Canonical decomposition in $\mathbb{r}_+^*$ of a convergent natural number by the collatz iterations

TL;DR

This paper introduces a canonical decomposition for natural numbers that converge under Collatz iterations, using a function that captures the convergence value for all such sequences.

Contribution

It constructs a function representing the convergence value and formalizes a canonical decomposition for numbers converging under Collatz iterations.

Findings

A function $f_{X,Y}$ representing convergence values is defined.

A canonical decomposition for convergent numbers is established.

The approach provides a new perspective on Collatz convergence patterns.

Abstract

The Collatz variations pattern seems not to have any recurrence relation between numbers. But knowing that there is at least a natural number that converges after several iterations we construct a function $f_{X,Y}$ that is equal to the value of convergence for all convergent sequences. A canonical decomposition can be expressed for such numbers.

Equations

$$g(n)=\{\begin{array}{ll}\frac{n}{2},& \text{mbox}ifn\text{mbox}iseven\\ 3n+1,& \text{mbox}ifn\text{mbox}isodd\end{array}$$

$$C_p=\{\begin{array}{ll}n,& \text{mbox}forp=0\\ g\left(C_{p-1}\right),& \text{mbox}forp>0\end{array}$$

$$E=\left\{C_0,C_1,.......,C_{k-1}\right\}.$$

$$Card\left\{E_1\right\}=X$$

$$Card\left\{E_2\right\}=Y.$$

$$\underset{p\to {\sigma }_{\infty }}{\lim }(C_p)=1.$$

$$f_{X,Y}:\mathcal{N}\to \mathbb{N}$$

$$f_{X,Y}\left(n\right)=\big{\lceil}\dfrac{3^{X}\left(2n+1\right)-1}{2^{Y+1}}\big{\rceil}.$$

$$Z_n=1-\epsilon .$$

$$3Z_n+1$$

$$=(3L+1)-3{\epsilon }^{\prime \prime }$$

$$3Z_n+1$$

$$=4-3{\epsilon }^{\prime \prime }$$

$$\forall n=n(X,Y)\in \mathcal{N},f_{X,Y}(n)=1.$$

$$Z_n=1-\epsilon$$

$$\displaystyle f_{X,Y}\left(n\right)=\big{\lceil}Z_{n}\big{\rceil}$$

$$n^{\prime }=2^Y(3^{-X}-{\epsilon }_n);\text{mbox}$$

$$0$$

$$-1$$

$$0$$

$$\big{\lceil}Z_{n}\big{\rceil}=1\Longleftrightarrow Z_{n}=1-\varepsilon$$

$$\frac{3^X\left(2n+1\right)-1}{2^{Y+1}}$$

$$3^X\left(2n+1\right)-1$$

$$=2^{Y+1}-2^{Y+1}\epsilon$$

$$2\cdot 3^{X}n+3^X-1$$

$$2\cdot 3^{X}n$$

$$n$$

$$=\frac{1}{2}(3^{-X}-1)+3^{-X}\cdot 2^Y-2^Y{\epsilon }_n$$

$$=\frac{1}{2}(3^{-X}-1)+2^Y(3^{-X}-{\epsilon }_n)$$

$$=2^Y(3^{-X}-{\epsilon }_n)-\frac{1}{2}(1-3^{-X})$$

$$=n^{\prime }-\frac{1}{2}(1-3^{-X})$$

$$n=n^{\prime }-\frac{1}{2}(1-3^{-X}).$$

$$n=\lfloor n^{\prime }\rfloor .$$

$$n$$

$$n^{\prime }$$

$$\lfloor n^{\prime }\rfloor$$

Full text

$\mathbf{Keywords}$: Syracuse problem, 3N+1 problem, Collatz conjecture.

2010 Mathematics Subject Classification: primary $\mathbf{11B99}$ ; secondary $\mathbf{11T30}$ .

1 Introduction

The Collatz conjecture has many denominations. It is also known as the Syracuse problem or the 3N+1 problem. The problem was first stated by the German mathematician Lothar Collatz in the 1930’s [1]. The conjecture is sumarized as follows. Take any natural number $n$ not equal to zero. If $n$ is even divide by 2. If $n$ is odd multiply it by 3 and add 1. Repeat the process to infinity. Does the sequence created reaches 1 for every initial number $n$? The Collatz sequence $(C_{p})_{p\in\mathbb{N}}$ started with a natural number $n$ different of zero is called convergent when after $k$ iterations the sequence is equal to 1. The total stopping time $\sigma_{\infty}(n)=inf\left\{k:T^{k}(n)=1\right\}$ [2]; $k$ is the finite least iterations before $(C_{p})$ converges. Condider the function:

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Form the sequence by performing an infnite operation of the fonction. Notation:

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$C_{p}$ is the value of $g$ applied to $n$ recursively $p$ times: in notation $C_{p}=g^{p}\left(n\right)$. The smallest $p$ such that $C_{p}=1$ is nothing than $\sigma_{\infty}(n)$ defined earlier as the total stopping time ($p=k$).

