{a}}

by Proposition 19 we have

\begin{matrix} \frac{3 γ x + r}{2^{a}} = σ_{3 γ}^{a} (r) = σ_{3 γ}^{a} (r_{1}) + σ_{3 γ}^{a} (r_{2}) - 3^{3 γ} u with u \in Z \end{matrix}

Then

\begin{matrix} x = \frac{2^{a} σ_{3 γ}^{a} (r_{1}) - r_{1}}{3 γ} + \frac{2^{a} σ_{3 γ}^{a} (r_{2}) - r_{2}}{3 γ} - 2^{a} u \end{matrix}

Let

k \in N

such that

r < 2^{2 γ k}

and

2 γ k > a

and let

δ_{j}, ε_{j}, η_{j} \in {0, 1}

such that

\begin{matrix} r_{1} \{\frac{2^{2 γ k} - 1}{3 γ}\} & = \sum_{j = 0}^{2 γ k - 1} δ_{j} 2^{j} \\ r_{2} \{\frac{2^{2 γ k} - 1}{3 γ}\} & = \sum_{j = 0}^{2 γ k - 1} ε_{j} 2^{j} and \\ r \{\frac{2^{2 γ k} - 1}{3 γ}\} & = \sum_{j = 0}^{2 γ k - 1} η_{j} 2^{j} \end{matrix}

by Proposition 25 we have

\begin{matrix} (\frac{2^{a} σ_{3 γ}^{a} (r_{1}) - r_{1}}{3 γ}) (Base 2) & = {δ_{j}}_{j = 0}^{a - 1} \\ (\frac{2^{a} σ_{3 γ}^{a} (r_{2}) - r_{2}}{3 γ}) (Base 2) & = {ε_{j}}_{j = 0}^{a - 1} and \\ (\frac{2^{a} σ_{3 γ}^{a} (r) - r}{3 γ}) (Base 2) & = {η_{j}}_{j = 0}^{a - 1} \end{matrix}

Then we have

\begin{matrix} \frac{2^{a} σ_{3 γ}^{a} (r) - r}{3 γ} = \frac{2^{a} σ_{3 γ}^{a} (r_{1}) - r_{1}}{3 γ} + \frac{2^{a} σ_{3 γ}^{a} (r_{2}) - r_{2}}{3 γ} - 2^{a} u \\ if and only if \sum_{j = 0}^{a - 1} η_{j} 2^{j} = \sum_{j = 0}^{a - 1} δ_{j} 2^{j} + \sum_{j = 0}^{a - 1} ε_{j} 2^{j} - 2^{a} u \end{matrix}

Then we have

\sum_{j = 0}^{a - 1} η_{j} 2^{j} = \sum_{j = 0}^{a - 1} δ_{j} 2^{j} + \sum_{j = 0}^{a - 1} ε_{j} 2^{j} mod 2^{a}

, therefore

\begin{matrix} C o d σ_{3 γ}^{a} (r) = C o d σ_{3 γ}^{a} (r_{1}) + C o d σ_{3 γ}^{a} (r_{2}) mod 2^{a} \end{matrix}

QED.

Proof: By Corollary 4 we have

C o d σ_{3^{b}}^{n} (r_{1} + r_{2}) = C o d σ_{3^{b}}^{n} (r_{1}) + C o d σ_{3^{b}}^{n} (r_{2}) mod 2^{n}

for all

n \in N

, this is equivalent to

\begin{matrix} | | C o d σ_{3^{b}}^{n} (r_{1} + r_{2}) - C o d σ_{3^{b}}^{n} (r_{1}) + C o d σ_{3^{b}}^{n} (r_{2}) {| |}_{2} < 2^{- n} for all n \in N \end{matrix}

Therefore

\begin{matrix} C o d σ_{3^{b}} (r_{1} + r_{2}) = C o d σ_{3^{b}} (r_{1}) + C o d σ_{3^{b}} (r_{2}) \in Z_{2} \end{matrix}

QED.

Example 11.

Let $S (x) = \frac{9 x + 5}{2^{4}}$

$\begin{matrix} σ_{9}^{4} (5) & = σ_{9}^{4} (3) + σ_{9}^{4} (2) mod 9 \end{matrix}$

(a)

$O σ_{9}^{4} (3) = {3, 6, 3, 6, 3}$ then $C o d σ_{9}^{4} (3) = 0101$ .

(b)

$O (σ_{9}^{4} (2)) = {2, 1, 5, 7, 8}$ then $C o d σ_{9}^{4} (2) = 1110$ .

$\begin{matrix} C o d σ_{9}^{4} (5) & = C o d σ_{9}^{4} (3) + C o d σ_{9}^{4} (2) mod 2^{4} \\ = 1110 + 0101 mod 2^{4} \\ = 0011 \end{matrix}$

$O (σ_{9}^{4} (5)) = {5, 7, 8, 4, 2}$ then $C o d σ_{9}^{4} (5) = 0011$ , Then we have $ρ_{0} (S) = 1 + 2 = 3$

$\begin{matrix} S (3) = \frac{9 \cdot 3 + 5}{2^{4}} = 2 \end{matrix}$
Let $S (x) = \frac{9 x + 3 + 2^{θ_{1}}}{2^{θ_{1} + θ_{2}}}$ with $θ_{1}$ even. We have

$\begin{matrix} σ_{9}^{θ_{1} + θ_{2}} (3 + 2^{θ_{1}}) & = σ_{9}^{θ_{1} + θ_{2}} (3) + σ_{9}^{θ_{1} + θ_{2}} (2^{θ_{1}}) mod 9 \\ = 3 σ_{3}^{θ_{1} + θ_{2}} + σ_{9}^{θ_{1} + θ_{2}} (2^{θ_{1}}) mod 9 \end{matrix}$

Then

$\begin{matrix} C o d σ_{9}^{θ_{1} + θ_{2}} (3 + 2^{θ_{1}}) & = C o d σ_{9}^{θ_{1} + θ_{2}} (3) + C o d σ_{9}^{θ_{1} + θ_{2}} (2^{θ_{1}}) mod 2^{θ_{1} + θ_{2}} \\ = C o d σ_{3}^{θ_{1} + θ_{2}} + C o d σ_{9}^{θ_{1} + θ_{2}} (2^{θ_{1}}) mod 2^{θ_{1} + θ_{2}} \end{matrix}$

(a)

$O (σ_{3}^{θ_{1} + θ_{2}}) = {1, 2, 1, 2, 1, 2, \dots}$ then $C o d σ_{3}^{θ_{1} + θ_{2}} = {\underset{︸}{\dots 010101}}_{θ_{1} + θ_{2}}$ .

(b)

$O (σ_{9}^{θ_{1} + θ_{2}} (2^{θ_{1}})) = {2^{θ_{1}}, 2^{θ_{1} - 1}, \dots, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, \dots}$ then

$\begin{matrix} C o d σ_{9}^{θ_{1} + θ_{2}} (2^{θ_{1}}) = \underset{θ_{2}}{\underset{︸}{\dots 000111000111}} \underset{θ_{1}}{\underset{︸}{\dots 00000}} \end{matrix}$

.

Then we have

$\begin{matrix} C o d σ_{9}^{θ_{1} + θ_{2}} (3 + 2^{θ_{1}}) & = \underset{θ_{2}}{\underset{︸}{\dots 000111000111}} \underset{θ_{1}}{\underset{︸}{\dots 00000}} + \underset{θ_{1} + θ_{2}}{\underset{︸}{\dots 010101}} mod 2^{θ_{1} + θ_{2}} \\ = \underset{θ_{2}}{\underset{︸}{\dots 00011100011100011100}} \underset{θ_{1}}{\underset{︸}{\dots 1010101010}} mod 2^{θ_{1} + θ_{2}} \\ = C o d σ_{9}^{θ_{2} - 2} 00 C o d σ_{3}^{θ_{1}} mod 2^{θ_{1} + θ_{2}} \end{matrix}$

Then we have $ρ_{0} (S) (base 2) = C o d σ_{9}^{θ_{2} - 2} 00 C o d σ_{3}^{θ_{1}} mod 2^{θ_{1} + θ_{2}}$ .
Let $S_{n} (x) = \frac{3^{n} x + 4^{n} - 3^{n}}{4^{n}}$

$\begin{matrix} σ_{3^{n}}^{2 n} (4^{n} - 3^{n}) = σ_{3^{n}}^{2 n} (4^{n}) - σ_{3^{n}}^{2 n} (3^{n}) = 1 - 3^{n} σ_{1}^{2 n} = 1 mod 3^{n} \end{matrix}$

On the other hand

(a)

$O (σ_{3^{n}}^{2 n} (2^{2 n})) = {2^{2 n}, 2^{2 n - 1}, 2^{2 n - 2}, \dots, 1}$ then $C o d σ_{3^{n}}^{2 n} (2^{2 n}) = {\underset{︸}{\dots 00000}}_{2 n}$ .

(b)

$O (σ_{3^{n}}^{2 n} (3^{n})) = {3^{n}, 3^{n}, 3^{n}, 3^{n} \dots}$ then $C o d σ_{3^{n}}^{2 n} (3^{n}) = {\underset{︸}{\dots 11111111}}_{2 n}$ .

Then we have

$\begin{matrix} \underset{2 n}{\underset{︸}{0 \dots 0}} - \underset{2 n}{\underset{︸}{1 \dots 1}} mod 2^{2 n} \\ = & 1 \underset{2 n}{\underset{︸}{0 \dots 0}} - 0 \underset{2 n}{\underset{︸}{1 \dots 1}} mod 2^{2 n} \\ = & 1 mod 2^{2 n} \\ = & 1 \end{matrix}$

Then

$\begin{matrix} C o d σ_{3^{n}}^{2 n} (4^{n} - 3^{n}) = 00 \dots 001 \end{matrix}$

That is, $C o d σ_{3^{n}}^{2 n} (4^{n} - 3^{n})$ has a constant coding equal to 1. This is natural, since $S_{n}$ is stable.

Proof: Let

u = \bar{γ_{b}} \in Z_{2}

with

γ_{b} = C o d σ_{3^{b}}^{2 \cdot 3^{b - 1}}

. Let

u = x

then multiplying by

2^{2 \cdot 3^{b - 1}}

we have

u {\underset{︸}{0 \dots 0}}_{2 \cdot 3^{b - 1}} = 2^{2 \cdot 3^{b - 1}} x

and subtracting, we have

\begin{matrix} u - u \underset{2 \cdot 3^{b - 1}}{\underset{︸}{0 \dots 0}} & = - (2^{2 \cdot 3^{b - 1}} - 1) x \\ γ_{b} & = - (2^{2 \cdot 3^{b - 1}} - 1) x \end{matrix}

On the other hand we have that

γ_{b} = \frac{2^{2 \cdot 3^{b - 1}} - 1}{3^{b}}

, then

\begin{matrix} \frac{2^{2 \cdot 3^{b - 1}} - 1}{3^{b}} = - (2^{2 \cdot 3^{b - 1}} - 1) then we have x = - \frac{1}{3^{b}} \end{matrix}

Other way for proof it, is

\begin{matrix} u = \sum_{n = 0}^{\infty} u_{n} 2^{n} & = \{\sum_{n = 0}^{2 \cdot 3^{b - 1} - 1} u_{n} 2^{n}\} + \{\sum_{n = 2 \cdot 3^{b - 1}}^{4 \cdot 3^{b - 1} - 1} u_{n} 2^{n}\} + \{\sum_{n = 4 \cdot 3^{b - 1}}^{6 \cdot 3^{b - 1} - 1} u_{n} 2^{n}\} \dots \\ = C o d σ_{3^{b}}^{2 \cdot 3^{b - 1}} + 2^{2 \cdot 3^{b - 1}} C o d σ_{3^{b}}^{2 \cdot 3^{b - 1}} + {(2^{2 \cdot 3^{b - 1}})}^{2} C o d σ_{3^{b}}^{2 \cdot 3^{b - 1}} + \dots \\ = C o d σ_{3^{b}}^{2 \cdot 3^{b - 1}} \{\sum_{n = 0}^{\infty} {(2^{2 \cdot 3^{b - 1}})}^{n}\} \\ = C o d σ_{3^{b}}^{2 \cdot 3^{b - 1}} \{\frac{1}{1 - 2^{2 \cdot 3^{b - 1}}}\} \\ = - \{\frac{1 - 2^{2 \cdot 3^{b - 1}}}{3^{b}}\} \{\frac{1}{1 - 2^{2 \cdot 3^{b - 1}}}\} \\ = - \frac{1}{3^{b}} \end{matrix}

Now let’s try the second part. Let

v \in N

\begin{matrix} C o d σ_{3^{b}}^{k} (v) = - v \{\frac{1 - 2^{2 \cdot 3^{b - 1} k}}{3^{b}}\} = - \frac{v}{3^{b}} + \frac{v 2^{2 \cdot 3^{b - 1} k}}{3^{b}} \end{matrix}

On the other hand

\begin{matrix} {||C o d σ_{3^{b}}^{k} (v) - (- \frac{v}{3^{b}})||}_{2} \leq 2^{- 2 \cdot 3^{b - 1} k} \to 0 as k \to \infty \end{matrix}

Therefore

C o d σ_{3^{b}} (v) = - \frac{v}{3^{b}}

QED.

