Furstenberg Topology and Collatz Problem

Abstract

The aims of this paper are two-fold. First, we present the result of the decomposition on the iterations of a Collatz transform into arithmetic sequences. With this, we prove that in Furstenberg topology, the set of (odd) integers with an infinite stopping time is closed and nowhere dense. Then, we move our considerations to some monoids $\mathbb{L}$ in $\mathbb{N}$, where we define a suitably modified Collatz transform, and we present some results of numerical investigations on the behaviour of these modified transforms.

1. Introduction

1.1. Notation

Throughout this paper, $\mathbb{N}=\{0,1,2,\dots \}$ is the set of non-negative integers (natural numbers), ${\mathbb{N}}^{*}$ is the set of positive integers and X denotes the set of odd positive integers. The classical Collatz function $C:{\mathbb{N}}^{*}\to {\mathbb{N}}^{*}$ is defined by the formula

$$C\left(x\right)=\left\{\begin{array}{c}3x+1\mathrm{if}x\mathrm{is}\mathrm{odd}\hfill \\ {\displaystyle \frac{x}{2}}\mathrm{if}x\mathrm{is}\mathrm{even}.\hfill \end{array}\right.$$

(1)

The so-called $3x+1$ function $T:{\mathbb{N}}^{*}\to {\mathbb{N}}^{*}$ is defined as

$$T\left(x\right)=\left\{\begin{array}{c}{\displaystyle \frac{3x+1}{2}}\mathrm{if}x\mathrm{is}\mathrm{odd}\hfill \\ {\displaystyle \frac{x}{2}}\mathrm{if}x\mathrm{is}\mathrm{even}.\hfill \end{array}\right.$$

(2)

For future use, we also define a version of the map T. Let $n\in \mathbb{N}$. We may uniquely write $n=2^k·t\left(n\right)$, where $t\left(n\right)$ is odd and $k\ge 0$. The number $2^{-k}$ (i.e., the $2-$adic norm of n) is denoted by ${\left|n\right|}_2$.

Definition 1. 

The map $\mathcal{T}:X\to X$ given by the formula

$$\mathcal{T}\left(n\right)=(3n+1)\left(\right|3n+1{|}_2)$$

is called the Collatz transform.

1.2. On the Collatz Problem

Here, we recall only basic facts about the Collatz problem. For much more information on the subject and a comprehensive bibliography, we refer the reader to [1,2,3,4,5].

With function T defined above and the integer $n\in {\mathbb{N}}^{*}$, the trajectory of n under the function T is an infinite sequence: $\widehat{T}\left(n\right):={\left(T^k\left(n\right)\right)}_{k=0}^{\infty },$ where $T^0\left(n\right)=n$, $T^1\left(n\right)=T\left(n\right)$ and $T^{k+1}\left(n\right)=T\left(T^k\left(n\right)\right)$.

For a given n, we have a priori two possibilities: the correspondence $k\to T^k\left(n\right)$ is injective or it is not. If the map $k\to T^k\left(n\right)$ is injective, it must tend to infinity, ${lim}_{k\to \infty }{T}^k\left(n\right)=+\infty$. Such trajectories are said to be divergent. If the map $k\to T^k\left(n\right)$ is not injective, then there exists $p\in \mathbb{N}$, $p>1$ such that for all sufficiently large $k\in \mathbb{N}$, we have $T^k\left(n\right)=T^{k+p}\left(n\right)$. Such a trajectory ${\left(T^k\left(n\right)\right)}_{k=0}^{\infty }$ is said to be cyclic. For a cyclic trajectory, let $k_0=min\{k:T^k\left(n\right)=T^{k+p}\left(n\right)\}$. The finite sequence $(T^{k_0}\left(n\right),T^{k_0+1}\left(n\right),\dots ,T^{k_0+p-1}\left(n\right))$ is called a cycle of the trajectory ${\left(T^k\left(n\right)\right)}_{k=0}^{\infty }$. The smallest such number p is the length of the cycle. When the length of the cycle equals 2 (i.e., $p=2$), then it is easy to see that the cycle equals $(1,2)$. In this case, it is called a trivial cycle, and the trajectory is said to be convergent. When $p>2$, the cycle is said to be non-trivial.

The Collatz problem, see [6], may be formulated as follows.

Conjecture 1 

([6]). For each $n\in \mathbb{N}$, the trajectory $\widehat{T}\left(n\right)$ is convergent.

Thus, the Collatz conjecture excludes the existence of divergent trajectories and non-trivial cycles.

This conjecture is a constant source of inspiration for mathematicians and computer scientists, who connect it with various fields of research: complex analysis, operator theory, number theory, probability, and many others; see, as an example [7,8,9,10,11,12,13,14].

The convergence of the trajectories was checked by computers, to our knowledge, for $n<2^{68}$ [15]. Terras, in 1976 [16], proved that a set of n such that for every $k\in \mathbb{N}$ there is $T^k\left(n\right)\ge n$, even if not empty, is relatively small (i.e., its density is zero; see Section 4). In 2019, Tao [17] proved that for any function $f:X\to \mathbb{R}$ with ${lim}_{n\to \infty }f\left(n\right)=+\infty$, the inequality $min\{T^k\left(n\right),k\in \mathbb{N}\}\le f\left(n\right)\}$ holds, in the sense of logarithmic density, for almost all $n\in {\mathbb{N}}^{*}$.

Additionally, we would like to remark that in this paper, we mainly focus on the transform $\mathcal{T}$ from Definition 1. The notions of cyclic or divergent trajectories for $\mathcal{T}$ are analogous to those for T. However, the lengths of the cycles may, of course, differ; e.g., a trivial cycle for $\mathcal{T}$ has a length of one.

1.3. Decomposition into Arithmetic Sequences

We show a, hopefully, interesting description of the k-th iteration of $\mathcal{T}$. We think this result is noteworthy and may provide new insight into the Collatz problem. The main idea behind this result is that the domain of $\mathcal{T}$ (that is, the set X) can be divided into separate arithmetic sequences, and for integers in any such sequence, the graph of ${\mathcal{T}}^{\left(k\right)}$ lies on a line with a given slope. Before we quote the result (Theorem 3), let us present Figure 1, showing the idea of the result.

Figure 1. The idea behind Theorem 3; a graph of ${\mathcal{T}}^{\left(4\right)}$.

Theorem 1. 

Let $k\in {\mathbb{N}}^{*}$ and let $\beta =(p_1,\dots p_k)\in {\mathbb{N}}^{*}k$. Let $\left|\beta \right|=p_1+\cdots +p_k$. Then there exists a family of subsets $K_{\beta }\subset X$ satisfying the following properties:

(1)

Each $K_{\beta }$ is an arithmetic sequence.

(2)

If ${\beta }_1\ne {\beta }_2$, then

$$K_{{\beta }_1}\cap K_{{\beta }_2}=\varnothing .$$

(3)

${\bigcup }_{\beta \in {\left({\mathbb{N}}^{*}\right)}^k}{K}_{\beta }=X.$

(4)

For each $\beta \in {\left({\mathbb{N}}^{*}\right)}^k$, there exist positive rational numbers $a\left(\beta \right)$ and $b\left(\beta \right)$ such that for each $x\in K_{\beta }$, we have

$${\mathcal{T}}^{\left(k\right)}\left(x\right)=a\left(\beta \right)·x+b\left(\beta \right).$$

(5)

$$a\left(\beta \right)={\displaystyle \frac{3^k}{2^{\left|\beta \right|}}},b\left(\beta \right)=\sum _{i=1}^k{\displaystyle \frac{3^{k-i}}{2^{p_i+p_{i+1}+..+p_k}}}.$$

(6)

The difference in the sequence $K_{\beta }$ equals $2^{\left|\beta \right|+1}$.

(7)

The difference in the sequence $T^{\left(k\right)}\left(K_{\beta }\right)$ equals $2·3^k$.