A divergent sequence isn’t yet found. The divergence would consist of a total stopping being infinity. In notation: $\sigma_{\infty}(n)=\infty$ [2]. Even though computational method had proven the convergence of all natural number $n<20\cdot 2^{58}$ [3], does not totally prove the Collatz conjecture. But it tells us the existence of several convergent numbers (The partition set of the convergent numbers in $\mathbb{N}$ is not empty).

This document is intented to prove that all convergent numbers have their convergence same as a function $f_{X,Y}$ . In general, $g^{k}\left(n\right)=f_{X,Y}=1$. This paper also includes properties of convergent numbers by the Collatz sequence and a generalisation of the idea that the set of convergent $n$ is never empty to an infinite set.

2 The odd and even iterations X and Y at convergence

2.1 The $k-tuple$ associated to a Collatz sequence at the total stopping time $\sigma_{\infty}(n)=k$

$\mathbf{Definition\mbox{ }2.1}$: The $k-tuple$ associated to $n$ after $k$ total Collatz iterations is the chain constitued of all values $C_{p}$ when $p$ varies from [math] to $p-1$ ($p=0,1,......,k-1$). Then the $k-tuple$ is $(C_{0},C_{1},.......,C_{k-1})$.

$\mathbf{Examples\mbox{ }2.1}$

For $n=6$, $C_{8}=1$. The $8-tuple$ associated to $6$ is $(6,3,10,5,16,8,4,2)$ .

Another example is $n=19$, it takes $20$ iterations before it gets to $1$. $C_{20}=1$ and its $20-tuple$ is $(19,58,29,88,44,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2)$ .

2.2 The smallest odd and even iterations $X$ and $Y$

Consider the $k-tuple$ $(C_{0},C_{1},.......,C_{k-1})$ associated to a convergent sequence of $n$. Let us make a set $E$ of all the element in the $k-tuple$ chain and 2 subsets $E_{1}$, $E_{2}$ defined respectively as set of all the odd and all even numbers of $E$.

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$\mathbf{Definition\mbox{ }2.2}$: The iterations on odd numbers X is the cardinal of the set $E_{1}$ and $Y$, the iterations on even number is the cardinal of the set $E_{2}$.

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and,

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$\mathbf{Remark\mbox{ }2.2}$

$Card\left\{{E}\right\}=Card\left\{E_{1}\right\}+Card\left\{E_{2}\right\}$,

$\left\{{E}\right\}=E_{1}\cup E_{2}$,

$\sigma_{\infty}(n)=k=X+Y$ .

By convenience we’ll note a convergent sequence of $n$ after $X$ and $Y$ iterations $n=n(X,Y)$, and we’ll denote by $\mathcal{N}$ the set of convergent natural numbers.

By definition of the Collatz conjecture $n\not=0$, so $\mathcal{N}\subseteq\mathbb{N}^{*}$.

3 The function $f_{X,Y}$ associated to the convergence value $C_{k}(n)$ of $n$

3.1 The value at the convergence

$\mathbf{Definition\mbox{ }3.1}$: The sequence $(C_{p})_{p\in\mathbb{N}}$ is called convergent when after $p=k$ iterations $C_{k}(n)=1$. The value at the convergence of any Collatz sequence started with $n$ non-zero positive integer is the limit taken at the total stopping time. In terms of limit notation:

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3.2 The function $f_{X,Y}$

$\mathbf{Definition\mbox{ }3.2}$: Let $\mathcal{N}$ be the set ol the convergent $n=n(X,Y)$, with the couple $(X,Y)$ associated to the its respective $n$. By a function at the convergence of $n$ we mean a map

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where:

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$\mathbf{Lemma\mbox{ }3.1}$: Let $Z_{n}=\dfrac{3^{X}\left(2n+1\right)-1}{2^{Y+1}}$. $\forall{n=n(X,Y)}\in\mathcal{N}$, $\exists\varepsilon$ such that $0\leq\varepsilon<\dfrac{1}{3}$ , for which,

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$Proof$: For $n(0,i)=2^{i}$ ($i\in\mathbb{Z_{+}^{*}}$), then $Z_{n}=\dfrac{2^{i}}{2^{i}}=1$ where $\varepsilon=0$.