Proof: Claim 1: Let

ξ \in Σ_{2}^{*}

and

k \in N

, the we have

C o d σ (π_{k}^{1} (ξ)) = - π_{k}^{1} (ξ) \in Z_{2}

Indeed. Let

k \in N

and

ξ \in Σ_{2}^{*}

. By Lemma 7, Proposition 26 and Lemma 8, we have

\begin{matrix} C o d σ (π_{k}^{1} (ξ)) & = C o d σ (\sum_{j = 1}^{b_{k}} \frac{2^{a_{j - 1}}}{3^{j}}) \\ = C o d σ_{3^{b_{k}}} (\sum_{j = 1}^{b_{k}} 2^{a_{j - 1}} 3^{b_{k} - j}) \\ = \sum_{j = 1}^{b_{k}} C o d σ_{3^{b_{k}}} (2^{a_{j - 1}} 3^{b_{k} - j}) \\ = \sum_{j = 1}^{b_{k}} C o d σ_{3^{j}} (2^{a_{j - 1}}) \\ = - \sum_{j = 1}^{b_{k}} \frac{2^{a_{j - 1}}}{3^{j}} \\ = - π_{k}^{1} (ξ) \end{matrix}

Since

- \frac{2^{a_{j - 1}}}{3^{j}} \in Z_{2}

then

- π_{k}^{1} (ξ) \in Z_{2}

Claim 2:

lim_{k \to \infty} π_{k}^{1} (ξ) = π^{1} (ξ) \in Z_{2}

Indeed. We have the following equivalence on

Q_{2}

\begin{matrix} π^{1} (ξ) = \sum_{n = 1}^{\infty} \frac{2^{a_{n - 1}}}{3^{n}} \in Q_{2} if and only if lim_{k \to \infty} \frac{2^{a_{k - 1}}}{3^{k}} = 0 on Q_{2} \end{matrix}

On the other hand

\begin{matrix} {∥\frac{2^{a_{k - 1}}}{3^{k}}∥}_{2} = {∥C o d σ_{3^{b}} (2^{a_{k - 1}})∥}_{2} = {∥C o d σ_{3^{b}} 0^{a_{k - 1}}∥}_{2} = 2^{- a_{k - 1}} \to 0 \sin ce a_{k - 1} it is increasing \end{matrix}

Therefore

π^{1} (ξ) = \sum_{n = 1}^{\infty} \frac{2^{a_{n - 1}}}{3^{n}} \in Q_{2}

. On the other hand, we have that

π_{k}^{1} (ξ)

is a Cauchy sequence on

Z_{2}

. Indeed

\begin{matrix} {∥π_{k + 1}^{1} (ξ) - π_{k}^{1} (ξ)∥}_{2} = {∥\frac{2^{a_{k - 1}}}{3^{k}}∥}_{2} = 2^{- a_{k - 1}} \to 0 as k \to \infty \end{matrix}

and we also have that

Z_{2}

is a complete metric space, then

lim_{k \to \infty} π_{k}^{1} (ξ) = π^{1} (ξ) \in Z_{2}

QED.

8. The Extension of Collatz Function on $Z_{2}$

In this section we will study the extension of the Collatz function on

Z_{2}

proposed by Lagaria in [3], and in an analogous way we will define the dyadic integer sets and the encoding set. We will prove that given an coding there exists a unique dyadic integer with this coding. We will show that this extension is topologically conjugate to the shift function in

Σ_{2}^{*}

and we will use this result to prove that codings in

G_{1}

are unstable.

8.1. Summary of Propositions in the Section

Lemma 9: Equivalence of the parity of fractions and their dyadic representation.
Definition 20: Extension of the Collatz function on the set of dyadic numbers and the definitions of dyadic integer set and coding set.
Proposition 28: Characterization of the dyadic integer set.
Proposition 29: Establishes that the Coding set and the Dyadic Integer Set are the same.
Proposition 30: It establishes that given a coding there is a unique dyadic number with said coding.
Theorem 5: The Collatz function on the set of dyadic numbers is topologically conjugate to the Shift function.
Corollary 5: The periodic points of the Collatz function in $Z_{2}$ are dense in $Z_{2}$ .
Proposition 31: The periodic sequences of $G_{0}$ correspond to positive periodic points of the Collatz functions and the periodic sequences of $G_{\infty}$ correspond to negative periodic points of the Collatz functions.
Lemma 10: $ρ (S_{k}) = C o d σ_{3^{k}}^{A_{k}} (N_{k})$ .
Proposition 32 $ρ_{0} (S_{k}) = C o d σ (π^{1} (ξ)) mod 2^{a_{k}}$ .
Theorem 6: The sequences in $G_{1}$ are positively and negatively unstable.

8.2. Extension of the Collatz Function on $Z_{2}$ .

Now we are going to extend the Collatz function to the set of

Z_{2}

. In order for the extension to be compatible with the results obtained in the previous sections, we will first show that the parity of the elements of

Q_{o d d}

is preserved in

Z_{2}

Proof: Let p a even number, so we have

\begin{matrix} \frac{p}{q} = \frac{2 k}{q} = δ_{0} + 2 M for any M \in Z_{2}, so 2 k = q δ_{0} + 2 M q then δ_{0} = 0 . Sin ce q is odd number \end{matrix}

Let p a odd number, so we have

\begin{matrix} \frac{p}{q} = \frac{2 k + 1}{q} = δ_{0} + 2 M for any M \in Z_{2}, so 2 k + 1 = q δ_{0} + 2 M q then δ_{0} = 1 . Sin ce q is odd number \end{matrix}

QED.

Now let us consider the following extension of the Collatz function on

Z_{2}

Next we will show the version in

Z_{2}

to the results seen in previous Sections. The following Proposition characterizes the set of dyadic integers of

S_{k} \in 〈 θ, ψ 〉

analogously to the entire set.

Proof: Let

β_{0} \in D (S_{k})

then we have

β_{0} + 2^{A_{k}} Z_{2} \subset D (S_{k})

. Indeed. Let

ξ \in Σ_{2}^{*}

and

S_{k} \in 〈 θ, ψ 〉

such that

C o d (S_{k}) = {ξ_{j}}_{j = 0}^{k - 1}

, then

\begin{matrix} S_{k} (x) = \frac{3^{b_{k}} x + N_{k}}{2^{A_{k}}} \end{matrix}

\begin{matrix} S_{k} (β_{0} + 2^{A_{k}} Z_{2}) = \frac{3^{b_{k}} (β_{0} + 2^{A_{k}} Z_{2}) + N_{k}}{2^{A_{k}}} = \frac{3^{b_{k}} β_{0} + N_{k}}{2^{A_{k}}} + Z_{2} \subset Z_{2} \end{matrix}

Now

D (S_{k}) \subset β_{0} + 2^{A_{k}} Z_{2}

. Let

β \in D (S_{k})

, so

3^{b_{k}} β + N_{k} = 0 mod 2^{A_{k}}

, so

\begin{matrix} 3^{b_{k}} β + N_{k} - (3^{b_{k}} β_{0} + N_{k}) = 0 mod 2^{A_{k}} \\ 3^{b_{k}} (β - β_{0}) = 0 mod 2^{A_{k}} \\ β - β_{0} = 0 mod 2^{A_{k}} \end{matrix}

Therefore, we have

\begin{matrix} β - β_{0} = \underset{2^{A_{k}}}{\underset{︸}{0 \dots 0}} α = 2^{A_{k}} α, with α \in Z_{2} \Rightarrow β = β_{0} + 2^{A_{k}} α \end{matrix}

QED.

The following proposition states that the entire dyadic set is equal to the coding set of

S_{k}

Proof: Let

β \in C o d^{k} (ξ)

, so by definition we have

C o d^{k} (β) = {ξ_{j}}_{j = 0}^{k - 1}

i.e

ξ_{i} = 0

C o d^{j} (β) = 0 mod 2

and

ξ_{i} = 10

C o d^{j} (β) = 1 mod 2

. We can rewrite as

β = β_{k} + 2^{A_{k}} α_{k}

for any

α_{k} \in Z_{2}

A_{k} =

numbers from 0 to

{ξ_{j}}_{j = 0}^{k - 1}

and

β_{k} = β mod 2^{A_{k}}

. We Claim that

β_{k} \in C o d^{k} (ξ)

. Indeed

\begin{matrix} C o l^{l} (β) = C o l^{l} (β_{k}) + 2^{A_{k} - l} α_{k} with l \leq A_{k}, \end{matrix}

We have that the parity on the right side only depends on

C o l^{l} (β_{k})

, then the latter must have the same

j -

coding as

C o l^{j} (β)

by Proposition 11 we have

β_{k} \in E (S_{k})

. So

S_{k} (β) = S_{k} (β_{k}) + α_{k} \in Z_{2}

, so

β \in D (S_{k})

Let

β \in D (S_{k})

and let

β_{k} = β mod 2^{A_{k}}

, then

β_{k} \in D (S_{k})

. On the other hand, since

β_{k}

trivially has a representation as a natural number, we have that

S_{k} (β_{k})

it is also a natural number, so

β_{k} \in E (S_{k})

by Proposition 11 we have that

β_{k} \in C o d^{k} (ξ)

. By hypothesis we have that

β_{k} = β mod 2^{A_{k}}

, then exist

α_{k} \in Z_{2}

such that

β = β_{k} + 2^{A_{k}} α_{k}

, applying

S_{k}

l -

times with

l \leq A_{k}

, we have

\begin{matrix} C o l^{l} (β) = C o l^{l} (β_{k}) + 2^{A_{k} - l} α_{k} \end{matrix}

Since

2^{A_{k} - l} α_{k}

is always even for all

l < A_{k}

, we have that the parity of each iteration only depends on

C o l^{l} (β_{k})

, then

β \in C o d^{k} (ξ)

QED.

In theorem 3 we saw that given

ξ \in Σ_{2}^{*}

if there exists a rational whose encoding is exactly

ξ

, then it is the only rational solution. However we could not guarantee the existence of such a number. The following Proposition guarantees us that there exists a solution in the set of dyadic numbers.

Proof: Let

ξ \in Σ_{2}^{*}

by Lemma 27 we have

- π^{1} (ξ) \in Z_{2}

. Now we are going to prove that:

Claim

- π^{1} (ξ) \in C o d^{k} (ξ)

for all

k \in N

. Indeed, let

S_{k} \in 〈 θ, ψ 〉

such that

C o d (S_{k}) = {ξ_{j}}_{j = 0}^{k}

, we have

S_{k} (- π_{k}^{1} (ξ)) = 0

for all

k \in N

, Then

\begin{matrix} - π^{1} (ξ) & = - π_{k}^{1} (ξ) - \sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} \\ so S_{k} (- π^{1} (ξ)) & = S_{k} (- π_{k}^{1} (ξ)) - \frac{3^{k}}{2^{a_{k}}} \sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} \\ = - \sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1} - a_{k}}}{3^{j - k}} \in Z_{2} \sin ce {∥\sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1} - a_{k}}}{3^{j - k}}∥}_{2} = 2^{- (a_{k} - a_{k})} = 1 \end{matrix}

By Proposition 29 we have

- π^{1} (ξ) \in C o d^{k} (ξ)

for all

k \in N

. Since

C o d^{k} (ξ) \subset C o d^{k - 1} (ξ)

, we have

\begin{matrix} - lim_{k \to \infty} π_{k}^{1} (ξ) \in ⋂_{k \in N} C o d^{k} (ξ) = C o d (ξ) \end{matrix}

Suppose there exists another dyadic integer

α

such that it is also in

C o d (ξ)

, then

\begin{matrix} α, - π^{1} (ξ) \in C o d (S_{k}) for all k \in N so {∥ π^{1} (ξ) - α ∥}_{2} < 2^{- A_{k}} \to 0 as k \to \infty \end{matrix}

Therefore

α = - π^{1} (ξ)

QED.

The existence of solutions to the equation

C o d (β) = ξ

in the dyadic numbers does not guarantee the existence of rational solutions. This will depend primarily on whether the dyadic solution can be represented as a rational number or, more generally, as a real number. Based on the nature of this solution, we can determine whether or not a rational solution exists.

8.3. Topological Conjugation

The Shift function

ω

Σ_{2}^{*}

is defined as the mapping that deletes the first term of the sequence. The following theorem states that the Collatz function on

Z_{2}

is dynamically equivalent to the Shift function on

Σ_{2}^{*}

and that the function

- π^{1}

is a homeomorphism between these two spaces. A similar result can be found in [3] where the Shift function is defined on

Z_{2}

instead of

Σ_{2}^{*}

Proof: Let’s first prove that the diagram is commutative.