1.4. LikeNs, Motivation and a Very Short Introduction

In this subsection, we present structures $\mathbb{L}\subset X$ named ‘like$\mathbb{N}$s’ described in [18,19]. We investigate the behaviour of a suitably modified Collatz transform on $\mathbb{L}$ in Section 5. The motivation for considering such structures is the fact that the map $\mathcal{T}:X\to X$ can be considered as well as the map $\mathcal{T}:Y:=X\setminus 3X\to X\setminus 3X,$ where $3X$ is a set of odd numbers divisible by 3. The problem is that $\mathcal{T}$ restricted, for example, to $Y\setminus 5X$ is not a map into $Y\setminus 5X$. It turns out that by defining the modified Collatz transform on some subsets of X, that is, on likens, we may omit this problem. Moreover, our approach makes the domain of the Collatz transform smaller and, at the same time, slightly changes the ‘affinity’ of this function. In this way, we obtain a source of interesting numerical results, obtained with low computer processing power. We hope that these results broaden our understanding of the Collatz problem.

Take ${\mathbb{N}}^{*}$, treated as a monoid. We define likens as follows:

Definition 2. 

Let $\mathbb{L}\subset {\mathbb{N}}^{*}$ be an infinite subset of ${\mathbb{N}}^{*}$ such that $1\in \mathbb{L}$, and $\mathbb{L}$ is closed with respect to the multiplication in ${\mathbb{N}}^{*}$, i.e., $\mathbb{L}$ is a monoid.

For future use, observe that we may analogously define $\mathbb{L}$ as a monoid in X.

As $\mathbb{L}$ is a strictly increasing sequence of natural numbers, $\mathbb{L}=(1=k_1<k_2<\cdots <k_i<k_{i+1}<\dots ).$ For any $n\in {\mathbb{N}}^{*}$ (or in X), we define a successor of n in $\mathbb{L}$:

Definition 3. 

$$s_{\mathbb{L}}\left(n\right)=min\left\{k_i:k_i>n\right\}.$$

If the context is clear, we write only $s\left(n\right)$.

Now, let $\mathbb{P}$ denote the sequence of prime numbers and let q be a fixed prime number. Then, the liken, denoted as ${\mathbb{L}}_q\subset {\mathbb{N}}^{*}$, is the monoid generated by the set $\mathbb{P}\setminus \left\{q\right\}$. For example, ${\mathbb{L}}_2$ is the monoid of all odd numbers, ${\mathbb{L}}_5=\{1,2,3,4,6,7,8,9,11,\dots \}$. More generally, ${\mathbb{L}}_A$ is the monoid in ${\mathbb{N}}^{*}$, generated by the set $\mathbb{P}\setminus A$, where A is a finite subset of $\mathbb{P}$.

Remark 1. 

In what follows, we are interested only in likens ${\mathbb{L}}_{2,q}$, or even ${\mathbb{L}}_{2,3,q}$, where $2<q\in \mathbb{P}$, so when the context is clear, we will write ${\mathbb{L}}_q\subset X$ instead of ${\mathbb{L}}_{2,q}$ or ${\mathbb{L}}_{2,3,q}$.

1.5. Paper Organization

The paper is organized as follows. In Section 2, we present the result of the decomposition of the iterations of a transform $\mathcal{T}$ into arithmetic sequences. In Section 3, we deal with the Furstenberg topology in X and prove that $\mathcal{T}$ is continuous in this topology. In Section 4, we prove that a set of $n\in X$, with infinite stopping time, is closed and nowhere dense (in the Furstenberg topology). In Section 5, we generalize $\mathcal{T}$ to likens, and in Section 6, we present the results of the numerical experiments.

2. Arithmetic Sequences

Let $\mathcal{T}\left(n\right)$ be the Collatz transform, as in Definition 1. The aim of this section is to construct an infinite sequence ${\left({\mathcal{F}}_k\right)}_{k=1}^{\infty }$ of families of sets, and then to study its properties. Each family ${\mathcal{F}}_k$ is a countable family of subsets of X. The elements of the family ${\mathcal{F}}_k$ are denoted by $K(k,p_1,p_2,\dots ,p_k)$, where $p_i\in {\mathbb{N}}^{*}.$ The sequence ${\mathcal{F}}_k$ is constructed recursively.

2.1. Construction of ${\mathcal{F}}_k$

Family ${\mathcal{F}}_1$ is constructed as follows. Take an arbitrary $p_1\in {\mathbb{N}}^{*}.$ Let $K(0,p_1)$ be given by

$$x\in K(0,p_1)\iff x\in X\mathrm{and}{\displaystyle \frac{3x+1}{2^{p_1}}}\in X.$$

Then,

$$K(1,p_1):=\mathcal{T}\left(K(0,p_1)\right).$$

Now, we describe how to construct ${\mathcal{F}}_{k+1}$ when ${\mathcal{F}}_k$ is given. The family ${\mathcal{F}}_{k+1}$ consists of sets of the form $K(k+1,p_1,p_2,\dots ,p_k,p_{k+1})$, so we choose an index $(\mathrm{k}+1,p_1,p_2,\dots ,p_k,p_{k+1})$. For this selected index, we consider the index $(k,p_1,p_2,\dots ,p_k)$. This index is associated with the set $K(k,p_1,p_2,\dots p_k)$ that is in ${\mathcal{F}}_k$, which has already been constructed. Define an auxiliary set $K(k,p_1,p_2,\dots ,p_k,p_{k+1})$ using the formula

$$x\in K(k,p_1,p_2,\dots ,p_k,p_{k+1})\iff$$

$$\iff x\in K(k,p_1,p_2,\dots ,p_k)\mathrm{and}{\displaystyle \frac{3x+1}{2^{p_{k+1}}}}\in X.$$

Finally, we put:

$$K(k+1,p_1,p_2,\dots ,p_k,p_{k+1})=\mathcal{T}\left(K(k,p_1,p_2,\dots ,p_k,p_{k+1})\right).$$

2.2. Some Properties of ${\mathcal{F}}_k$

Now, we study the properties of the constructed sequence ${\mathcal{F}}_k$.

Property 1. 

For each $p\in {\mathbb{N}}^{*}$, the set

$$K(0,p)=\{x\in X:{\displaystyle \frac{3x+1}{2^p}}\in X\}$$

is an arithmetic sequence with difference $2^{p+1}$.

Proof. 

Note that $x\in K(0,p)\Rightarrow x+2^{p+1}\in K(0,p)$.

Suppose now that there exists an even integer r such that $0<r<2^{p+1}$ and such that for some $x\in K(0,p)$, $x+r\in K(0,p)$. Then, write $r=2^s·t$ for some $1\le s\le p$ and $t\in X$. Hence, as ${\displaystyle \frac{3x+1}{2^p}}=m\in X,$ we have $3(x+r)+1=3x+1+3r=2^p·m+3·2^s·t.$ Since $x+r\in K(0,p)$, the left-hand side is divisible by $2^p$, so $s\ge p$, and so, $s=p$. Thus, we have $3(x+r)+1=2^p·m+3·2^p·t=2^p·(m+3t).$ As a consequence, since $m+3t$ is even, we see that $3(x+r)+1$ is divisible by $2^{p+1}$, which is, according to the definition of $K(0,p)$, impossible. □

Property 2. 

For each $p,q\in p,q$, if $p\ne q$, then $K(0,p)\cap K(0,q)=\varnothing .$

Proof. 

Suppose, on the contrary, that there exists $x\in K(0.p)\cap K(0,q)$. Hence, $x=2^p·m=2^q·n$, for some $m,n\in X$. This is impossible when $p\ne q$. □

Property 3. 

We have

$$\bigcup _{p\in X}K(0,p)=X.$$

Proof. 

Indeed, if $x\in X$, then $3x+1$ is even, so there exists $p\in {\mathbb{N}}^{*}$ and $m\in X$ such that $3x+1=2^p·m$. As a consequence, $x\in K(0,p)$. □

2.3. Arithmetic Sequences and ${\mathcal{T}}^{\left(1\right)}$

As the proof of Theorem 3 is by inductive argument, it will be useful to consider the graph of the first iteration $\mathcal{T}={\mathcal{T}}^{\left(1\right)}$ of the Collatz transform. As we have observed above, its domain is the set X, which can be decomposed into a countable family of arithmetic progressions ${\left\{K(0,p)\right\}}_{p\in X}$. Thus, ${\mathcal{T}}^{\left(1\right)}$ is the disjoint union of its restriction to the sets $K(0,p)$, and we have the following:

Property 4. 

$$\mathit{Graph}\left(\mathcal{T}\right)=\bigcup _{p\in X}\left\{\left(n,{\mathcal{T}}^{\left(1\right)}\left(n\right)\right)|n\in K(0,p)\right\}.$$

Property 5. 