Let $n\not=n(0,i)$. Proceed by ABSURD i.e we supose that $\exists n=n(X,Y)$, and $\varepsilon>\dfrac{1}{3}$ such that $Z_{n}\not=1-\varepsilon$

If $Z_{n}<1-\varepsilon$

$Z_{n}<1-\varepsilon\Longrightarrow\exists\varepsilon^{\prime}$ such that $\dfrac{1}{3}<\varepsilon^{\prime}<1$ and $Z_{n}=1-\varepsilon^{\prime}$. That’s ABSURD.

If $Z_{n}>1-\varepsilon$

$Z_{n}>1-\varepsilon\Longrightarrow\exists\varepsilon^{\prime}$ such that $0<\varepsilon^{\prime}<\varepsilon<\dfrac{1}{3}$ and $Z_{n}=1-\varepsilon^{\prime}$ (ABSURD),

or

$\exists L>1$ and $\varepsilon^{\prime\prime}$, for which $Z_{n}=L-\varepsilon^{\prime\prime}$ with $\dfrac{1}{3}<\varepsilon^{\prime\prime}<1$ ($L\in\mathbb{N}$).

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$n$ converges, $n=n(X,Y)\Longrightarrow Z_{n}=1-\varepsilon^{\prime\prime}$ and $0<\varepsilon^{\prime\prime}<\dfrac{1}{3}$ ($n\not=n(0,i)$) .

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For the same $n=n(X,Y)$, we have 2 values of $3Z_{n}+1$ where one’s function of $L$. Since a number cannot differ from itself, $L$ must be equal to $1$ and $0<\varepsilon^{\prime\prime}<\dfrac{1}{3}$. There is a contradiction meaning that there is no such $L$ greater than $1$ and there is no $\varepsilon^{\prime\prime}$, such that $\dfrac{1}{3}<\varepsilon^{\prime\prime}<1$ for which $Z_{n}=L-\varepsilon^{\prime\prime}$ .

$Conclusion$: $\forall{n=n(X,Y)}\in\mathcal{N}$, $\exists\varepsilon$ such that $0\leq\varepsilon<\dfrac{1}{3}$ for which $Z_{n}=1-\varepsilon$.

We can now prove the following theorem;

$\mathbf{Theorem\mbox{ }3.1}$: For all $n=n(X,Y)\in\mathcal{N}$, there is a function f_{X,Y}(n)=\big{\lceil}\dfrac{3^{X}\left(2n+1\right)-1}{2^{Y+1}}\big{\rceil} which is equal to the value of convergence $C_{k}$ of $n$.

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$Proof$: From Lemma 3.1 for all $n=n(X,Y)$ there is always $\varepsilon$ satisfying the condition $0\leq\varepsilon<\dfrac{1}{3}$ and we have $Z_{n}=1-\varepsilon$.

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$\mathbf{Corollary\mbox{ }3.1}$: $\forall{n=n(X,Y)}\in\mathcal{N}$, $f_{X,Y}(n)$ is a constant function.

4 Canonical decomposition $n^{\prime}$ of $n=n(X,Y)$

$\mathbf{Definition\mbox{ }4.1}$: For $n$ in $\mathcal{N}$ with its respective couple $(X,Y)$, the expression $n^{\prime}$ in $\mathbb{R}_{+}^{*}$ of $n$ is:

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$n^{\prime}$ is called by definition the canocical decomposition of $n$.

$\mathbf{Lemma\mbox{ }4.1}$: For all convergent $n=n(X,Y)$, $0<\dfrac{1}{2}(1-3^{-X})<\dfrac{1}{2}$ .

$Proof$: $\forall X\in\mathbb{N}$,

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$\mathbf{Lemma\mbox{ }4.2}$: If $\left\{n^{\prime}\right\}$ is the fractionnal part of $n^{\prime}$ then, $\left\{n^{\prime}\right\}=\dfrac{1}{2}(1-3^{-X})$ and $\varepsilon_{n}=3^{-X}\varepsilon$ .

$Proof$: Let $n(X,Y)\in\mathcal{N}$, then from (6) we know that \big{\lceil}Z_{n}\big{\rceil}=1 .

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The number $n$ is a natural number written as the difference of $2$ real numbers which are positive. Also $0<\dfrac{1}{2}(1-3^{-X})<\dfrac{1}{2}$ and $n^{\prime}>\dfrac{1}{2}(1-3^{-X})$. The relation $n=n^{\prime}-\dfrac{1}{2}(1-3^{-X})$ is true if and only if $\dfrac{1}{2}(1-3^{-X})$ is the fractionnal part of $n^{\prime}$; i.e $\left\{n^{\prime}\right\}=\dfrac{1}{2}(1-3^{-X})$.

$\mathbf{Lemma\mbox{ }4.3}$: If $n^{\prime}$ is the canocical decomposition of $n$ then $0\leq\varepsilon_{n}<\dfrac{1}{3^{X+1}}$ .