Let

{ξ_{j}}_{j = 0}^{\infty} = 10^{θ_{1}} 10^{θ_{2}} \dots = \prod_{j = 0}^{\infty} 10^{θ_{j}} \in Σ_{2}^{*}

with

θ_{j} \in N

. Writing this way, we get an explicit form for the function

a_{k} = \sum_{j = 0}^{k} θ_{j}

. If

a_{0} = 0

we have

\begin{matrix} C o l \circ (- π^{1}) (\prod_{j = 0}^{\infty} 10^{θ_{j}}) & = C o l (- \sum_{k = 1}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty})}}{3^{k}}) \\ = \frac{1}{2} (- 3 \sum_{k = 1}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty})}}{3^{k}} + 1) \\ = \frac{1}{2} (- \sum_{k = 2}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty})}}{3^{k - 1}} + 1) \\ = \frac{1}{2} (- 1 - \sum_{k = 2}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty})}}{3^{k - 1}} + 1) \\ = - \sum_{k = 2}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty}) - 1}}{3^{k - 1}} \\ = - \sum_{k = 1}^{\infty} \frac{2^{a_{k} ({ξ_{j}}_{j = 0}^{\infty}) - 1}}{3^{k}} \end{matrix}

and

\begin{matrix} - π^{1} \circ ω (\prod_{j = 0}^{\infty} 10^{θ_{j}}) = - π^{1} (0^{θ_{0} - 1} \prod_{j = 1}^{\infty} 10^{θ_{j}}) = - \sum_{k = 1}^{\infty} \frac{2^{a_{k} ({ξ_{i}}_{i = 1}^{\infty}) - 1}}{3^{k}} \end{matrix}

where both parts are equal. On the other hand. suppose that

{ξ_{j}}_{j = 0}^{\infty} = 0^{θ_{0}} 10^{θ_{1}} 1 \dots = \prod_{j = 0}^{\infty} 0^{θ_{j}} 1 \in Σ_{2}^{*}

with

θ_{j} \in N

. We have

\begin{matrix} C o l \circ (- π^{1}) (\prod_{j = 0}^{\infty} 0^{θ_{j}} 1) = C o l (- \sum_{k = 1}^{\infty} \frac{2^{a_{k - 1} ({ξ_{i}}_{i = 0}^{\infty})}}{3^{k}}) = - \sum_{k = 1}^{\infty} \frac{2^{a_{k - 1} ({ξ_{i}}_{i = 0}^{\infty}) - 1}}{3^{k}} \end{matrix}

and

\begin{matrix} (- π^{1}) \circ ω (\prod_{j = 0}^{\infty} 0^{θ_{j}} 1) = (- π^{1}) \circ (θ^{θ_{0} - 1} \prod_{j = 1}^{\infty} 0^{θ_{j}} 1) = - \sum_{k = 1}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty}) - 1}}{3^{k}} \end{matrix}

where again both parts are equal. Then we conclude that the diagram is commutative.

Now we are going to prove that

- π^{1}

is a bijection.

Let

C o d : Z_{2} \to Σ_{2}^{*}

given by

ξ_{j} = 0

C o l^{j} (β) = 0 mod 2

and

ξ_{j} = 10

C o l^{j} (β) = 1 mod 2

. Let’s prove that

C o d \circ - π^{1} (ξ) = ξ

and

- π^{1} \circ C o d (β) = β

$C o d \circ - π^{1} = I d_{Σ_{2}^{*}}$ : By Corollary 30 we have $C o d (- π^{1} (ξ)) = ξ$ .
$- π^{1} \circ C o d = I d_{Z_{2}}$ : Let $β \in Z_{2}$ and $ξ \in Σ_{2}^{*}$ such that $ξ = C o d (β)$ , so $β \in C o d^{k} (ξ)$ for all $k \in N$ . On the other hand we have $- π^{1} (ξ) \in C o d^{k} (ξ)$ for all $k \in N$ . Exists $γ_{k} \in Z_{2}$ for all $k \in N$ such that $β = - π^{1} (ξ) + 2^{A_{k}} γ_{k}$ so $β = - π^{1} (ξ) mod 2^{a_{k}}$ then $∥ β - {- π^{1} (ξ)} ∥_{2} \leq 2^{- A_{k}} \to 0$ as $k \to \infty$ Thus $β = - π^{1} (C o d (β))$ .

Let us show that the applications

- π^{1}

and

C o d

are uniformly continuous. Let

D : Σ_{2}^{*} \times Σ_{2}^{*} \to [0, \infty)

the symbolic metric of two symbols given by

\begin{matrix} D (ξ, η) = \sum_{j = 1}^{\infty} \frac{1}{2^{j}} Δ (ξ_{j}, η_{j}), with ξ_{j}, η_{j} \in {0, 10} \end{matrix}

and

\begin{matrix} Δ (ξ_{j}, η_{j}) = \{\begin{matrix} 1 & if ξ_{j} \neq η_{j} \\ 0 & if ξ_{j} = η_{j} \end{matrix} \end{matrix}

The space

(Σ_{2}^{*}, D)

is a complete metric space with the property that if two sequences are arbitrarily close if and only if their first terms are equal.

\begin{matrix} D (ξ, η) < \frac{1}{2^{r}} if and only if ξ_{j} = η_{j} for all j < r . \end{matrix}

$π^{1}$ is uniformly continuous: Let

ε > 0

and

A \in N

such that

2^{- A} < ε

and

ξ, η \in Σ_{2}^{*}

. So let

δ = \frac{1}{2^{A + 2}}

\begin{matrix} D (ξ, η) < δ = \frac{1}{2^{A + 2}} then we have ξ_{j} = η_{j} for j \leq A + 1 \end{matrix}

Let

b_{A + 1}

the number from 1 to the

A + 1

-th term of

ξ

, then

a_{k - 1} (ξ) = a_{k - 1} (η)

for

k \leq b_{A + 1}

- \sum_{k = 1}^{b_{A + 1}} \frac{2^{a_{k - 1} (ξ)}}{3^{k}} = - \sum_{k = 1}^{b_{A + 1}} \frac{2^{a_{k - 1} (η)}}{3^{k}} \in C o d^{A + 1} (ξ)

, then in particular

\begin{matrix} - \sum_{k = 1}^{b_{A + 1}} \frac{2^{a_{k - 1} (ξ)}}{3^{k}} = - \sum_{k = 1}^{b_{A + 1}} \frac{2^{a_{k - 1} (η)}}{3^{k}} mod 2^{A + 1} \end{matrix}

\begin{matrix} - π^{1} (ξ) = - π^{1} (η) mod 2^{A + 1} then we have {∥ π^{1} (ξ) - π^{1} (η) ∥}_{2} \leq 2^{- A - 1} < 2^{- A} < ε . \end{matrix}

$C o d$ is uniformly continuous: Let

ε > 0

and

A \in N

such that

δ = 2^{- A} < ε

and

α, β \in Z_{2}

such that

{∥ α - β ∥}_{2} < δ = 2^{- A}

then

α = β mod 2^{A}

, so

C o d^{A} (α) = C o d^{A} (β)

, so

D (C o d (α), C o d (β)) < \frac{1}{2^{A}} < ε

Therefore

π^{1}

is a homemorphism and

C o l

is topologically conjugate to

ω

QED.

As a first consequence of topological conjugation, we have that the set of periodic points of the Collatz function is dense in

Z_{2}

Proof: consequence of the continuity of the function

- π^{1}

and the fact that the periodic sequences of the Shift function are dense in

Σ_{2}^{*}

QED.

Now we are going to characterize the rational representation of the periodic points. By Corollary 5 we have that the set of the periodic numbers is dense in the set of the dyadic numbers, we will use these results to show that there exists no rational number (in general real) whose encoding is in

G_{1}

Proof: Let

ξ \in Σ_{2}^{*}

periodic. Without loss of generality we can assume that

ξ = \prod_{j = 0}^{\infty} 10^{θ_{j}}

, then exit

K \in N

such that

a_{r K + j} = r a_{K} + a_{j - 1}

for all

R \in N

. Indeed. Since

ξ

is periodic, exist

K \in N

such that

\begin{matrix} 10^{θ_{0}} 10^{θ_{1}} \dots 10^{θ_{K}} 10^{θ_{0}} 10^{θ_{1}} \dots 10^{θ_{K}} 10^{θ_{0}} 10^{θ_{1}} \dots 10^{θ_{K}} \dots \end{matrix}

So, by induction we have:

\begin{matrix} a_{K + 1} & = θ_{0} + θ_{1} + \dots + θ_{K} + θ_{K + 1} = θ_{0} + θ_{1} + \dots + θ_{K} + θ_{0} = a_{K} + a_{0} \\ a_{K + 2} & = θ_{0} + θ_{1} + \dots + θ_{K} + θ_{K + 1} + θ_{K + 2} = θ_{0} + θ_{1} + \dots + θ_{K} + θ_{0} + θ_{1} = a_{K} + a_{1} \\ ⋮ \\ a_{2 K + 1} & = θ_{0} + θ_{1} + \dots + θ_{K} + θ_{K + 1} + θ_{K + 2} + \dots + θ_{2 K + 1} = θ_{0} + θ_{1} + \dots + θ_{K} + θ_{0} + θ_{1} + \dots + θ_{K} + θ_{0} \\ = 2 a_{K} + a_{0} \\ ⋮ \\ a_{r K + j} & = r a_{K} + a_{j - 1} \\ ⋮ \end{matrix}

We are going to show that

- π^{1} (ξ)

is rational

\begin{matrix} - \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{k}} & = - \{\frac{2^{a_{0}}}{3} + \frac{2^{a_{1}}}{3^{2}} + \dots + \frac{2^{a_{K - 1}}}{3^{K}}\} - \{\frac{2^{a_{K - 1} + a_{0}}}{3^{K + 1}} + \frac{2^{a_{K - 1} + a_{1}}}{3^{K + 2}} - \dots + \frac{2^{a_{K - 1} + a_{K - 1}}}{3^{2 K}}\} \\ - \dots - \{\frac{2^{r a_{K - 1} + a_{0}}}{3^{r K + 1}} + \frac{2^{r a_{K - 1} + a_{1}}}{3^{r K + 2}} + \dots + \frac{2^{r a_{K - 1} + a_{K - 1}}}{3^{(r + 1) K}}\} - \dots \\ = - \{\frac{2^{a_{0}}}{3} + \frac{2^{a_{2}}}{3^{2}} + \dots + \frac{2^{a_{K - 1}}}{3^{K}}\} - \frac{2^{a_{K - 1}}}{3^{K}} \{\frac{2^{a_{0}}}{3} + \frac{2^{a_{1}}}{3^{2}} + \dots + \frac{2^{a_{K}}}{3^{K}}\} \\ - \dots - {(\frac{2^{a_{k - 1}}}{3^{K}})}^{r} \{\frac{2^{a_{0}}}{3} + \frac{2^{a_{1}}}{3^{2}} + \dots + \frac{2^{a_{K - 1}}}{3^{K}}\} - \dots \\ = - \{\frac{2^{a_{0}}}{3} + \frac{2^{a_{1}}}{3^{2}} + \dots + \frac{2^{a_{K - 1}}}{3^{K}}\} \sum_{r = 1}^{\infty} {(\frac{2^{a_{k - 1}}}{3^{K}})}^{r} \\ = - \{\frac{2^{a_{0}}}{3} + \frac{2^{a_{1}}}{3^{2}} + \dots + \frac{2^{a_{K - 1}}}{3^{K}}\} (\frac{1}{1 - \frac{2^{a_{K - 1}}}{3^{K}}}) \in Z_{2} \end{matrix}

which corresponds to a rational number, where its sign depends on the denominator

1 - \frac{2^{a_{K}}}{3^{K}}

, if

ξ

are in

G_{0}

then we necessarily have that

\frac{3^{K}}{2^{a_{K}}}

must be less than 1, so the denominator is negative, so

- π^{1} (ξ)

is positive, analogously when

ξ

is in

G_{\infty}

we have that

\frac{3^{K}}{2^{a_{K}}}

must be greater than 1, so the denominator is positive, so

- π^{1} (ξ)

is negative.

QED.

We will now show a connection between the minimum positive integer value and the encoding of the sigma function.

Proof: Let

S_{k} \in 〈 θ, ψ 〉

. Then

\begin{matrix} ρ_{0} (S_{k}) = \frac{2^{A_{k}}}{3^{k}} σ_{3^{k}}^{A_{k}} (N_{k}) - \frac{N_{k}}{3^{k}} = \frac{2^{A_{k}}}{3^{k}} \{\frac{N_{k}}{2^{A_{k}}} + \frac{3^{k} C o d σ_{3^{k}}}{2^{A_{k}}}\} - \frac{N_{k}}{3^{k}} = C o d σ_{3^{k}}^{A_{k}} (N_{k}) = C o d σ^{A_{k}} (\frac{N_{k}}{3^{k}}) . \end{matrix}

QED.

As a consequence of the next proposition, we have that if

- π^{1}

is a negative or non-integer number, the minimum value diverges, since we have that the dyadic representation of these numbers always has an infinite amount of numbers.

Proof: Let

ξ = C o d ({S_{k}}_{k \in N}

. So

\begin{matrix} {∥C o d σ (π^{1} (ξ)) - C o d σ^{A_{k}} (\frac{N_{k}}{3^{k}})∥}_{2} = {∥π^{1} (ξ) - π_{k}^{1} (ξ)∥}_{2} \leq 2^{- a_{k}}, \sin ce π_{k}^{1} (ξ) = π^{1} (ξ) mod 2^{a_{k}} \end{matrix}

Then

C o d σ^{A_{k}} (\frac{N_{k}}{3^{k}}) = C o d σ (π^{1} (ξ)) mod 2^{a_{k}} .