$K(1,p)$ is an arithmetic sequence with the difference $r=6.$

We have $K(1,p)=\mathcal{T}\left(K\right(0,p\left)\right)$. From Property 1, $K(0,p)$ is an arithmetic sequence with the difference $2^{p+1}$. Take two consecutive terms of $K(0,p)$, say, x and $x+2^{p+1}$. We have

$$\mathcal{T}(x+2^{p+1})-\mathcal{T}\left(x\right)={\displaystyle \frac{3(x+2^{p+1})+1}{2^p}}-{\displaystyle \frac{3x+1}{2^p}}={\displaystyle \frac{3·2^{p+1}}{2^p}}=6.$$

As a consequence of the above properties, we obtain the following:

Theorem 2. 

Let $p\in X$. Then, the points $(n,\mathcal{T}(n\left)\right)$ of the graph of ${\mathcal{T}|}_{K(0,p)}$ satisfy $\mathcal{T}\left(n\right)={\displaystyle \frac{3}{2^p}}·n+{\displaystyle \frac{1}{2^p}}$. Hence, they lie on the (real) line $y={\displaystyle \frac{3}{2^p}}·x+{\displaystyle \frac{1}{2^p}}.$

The next property concerns some arithmetic sequences in X.

Property 6. 

Let $U(x_0,r)=(\left\{x_i\right\}:i\in \mathbb{N})$ be an arithmetic sequence in X such that $x_i=x_0+i·2r$, where r is odd. Take $p\in {\mathbb{N}}^{*}$.

Then, the set

$$K_p\left(U\right)=\{x_i\in U:{\displaystyle \frac{3x_i+1}{2^p}}\in X\}$$

is an arithmetic sequence.

Proof. 

It is enough to show that the set $V_p=\{i\in \mathbb{N}:{\displaystyle \frac{3x_i+1}{2^p}}\in X\}$ is an arithmetic sequence.

Any i such that ${\displaystyle \frac{3x_i+1}{2^p}}\in X$ must satisfy

$$3x_0+1+i·3·2r=0mod{2}^p.$$

Let $3x_0+1=-ymod{2}^p.$ As $3x_0+1$ is even, $-y$ is also even, so $y=2z$ for some z. Thus,

$$i\in V_p\iff i·3·2r=2zmod{2}^p.$$

As $3r$ is odd, it has an inverse in the ring ${\mathbb{Z}}_{2^p}$; hence,

$$2i=2z·{\left(3r\right)}^{-1}mod{2}^p.$$

Denote ${\left(3r\right)}^{-1}mod{2}^p$ as t.

Thus, $i=z·tmod{2}^{p-1}$, so $i\in V_p$ iff

$$i=z·t+j·2^{p-1},$$

and thus $V_p$ is an arithmetic sequence. □

Finally, let us observe the following:

Property 7. 

Keeping the notation as above, given a sequence $U=U(x_0,r)$, we obtain a sequence of arithmetic progressions ${\left(K_p\right)}_{p\in {\mathbb{N}}^{*}}$ such that $K_p\cap K_q=\varnothing$ for $p\ne q$ and ${\bigcup }_{p\in {\mathbb{N}}^{*}}{K}_p=U$.

Proof. 

Indeed, for a given p and q, suppose that $x_i$ is the common element of $K_p$ and $K_q$. Then, for some k and m, we have ${\displaystyle \frac{3x_i+1}{2^p}}=2k+1$ and ${\displaystyle \frac{3x_i+1}{2^q}}=2m+1$, so $2^p(2k+1)=2^q(2m+1)$, and this is impossible.

Then, take $x\in U$. As $3x+1$ is even, there exists (only one) p such that ${\displaystyle \frac{3x+1}{2^p}}$ is odd, so ${\bigcup }_{p\in {\mathbb{N}}^{*}}{K}_p=U$. □

2.4. Description of the Second Iteration ${\mathcal{T}}^{\left(2\right)}$

Section 2.3 gives a description of the graph of $\mathcal{T}$. This section describes the graph of ${\mathcal{T}}^{\left(2\right)}$. In fact, it is the second step in the inductive proof of Theorem 3. We include it here to simplify the reading of the (rather technical) proof of Theorem 3.

Recall the construction of the family of sets $K(2,p_1,p_2)$.

The family ${\mathcal{F}}_2$ consists of sets of the form $K(2,p_1,p_2)$. So we choose the index $(p_1,p_2)$ and take the index $(1,p_1)$. This index is associated with the set $K(1,p_1)$ in ${\mathcal{F}}_1$, which has already been constructed. Define auxiliary sets $K(1,p_1,p_2)$ as $x\in K(1,p_1,p_2)\iff x\in K(1,p_1)\mathrm{and}{\displaystyle \frac{3x+1}{2^{p_2}}}\in X.$ Then,

$$K(2,p_1,p_2)=\mathcal{T}\left(K(k,p_1,p_2)\right).$$

From Properties 6 and 7 we see that the set $K(1,p_1,p_2)$ is an arithmetic sequence, consisting of such $x\in K(1,p_1)$ for which ${\displaystyle \frac{3x+1}{2^{p_2}}}$ is odd; moreover, changing $p_2\in X$, we obtain a disjoint decomposition of $K(1,p_1)$ into arithmetic sequences $K(1,p_1,p_2)$, i.e., such that $p\ne q\Rightarrow K(1,p_1,p)\cap K(1,p_1,q)=\varnothing$ and ${\bigcup }_{p_2}K(1,p_1,p_2)=K(1,p_1).$

Changing both $p_1$ and $p_2$, we obtain a decomposition of

$$\mathcal{T}\left(X\right)={\mathcal{T}}^{\left(1\right)}\left(X\right)=\bigcup _{p_1,p_2}K(1,p_1,p_2)$$

into arithmetic sequences.

Now, take $K(2,p_1,p_2)=\mathcal{T}\left(K(1,p_1,p_2)\right).$ The set $K(2,p_1,p_2)$ is the set of values of the first iteration $\mathcal{T}$ on the arithmetic sequence $K(1,p_1)$. Now, take into account an arbitrary point $x\in X.$ For this point, there is exactly one integer $p_1$ such that $x\in K(0,p_1)$. Thus, $x_1:=\mathcal{T}\left(x\right)$ belongs to $K(1,p_1)$ and $\mathcal{T}\left(x\right)=x_1={\displaystyle \frac{3}{2^{p_1}}}·x+{\displaystyle \frac{1}{2^{p_1}}}$. Since $x_1\in K(1,p_1)$, there exists exactly one integer $p_2$ such that $T\left(x\right)=x_1$ belongs to the set $K(1,p_1,p_2)$ and $x_2:=\mathcal{T}\left(x_1\right)={\displaystyle \frac{3}{2^{p_2}}}·x_1+{\displaystyle \frac{1}{2^{p_2}}}.$ Thus, ${\mathcal{T}}^{\left(2\right)}\left(x\right)={\displaystyle \frac{3}{2^{p_2}}}·({\displaystyle \frac{3}{2^{p_1}}}·x+{\displaystyle \frac{1}{2^{p_1}}})+{\displaystyle \frac{1}{2^{p_2}}}.$

As a consequence, the point $(x,{\mathcal{T}}^{\left(2\right)}\left(x\right))=(x,\mathcal{T}\left(\mathcal{T}\left(x\right)\right))$ lies on the graph of an affine function $y=a(p_1,p_2)·x+b(p_1,p_2)$, where $a(p_1,p_2)={\displaystyle \frac{3^2}{2^{p_1+p_2}}}$ and $b(p_1,p_2)={\displaystyle \frac{3}{2^{p_1+p_2}}}+{\displaystyle \frac{1}{2^{p_2}}}.$

2.5. Description of the k-th Iteration ${\mathcal{T}}^{\left(k\right)}$

The description of ${\mathcal{T}}^{\left(k\right)}$ is one of the essential results of this paper. Perhaps this result is already known to specialists on the subject, but we were unable to find it. Thus, we present it as the following Theorem. To simplify the notation, we slightly change the indexing of the sets in the family ${\mathcal{F}}_k$.

Theorem 3. 

Let $k\in {\mathbb{N}}^{*}$ and let $\beta =(p_1,\dots p_k)\in {\left({\mathbb{N}}^{*}\right)}^k$. Let $\left|\beta \right|=p_1+\dots +p_k$. Then, there exists a family of subsets $K_{\beta }\subset X$ satisfying the following properties:

(1)

Each $K_{\beta }$ is an arithmetic sequence.