$Proof$: $\varepsilon_{n}=3^{-X}\varepsilon$ and $0\leq\varepsilon<\dfrac{1}{3}$ .

$\mathbf{Remarks\mbox{ }4.1}$: For the same iterations $X$ and $Y$ at the convergence of $n$ and $m$, $\varepsilon_{n}=\varepsilon_{m}$ ,

For $n=n(0,i)$ (or $n=2^{i}$, $i\in\mathbb{Z_{+}^{*}}$), $\varepsilon_{n}=0$ .

We can now state the following theorem;

$\mathbf{Theorem\mbox{ }4.1}$: Let $n^{\prime}$ be the canonical decomposition of $n=n(X,Y)$ in $\mathbb{R}_{+}^{*}$. The expression of $n$ in function of $n^{\prime}$ is:

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$Proof$: From (8) we have the equality $n=n^{\prime}-\dfrac{1}{2}(1-3^{-X})$

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$\mathbf{Corollary\mbox{ }4.1}$: $\forall{n}\in\mathcal{N}$, $n=n(X,Y)\Longleftrightarrow n=\lfloor{n^{\prime}}\rfloor$ .

4.1 Properties

Consider $\mathcal{N}$, $n^{\prime}$ the canocical decomposition of $n=n(X,Y)$, and $a$ and $b$ be 2 elements of $\mathcal{N}$. We consider the following strong properties arising from the canonical decomposition:

$\mathbf{Unicity\mbox{ }of\mbox{ }the\mbox{ }couple\mbox{ }(X,Y)}$: $a=a(X,Y)$ and $b=b(X,Y)$ iff $a=b$ .

$Proof$: Let $a=a(X,Y)$ and $b=b(X,Y)$ then $a=\lfloor{a^{\prime}}\rfloor$ and $b=\lfloor{b^{\prime}}\rfloor$

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It’s known from remark 4.1 that for the same iterations $X$ and $Y$ at the convergence of $n$ and $m$, $\varepsilon_{n}=\varepsilon_{m}$ then:

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$\mathbf{The\mbox{ }a+b\mbox{ }addition}$: If $a$ and $b$ converge so does $a+b$: i.e $a=a(X,Y)$ and $b=b(X^{\prime},Y^{\prime})$, then $\exists(X^{\prime\prime},Y^{\prime\prime})$ such that $a+b=[a+b](X^{\prime\prime},Y^{\prime\prime})$

$Proof$: Let $a=a(X,Y)$ and $b=b(X,Y)$ then $a=\lfloor{a^{\prime}}\rfloor$ and $b=\lfloor{b^{\prime}}\rfloor$

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Because $\left\{a^{\prime}\right\}$ and $\left\{b^{\prime}\right\}$ are both less than $\dfrac{1}{2}$ . (See Lemma 4.1).

5 Algebra of the set $\mathcal{N}$

5.1 Equipotence to $\mathbb{N}$

$\mathbf{Definition\mbox{ }5.1}$: $\mathcal{N}$ is equipotent to $\mathbb{N}$ or countably infinite when there exist a function bijective from $\mathcal{N}$ to $\mathbb{N}$.

$\mathbf{Lemma\mbox{ }{5.1}}$: There is a bijection from $\mathbb{N}$ to its infinite subsets especially to $\mathcal{N}$.

$Proof$: $\mathcal{N}\subset\mathbb{N}$ ; Let consider an order relation $\leq$ on $\mathcal{N}$ and let the set be finite,

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From properties above the addition of 2 convergent numbers is convergent so must $M+1$ also be in $\mathcal{N}$ i.e also convergent. We arrive at a contracdition. $\mathcal{N}$ is not majored and not a finite set.

The application which to every single element of $\mathcal{N}$ associate their perfect square in $\mathbb{N}$ is bijective.

5.2 Total order relation in $\mathcal{N}$

The order relation $\leq$ is total in $\mathcal{N}$. By definition $\leq$ is a total relation order when $\forall a$ and $b$ in the set such that $a\leq b$, there is also $c$ in the set such $a+c=b$.

In $\mathcal{N}$ this relation is verified. In fact, if there is $M$ in $\mathcal{N}$, $M+1$ also is in $\mathcal{N}$ leading to state that two elements $a$ and $b$ in the set are always comparable: $a\leq b$ or $b\leq a$.

5.3 Conclusion

$We\mbox{ }recall\mbox{ }some\mbox{ }basic\mbox{ }properties$: Any partition of $\mathbb{N}$ different from the empty set ($\emptyset$) has a least element. The least element to converge in $\mathcal{N}$ is $1$ .

The addition in $\mathcal{N}$ is an internal law of composition.

So we can assure these following inclusions:

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