QED.

The above proposition tells us that for an coding

ξ

to be positive, it is necessary and sufficient that the dyadic expansion of

- π^{1} (ξ)

has a finite amount of 1’s.

Now, we are going to show that for all

ξ \in G_{1}

, must necessarily be unstable. For this we will show that the encoding of

- π^{1} (ξ)

must necessarily have an infinite amount of 1

Proof: Let

ξ \in G_{1}

and

{ξ_{k}}_{k \in N}

such that

ξ_{k}

is periodic with period k and

ξ_{k} \to ξ

k \to \infty

. Due to periodicity, we have

ξ_{k} \in G_{0} \cup G_{\infty}

. By Proposition 31, we have that

- π^{1} (ξ_{k})

admits a rational representation

- π^{1} (ξ_{k}) = - \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{k}} = - \{\frac{2^{a_{0}}}{3} + \frac{2^{a_{1}}}{3^{2}} + \dots + \frac{2^{a_{k - 1}}}{3^{k}}\} (\frac{1}{1 - \frac{2^{a_{k - 1}}}{3^{k}}})

= - \{\frac{2^{a_{0}}}{3} + \frac{2^{a_{1}}}{3^{2}} + \dots + \frac{2^{a_{k - 1}}}{3^{k}}\} (\frac{1}{1 - {(\frac{2^{\frac{a_{k - 1}}{k}}}{3})}^{k}})

By the continuity of the function

- π^{1}

Z_{2}

, we have that

\begin{matrix} - π^{1} (ξ) & = - lim_{k \to \infty} π^{1} (ξ_{k}) \\ = - lim_{k \to \infty} \{\frac{2^{a_{0}}}{3} + \frac{2^{a_{1}}}{3^{2}} + \dots + \frac{2^{a_{k - 1}}}{3^{k}}\} (\frac{1}{1 - {(\frac{2^{\frac{a_{k - 1}}{k}}}{3})}^{k}}) \end{matrix}

This means that as k increases, the dyadic expansion on the right approaches the dyadic expansion of r, since

ξ_{k}

converges to

ξ

. On the other hand, we have that

lim_{k \to \infty} \frac{a_{k - 1}}{k} = \frac{ln (3)}{ln (2)}

This means that the rational representation of

- π^{1} (ξ_{k})

increases (in absolute value) as k increases. Then its dyadic expansion of

- π^{1} (ξ)

cannot have a finite amount of 1. By Proposition 32 we have that

ρ_{0} (S_{k}) = C o d σ (π^{1} (ξ)) mod 2^{a_{k}} \to \infty .

k \to \infty

QED.

Let

ξ \in G_{\infty}

, if

- π^{1} (ξ) \in Z_{2}

admits real representation then, this representation is negative? Unfortunately, we cannot obtain an answer for dyadic numbers in this way, as the sign of a representation is not determined through approximation as it is for real numbers. One might be tempted to argue that since the function

- π^{1}

converges in the real numbers, its representation in

Z_{2}

must coincide, but this is not always the case. However, we will show in the next section that they do indeed coincide, since in reality

- π^{1}

is a solution when it converges.

9. The Coding of $π^{1}$

Now we are going to prove that there is a complete metric on

G_{\infty}

. We will use this result to prove that if

π^{1} ({ξ_{j}}_{j = 0}^{\infty}) \in Q_{o d d}

then

C o d (π^{1} ({ξ_{j}}_{j = 0}^{\infty})) = {ξ_{j}}_{j = 0}^{\infty}

and in the case that

π^{1} ({ξ_{j}}_{j = 0}^{\infty}) \in R ∖ Q

, then there is no rational r such that

C o d (r) = {ξ_{j}}_{j = 0}^{\infty}

. We also demonstrate that the parity of the Collatz function on

π^{1} (G_{\infty})

depends solely on the first term. Building upon this insight, we extend the Collatz function to

π^{1} (G_{\infty})

and conclude the section by showing that the Collatz function is topologically conjugate to the Shift function on

Σ_{2}^{*}

. We will use this result to establish that the set of periodic orbits is dense.

9.1. Summary of Propositions in the Section

Lemma 11: Established that when the function $d (π^{1} (ξ), π^{1} (η)) = \sum_{j \in N} \frac{1}{3^{j}} | 2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)} | \leq \frac{1}{3^{r}}$ then $ξ$ and $η$ share at least the first $k - 1$ terms.
Proposition 33: Established that $(π^{1} (G_{\infty}), d)$ is a complete metric space.
Corollary 6: Established that the $k -$ coding set is an open set.
Theorem 7: Established that the full coding set is either as a singleton set or as an empty set depending on whether $π^{1} (ξ)$ is rational or not.
Proposition 34: It establishes that the parity of $π^{1} (ξ)$ depends only on the first term of the series.
Definition 21: Defines an extension of the Collatz function over all $π^{1} (G_{\infty})$ .
Proposition 35: The Collatz functions are continuous.
Proposition 36: The Collatz function on $π^{1} (G_{\infty})$ is topological conjugacy to Shift map on $Σ_{2}^{*}$
Corollary 7: It is stable that the periodic points of the Collatz function in $(π^{1} (G_{\infty}), d)$ are dense.

9.2. $π^{1} (G_{\infty})$ as Complete Metric Space

To ensure the coherent definition of a metric in

π^{1} (G_{\infty})

, we need to "complete" the missing terms of the series to enable the calculation of the difference

| 2^{a_{k - 1} (ξ)} - 2^{a_{k - 1} (η)} |

for all

k \in N

, irrespective of whether

ξ

η

has a null tail. To accomplish this, we define that when the sequence of 1s in

ξ

terminates, the function

a_{k}

will take on the value

- \infty

. Hence, we have

| 2^{a_{k - 1} (ξ)} - 2^{a_{k - 1} (η)} | = | 2^{- \infty} - 2^{a_{k - 1} (η)} | = 2^{a_{k - 1} (η)}

from the index of

ξ

. Let

ξ \in Σ_{2}^{*}

with null tail with index J. We will write

\begin{matrix} π^{1} ({ξ}_{j = 1}^{\infty}) : = \sum_{j = 1}^{J} \frac{2^{a_{k - 1}}}{3^{j}} + \sum_{j = J + 1}^{J} \frac{2^{- \infty}}{3^{j}} \end{matrix}

In the following lemma, we are going to introduce a new function, which, as we will see later, corresponds to a metric in the space

π^{1} (G_{\infty})

. Additionally, we will present another result that we will examine more closely in this section, and essentially indicates to us that, since the parity of

π^{1} (ξ)

depends only on the first term, we can interpret this in the following way: if two sequences are arbitrarily close, then they share the first terms of their encoding. This is of great importance for understanding the behavior of the orbits of the Collatz function, as, if we consider the Euclidean metric in

Q

or that of the absolute value, we observe the phenomenon that even though two numbers are arbitrarily close, their dynamics are completely different. One may converge to a cycle in a few iterations, while the other may take a very large amount of time.

Proof: Let

ξ, η \in G_{\infty}

, then we have

Claim Let,

a, b \in N

then

| 2^{a} - 2^{b} | < 2^{max {a, b}}

a = b

then

0 < 2^{a}

. Suppose that

a > b

then

\begin{matrix} | 2^{a} - 2^{b} | = 2^{a} |1 - \frac{1}{2^{a - b}}| < 2^{a} . \end{matrix}

QED of the Claim.

Now we prove that it is well-defined, by Claim we have:

\begin{matrix} d (\sum_{j = 1}^{\infty} \frac{2^{a_{j - 1} (ξ)}}{3^{j}}, \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1} (η)}}{3^{j}}) = \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| < \sum_{j = 1}^{\infty} \frac{2^{max {a_{j - 1} (ξ), a_{j - 1} (η)}}}{3^{j}} \end{matrix}

\frac{2^{a_{j - 1} (ξ)}}{3^{j}}

and

\frac{2^{a_{j - 1} (η)}}{3^{j}}

converge to 0, then for

n > N

exist

J \in N

such that if

j > J

we have

\begin{matrix} |\frac{2^{a_{j - 1} (ξ)}}{3^{j}}|, |\frac{2^{a_{j - 1} (η)}}{3^{j}}| < \frac{1}{n} \end{matrix}

Then for

j > J

we have too

\begin{matrix} \frac{2^{max {a_{j - 1} (ξ), a_{j - 1} (η)}}}{3^{j}} < \frac{1}{n} \end{matrix}

Then we have that

\frac{2^{max {a_{j - 1} (ξ), a_{j - 1} (η)}}}{3^{j}}

also converges to 0. Then by Proposition 16 we have

\sum_{j = 1}^{\infty} \frac{2^{max {a_{j - 1} (ξ), a_{j - 1} (η)}}}{3^{j}} < \infty

. To prove the statement, we will consider whether the sequences

ξ

and

η

Σ_{2}^{*}

have a null tail or not.

Let us first assume that the sequence does not have a null tail, then if we have

\begin{matrix} \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| \leq \frac{1}{3^{r}} \end{matrix}

All terms less than r must be null. Suppose there exists some non-zero term, then we have that

\begin{matrix} \frac{1}{3^{r - 1}} \leq \sum_{j = 1}^{r - 1} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| \leq \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| \leq \frac{1}{3^{r}} Then r - 1 \geq r \end{matrix}

which is absurd. Then we have that

\sum_{j = 1}^{r - 1} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| = 0

. Which implies that

\begin{matrix} a_{j - 1} (ξ) = a_{j - 1} (η) for all j < r . \end{matrix}

Now we will prove that the sequences coincide up to r.

\begin{matrix} ξ = 0^{θ_{0}^{1}} \prod_{i = 1}^{\infty} 10^{θ_{i}^{1}} and η = 0^{θ_{0}^{2}} \prod_{i = 1}^{\infty} 10^{θ_{i}^{2}} \end{matrix}

writing this way, we have to

\begin{matrix} a_{j} (ξ) = \sum_{i = 0}^{j} θ_{i}^{1} and a_{j} (η) = \sum_{i = 0}^{j} θ_{i}^{2} \end{matrix}

then

\begin{matrix} θ_{j - 1}^{1} = a_{j - 1} (ξ) - a_{j - 2} (ξ) = a_{j - 1} (η) - a_{j - 2} (η) = θ_{j - 1}^{2} for all 0 \leq j < r \end{matrix}

which means that

ξ

and

η

share the first

(r - 1) - 10^{θ_{j}}

blocks.

Now suppose that

ξ

has a null tail of index I and

η

has no tail null. Then we have

\begin{matrix} \sum_{j = 1}^{I} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| + \sum_{j = I + 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (η)}| \leq \frac{1}{3^{r}} \end{matrix}

r \leq I

then we have the previous case, then

a_{j - 1} (ξ) = a_{j - 1} (η) for all j < r

. Now if

r > I

we have

\begin{matrix} \sum_{j = 1}^{I} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| + \sum_{j = I + 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (η)}| \leq \frac{1}{3^{r}} \\ \Rightarrow \sum_{j = 1}^{I} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| + \sum_{j = I + 1}^{r - 1} \frac{1}{3^{j}} |2^{a_{j - 1} (η)}| = 0 \end{matrix}

The latter makes sense if

η

also has a null tail of index I, then

\begin{matrix} ξ_{j - 1} = η_{j - 1} for all j \in N \end{matrix}

In particular

ξ_{j - 1} = η_{j - 1} for all j < r

. Finally, suppose that

ξ

and

η

have a null tail of index I and L respectively, without loss of generality we can assume that

I \leq L

. Then

\begin{matrix} \sum_{j = 1}^{I} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| + \sum_{j = I + 1}^{L} \frac{1}{3^{j}} |2^{a_{j - 1} (η)}| + \sum_{j = L + 1}^{\infty} \frac{1}{3^{j}} |2^{- \infty}| \leq \frac{1}{3^{r}} \end{matrix}

If $1 \leq r \leq I$ then all terms with an index less than r are null and in particular we have $a_{j - 1} (ξ) = a_{j - 1} (η)$ for $j < r$ . and as we already saw in the proofs above, this implies that $ξ_{j - 1} = η_{j - 1}$ for all $j < r$ .
If $r \geq I$ . Then $\sum_{j = I + 1}^{L} \frac{1}{3^{j}} |2^{a_{j - 1} (η)}| = 0$ we have that $I = J$ therefore $ξ = η$ . particular we have $a_{j - 1} (ξ) = a_{j - 1} (η)$ for $j < r$ .

QED.

Now we are going to show that the function we defined above is a complete metric on

π^{1} (G_{\infty}) .