(2)

If ${\beta }_1\ne {\beta }_2$, then

$$K_{{\beta }_1}\cap K_{{\beta }_2}=\varnothing .$$

(3)

${\bigcup }_{\beta \in {\left({\mathbb{N}}^{*}\right)}^k}{K}_{\beta }=X.$

(4)

For each $\beta \in {\left({\mathbb{N}}^{*}\right)}^k$, there exist positive rational numbers $a\left(\beta \right)$ and $b\left(\beta \right)$ such that for each $x\in K_{\beta }$, we have

$${\mathcal{T}}^{\left(k\right)}\left(x\right)=a\left(\beta \right)·x+b\left(\beta \right).$$

(5)

$$a\left(\beta \right)={\displaystyle \frac{3^k}{2^{\left|\beta \right|}}}.$$

$$b\left(\beta \right)=\sum _{i=1}^k{\displaystyle \frac{3^{k-i}}{2^{p_i+p_{i+1}+..+p_k}}}.$$

(6)

The difference in the sequence $K\left(\beta \right)$ equals $2^{\left|\beta \right|+1}$.

(7)

The difference in the sequence $K(k,\beta )=T^{\left(k\right)}\left(K_{\beta }\right)$ equals $2·3^k$.

Before starting the proof, we present a picture illustrating the idea of the Theorem. Figure 2 shows $(n,{\mathcal{T}}^{\left(19\right)}\left(n\right))$ for all odd n up to 500.

Figure 2. Graph of ${\mathcal{T}}^{\left(19\right)}\left(n\right)$.

Proof. 

For $k=1$n the proof follows from the construction given in Section 2.1, Properties 1–5 and Theorem 2.

Next, assume that $k\ge 1$, $\beta \in {\left({\mathbb{N}}^{*}\right)}^k$ and there exist sets $K\left(\beta \right)$ satisfying Properties 1–7. We want to prove that for each $\alpha \in {\left({\mathbb{N}}^{*}\right)}^{k+1}$, there exist sets $K_{\alpha }$ satisfying (1)–(7).

Each such $\alpha$ is of the form $(\beta ,p_{k+1})$, for some $\beta \in {\left({\mathbb{N}}^{*}\right)}^k$, $p_{k+1}\in {\mathbb{N}}^{*}.$ We define $K_{\alpha }$ using Property 6, which we are able to do as, according to the inductive assumption, the sequence ${\mathcal{T}}^k\left(K_{\beta }\right)$ is arithmetic. Take, as in this property, $K\left(U\right):={\mathcal{T}}^{\left(k\right)}\left(K_{\beta }\right)$ and $p:=p_{k+1}$. Define

$$K_{\alpha }:={\left({\mathcal{T}}^{\left(k\right)}\right)}^{-1}\left({\mathcal{T}}^{\left(k\right)}{\left(K_{\beta }\right)}_{p_{k+1}}\right).$$

Thus, the following are true:

(1)

The below follows from Property 6.

(2)

If ${\alpha }_1=({\beta }_1,p_1)\ne {\alpha }_2=({\beta }_2,p_2)$, then ${\beta }_1\ne {\beta }_2$ or $p_1\ne p_2$. If ${\beta }_1\ne {\beta }_2$, then $K_{{\beta }_1}\cap K_{{\beta }_2}$ by inductive assumption, and as $K_{{\alpha }_i}\subset K_{{\beta }_i}$, we are finished. If ${\beta }_1={\beta }_2$ but $p_1\ne p_2$, we obtain the claim from Property 7.

(3)

Take ${\bigcup }_{\alpha \in {\left({\mathbb{N}}^{*}\right)}^{k+1}}{K}_{\alpha }$. We have

$$\bigcup _{\alpha \in {\left({\mathbb{N}}^{*}\right)}^{k+1}}{K}_{\alpha }=\bigcup _{\beta \in {\left({\mathbb{N}}^{*}\right)}^k}\bigcup _{p\in {\mathcal{N}}^{*}}{\left({\mathcal{T}}^{\left(k\right)}\right)}^{-1}\left({\mathcal{T}}^{\left(k\right)}{\left(K_{\beta }\right)}_p\right)=\bigcup _{\beta \in {\left({\mathbb{N}}^{*}\right)}^k}{\left({\mathcal{T}}^{\left(k\right)}\right)}^{-1}\left(\bigcup _{p\in {\mathcal{N}}^{*}}{\mathcal{T}}^{\left(k\right)}{\left(K_{\beta }\right)}_p\right).$$

From Property 7,

$$\bigcup _{p\in {\mathcal{N}}^{*}}{\mathcal{T}}^{\left(k\right)}{\left(K_{\beta }\right)}_p={\mathcal{T}}^{\left(k\right)}\left(K_{\beta }\right).$$

Thus, we obtain

$$\bigcup _{\beta \in {\left({\mathbb{N}}^{*}\right)}^k}{\left({\mathcal{T}}^{\left(k\right)}\right)}^{-1}\left({\mathcal{T}}^{\left(k\right)}\left(K_{\beta }\right)\right)=X,$$

by inductive assumption.

(4)

We want to prove that if $x\in K_{\alpha }$, then ${\mathcal{T}}^{(k+1)}\left(x\right)=a\left(\alpha \right)x+b\left(\alpha \right)$, for some numbers ($a\left(\alpha \right)$ and $b\left(\alpha \right)$). Observe that on ${\mathcal{T}}^{\left(k\right)}{\left(K_{\beta }\right)}_p$, the transform $\mathcal{T}$ acts as $\mathcal{T}\left(z\right)={\displaystyle \frac{3}{2^p}}z+{\displaystyle \frac{1}{2^p}}.$ Replacing z with ${\mathcal{T}}^{\left(k\right)}\left(x\right)$ and using the inductive assumption, we are finished.

(5)

The proof is a direct consequence of the above.

(6)

Let $x,y\in K\left(\beta \right)$. We have

$${\mathcal{T}}^{k+1}\left(x\right)=a\left(\beta \right))·x+b\left(\beta \right))$$

and

$${\mathcal{T}}^{k+1}\left(y\right)=a\left(\beta \right))·y+b\left(\beta \right)),$$

so

$$T^k\left(x\right)-T^k\left(y\right)=a\left(\beta \right)·(x-y).$$

From (5), we have $a_{p_1,...,p_k,p_{k+1}}={\displaystyle \frac{3^{k+1}}{2^{\left|\beta \right|}}},$ and so,

$$({\mathcal{T}}^k\left(x\right)-{\mathcal{T}}^k\left(y\right))·2^{\left|\beta \right|}=3^k(x-y).$$

As $({\mathcal{T}}^k\left(x\right)-{\mathcal{T}}^k\left(y\right))$ is even, $(x-y)$ is divisible by $2^{\left|\beta \right|+1}$.

On the other hand, analogously, as in the proof of Property 1, we check that the difference in the sequence cannot be less than $2^{\left|\beta \right|+1}$.

(7)

Take two consecutive elements of $K_{\beta }$. From (6), we know that they are given by $x\in K_{\beta }$ and $x+2^{\left|\beta \right|+1}$.

Thus, two consecutive terms of ${\mathcal{T}}^{\left(k\right)}\left(K_{\beta }\right)$ are given by ${\mathcal{T}}^k\left(x\right)$ and ${\mathcal{T}}^k(x+2^{\left|\beta \right|+1})$. From this, we have

$${\mathcal{T}}^k(x+2^{\left|\beta \right|+1})-{\mathcal{T}}^k\left(x\right)={\displaystyle \frac{3^k·2^{\left|\beta \right|+1}}{2^{\left|\beta \right|}}}=2·3^k.$$

□

3. Furstenberg Topology on Odd Positive Integers

In 1955, Furstenberg, in [20], defined a topology on integers. We will use his definition, restricted to X.

Let X be a set of all odd positive integers.

Definition 4. 

Let $U\subset X$. Then, $U\in {\tau }_{ar}$ if and only if for any $a\in U$, there exists $r>0$ such that for all $i\in \mathbb{N}$, we have $a+i·2r\in U$.

Thus, $U\subset X$ is open iff every point $a\in U$ is contained in U together with an arithmetic sequence, or $U=\varnothing .$

Remark 2. 

(1) ${\tau }_{ar}$ is a topology in X.