Proof: Let’s prove that d is a metric through the axioms of metric:

$d (π^{1} (ξ), π^{1} (η)) = 0$ if and only if $π^{1} (ξ) = π^{1} (η)$ for all $π^{1} (ξ), π^{1} (η) \in π^{1} (G_{\infty})$ : Trivially we have that if $π^{1} (ξ) = π^{1} (η)$ , then

$\begin{matrix} d (\sum_{j = 1}^{\infty} \frac{2^{a_{j - 1} (ξ)}}{3^{j}}, \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1} (ξ)}}{3^{j}}) = \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (ξ)}| = 0 . \end{matrix}$

Let $π^{1} (ξ), π^{1} (η) \in π^{1} (G_{\infty})$ such that

$\begin{matrix} d (\sum_{j = 1}^{\infty} \frac{2^{a_{j - 1} (ξ)}}{3^{j}}, \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1} (η)}}{3^{j}}) = \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| = 0 \end{matrix}$

by lemma 11 we have

$\begin{matrix} \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| \leq \frac{1}{3^{r}} \Rightarrow a_{j - 1} (ξ) = a_{j - 1} (η) for j < r \end{matrix}$

In particular, for $r \to \infty$ we have $ξ = η$ .
$d (π^{1} (ξ), π^{1} (η)) = d (π^{1} (η), π^{1} (ξ))$ for all $π^{1} (ξ), π^{1} (η) \in π^{1} (G_{\infty})$ :

$\begin{matrix} \sum_{j = 1}^{k} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| = \sum_{j = 1}^{k} \frac{1}{3^{j}} |2^{a_{j - 1} (η)} - 2^{a_{j - 1} (ξ)}| \end{matrix}$

then

$\begin{matrix} \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| = \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (η)} - 2^{a_{j - 1} (ξ)}| \end{matrix}$
$d (π^{1} (ξ), π^{1} (η)) \leq d (π^{1} (ξ), π^{1} (κ)) + d (π^{1} (κ), π^{1} (η))$ for all, $π^{1} (ξ), π^{1} (η), π^{1} (κ) \in π^{1} (G_{\infty})$

$\begin{matrix} \sum_{j = 1}^{k} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| \leq \sum_{j = 1}^{k} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (κ)}| + \sum_{j = 1}^{k} \frac{1}{3^{j}} |2^{a_{j - 1} (κ)} - 2^{a_{j - 1} (η)}| \end{matrix}$

then

$\begin{matrix} \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (η)}| \leq \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (κ)}| + \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (κ)} - 2^{a_{j - 1} (η)}| \end{matrix}$

then

(π^{1} (G_{\infty}), d)

is a metric space. Now we are going to prove that it is a complete metric space. Let

{{π^{1} (ξ_{j}^{k})}_{j = 1}^{\infty}}_{k = 1}^{\infty}

be a Cauchy sequence on

π^{1} (G_{\infty})

then for any

ε > 0

exist

K > 0

such that

\begin{matrix} d ({π^{1} (ξ_{j}^{n})}_{j = 1}^{\infty}, {π^{1} (ξ_{j}^{m})}_{j = 1}^{\infty}) = \sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ^{n})} - 2^{a_{j - 1} (ξ^{m})}| < ε for all n, m > K \end{matrix}

Let

r > 0

such that

\sum_{j = 1}^{\infty} \frac{1}{3^{j}} |2^{a_{j - 1} (ξ^{n})} - 2^{a_{j - 1} (ξ^{m})}| < \frac{1}{3^{r}} < ε

by lemma 11 we have

\begin{matrix} ξ_{j}^{n} = ξ_{j}^{m} for all j < r \end{matrix}

On the other hand let

D : Σ_{2}^{*} \times Σ_{2}^{*} \to [0, \infty)

the symbolic metric of two symbols given by

\begin{matrix} D (ξ, η) = \sum_{j = 1}^{\infty} \frac{1}{2^{j}} Δ (ξ_{j}, η_{j}) \end{matrix}

with

\begin{matrix} Δ (ξ_{j}, η_{j}) = \{\begin{matrix} 1 & if ξ_{j} \neq η_{j} \\ 0 & if ξ_{j} = η_{j} \end{matrix} \end{matrix}

The space

(Σ_{2}^{*}, D)

is a complete metric space with the property that if two sequences are arbitrarily close if and only if their first terms are equal.

\begin{matrix} D (ξ, η) < \frac{1}{2^{r}} if and only if ξ_{j} = η_{j} for all j < r . \end{matrix}

then given a Cauchy sequence in

π_{1} (G_{\infty})

by the observation above we obtain a Cauchy sequence in

Σ_{2}^{*}

and the latter being complete there is a

ξ \in Σ_{2}^{*}

such that

\begin{matrix} ξ^{n} \to ξ as n \to \infty \end{matrix}

We will now prove that

ξ

is in

G_{\infty}

and that the sequence

π_{1} (ξ^{n})

converges to

π_{1} (ξ)

Suppose that $ξ$ is not in $G_{\infty}$ , so $\frac{2^{a_{j - 1} (ξ)}}{3^{j}}$ does not converge to 0, then exist $L, k > 0$ and subsequence $a_{j_{k} - 1} (ξ)$ such that $\frac{2^{a_{j_{k} - 1} (ξ)}}{3^{j_{k}}} > L$ for all $k > K$ . Since $ξ^{n}$ converges to $ξ$ . Exist $N > 0$ such that

$\begin{matrix} D (ξ^{n}, ξ) < \frac{1}{2^{r}} for n > N \end{matrix}$

then the first $r -$ terms begin to equal, then

$\begin{matrix} |\frac{2^{a_{j - 1} (ξ)}}{3^{j}} - \frac{2^{a_{j - 1} (ξ^{n})}}{3^{j}}| = 0 for all j \leq r \end{matrix}$

Since $π^{1} (ξ^{n})$ is a sequence in $π_{1} (G_{\infty})$ for all $k \in N$ then $\frac{2^{a_{j - 1} (ξ^{n})}}{3^{j}} \to 0$ as $j \to \infty$ . In particular for L exist $J > K > 0$ such that $\frac{2^{a_{j - 1} (ξ^{n})}}{3^{j}} < L$ so $\frac{2^{a_{j_{k} - 1} (ξ)}}{3^{j_{k}}} < L$ this is a contradiction, then $ξ$ is in $G_{\infty}$ .
Let us suppose, for absurdity, that $D (ξ, ξ^{n}) \to 0$ but $d (π_{1} (ξ), π_{1} (ξ^{n}))$ does not converge to 0, then exist $k, N > 0$ such that

$\begin{matrix} d (π^{1} (ξ), π^{1} (ξ^{n})) \geq \frac{1}{3^{k}} for all n > N \end{matrix}$

In particular, exists in $N_{0}$ such that the $a_{j}$ of both sequences are no longer equal, otherwise, we would have that for N that is large enough the distance would be less than $\frac{1}{3^{k}}$ . On the other hand, the fact that they $a_{j - 1}$ are different implies that the terms of $ξ$ and $ξ^{n}$ must be different for $N > N_{0}$ , but this contradicts the fact that the sequences $ξ$ and $ξ^{n}$ come closer, since with the metric D getting closer is the same as having the first terms of the sequences become equal.

then we can conclude that the metric space

(π^{1} (G_{\infty}), d)

is complete.

QED.

With this metric, we have that the coding sets are open sets.

Proof: Let

ξ \in G_{\infty}

and

u \in π^{1} (G_{\infty})

such that

u \in C o d^{k} (ξ)

. Let us consider v in

π^{1} (G_{\infty})

such that

d (u, v) < \frac{1}{3^{k + 1}}

, by definition, exist

μ, τ \in G_{\infty}

such that

π^{1} (μ) = u

and

π^{1} (τ) = v

. By Lemma 11 have

μ_{j} = τ_{j}

for

j \leq k

, then we have that

v \in C o d^{k} (ξ)

. Therefore, then

B (u, \frac{1}{3^{k + 1}})

i.e. the ball of radius

\frac{1}{3^{k + 1}}

and center u is a subset of

C o d^{k} (ξ)

, therefore

C o d^{k} (ξ)

is an open set.

QED.

9.3. Characterization of the Full Coding Sets through the Function $π^{1}$

Proof: Let

ξ \in G_{\infty}

, We have by definition that

π^{1} (ξ) = lim_{k \to \infty} π_{k}^{1} (ξ) \in Q

. First, we will prove that this limit also makes sense in

(π^{1} (G_{\infty}), d)

. We have:

Claim 1:

d (π_{k}^{1} (ξ), π^{1} (ξ)) \to 0

k \to \infty

Indeed, we have

\begin{matrix} d (π_{k}^{1} (ξ), π^{1} (ξ)) = \underset{0}{\underset{︸}{\sum_{j = 1}^{n} \frac{1}{3^{j}} | 2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (ξ)} |}} + \sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1} (ξ)}}{3^{j}} = \sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1} (ξ)}}{3^{j}} \end{matrix}

By Proposition 16, we have

\sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1} (ξ)}}{3^{j}} \to 0

k \to \infty

, then

d (π_{k}^{1} (ξ), π^{1} (ξ)) \to 0

k \to \infty

QED of the Claim.

We will now prove, using the completeness of

(π^{1} (G_{\infty}), d)

, that

π^{1} (ξ) \in π^{1} (G_{\infty})

Claim 2:

π^{1} (ξ) \in π^{1} (G_{\infty})

We can be rewritten,

\begin{matrix} π_{k}^{1} (ξ) = π^{1} ({ξ_{j}}_{j = 1}^{k} 000 \dots) \end{matrix}

let’s prove that

π_{k}^{1} (ξ)

is a Cauchy sequence in

π^{1} (G_{\infty})

. Let

ε > 0

and

n, m \in N

with

n < m

, then

\begin{matrix} d (π_{n}^{1} (ξ), π_{m}^{1} (ξ)) = \underset{0}{\underset{︸}{\sum_{j = 1}^{n} \frac{1}{3^{j}} | 2^{a_{j - 1} (ξ)} - 2^{a_{j - 1} (ξ)} |}} + \sum_{j = n + 1}^{m} \frac{2^{a_{j - 1} (ξ)}}{3^{j}} = \sum_{j = n + 1}^{m} \frac{2^{a_{j - 1} (ξ)}}{3^{j}} \end{matrix}

Since

ξ \in G_{\infty}

we have by Proposition 16

π^{1} (ξ)

is convergent, then

\sum_{j = n + 1}^{m} \frac{2^{a_{j - 1} (ξ)}}{3^{j}} \to 0

n, m \to \infty

. Therefore, exists

N \in N

such that

n, m > N

we have

\sum_{j = n + 1}^{m} \frac{2^{a_{j - 1} (ξ)}}{3^{j}} < ε

. Then the sequence is Cauchy and since

π^{1} (G_{\infty})

is a complete metric space, we have that

π^{1} (ξ) = lim_{k \to \infty} π_{k}^{1} (ξ) \in π^{1} (G_{\infty})

QED of the Claim.

Claim 3:

- π_{k}^{1} (ξ) \in C o d^{k} (ξ)

Indeed, let

S_{k} \in 〈 θ, ψ 〉

such that

C o d (S_{k}) = {ξ_{j}}_{j = 1}^{k}

by Proposition 12 we have

\begin{matrix} S_{k} (- π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty}) = \frac{3^{b_{k}} (- π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty}) + N_{k}}{2^{A_{k}}} = \frac{3^{b_{k}} (- \frac{N_{k}}{3^{b_{k}}}) + N_{k}}{2^{A_{k}}} = 0 . \end{matrix}

In other hand, we have

\begin{matrix} \frac{3^{b_{k}} (- \frac{N_{k}}{3^{b_{k}}}) + N_{k}}{2^{A_{k}}} = \frac{3^{b_{k}} (- N_{k}) + 3^{b_{k}} (N_{k})}{3^{b_{k}} 2^{A_{k}}} = \frac{1}{3^{b_{k}}} (\frac{3^{b_{k}} (- N_{k}) + 3^{b_{k}} (N_{k})}{2^{A_{k}}}) = \frac{1}{3^{b_{k}}} S^{3^{b_{k}}} \end{matrix}

then

- N_{k} \in E (S^{3^{b_{k}}})

, by Proposition 8 we have

C o d_{3^{b_{k}}} (S_{k}^{3^{b_{k}}}) = C o d (S_{k})

. Therefore

C o d^{k} (\frac{- N_{k}}{3^{b_{k}}}) = C o d^{k} (- π_{k}^{1} (ξ)) = C o d_{3^{b_{k}}} (S^{3^{b_{k}}}) = C o d (S_{k}) = {ξ_{j}}_{j = 1}^{k}

QED of the Claim.

Claim 4:

- π^{1} (ξ) \in C o d^{k} (ξ)

for all

k \in N

Let

L + 1 \in N

, we have by Claim 1, exist

K \in N

such that if

k > K > L + 1

d (π_{k}^{1} (ξ), π^{1} (ξ)) < \frac{1}{3^{L + 1}}

, By Lemma 11 we have to share the first L terms of the coding. On the other hand, by Claim 3, we have that

- π_{k}^{1} (ξ) \in C o d^{k} (ξ)

and since we have

k > K > L + 1

, due to the monotony of

C o d

, we have

- π_{k}^{1} (ξ) \in C o d^{L} (ξ)

. Therefore

- π^{1} (ξ) \in C o d^{L} (ξ)

QED of the Claim.

Claim 5:

π^{1} (ξ) \in C o d (ξ)

Since

- π^{1} (ξ) \in C o d^{k} (ξ)

for all

k \in N

, we have to

\begin{matrix} - π^{1} (ξ) \in ⋂_{k \in N} C o d^{k} (ξ) = C o d (ξ) . \end{matrix}

QED of the Claim.

Claim 6:

C o d (- π^{1} ({ξ_{j}}_{j = 0}^{\infty})) = {ξ_{j}}_{j = 0}^{\infty}

We first show that

π^{1} (ξ) \in Q_{o d d}

. Let’s assume that

π^{1} ({ξ_{j}}_{j = 0}^{\infty})

converges to a fraction with an even denominator; then its coding is

1111 \dots

. However,

1111 \dots

is not an element of

Σ_{2}^{*}

, which leads to a contradiction with claim 4. Then we have

π^{1} (ξ) \in Q_{o d d}

and, by Theorem 3, we have

C o d (- π^{1} ({ξ_{j}}_{j = 0}^{\infty})) = {ξ_{j}}_{j = 0}^{\infty}

QED of the Claim.

For the next part of the proposition, we will leverage the results presented in Section 7. In this section, we introduce the Sigma function along with its main properties and applications in solving linear Diophantine equations. It serves as an alternative to classical methods for solving this type of equation.

Claim 7: If

π^{1} ({ξ_{j}}_{j = 0}^{\infty})

it is irrational, then there is no rational then there is no rational

\frac{p}{q}

such that

c o d (\frac{p}{q}) = {ξ_{j}}_{j = 0}^{\infty}

We will prove that there is no rational solution, By the theorem 2 we have that if there is another rational solution it must be a minimum positive integer value or a maximum negative integer value for

S_{k} (x) \in 〈 θ, ψ^{q} 〉

such that

C o d (S_{k}) = {ξ_{j}}_{j = 1}^{k}

for unique q not null, by proposition 7 we have

\begin{matrix} S_{k} (x) = \frac{3^{b_{k}} x + q N_{k} ({ξ_{j}}_{j = 0}^{\infty})}{2^{A_{k}}} \end{matrix}

by Propositions 4 and 21 the minimum positive integer value is

\begin{matrix} ρ_{0} (S_{k}) = \frac{1}{3^{b_{k}}} (2^{a_{k}} σ_{3^{b_{k}}}^{a_{k}} (q N_{k}) - q N_{k}) = 2^{a_{k}} σ^{a_{k}} (q π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty}) - q π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty}) = C o d σ^{a_{k}} (q π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty}) \end{matrix}

and the maximum negative integer value is

\begin{matrix} ρ_{1} (S_{k}) = \frac{1}{3^{b_{k}}} (2^{a_{k}} σ_{- 3^{b_{k}}}^{a_{k}} (q N_{k}) - q N_{k}) & = \frac{1}{3^{b_{k}}} (2^{a_{k}} σ_{3^{b_{k}}}^{a_{k}} (q N_{k}) - q N_{k}) - 2^{a_{k}} \\ = C o d σ^{a_{k}} (q π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty}) - 2^{a_{k}} \end{matrix}

when

\begin{matrix} C o d σ^{a_{k}} (q π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty}) = \sum_{j = 0}^{a_{k} - 1} δ_{j} 2^{j} with δ_{j} \in {0, 1} \end{matrix}

On the other hand, by Proposition 32 we have that the coding of the

σ^{a_{k}} (q π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty})

corresponds to the dyadic expansion of

- q π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty})

. Since

π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty})

is irrational, then the dyadic expansion of

q π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty})

will never have a tail of 0 or 1 for all

q \in Z

, so

C o d σ^{a_{k}} (q π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty})

does not converge for all

q \in Z

. In particular we have to

ρ_{0} (S_{k}) \to \infty

. Now for

ρ_{1} (S_{k})

, we have

\begin{matrix} ρ_{1} (S_{k}) = \sum_{j = 0}^{a_{k} - 1} δ_{j} 2^{j} - 2^{a_{k}} = \sum_{j = 0}^{a_{k} - 1} δ_{j} 2^{j} - \sum_{j = 0}^{a_{k} - 1} 2^{j} = \sum_{j = 0}^{a_{k} - 1} (δ_{j} - 1) 2^{j} \end{matrix}

for this sum to be finite it is necessary exist

J > 0

such that

δ_{j}

are all 1 for

j > J

, however as

q π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty})

this is impossible, then this sum is divergent.

Then if

π_{k}^{1} ({ξ_{j}}_{j = 0}^{\infty})

is irrational then there is no rational

\frac{p}{q}

such that

c o d (\frac{p}{q}) = {ξ_{j}}_{j = 0}^{\infty}

QED of the Claim.

QED.

Example 12.

Let $ξ_{1} = 101010 \dots \in Σ_{2}^{*}$ , then $π^{1} (ξ) = 1$ (see Example 7). Therefore $C o d (- 1) = 101010 \dots$ .
Let $ξ_{2} = 1010010100 \dots \in Σ_{2}^{*}$ , then

$\begin{matrix} π^{1} (ξ_{2}) & = \frac{1}{3} + \frac{2}{3^{2}} + \frac{2^{3}}{3^{3}} + \frac{2^{4}}{3^{4}} + \frac{2^{6}}{3^{5}} + \frac{2^{7}}{3^{6}} + \frac{2^{9}}{3^{7}} + \dots \\ = \frac{1}{3} + \{\frac{2}{3^{2}} + \frac{2^{3}}{3^{3}}\} + \frac{2^{3}}{3^{2}} \{\frac{2}{3^{2}} + \frac{2^{3}}{3^{3}}\} + {(\frac{2^{3}}{3^{2}})}^{2} \{\frac{2}{3^{2}} + \frac{2^{3}}{3^{3}}\} + \dots \\ = \frac{1}{3} + \{\frac{2}{3^{2}} + \frac{2^{3}}{3^{3}}\} \sum_{j = 0}^{\infty} {(\frac{2^{3}}{3^{2}})}^{j} \\ = \frac{1}{3} + \{\frac{2}{3^{2}} + \frac{2^{3}}{3^{3}}\} \frac{1}{1 - \frac{2^{3}}{3^{2}}} = 5 \end{matrix}$

Therefore $C o d (- 5) = 1010010100 \dots$

9.4. Extension of the Collatz Function on $π^{1} (G_{\infty})$

Proof Let’s prove that the parity of

π^{1} ({ξ_{j}}_{j = 0}^{\infty})

only depends on the first term

\begin{matrix} π^{1} ({ξ_{j}}_{j = 0}^{\infty}) = \frac{2^{a_{0}}}{3} + \sum_{j = 2}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} = \frac{2^{a_{0}}}{3} + 2 (\sum_{j = 2}^{\infty} \frac{2^{a_{j - 1} - 1}}{3^{j}}) \end{matrix}

Claim: The series

\sum_{j = 2}^{\infty} \frac{2^{a_{j - 1} - 1}}{3^{j}}

cannot converge to a fraction with an even denominator.

Let us assume by contradiction that have

\sum_{j = 2}^{\infty} \frac{2^{a_{j - 1} - 1}}{3^{j}} = \frac{p}{2 q}

with

(q, p) = (p, 2) = 1

. Let

{ξ_{j}}_{j = 0}^{l} 0 \dots \in G_{\infty}

such that

π^{1} {ξ_{j}}_{j = 0}^{l} 0 \dots = \frac{2^{a_{1}}}{3}

. Let

η \in G_{\infty}

given by

η = {ξ_{j}}_{j = l + 1}^{\infty}

π^{1} (η) = π^{1} {ξ_{j}}_{j = l + 1}^{\infty} = \sum_{j = 2}^{\infty} \frac{2^{a_{j - 1} - 1}}{3^{j}} = \frac{p}{2 q}

and since

η

are in

G_{\infty}

, this generates a contradiction to the Theorem 7.

QED of the Claim.

Let

\sum_{j = 2}^{\infty} \frac{2^{a_{j} - 1}}{3^{j}} = \frac{p}{q}

with

(p, q) = (q, 2) = 1

. We have:

\begin{matrix} π^{1} ({ξ_{j}}_{j = 0}^{\infty}) = \frac{2^{a_{0}}}{3} + \sum_{j = 2}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} = \frac{2^{a_{0}}}{3} + 2 (\sum_{j = 2}^{\infty} \frac{2^{a_{j - 1} - 1}}{3^{j}}) = \frac{2^{a_{0}}}{3} + \frac{2 p}{q} = \frac{2^{a_{0}} q + 6 p}{3 q} \end{matrix}

a_{0} = 0

then

2^{0} q + 6 p = q + 6 p

is odd, since q is odd. Then

π^{1} ({ξ_{j}}_{j = 0}^{\infty})

is odd, and if

a_{0} > 0

then

2^{a_{1}} q + 6 p

is even. Then

π^{1} ({ξ_{j}}_{j = 0}^{\infty})

is even. In the case that

a_{0} = - \infty

we have that

ξ = 0

or equivalent

ξ_{j} = 0

for all

j \in N_{0}

, so

π^{1} (0) = 0

, then

C o l (0) = \frac{0}{2} = 0

QED.

From the result above, we can extend the Collatz function over the entire set

G_{\infty}

, thus obtaining a way of defining Collatz for the cases where the function

π^{1}

is irrational.

The extension of the Collatz function on -

π^{1} (G_{\infty})

is continuous.

Proof: Let us consider the metric induced in

- π^{1} (G_{\infty})

- i d : π^{1} (G_{\infty}) \to - π^{1} (G_{\infty})

, that is,

d_{- π^{1} (G_{\infty})} (- u, - v) = d (u, v)

π^{1} (G_{\infty})

, we will use the same notation for both metrics.

Let

u \in - π^{1} (G_{\infty})

and

{u_{j}}_{j \in N}

sequence of

- π^{1} (G_{\infty})

such that

u_{j} \to u

. Let

ξ, ξ_{j} \in G_{\infty}

such that

C o d (u) = ξ

and

C o d (ξ_{j}) = u_{j}

. Let

ε > 0

and

r \in N

such that

r > 2

and

\frac{1}{3^{r - 1}} < ε

\begin{matrix} d (u_{j}, u) = \sum_{k \in N} \frac{| 2^{a_{k - 1} (ξ_{j})} - 2^{a_{k - 1} (ξ)} |}{3^{k}} < \frac{1}{3^{r - 1}} \end{matrix}

Since

r > 3

then at least the first term must coincide.

Suppose that,

a_{0} > 0

then

\begin{matrix} d (C o l (u_{j}), C o l (u)) & = \sum_{k \in N} \frac{| 2^{a_{k - 1} (ξ_{j}) - 1} - 2^{a_{k - 1} (ξ) - 1} |}{3^{k}} \\ = \frac{2}{2} \sum_{k \in N} \frac{| 2^{a_{k - 1} (ξ_{j}) - 1} - 2^{a_{k - 1} (ξ) - 1} |}{3^{k}} \\ = \frac{1}{2} \sum_{k \in N} \frac{| 2^{a_{k - 1} (ξ_{j})} - 2^{a_{k - 1} (ξ)} |}{3^{k}} \\ < \frac{1}{3^{r - 1} 2} = \frac{ε}{2} < ε \end{matrix}

Suppose now that

a_{0} = 0

\begin{matrix} d (C o l (u_{j}), C o l (u)) & = \sum_{k \in N} \frac{| 2^{a_{k} (ξ_{j}) - 1} - 2^{a_{k} (ξ) - 1} |}{3^{k}} \\ = \frac{2}{2} \frac{3}{3} \sum_{k > 1} \frac{| 2^{a_{k - 1} (ξ_{j}) - 1} - 2^{a_{k - 1} (ξ) - 1} |}{3^{k - 1}} \\ = \underset{0}{\underset{︸}{\frac{1}{3} | 2^{a_{0} (ξ)} - 2^{a_{0} (ξ_{j})} |}} + \frac{3}{2} \sum_{k > 1} \frac{| 2^{a_{k - 1} (ξ_{j})} - 2^{a_{k - 1} (ξ)} |}{3^{k}} \\ = \frac{3}{2} \sum_{k \in N} \frac{| 2^{a_{k - 1} (ξ_{j})} - 2^{a_{k - 1} (ξ)} |}{3^{k}} \\ < \frac{3}{3^{r - 1} 2} = \frac{1}{3^{r} 2} < \frac{1,}{3^{r - 1} 2} < ε \end{matrix}

Therefore

C o l

is continuous.

QED.

Now we will prove that the extension of the function Collatz on

G_{\infty}

is topologically conjugate to the shift function in

Σ_{2}^{*}

9.5. Topological Conjugation

In a dynamic system, there is a well-studied dynamics in the space of sequences of two symbols, known as the Shift map. This map acts on the sequences by eliminating the first term. It is known that with the metric D, this map is continuous, and its periodic orbits form a dense set. In the following proposition, we will show that the extension of the Collatz function on

- π^{1} (G_{\infty})

is, in fact, topologically conjugate to the dynamics of the Shift map.

Proof: We are going to prove that this diagram is commutative

and that

- π^{1}

is a homeomorphism.