(2)

The set $\{a+i·2r,i\in \mathbb{N}\cup \{0\left\}\right\}$ is a neighbourhood of a.

(3)

Let $a\in X$ and $r\in {\mathbb{N}}^{*}$. Then, the set (arithmetic sequence) $U=U(a,r):=\{a+i·2r,i\in \mathbb{N}\cup \{0\left\}\right\}$ is closed in ${\tau }_{ar}.$

Proof. 

(1) and (2) are an immediate consequence of the definition.

(3) is also easy, but we put it here for the sake of completeness.

Take $V:=X\setminus U$. We have to show that V is an open set. Observe that $V=\{x\in X:x-a\ne 0mod(2r\left)\right\}$. For an integer $0<s<r$, let $V_s:=\{x\in X:x-a=2smod\left(2r\right)\}$. Thus, $V_s$ is an open set, as it is in fact $\{a+i·2s,i\in \mathbb{N}\}$. As $V={\bigcup }_{0<s<r}{V}_s$, we are finished. □

We recall below some properties of the Furstenberg topology. They are proved in [20].

Proposition 1. 

(1) $(X,{\tau }_{ar})$ is a Hausdorff space.

(2)

($X,{\tau }_{ar})$ is a regular space.

(3)

($X,{\tau }_{ar})$ has a countable basis.

(4)

($X,{\tau }_{ar})$ is (from the Uryhson theorem) metrizable.

The following theorem is, to the best of our knowledge, new.

Theorem 4. 

$\mathcal{T}:X\ni n\to \mathcal{T}\left(n\right)\in X$ is continuous as a map from $(X,{\tau }_{ar})$ to itself.

Proof. 

We have to show that $\mathcal{T}$ is continuous in every $x\in X$. This means that for every neighbourhood V of $\mathcal{T}\left(x\right)$, there exists a neighbourhood U of x, such that $\mathcal{T}\left(U\right)\subset V$.

Without loss of generality, we may assume that $V=\left\{\mathcal{T}\right(x)+j·2r,j\in \mathbb{N}\},$ where r is a positive integer. Let us also define an auxiliary set $V^{\prime }\subset V$ as $V^{\prime }=\{\mathcal{T}\left(x\right)+\left(3j\right)·2r,j\in \mathbb{N}\cup \left\{0\right\}\}$. Now, let p be such an integer that $\mathcal{T}\left(x\right)={\displaystyle \frac{3x+1}{2^p}}$ is an odd number (thus, $2^{-p}={|3x+1|}_2$). Let

$$U:=\left\{x+j·2^p·\left(2r\right)\right\}.$$

U is a neighbourhood of x, as p is fixed. Take a number $n\in U$. Thus, $n=x+j·2^p·\left(2r\right)$ for some j. We have

$$\mathcal{T}\left(n\right)={(3n+1)\left(\right|3n+1|}_2{)=(3x+1)(|3n+1|}_2)+(3j·2^p·\left(2r\right)){\left(\right|3n+1|}_2)=$$

$$=2^p{·\mathcal{T}\left(x\right)|3n+1|}_2+(3j·2^p·\left(2r\right)){\left(\right|3n+1|}_2).$$

Let ${|3n+1|}_2=2^{-s}$.

Then, $\mathcal{T}\left(n\right)=2^p\mathcal{T}\left(x\right)2^{-s}+(3j·2r)2^p{2}^{-s}.$ As $\mathcal{T}\left(n\right)$ must be odd and $\mathcal{T}\left(x\right)$ is odd, the only possibility is $s=p$. So,

$$\mathcal{T}\left(n\right)=\mathcal{T}\left(x\right)+\left(3j\right)·2r.$$

So, $\mathcal{T}\left(n\right)\in V^{\prime }\subset V$, and we are finished. □

Remark 3. 

From this Theorem, we immediately determine that any iteration of $\mathcal{T}$ is continuous in $(X,{\tau }_{ar})$.

4. Stopping Time

The main result of this Section is Theorem 6, which describes, in $(X,{\tau }_{ar})$, a property of the set of odd integers with a finite stopping time.

Recall the notion of a (finite) stopping time.

Definition 5. 

We say that $n\in X\setminus \left\{1\right\}$ has a finite stopping time if and only if there exists a positive integer k such that ${\mathcal{T}}^k\left(n\right)<n$. The least such k, if it exists, is called $\mathit{st}\left(n\right)$, the stopping time of n. If a given n does not have a finite stopping time, we say $\mathit{st}\left(n\right)=\infty$.

For a natural number k, define a set $V_k=\left\{n\in X:\mathrm{st}\left(n\right)=k\right\}\cup \left\{1\right\}.$ Let $V={\bigcup }_{k\in \mathbb{N}}{V}_k$ and take the complement of V in X, namely the set $S=X\setminus V$. Then, for $n\in S$, there is $\mathrm{st}\left(n\right)=\infty .$ Note that in a nontrivial cycle (if one exists), with $n_{+}$ being the maximum and $n_{-}$ the minimum of the cycle values, $n_{+}$ does have a finite stopping time and $n_{-}$ does not. However, it is not difficult to see the following:

Remark 4. 

Collatz Conjecture is true if and only if S is empty.

Unfortunately, so far, there is no proof that S is empty. However, Terras, in his paper [16], proved the following theorem.

Theorem 5. 

The density of S is 0.

Recall that the (natural) density of a set $A\subset \mathbb{N}$ is defined as follows:

Definition 6. 

Let $A\subset \mathbb{N}$. Let $a\left(n\right):=|A\cap \{1,2,\dots ,n\left\}\right|$. The upper asymptotic density of A is $\overline{d}\left(A\right)={lim\; sup}_{n\to \infty }{\displaystyle \frac{a\left(n\right)}{n}}$, and the lower asymptotic density of A is $\underline{d}\left(A\right)={lim\; inf}_{n\to \infty }{\displaystyle \frac{a\left(n\right)}{n}}$. If $\overline{d}\left(A\right)=\underline{d}\left(A\right)=:d\left(A\right)$, then $d\left(A\right)$ is the density of A.

Remark 5. 

Observe that if A contains an (infinite) arithmetic sequence, then $d\left(A\right)>0.$

The main result of this Section is a theorem about the topological properties of the set of $x\in X$ with an infinite stopping time. We prove that this set is closed and nowhere dense in $(X,{\tau }_{ar})$; this gives a result densewise weaker than Theorem 5, but perhaps interesting on its own.

Theorem 6. 

The set S (of n with $\mathit{st}\left(n\right)=\infty$) is closed and nowhere dense in $(X,{\tau }_{ar})$.

Proof. 

Let $U(x_0,R)$ denote an arithmetic sequence in X, with $x_i=x_0+i·R$. As R must be even, there exist (unique) integers, $s\ge 0$ and $r\in X$, such that $R=2^s·2r.$ Denote $s=c\left(U(x_0,R)\right)$.

From the definition of $\mathcal{T}$, it follows that there exists a unique integer $p>0$, such that $\mathcal{T}\left(x_0\right)·2^p=3x_0+1.$ We write $p=\left|U\right(x_0,R\left)\right|$.

First, we prove the following:

Lemma 1. 

For every $U(x_0,R)$ with $c\left(U(x_0,R)\right)>0$, there exists a smaller open set $U^{\prime }(x_0,R^{\prime })\subset U(x_0,R)$ such that $\mathcal{T}\left(U^{\prime }(x_0,R^{\prime })\right)\ni \mathcal{T}\left(x_0\right)$ and $c\left(U^{\prime }(x_0,R^{\prime })\right)<c\left(U(x_0,R)\right)$.

Proof. 

Let $p_0:=\left|U(x_0,R)\right|$ and $s_0:=c\left(U(x_0,R)\right)$.

Case 1. $p_0<s_0$. Recall that $U(x_0,r)=\{x_i=x_0+i·R=x_0+i·2^{s_0}\left(2r\right),i\in \mathbb{N}\}$, where $r\in X.$ Observe that $|3x_i{+1|}_2={\displaystyle \frac{1}{2^{p_0}}}$, i.e., ${\displaystyle \frac{3x_i+1}{2^{p_0}}}\in X$. Indeed,

$${\displaystyle \frac{3x_i+1}{2^{p_0}}}={\displaystyle \frac{3x_0+1}{2^{p_0}}}+{\displaystyle \frac{i·2^{s_0}2·3r}{2^{p_0}}}=\mathcal{T}\left(x_0\right)+i·2^{s_0-p_0}·2·3r.$$

In this case, establish $U(x_0,R^{\prime })=U(x_0,R)$. Then, $\mathcal{T}\left(U(x_0,R^{\prime })\right)=\mathcal{T}\left(x_0\right)+i·2^{s_0-p_0}·2·\left(3r\right)$, and so, $\mathcal{T}\left(U(x_0,R^{\prime })\right)$ is a neighbourhood of $\mathcal{T}\left(x_0\right)$ with $c\left(\mathcal{T}\left(U(x_0,R^{\prime })\right)\right)=s_0-p_0<s_0$, as $p_0\ge 1$.

Case 2. $s_0\le p_0.$ Take $m_0:=p_0-s_0\ge 0$ and take a subsequence of $U(x_0,R)$ given by $U(x_0,R^{″})=\left\{x_{i_j}\right\}$, where

$$x_{i_j}:=x_0+j·2^{m_0}·2^{s_0}·\left(2r\right)=x_0+j·2^{p_0}·\left(2r\right).$$

We compute, as above, that $|3x_{i_j}{+1|}_2={\displaystyle \frac{1}{2^{p_0}}}$. Thus, $\mathcal{T}\left(U(x_0,R^{″})\right)=\mathcal{T}\left(x_0\right)+j·2\left(3r\right)$ is a neighbourhood of $\mathcal{T}\left(x_0\right)$ with $c(\mathcal{T}\left(U(x_0,R^{″})\right)=0<s_0$, as $s_0>0.$ □

From Lemma 1 we have the following:

Corollary 1. 

Every neighbourhood $U_i({\mathcal{T}}^{\left(i\right)}\left(x_0\right),R_i)$ with $c(U_i({\mathcal{T}}^{\left(i\right)}\left(x_0\right),R_i)>0$ is an image by ${\mathcal{T}}^{\left(i\right)}$ of some $V_i(x_0,R_{V_i})\subset U(x_0,R)$.

Proof. 

Indeed, applying the lemma to $U_i({\mathcal{T}}^{\left(i\right)}\left(x_0\right),R_i)$ instead of $U(x_0,R)$, we obtain $U_i({\mathcal{T}}^{\left(i\right)}\left(x_0\right),R_i)=\mathcal{T}\left(U_{i-1}^{\prime }({\mathcal{T}}^{(i-1)}\left(x_0\right),R_{i-1}^{\prime })\right)$. We have, as in the lemma, $U_i({\mathcal{T}}^{\left(i\right)}\left(x_0\right),R_i)=\mathcal{T}\left(W({\mathcal{T}}^{(i-1)}\left(x_0\right),R_W)\right)$ for some neighbourhood $W({\mathcal{T}}^{(i-1)}\left(x_0\right),R_W)$ of ${\mathcal{T}}^{(i-1)}\left(x_0\right)$; then, arguing inductively, we obtain

$$U_i({\mathcal{T}}^{\left(i\right)}\left(x_0\right),R_i)=\mathcal{T}({\mathcal{T}}^{(i-1)}\left(V_W(x_0,R_{V_W})\right)={\mathcal{T}}^{\left(i\right)}(V_i(x_0,R_{V_i}).$$

□

Next, we need

Lemma 2. 

For every neighbourhood $U_0(x_0,R_0)$ of $x_0$, there exists a (finite) sequence of neighbourhoods $U_i({\mathcal{T}}^{\left(i\right)}\left(x_0\right),R_i)$ of ${\mathcal{T}}^{\left(i\right)}\left(x_0\right)$, $i=0,\dots ,k$ with

1. 

$U_i({\mathcal{T}}^{\left(i\right)}\left(x_0\right),R_i)$ is a neighbourhood of ${\mathcal{T}}^{\left(i\right)}\left(x_0\right)$ with $c\left(U_i({\mathcal{T}}^{\left(i\right)}\left(x_0\right),R_i)\right)=:s_i$.

2. 

$s_k=0$.

3. 

$U_k({\mathcal{T}}^{\left(k\right)}\left(x_0\right),R_k)={\mathcal{T}}^{\left(k\right)}(V_k(x_0,R_{V_k})$, with $V_k(x_0,R_{V_k})\subset U_0(x_0,R_0).$

Proof. 

We proceed as follows. If $s_0=0$, then we are finished. If $s_0>0$, then apply Lemma 1 to this $U_0(x_0,R_0)$. This way, we obtain $U_1(\mathcal{T}\left(x_0\right),R_1)$ with $s_1<s_0$. We proceed in this way until for some (finite) k, we obtain $s_k=0$. The last statement follows from Corollary 1. □

The last Lemma we need is as follows:

Lemma 3. 

For any $k\in {\mathbb{N}}^{*}$, the set $V_k=\left\{n\in X:st\left(n\right)=k\right\}$ is open.

Proof. 

We proceed by induction. Take $n_0\in V_1$, so $\mathcal{T}\left(n_0\right)<n_0$. From Theorem 3, we know that there exists an arithmetic sequence $n_i=n_0+ir,i\in \mathbb{N}$ such that all points $(n_i,\mathcal{T}\left(n_i\right)$ are on a line with a slope ${\displaystyle \frac{3}{2^{1+j}}}$ (for some natural j). As $\mathcal{T}\left(n_0\right)<n_0$, this slope must be less than one, so at most, ${\displaystyle \frac{3}{4}}$. As a consequence, all the points $(n_i,\mathcal{T}\left(n_i\right))$ are “under” the line $y=x$. Thus, for all i, $\mathcal{T}\left(n_i\right)<n_i$; the sequence $\left(n_i\right)$ is contained in $V_1$, so $V_1$ is open.

Let $k\ge 2$ and let $n\in V_k$. Thus, $\mathrm{st}\left(n\right)=k$, so ${\mathcal{T}}^{\left(j\right)}\left(n\right)>n$ for $j=1,2,\dots k-1$ and ${\mathcal{T}}^{\left(k\right)}\left(n\right)<n$; in other words, $\mathcal{T}\left({\mathcal{T}}^{(k-1)}\left(n\right)\right)<n<{\mathcal{T}}^{(k-1)}\left(n\right).$ This implies that $\mathrm{st}\left({\mathcal{T}}^{k-1}\left(n\right)\right)=1$, so ${\mathcal{T}}^{k-1}\left(n\right)\in V_1.$ Thus, $n\in {\left({\mathcal{T}}^{(k-1)}\right)}^{-1}\left(V_1\right)$. As $V_1$ is open and ${\mathcal{T}}^{(k-1)}$ is continuous, the set ${\left({\mathcal{T}}^{(k-1)}\right)}^{-1}\left(V_1\right)$ is open, which ends the proof. □

Let us now prove the theorem. As $V={\bigcup }_{k\in \mathbb{N}}{V}_k$ is open and $S=X\setminus V$, S is closed. Assume that S is not nowhere dense. Then S has an interior point (in topology ${\tau }_{ar}$). Thus, there exists $x_0\in S$ such that for some even $R_0$, there is $U(x_0,R_0)\in S$. Using Lemmas 1 and 2, we obtain a neighbourhood $U_k({\mathcal{T}}^{\left(k\right)}\left(x_0\right),R_k)$ with $c\left(U_k({\mathcal{T}}^{\left(k\right)}\left(x_0\right),R_k)\right)=0$ and form Corollary 1, $U_k({\mathcal{T}}^{\left(k\right)}\left(x_0\right),R_k)={\mathcal{T}}^{\left(k\right)}\left(V_k(x_0,R_k)\right)$, $V_k(x_0,R_k)\subset U_0(x_0,R_0)$. Thus, we may apply Property 6 to $U_k({\mathcal{T}}^{\left(k\right)}\left(x_0\right),R_k)$, so we may decompose $U_k({\mathcal{T}}^{\left(k\right)}\left(x_0\right),R_k)$ into (nonempty for every $p\in {\mathbb{N}}^{*}$) arithmetic sequences, say, ${\left(U_k\right)}_p,$ for short, such that for every $z\in {\left(U_k\right)}_p$, we have

$$\mathcal{T}\left(z\right)={\displaystyle \frac{3}{2^p}}z+1/2^p.$$

As ${\mathcal{T}}^k:V_k(x_0,R_{V_k})\to U_k({\mathcal{T}}^k\left(x_0\right),R_k)$ is bijective, there exist $V_{k,p}\subset V_k(x_0,R_{V_k})$ satisfying $U_{k,p}={\mathcal{T}}^k\left(V_{k,p}\right).$