Claim 1: The diagram is commutative

Suppose that

{ξ_{j}}_{j = 0}^{\infty} = 10^{θ_{1}} 10^{θ_{2}} \dots = \prod_{j = 1}^{\infty} 10^{θ_{j}} \in G_{\infty}

with

θ_{j} \in N

. Writing this way, we get an explicit form for the function

a_{k} = \sum_{j = 0}^{k} θ_{j}

. If

a_{0} = 0

we have

\begin{matrix} C o l \circ (- π^{1}) (\prod_{j = 1}^{\infty} 10^{θ_{j}}) = C o l (- \sum_{j = 1}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty})}}{3^{k}}) & = \frac{1}{2} (- 3 \sum_{k = 1}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty})}}{3^{k}} + 1) \\ = \frac{1}{2} (- \sum_{k = 2}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty})}}{3^{k - 1}} + 1) \\ = \frac{1}{2} (- 1 - \sum_{k = 2}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty})}}{3^{k - 1}} + 1) \\ = - \sum_{k = 2}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty}) - 1}}{3^{k - 1}} \\ = - \sum_{k = 1}^{\infty} \frac{2^{a_{k} ({ξ_{j}}_{j = 0}^{\infty}) - 1}}{3^{k}} \end{matrix}

and

\begin{matrix} - π^{1} \circ ω (\prod_{j = 1}^{\infty} 10^{θ_{j}}) = - π^{1} (0^{θ_{1} - 1} \prod_{j = 2}^{\infty} 10^{θ_{j}}) = - \sum_{j = 1}^{\infty} \frac{2^{a_{k} ({ξ_{i}}_{i = 1}^{\infty}) - 1}}{3^{k}} \end{matrix}

where both parts are equal. On the other hand. suppose that

{ξ_{j}}_{j = 0}^{\infty} = 0^{θ_{0}} 10^{θ_{1}} 1 \dots = \prod_{j = 0}^{\infty} 0^{θ_{j}} 1 \in G_{\infty}

with

θ_{j} \in N

. We have

\begin{matrix} C o l \circ (- π^{1}) (\prod_{j = 0}^{\infty} 0^{θ_{j}} 1) = C o l (- \sum_{j = 0}^{\infty} \frac{2^{a_{k - 1} ({ξ_{i}}_{i = 0}^{\infty})}}{3^{k}}) = - \sum_{j = 1}^{\infty} \frac{2^{a_{k - 1} ({ξ_{i}}_{i = 0}^{\infty}) - 1}}{3^{k}} \end{matrix}

and

\begin{matrix} (- π^{1}) \circ ω (\prod_{j = 1}^{\infty} 0^{θ_{j}} 1) = (- π^{1}) \circ (θ^{θ_{0} - 1} \prod_{j = 1}^{\infty} 0^{θ_{j}} 1) = - \sum_{j = 1}^{\infty} \frac{2^{a_{k - 1} ({ξ_{j}}_{j = 0}^{\infty}) - 1}}{3^{k}} \end{matrix}

where again both parts are equal. Then we conclude that the diagram is commutative.

QED of the Claim.

Claim 2:

- π^{1} : G_{\infty} \to - π^{1} (G_{\infty})

with It is a bijective function.

Let

C o d : - π^{1} (G_{\infty}) \to G_{\infty}

let’s prove that

C o d \circ - π^{1} = I d_{G_{\infty}}

and

- π^{1} \circ C o d = I d_{- π^{1} (G_{\infty})}

$C o d \circ - π^{1} = I d_{G_{\infty}}$ :

Let $ξ = ξ_{0} ξ_{1} ξ_{2} \dots \in G_{\infty}$ with $ξ_{j} \in {0, 10}$ . Since the parity of $- \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}}$ depends only on the first term, if $ξ$ starts with 0 then $a_{0} > 0$ , then $- \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}}$ is even then the first term of its coding is 0, and if $ξ$ starts with 1 then $a_{0} = 0$ then $- \sum_{j = 1}^{\infty} \frac{2^{a_{j}}}{3^{j}}$ is odd then the first term of coding is 10. By applying the Collatz function we obtain the same result as applying a translation of the terms of $ξ$ . Indeed

(a)

if $a_{0} = 0$

$\begin{matrix} C o l (- \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}}) = \sum_{j = 1}^{\infty} \frac{2^{a_{j} - 1}}{3^{j}} = - π^{1} (ξ_{1} ξ_{2} \dots) . \end{matrix}$

.

(b)

if $a_{0} > 0$

$\begin{matrix} C o l (- \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}}) = \sum_{j = 1}^{\infty} \frac{2^{a_{j - 1} - 1}}{3^{j}} = - π^{1} (ξ_{1} ξ_{2} \dots) . \end{matrix}$

Then applying the function $- π^{1}$ . Then we can repeat the same procedure and we recover $ξ$ . Therefore

$\begin{matrix} C o d (- π^{1} (ξ)) = ξ \end{matrix}$
$- π^{1} \circ C o d = I d_{- π^{1} (G_{\infty})}$ :

Let $u \in - π^{1} (G_{\infty})$ and $ξ \in G_{\infty}$ such that $- π^{1} (ξ) = u$ . On the other hand we have $C o d (- π^{1} (ξ)) = C o d (u)$ and $C o d (- π^{1} (ξ)) = ξ$ then $C o d (u) = ξ$ , applying $- π^{1}$ on both sides we have $- π^{1} (C o d (u)) = - π^{1} (ξ) = u$ .

QED of the Claim.

Claim 3:

- π^{1} : G_{\infty} \to - π^{1} (G_{\infty})

is continuous.

Let

ξ \in G_{\infty}

and

{η_{j}}_{l \in N}

a sequence on

G_{\infty}

such that

η_{j} \to ξ

j \to \infty

. Let

r \in N

then if

D (η_{j}, ξ) < \frac{1}{2^{r}}

implies that the first elements are equal, then:

\begin{matrix} d (π^{1} (η_{j}), π^{1} (ξ)) = \sum_{k = r}^{\infty} \frac{| 2^{a_{k - 1} (η_{j}))} - 2^{a_{k - 1} (η))} |}{3^{k}} \to 0 as r \to \infty \end{matrix}

QED of the Claim.

Claim 4:

C o d : - π^{1} (G_{\infty}) \to G_{\infty}

is continuous.

Let

u \in - π^{1} (G_{\infty})

and let

ε > 0

. Take

r \in N

such that

\frac{1}{2^{r}} < ε

. then by Lemma 11 we have

\begin{matrix} d (u, v) < \frac{1}{3^{r}} \Rightarrow C o d {(u)}_{j} = C o d {(v)}_{j} for all j < r \Rightarrow D (C o d (u), C o d (v)) < \frac{1}{2^{r}} < ε \end{matrix}

QED of the Claim.

Therefore,

- π^{1}

is a homeomorphism and therefore a topological conjugation.

QED.

Proof: Direct consequence of the proposition 36

QED.

Proof Consequence of uniqueness, since

- lim_{k \to \infty} π^{1} (ξ) \in R .

is the only solution in

R

whose coding is

ξ

and we know that

Γ

is the only dyadic solution whose coding is

ξ

QED.

10. The Problem of Divergence

In this section, we address the fundamental aspects of divergence of the Collatz function. The primary focus is on the behavior of sequences and orbits, especially those with divergent slopes and their stability properties. The main results are summarized in the following key theorems:

Theorem 8: This theorem states that all sequences $S_{k}$ with a divergent slope are positively unstable, defining the sufficient condition under which a sequence becomes unstable.
Theorem 9: This theorem shows that all orbits with codings in $G_{0}$ are bounded.
Theorem 10: This theorem shows that all orbits with codings in $G_{\infty}$ are bounded.Consequently, the orbits of negative numbers necessarily fall into periodic orbits.
Theorem 11: This theorem concludes that all natural numbers have bounded orbits, implying the non-existence of divergent orbits for natural numbers.

First, we examine the conditions under which the slope of a function

S_{k}

diverges, leading to instability. Next, we explore the boundedness of orbits coded within

G_{0}

and

G_{\infty}

, providing proofs and corollaries to support these findings. Finally, we demonstrate the non-existence of divergent orbits for natural numbers.

10.1. Summary of Propositions in the Section

Theorem 8: It is stated that all sequences $S_{k}$ with a divergent slope are positively unstable.
Theorem 9: It is stated that all orbits with codings in $G_{0}$ are bounded.
Corollary 9: If exist a sub-sequence such that $lim_{j \to \infty} \frac{a_{k_{l}} (ξ)}{k_{l}} > \frac{ln (3)}{ln (2)}$ . Then exist $Ω > 0$ such that $π^{2} (ξ) (k_{l}) < Ω$ .
Theorem 10: Let $ξ \in G_{\infty}$ . Then we have exits $Ω > 0$ such that $C o l^{a_{k}} (- π^{1} (ξ)) > - Ω$ for all $k \in N$ . In particular if $- π^{1} (ξ) \in Z$ then its orbit necessarily falls into a cycle.
Theorem 11: It is stated that all natural numbers have bounded orbits.

10.2. The Problem of Divergence

The following theorem shows that if the slope of the function

S_{k}

diverges, then so does the minimum value, this is because the only value that satisfies the encoding of

S_{k}

is negative.

Proof: Let

ξ = C o d ({S_{j}}_{j = 1}^{\infty})

. Since

\frac{3^{b_{j}}}{2^{a_{j}}} \to \infty

then

ξ \in G_{\infty}

by Proposition 16, 30 and 32 and Theorem 7. We have

- π^{1} (ξ) = - \sum_{j = 1}^{\infty} \frac{2^{a_{j} - 1}}{3^{j}} \in R

is the only value whose coding is

ξ

. On the other hand, regardless of rationality, this number is always negative. Therefore, the minimum value must necessarily be divergent.

QED.

We are going to show a series of results referring to the bounds of the orbits of numbers whose coding is in

G_{0}

and

G_{\infty}

Proof Let

{S_{j}}_{j \in N}

〈 θ, ψ 〉

such that

C o d ({S_{j}}_{j \in N}) = C o d (n)

. Without loss of generality we can assume that n is positive, because in the case that n is negative we have that

\begin{matrix} lim_{j \to \infty} C o l^{j} (n) = lim_{j \to \infty} S_{j} (n) = lim_{j \to \infty} \frac{3^{j} n + N_{j}}{2^{a_{j}}} = lim_{j \to \infty} \frac{3^{j}}{2^{a_{j}}} n + lim_{j \to \infty} π^{2} (C o d (n)) (j) = lim_{j \to \infty} π^{2} (C o d (n)) (j) \geq 0 \end{matrix}

then eventually its orbit will fall into a non-negative number.

If the coding of

{S_{j}}_{j \in N}

has a null tail, the result is trivial. Suppose

{S_{j}}_{j \in N}

has no null tail, then:

\begin{matrix} C o l^{j} (n) = S_{j} (n) = \frac{3^{j} n + N_{j}}{2^{a_{j}}} = \frac{3^{j}}{2^{a_{j}}} n + π^{2} (C o d (n)) (j) \end{matrix}

Since

\frac{3^{j}}{2^{a_{j}}} \to 0

we have by Proposition that

π^{2} (C o d (n)) (j)

is bounded and let

M > 0

such that

π^{2} (C o d (n)) (j) < M

for all

j \in N

. On the other hand, since

\frac{3^{j}}{2^{a_{j}}} \to 0

we have

\frac{3^{j}}{2^{a_{j}}}

is bounded and let

H > 0

such that

\frac{3^{j}}{2^{a_{j}}} < H

for all

j \in N

, then

\begin{matrix} \frac{3^{j}}{2^{a_{j}}} n + π^{2} (C o d (n)) (j) < H n + M \end{matrix}

Therefore

C o l^{j} (n) < \infty

QED.

The following result is the version of the previous theorem for sub-sequence.

Proof: Let

j < k_{l}

then

\begin{matrix} lim_{k_{l} - j \to \infty} \frac{a_{k_{l}} - a_{j - 1}}{k_{j} - j} = lim_{k_{l} - 1 \to \infty} \frac{a_{k_{l}} - a_{0}}{k_{j} - 1} = lim_{l \to \infty} \frac{a_{k_{l}}}{k_{l}} > \frac{ln (3)}{ln (2)} \end{matrix}

then exist

T, ε > 0

such that if

k_{l} - j > T

we have

\begin{matrix} \frac{a_{k_{l}} - a_{j - 1}}{k_{l} - j} > \frac{ln (3)}{ln (2)} + ε \end{matrix}

Using the lemma 5 we have

\begin{matrix} π^{2} (ξ) (k_{l}) & = \frac{3^{k_{l}}}{2^{a_{k_{l}}}} \sum_{j = 1}^{k_{l}} \frac{2^{a_{j - 1}}}{3^{j}} \\ = \sum_{j = 1}^{k_{l}} \frac{2^{a_{j - 1} - a_{k_{l}}}}{3^{j - k_{l}}} \\ = \sum_{j = 1}^{k_{l}} {(\frac{3}{2^{\frac{a_{k_{l}} - a_{j - 1}}{k_{l} - j}}})}^{k_{l} - j} \\ = \sum_{j = 1}^{k_{l} - T} {(\frac{3}{2^{\frac{a_{k_{l}} - a_{j - 1}}{k_{l} - j}}})}^{k_{l} - j} + \sum_{j = k_{l} - T + 1}^{k_{l}} {(\frac{3}{2^{\frac{a_{k_{l}} - a_{j - 1}}{k_{l} - j}}})}^{k_{l} - j} \\ < \sum_{j = 1}^{k_{l} - T} {(\frac{3}{2^{\frac{ln (3)}{ln (2)} + ε}})}^{k_{l} - j} + \sum_{j = k_{l} - J + 1}^{k_{l}} {(\frac{3}{2})}^{k_{l} - j} \\ < \sum_{j = 1}^{k_{l}} {(\frac{3}{2^{\frac{ln (3)}{ln (2)} + ε}})}^{k_{l} - j} + \sum_{j = k_{l} - J + 1}^{k_{l}} {(\frac{3}{2})}^{k_{l} - j} \\ < \frac{1}{2^{ε k_{l}}} \sum_{j = 1}^{k_{l}} 2^{ε j} + {(\frac{3}{2})}^{k_{l}} \sum_{j = k_{l} - J + 1}^{k_{l}} {(\frac{2}{3})}^{j} \\ = \frac{1}{2^{ε k_{l}}} (\frac{1 - 2^{ε (k_{l} + 1)}}{1 - 2^{ε}}) + {(\frac{3}{2})}^{k_{l}} \{\frac{{(\frac{2}{3})}^{k_{l}} - {(\frac{2}{3})}^{k_{l} - T + 1 - 1}}{\frac{2}{3} - 1}\} \\ < \frac{\frac{1}{2^{ε k_{l}}} - 2^{(\frac{k_{l} + 1}{k_{l}})}}{1 - 2^{ε}} + \frac{1 - {(\frac{3}{2})}^{T}}{\frac{2}{3} - 1} \\ \leq \frac{4}{2^{ε} - 1} + 3 \{{(\frac{3}{2})}^{T} - 1\} = Ω \end{matrix}

Therefore

π^{2} (ξ) (k_{l}) < Ω

QED.