From Theorem 3, it follows that there exist $j\in \mathbb{N}$ and $b_{k,j}\in {\mathbb{N}}^{*}$ such that

$$x\in V_{k,p}\Rightarrow {\mathcal{T}}^k\left(x\right)={\displaystyle \frac{3^k}{2^{k+j}}}·x+b_{k,j}.$$

Thus,

$${\mathcal{T}}^k\left(z\right)={\mathcal{T}}^{k+1}\left(x\right)={\displaystyle \frac{3}{2^p}}\left({\displaystyle \frac{3^k}{2^{k+j}}}·x+b_{k,j}\right)+{\displaystyle \frac{1}{2^p}}={\displaystyle \frac{3^{k+1}}{2^{k+j+p}}}·x+b_{k,j,p}.$$

For big enough p, the coefficient ${\displaystyle \frac{3^{k+1}}{2^{k+j+p}}}$ is smaller than 1. So, there exist (infinitely many) points $x\in V_{k,p}$ with $(x,{\mathcal{T}}^{k+1}\left(x\right))$ “under” the line $y=x$, i.e., $x>{\mathcal{T}}^{k+1}\left(x\right)$, so they have a finite stopping time, which is a contradiction. □

Remark 6. 

The proof of Theorem 6 may be obtained in an easier way if we use Theorem 5.

Proof. 

As above, S is closed. Assume that S is not nowhere dense. Then, S has an interior point (in ${\tau }_{ar}$). Thus, this point is contained in S with an arithmetic sequence; see Remark 2. But then, $d\left(S\right)>0$, from Remark 5, and this is a contradiction with Theorem 5. □

Remark 7. 

Let $\mathcal{R}$ be a family of all arithmetic progressions of the form $n·2t,t\in {\mathbb{N}}^{*},n\in \mathbb{N}$. Let $\mathcal{S}\subset \mathcal{R}$ be a subfamily which is a filter basis. For such S, we have the topology ${\tau }_S$, analogous to the Furstenberg topology. We can then investigate the properties of $\mathcal{T}$ in $(X,{\tau }_S)$. As far as nowhere denseness is considered, the Furstenberg topology is the best, as it is the strongest one, so the nowhere dense set is the smallest.

5. Generalized Collatz Transform

The Collatz problem has been generalized or reformulated in many ways. One possible group of generalizations of the Collatz problem preserves the set of positive integers ${\mathbb{N}}^{*}$ as the domain of the Collatz function, but the formula defining T is replaced by a formula similar to it. For example, instead of $3x+1$ one may consider the functions $3x-1$ or $5x+1$ or, more generally, affine functions of type $px+q$. The Collatz functions of this type may not seem to differ, from a numerical point of view, from the $3x+1$ function. However, the numerical data obtained, for example, for the function $5x+1$, suggest that in this case, almost all orbits escape to infinity, that is, are divergent; see [2].

Another method of changing the problem is to reduce it to residue classes. A subset $\mathbb{S}\subset {\mathbb{N}}^{*}$ is sufficient if the convergence of all trajectories for $n\in \mathbb{S}$ implies the convergence of all trajectories for $n\in {\mathbb{N}}^{*}$. Trivially, the set $\mathbb{S}$ of all natural numbers that are not divisible by 3 is sufficient. A less trivial example of a sufficient set is the set $\mathbb{S}=\left\{x\in {\mathbb{N}}^{*}:x\equiv 1mod16\right\}$; see [1,21] for a generalization.

Finally, we may generalize both the domain of T and T itself. In [1,3], we can find information on the possibility of such modifications of the Collatz problem in $\mathbb{R}$ or in $\mathbb{C}$. There are many such extensions; for example, for $x\in \mathbb{R}$, let us denote $f\left(x\right)={\displaystyle \frac{1}{2}}\left(3^{{sin}^2(\pi x/2)}\right)x+{sin}^2(\pi x/2)$. One may easily observe that if $x\in \mathbb{N}$, then $f\left(x\right)=T\left(x\right)$; see [1,22].

Our approach to generalization of the Collatz problem, on the one hand, restricts the domain of the function, and on the other hand, leaves the Collatz transform intact as much as possible.

We restrict our considerations to likens. Let us start with some remarks, showing that likens have interesting topological properties.

Remark 8. 

The Furstenberg topology may also be defined on likens ${\mathbb{L}}_q\subset X.$ The small difference is as follows: Let $U\subset {\mathbb{L}}_q$. Then, $U\in {\tau }_{ar}\subset \mathcal{P}\left({\mathbb{L}}_q\right)$ if and only if for any $a\in U$, there exists an integer $r>0$ such that for all $i\in \mathbb{N}$, we have $a+i·2q·r\in U$.

Remark 9. 

For any primes $q_1,\cdots ,q_s$, the likens ${\mathbb{L}}_{q_1,\cdots ,q_s}$ are open in Furstenberg topology.

Indeed, arithmetic sequences $A_{q_j}:=\{0+i·2q_j,i\in \mathbb{N}\}$ are closed in X; thus, ${\mathbb{L}}_{q_1,\cdots ,q_s}=X\setminus {\sum }_{j=1}^s{A}_{q_j}$ is open.

From now on, we restrict our considerations to likens ${\mathbb{L}}_{2,3,q}$, where $q>3$ is a prime number. As the context is clear, we still denote ${\mathbb{L}}_{2,3,q}$ as ${\mathbb{L}}_q$.

We define a generalized Collatz transform as a function from ${\mathbb{L}}_q\to {\mathbb{L}}_q$, defined as follows:

Definition 7. 

Let $n\in {\mathbb{L}}_q$.

$${\mathcal{T}}_q\left(n\right)=s\left(3n\right){\left|s\left(3n\right)\right|}_2,$$

(3)

where $s\left(n\right)$ is the successor of $3n$ in ${\mathbb{L}}_q$; see Definition 3. In other words, if $3n+1$ is divisible by q, we replace it with $3n+2$.

Remark 10. 

(1)

Strictly speaking, the first iteration of the transforms ${\mathcal{T}}_q$ (i.e., ${\mathcal{T}}_q$ itself) is a map from X to ${\mathbb{L}}_q,$ and only starting from the second iteration may we speak of a map from ${\mathbb{L}}_q$ to ${\mathbb{L}}_q$. This is also the case for $\mathcal{T}$, treated as $\mathcal{T}:X\to {\mathbb{L}}_3$.

(2)

We will not prove it here, but, as it visible in Figure 3, this version of Theorem 3 holds also for ${\mathbb{L}}_q$ and ${\mathcal{T}}_q$.

Figure 3. Decomposition of ${\mathcal{T}}_{19}^{\left(3\right)}$ and ${\mathcal{T}}_7^{\left(4\right)}$ into arithmetic sequences.

6. Some Numerical Results and Questions

Numerically investigating ${\mathcal{T}}_q$ transformations in ${\mathbb{L}}_q$ for several initial values of q, we discovered some phenomena not detected (so far) in the classical case. In what follows, we present a number of numerical results for ${\mathcal{T}}_q$, where $3<q$ represents prime numbers less than 100. This investigation was carried out with Mathematica, version 13.0 [23].

6.1. Cycles

Observation 1. 

The Collatz function ${\mathcal{T}}_5$ admits a non-trivial cycle.

This cycle was found by performing a numerical search. We list below all the elements of the trajectory ${\left(T_5^{\left(k\right)}\left(71\right)\right)}_{k=0}^{\infty }$:

Cycle for ${\mathcal{T}}_5$
i	0	1	2	3	4	5	6	7*	8	9	10*	11	12	13
$T_{\left(5\right)}(k+i)$	71	107	161	121	91	137	103	311	467	701	263	791	1187	1781
i	14	15	16	17	18*	19	20	21	22	23	24	25	26	27
$T_{\left(5\right)}(k+i)$	167	251	377	283	851	1277	479	719	1079	2429	1619	1367	2051	3077
i	28	29	30*	31	32	33*
$T_{\left(5\right)}(k+i)$	577	433	1301	61	23	71

Remark 11. 

A comment is needed. We have $3·103=309$. In ${\mathbb{N}}^{*}$, there is $s_{{\mathbb{N}}^{*}}\left(309\right)=310$, but $310\notin {\mathbb{L}}_5$; hence, $s_{{\mathbb{L}}_5}\left(309\right)=311$.

We use the notation ${(k+i)}^{*}$ to denote a case when $s_{{\mathbb{L}}_q}\left({\mathcal{T}}^{\left(i\right)}(k+i)\right)>s_{{\mathbb{N}}^{*}}\left({\mathcal{T}}^{\left(i\right)}(k+i)\right)$.

Figure 4 shows the behaviour of ${\mathcal{T}}_5\left(71\right)$ in the logarithmic scale (ListLogPlot command in Mathematica [23]).

Figure 4. Cycle for 71 in ${\mathbb{L}}_5$.

Observation 2. 

Non-trivial cycles were also found to exist in ${\mathbb{L}}_5$ for primes $7,11,13,17.$

Non-trivial cycles were also found for other ${\mathbb{L}}_q$ values.

Observation 3. 

The following tables give non-trivial cycles: ${\mathbb{L}}_7\left(103\right)$, ${\mathbb{L}}_{11}\left(67\right),$${\mathbb{L}}_{13}\left(137\right),$${\mathbb{L}}_{17}\left(71\right)$, see Figure 5, Figure 6 and Figure 7.

Cycle for ${\mathcal{T}}_7$
i	0	1	2	3*	4	5	6	7	8	9
$T_{\left(7\right)}(k+i)$	103	155	233	701	263	395	593	445	167	251
i	10	11	12	13	14	15	16	17	18*	19
$T_{\left(7\right)}(k+i)$	377	283	425	319	479	719	1079	1619	2429	7289
i	20*	21	22	23	24	25*	26	27	28	29
$T_7(k+i)$	21,869	8201	6151	9227	13,841	41,525	3893	365	137	103

Figure 5. Cycle for 103 in ${\mathbb{L}}_7$.

Figure 5. Cycle for 103 in ${\mathbb{L}}_7$.

The shortest nontrivial cycle we found is for ${\mathcal{T}}_{11}$ in ${\mathbb{L}}_{11}$.

Cycle for ${\mathcal{T}}_{11}$
i	0	1	2	3	4*	5
$T_{11}(k+i)$	67	101	19	29	89	67

Figure 6. Cycle for 67 in ${\mathbb{L}}_{11}$.

Figure 6. Cycle for 67 in ${\mathbb{L}}_{11}$.

Cycle for ${\mathcal{T}}_{13}$
i	0	1	2	3	4	5	6	7	8	9	10	11
$T_{\left(13\right)}(k+i)$	137	103	155	233	175	395	263	395	593	445	167	251
i	12*	13	14	15	16	17	18*	19	20	21	22	23
$T_{13}(k+i)$	755	1133	425	319	479	719	2159	3239	4859	7289	5467	8201
i	24	25	26	27	28	29	30
$T_{13}(k+i)$	6151	9227	13,841	10,381	3893	365	137

The longest cycle we found was for 71 in ${\mathbb{L}}_{17}$.

Cycle for ${\mathcal{T}}_{17}$
i	0	1	2	3	4	5	6	7	8	9	10	11
$T_{17}(k+i)$	71	107	161	121	91	137	103	155	233	175	263	395
i	12	13	14	15	16	17	18*	19	20	21	22	23
$T_{\left(17\right)}(k+i)$	593	445	167	251	377	283	851	1277	479	719	1079	1619
i	24	25	26	27	28*	29	30	31	32	33*	34	35
$T_{17}(k+i)$	2429	911	1367	2051	6155	9233	6925	2597	487	1463	2195	3293
i	36	37*	38	39	40	41	42	43	44	45	46
$T_{17}(k+i)$	1235	3707	5561	4171	6257	4693	55	83	125	47	71

Figure 7. Cycle for 71 in ${\mathbb{L}}_{17}$.

Figure 7. Cycle for 71 in ${\mathbb{L}}_{17}$.

Observation 4. 

We searched for cycles for ${\mathcal{T}}_{19}$, but to no avail. If ${\mathcal{T}}_{19}$ allows a cycle, then it is “far”. Also, for the successive prime numbers from 23 to 47 in the studied ranges, all trajectories were convergent. But, rather unexpectedly, we found a cycle for ${\mathcal{T}}_{53}$.

Cycle for ${\mathcal{T}}_{53}$
i	0	1*	2	3	4	5	6	7	8	9	10
$T_{53}(k+i)$	35	107	161	121	91	137	103	155	233	175	263
i	11	12	13	14	15	16	17	18	19	20	21
$T_{53}(k+i)$	395	593	445	167	251	377	433	325	61	23	35

Observation 5. 

The biggest prime $q<100$ for which we found a cycle in ${\mathbb{L}}_q$ is 83.

Cycle for ${\mathcal{T}}_{83}$
i	0	1	2	3	4	5	6	7	8	9	10
$T_{17}(k+i)$	103	155	233	175	263	395	593	445	167	251	377
i	11	12	13	14	15	16*	17	18	19	20	21
$T_{\left(83\right)}(k+i)$	283	425	319	479	719	2159	3239	4859	7289	5467	8201
i	22	23	24	25	26	27	28	29
$T_{83}(k+i)$	6151	9227	13,841	10,381	3893	365	137	103

6.2. Divergent Trajectory?

Let us go back to ${\mathbb{L}}_7$. Our numerical experiments on the generalized Collatz problem in ${\mathbb{L}}_q$ allowed us to discover non-trivial cycles of Collatz transforms ${\mathcal{T}}_q$ for a few initial primes. The number of detected cycles is not too large, and it seems that there is no reason to conclude that the cycles listed above are all cycles for ${\mathcal{T}}_q$ (given q), or even that the number of cyclic trajectories is finite. We think that more extensive numerical research will detect further cycles in these ${\mathbb{L}}_q$s where we found a cycle, and further prime numbers q for which ${\mathcal{T}}_q$ admit a cycle.

The problem of the existence of divergent trajectories is more complicated as it cannot be solved numerically. However, to our surprise, we found a phenomenon in ${\mathbb{L}}_7$, which suggests the existence of a divergent trajectory in ${\mathbb{L}}_7$. For $k\le 126$, all trajectories ${\mathcal{T}}_7\left(k\right)$ are convergent or cyclic. But our numerical results suggest that ${\mathcal{T}}_7\left(127\right)$ is divergent.

Below there are some results of the computations and in Figure 8, a plot of the beginning of the orbit.

Figure 8. Graph showing possible divergence.

(1)

For $j={10}^2$, we have ${\mathcal{T}}_7^{\left(j\right)}\left(127\right)\sim 2·{10}^{13}$.

(2)

For $j={10}^3$, we have ${\mathcal{T}}_7^{\left(j\right)}\left(127\right)\sim 2·{10}^{23}$.

(3)

For $j={10}^4$, we have ${\mathcal{T}}_7^{\left(j\right)}\left(127\right)\sim 3·{10}^{353}$.

(4)

For $j={10}^5$, we have ${\mathcal{T}}_7^{\left(j\right)}\left(127\right)\sim 3·{10}^{2538}$.

(5)

For $j={10}^6$, we have ${\mathcal{T}}_7^{\left(j\right)}\left(127\right)\sim 8·{10}^{26133}$.

(6)

For $j={10}^7$, we have ${\mathcal{T}}_7^{\left(j\right)}\left(127\right)\sim 4·{10}^{256801}$.

…

(7)

For $j=2·{10}^8$, we have ${\mathcal{T}}_7^{\left(j\right)}\left(127\right)\sim {10}^{5113715}$.

(8)

For $j=2.5·{10}^8$, we have ${\mathcal{T}}_7^{\left(j\right)}\left(127\right)\sim {10}^{6395754}$.

(9)

For $j=3·{10}^8$, we have ${\mathcal{T}}_7^{\left(j\right)}\left(127\right)\sim {10}^{7675803}$.

(10)

For $j=4.5·{10}^8$, we have ${\mathcal{T}}_7^{\left(j\right)}\left(127\right)\sim {10}^{11515252}$.

…

(11)

Computations in progress.

We finish with the following questions.

Question 1. 

(1)

Is ${\mathcal{T}}_7\left(127\right)$ really divergent?

(2)

If yes, is ${\mathbb{L}}_7$ the only ${\mathbb{L}}_q$ where a divergent trajectory exists?

(3)

If yes, why?