We will now show that the orbits of numbers whose coding is in

G_{\infty}

are bounded below, in particular the orbits of integers whose coding is in

G_{\infty}

, so they must necessarily fall into a cycle.

Proof: Let

ξ \in G_{\infty}

, then exist

1 > ε > 0

such that

\underset{k \to \infty}{lim sup} \frac{a_{k}}{k} < \frac{ln (3)}{ln (2)} - ε

. Exist

J > 0

such that

\frac{a_{j - 1} - a_{k}}{j - k} < \frac{ln (3)}{ln (2)} - ε

for all

k > J

. so

\begin{matrix} C o l^{a_{k}} (- π^{1} (ξ)) = C o l^{a_{k}} (- \sum_{j \in N} \frac{2^{a_{j - 1}}}{3^{j}}) = - \frac{3^{k}}{2^{a_{k}}} \sum_{j \in N} \frac{2^{a_{j - 1}}}{3^{j}} + \frac{N_{k}}{2^{a_{k}}} \\ = - \frac{3^{k}}{2^{a_{k}}} \sum_{j = 1}^{k} \frac{2^{a_{j - 1}}}{3^{j}} - \frac{3^{k}}{2^{a_{k}}} \sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} + \frac{N_{k}}{2^{a_{k}}} = - \frac{N_{k}}{2^{a_{k}}} - \frac{3^{k}}{2^{a_{k}}} \sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} + \frac{N_{k}}{2^{a_{k}}} \\ = - \frac{3^{k}}{2^{a_{k}}} \sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1}}}{3^{j}} = - \sum_{j = k + 1}^{\infty} \frac{2^{a_{j - 1} - a_{k}}}{3^{j - k}} = - \sum_{j = k + 1}^{\infty} {(\frac{2^{\frac{a_{j - 1} - a_{k}}{j - k}}}{3})}^{j - k} \\ > - \sum_{j = k + 1}^{\infty} {(\frac{2^{\frac{ln (3)}{ln (2)} - ε}}{3})}^{j - k} = - \sum_{j = k + 1}^{\infty} 2^{- ε (j - k)} = - 2^{ε k} \sum_{j = k + 1}^{\infty} 2^{- ε j} \\ = - 2^{ε k} \{\frac{2^{- ε (k + 1)}}{1 - 2^{- ε}}\} = - \frac{2^{- ε}}{1 - 2^{- ε}} \end{matrix}

Let

Ω = max \{\frac{2^{- ε}}{1 - 2^{- ε}}, C o l^{a_{k}} (- π^{1} (ξ)) with k < J\}

, so

C o l^{a_{k}} (- π^{1} (ξ)) > - Ω

for all

k \in N .

In particular, if

- π^{1} (ξ) \in Z

, then

C o l^{a_{k}} (- π^{1} (ξ))

only takes negative integer values and being bounded from below, must necessarily repeat some value, then the orbit must fall into a cycle.

QED.

Monks and Yazinski also extend the results of Eliahou [4] (1993) and Lagarias [3] (1985) concerning the density of "odd" points in an orbit. Let

k_{n} (x)

denote the number of ones in the first n digits of the parity vector x. If

x \in Q_{o d d}

eventually enters an n-periodic orbit, then

\frac{ln (2)}{ln (3 + 1 / m)} \leq lim_{n \to \infty} \frac{k_{n} (x)}{n} \leq \frac{ln (2)}{ln (3 + 1 / M)}

where

m, M

are the least and greatest cyclic elements in the eventual cycle. If

x \in Q_{o d d}

diverges, then

\frac{ln (2)}{ln (3)} \leq \underset{n \to \infty}{lim inf} \frac{k_{n} (x)}{n} .

We will now show the main theorem of this work, where we finally show the nonexistence of divergent orbits for every positive integer. We will show that the necessary and sufficient condition for an orbit to be divergent is

\begin{matrix} \underset{j \to \infty}{lim sup} \frac{a_{j}}{j} < \frac{ln (3)}{ln (2)} \end{matrix}

which implies that the only solution if it exists must be

\begin{matrix} - \infty < - π^{1} (ξ) < 0 . \end{matrix}

Proof: Let

n \in N

such that

C o l^{j} (n) \to \infty

j \to \infty

and

{S_{j}}_{j = 1}^{\infty} \subset 〈 θ, ψ 〉

such that

C o d {S_{j}}_{j = 1}^{\infty} = C o d (n)

. We are going to prove that the necessary and sufficient condition for an orbit to be divergent is that the coding does not have a null tail and

\frac{3^{j}}{2^{a_{j}}} \to \infty

j \to \infty

C o d {S_{j}}_{j = 1}^{\infty}

has a null tail, it means that from a certain iteration, the orbit of n must always be even, which implies that this orbit must be decreasing. This contradicts the fact that we have assumed that

C o l^{j} (n) \to \infty

Now without loss of generality, we can assume that

C o d {S_{j}}_{j = 1}^{\infty}

does not have a null tail

\begin{matrix} C o l^{j} (n) = \frac{3^{j} n + N_{j}}{2^{a_{j}}} \to \infty then \frac{3^{j}}{2^{a_{j}}} \to \infty or π_{k}^{2} (C o d {S_{j}}_{j = 1}^{\infty}) = \frac{N_{j}}{2^{a_{j}}} \to \infty \end{matrix}

If $\frac{3^{j}}{2^{a_{j}}} \to \infty$ then $C o d {S_{j}}_{j = 1}^{\infty} \in G_{\infty}$ by Proposition 16 we have $π_{k}^{1} (C o d {S_{j}}_{j = 1}^{\infty}) < \infty$ this implies that

$\begin{matrix} π_{k}^{2} (C o d {S_{j}}_{j = 1}^{\infty}) = \frac{3^{j}}{2^{a_{j}}} π_{k}^{1} (C o d {S_{j}}_{j = 1}^{\infty}) \to \infty . \end{matrix}$
if $π_{k}^{2} (C o d {S_{j}}_{j = 1}^{\infty}) \to \infty$ , by Proposition 16 we have that $C o d {S_{j}}_{j = 1}^{\infty} \notin G_{0}$ then $\underset{j \to \infty}{lim inf} \frac{a_{j}}{j} \leq \frac{ln (3)}{ln (2)}$ . Let’s show now in fact $\underset{k \to \infty}{lim sup} \frac{a_{k}}{k} \leq \frac{ln (3)}{ln (2)}$ . Suppose that exist ${j_{k}}_{k \in N}$ such that $lim_{j \to \infty} \frac{a_{j_{k}}}{j_{k}} < \frac{ln (3)}{ln (2)} + ε$ with $ε > 0$ , so using the estimated bound in the demonstration of the Lemma 9, we have

$\begin{matrix} π^{2} (ξ) (j_{k}) < \frac{4}{2^{ε} - 1} + 3 \{{(\frac{3}{2})}^{T} - 1\} = Ω \end{matrix}$

Since $\frac{3^{j_{k}}}{2^{a_{j_{k}}}} \to 0$ as $j \to \infty$ . So $max_{k} \{\frac{3^{j_{k}}}{2^{a_{j_{k}}}}\} < \infty$ then we have

$\begin{matrix} C o l^{j_{k}} (n) = \frac{3^{j_{k}}}{2^{a_{j_{k}}}} n + \frac{N_{j_{k}}}{2^{j_{k}}} < max_{k} \{\frac{3^{j_{k}}}{2^{a_{j_{k}}}}\} n + Ω \end{matrix}$

Since $C o l^{j} (n) \in N$ for all $j \in N$ , then exist $K \in N$ such that $C o l^{j_{k}} (n) = C o l^{j_{k + K}} (n)$ . So we have that the orbit of n must fall into a cycle, which implies that $lim_{j \to \infty} \frac{3^{j}}{2^{a_{j}}} = 0$ which implies that $π^{2} (ξ) (j)$ is bounded, which contradicts the hypothesis that $C o l^{j} (n) \to \infty$ . Therefore $\underset{j \to \infty}{lim sup} \frac{a_{j}}{j} \leq \frac{ln (3)}{ln (2)}$ . This is equivalent to $\frac{3^{j}}{2^{a_{j}}} \to {1, \infty}$ as $j \to \infty$ .

On the other hand if $C o d {S_{j}}_{j = 1}^{\infty} \in G_{1}$ by Theorem 6 we have $ρ_{0} {S_{j}}_{j = 1}^{\infty} \to \infty$ which contradicts the hypothesis that ${S_{j}}_{j = 1}^{\infty}$ is positively stable, so $C o d {S_{j}}_{j = 1}^{\infty} \in G_{\infty}$

Since the Theorem 9 above, to have

C o l^{j} (u)

divergent it is necessary that

\frac{3^{k}}{2^{a_{k}}} \to \infty

, by Theorem 8 we have

{S_{j}}_{j = 1}^{\infty}

is positively unstable, by Theorem, 2 we have

E ({S_{j}}_{j = 1}^{\infty}) \cap N = \emptyset

. Therefore, cannot exist

n \in N

such that its orbit is divergent.

QED.

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The Collatz Conjecture "The Nonexistence of Divergent Orbits, through a Dynamic Diophantine Linear Analysis of Codification

Abstract

Keywords:

Subject:

1. Notations and Conventions

2. Introduction

2.1. Collatz’s Conjecture

2.2. Extension of the Collatz Function

3. Set Generate by θ and ψ q

3.1. Summary of Propositions in the Section

3.2. Set Generate by θ and ψ q

4. Stability and Instability of Integer Set

4.1. Summary of Propositions in the Section

4.2. Stability and Instability of Integer Set

5. Coding of the Orbits

5.1. Summary of Propositions in the Section

5.2. Coding of the Orbits

6. The π 1 and π 2 Functions

6.1. Summary of Propositions in the Section

6.2. The π 1 and π 2 Functions

7. The Sigma Function

7.1. Summary of Propositions in the Section

7.2. The Sigma function

7.3. Periodicity of the Sigma Function

7.4. Linearity of the Sigma Function Modulo a

7.5. Extension on the Q o d d of the Sigma Function

7.6. Properties of the Extension of the Sigma Function

7.7. Coding of Sigma Function

7.7.1. p − Adic Numbers

7.7.2. Coding of Sigma Function

8. The Extension of Collatz Function on Z 2

8.1. Summary of Propositions in the Section

8.2. Extension of the Collatz Function on Z 2 .

8.3. Topological Conjugation

9. The Coding of π 1

9.1. Summary of Propositions in the Section

9.2. π 1 ( G ∞ ) as Complete Metric Space

9.3. Characterization of the Full Coding Sets through the Function π 1

9.4. Extension of the Collatz Function on π 1 ( G ∞ )

9.5. Topological Conjugation

10. The Problem of Divergence

10.1. Summary of Propositions in the Section

10.2. The Problem of Divergence

References

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3. Set Generate by $θ$ and $ψ^{q}$

3.2. Set Generate by $θ$ and $ψ^{q}$

6. The $π^{1}$ and $π^{2}$ Functions

6.2. The $π^{1}$ and $π^{2}$ Functions

7.5. Extension on the $Q_{o d d}$ of the Sigma Function

7.7.1. $p -$ Adic Numbers

8. The Extension of Collatz Function on $Z_{2}$

8.2. Extension of the Collatz Function on $Z_{2}$ .

9. The Coding of $π^{1}$

9.2. $π^{1} (G_{\infty})$ as Complete Metric Space

9.3. Characterization of the Full Coding Sets through the Function $π^{1}$

9.4. Extension of the Collatz Function on $π^{1} (G_{\infty})$