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\TopicFR{Combinatoire, ThÃ©orie des nombres}
\TopicEN{Combinatorics, Number theory}

\addbibresource{CRMATH_Eliahou_20240323.bib}

\usepackage{mathtools}
\usepackage[autolanguage,np]{numprint}

\DeclareMathOperator{\sat}{sat}

\newcommand{\N}{\mathbb N}
\newcommand{\LL}{\mathcal{L}}

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\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
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\DeclarePairedDelimiter{\angles}{\langle}{\rangle}
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\title{The number system in rational base~$3/2$ and the $3x+1$ problem}
\alttitle{Le systÃ¨me de numÃ©ration en base rationnelle 3/2 et le problÃ¨me $3x+1$}

\author{\firstname{Shalom} \lastname{Eliahou}\CDRorcid{0000-0002-2305-6205}}
\address{Univ. Littoral C\^ote d'Opale, UR 2597 -- LMPA -- Laboratoire de Math\'ematiques Pures et Appliqu\'ees Joseph Liouville, F-62100 Calais, France}
\address{CNRS FR 2037, France}
\email{eliahou@univ-littoral.fr}

\author{\firstname{Jean-Louis} \lastname{Verger-Gaugry}\CDRorcid{0000-0002-5981-9841}}
\address{Universit\'e-Grenoble-Alpes, Universit\'e Savoie Mont Blanc, F-73000 Chamb\'ery, France}
\email{verger-gaugry.jean-louis@unsechat-math.fr}

\begin{abstract}
The representation of numbers in rational base $p/q$ was introduced in 2008 by Akiyama, Frougny \&amp;~Sakarovitch, with a special focus on the case $p/q=3/2$. Unnoticed since then, natural questions related to representations in that specific base turn out to intimately involve the Collatz $3x+1$ function. Our purpose in this note is to expose these links and motivate further research into them.
\end{abstract}

\begin{altabstract}
La reprÃ©sentation des nombres dans la base rationnelle $p/q$ a Ã©tÃ© introduite en 2008 par Akiyama, Frougny \&amp;~Sakarovitch, avec un accent particulier sur le cas $p/q=3/2$. Des questions naturelles liÃ©es aux reprÃ©sentations dans cette base spÃ©cifique, passÃ©es inaperÃ§ues jusqu'ici, s'avÃ¨rent impliquer intimement la fonction de Collatz $3x+1$. Le but de cette note est d'exposer ces liens et de motiver des recherches plus approfondies sur ceux-ci.
\end{altabstract}

\keywords{\kwd{Collatz conjecture} \kwd{iteration} \kwd{numeration} \kwd{odometer}}
\altkeywords{\kwd{Conjecture de Collatz} \kwd{itÃ©ration} \kwd{numÃ©ration} \kwd{odomÃ¨tre}}

\subjclass{11B83, 68R15}

\begin{document}

\maketitle

\section{Introduction}

The representation of numbers in rational base $p/q$ with $p&gt;q$ coprime positive integers was introduced in 2008 in~\cite{AFS}, with a focus on the case $p/q=3/2$ to study Mahler's $Z$-number problem and the Josephus problem.

Given $n \in \N$, its representation in rational base~$3/2$ is the unique expression of the form
\[
n = \sum_{i=0}^k \frac{a_i}2\parens[\Bigg]{\frac32}^i
\]
with digits $a_i \in \{0,1,2\}$ for all $i \ge 0$ and $a_k \neq 0$ if $n \ge 1$. We then denote the corresponding word in $\{0,1,2\}^*$ by
\[
\angles{n}=a_ka_{k-1}\cdots a_0.
\]
We set $\angles{0}=\epsilon$, the empty word, and we identify any word $0w \in \{0,1,2\}^*$ with $w$.

Denoting $\LL =\{\angles{n} \mid n \in \N\} \subset \{0,1,2\}^*$, here are two natural questions about $\LL$:
\begin{enumerate}
\item \label{question_1} If $w \in \LL$, which extensions among $w0,w1,w2$ still belong to $\LL$?
\item \label{question_2} Given $k \ge 1$, what is the lexicographically largest word of length $k$ in $\LL$? Equivalently, what is the largest $n \in \N$ such that $\angles{n}$ is of length $k$?
\end{enumerate}
Apparently unnoticed so far, answers to these questions turn out to intimately involve the Collatz $3x+1$ function, namely $T \colon \N \to \N$ defined by
\begin{equation}\label{eq T}
T(n) =
\begin{cases}
n/2 			&amp; \text{if $n \equiv 0 \bmod 2$}, \\
(3n+1)/2 	&amp; \text{if $n \equiv 1 \bmod 2$}.
\end{cases}
\end{equation}
The purpose of this note is to expose these links and motivate further research into them. Its contents are as follows. In Section~\ref{sec basic}, we recall basic properties of the numeration system in rational base~$3/2$, in particular the recursive construction of the word $\angles{n}$ in Section~\ref{sec word} and the odometer allowing to construct $\angles{n+1}$ from $\angles{n}$ in Section~\ref{sec odometer}. We also propose a variant of the Collatz conjecture in Section~\ref{sec collatz}. In Sections~\ref{sec extend} and~\ref{sec saturated}, we address the above questions~\eqref{question_1} and~\eqref{question_2}, respectively. Section~\ref{sec questions} concludes the note with two questions and two conjectures closely related to the Collatz conjecture and possibly of the same order of difficulty.

\section{Basics}\label{sec basic}

Following~\cite{AFS}, every $n\in \N$ has a unique expression of the form
\[
n=\frac12\parens[\big]{a_k(3/2)^k+a_{k-1}(3/2)^{k-1}+\dots+a_0}
\]
with $a_i \in \{0,1,2\}$ for all $i$ and $a_k \neq 0$ if $n \ge 1$. Accordingly, we associate to $n$ the word
\[
\langle n \rangle=a_ka_{k-1}\cdots a_0 \in \{0,1,2\}^*,
\]
called the \emph{representation of $n$ in rational base\/~$3/2$}. We set $\angles{0}=\epsilon$, the empty word, and we identify any word $0w \in \{0,1,2\}^*$ with $w$.

By~\cite[Lemma~4.1]{MWT}, the length of the word $\angles{n}$ is given by $\abs{\angles{n}}=\log_{3/2}(n) + O(1)$.

\begin{definition}
We say that a word $w \in \{0,1,2\}^*$ is \emph{admissible} if $w=\angles{n}$ for some $n \in \N$. We denote by $\LL \subset \{0,1,2\}^*$ the language of admissible words, i.e.
\[
\LL=\{\angles{n} \mid n \in \N\}.
\]
\end{definition}

As observed in~\cite{AFS}, \emph{the language $\LL$ is prefix-closed}, i.e.~stable under taking prefixes. That is, every prefix $w'$ of an admissible word $w=w'w''$ is itself admissible.

Conversely, given an admissible word $w \in \LL$, which extensions among $w0,w1,w2$ remain admissible? The answer involves the Collatz $3x+1$ function. See Proposition~\ref{prop link 1}.

\subsection{Construction of \texorpdfstring{$\angles{n}$}{the word $n$}}\label{sec word}

Given $n \in \N$, a recursive construction of $\angles{n} \in \{0,1,2\}^*$ may be described as follows. Set $\angles{0}=\epsilon$, the empty word.

\begin{proposition}\label{prop n1}
Let $n_0 \in \N \setminus \{0\}$. Denote $\angles{n_0}=a_k\cdots a_0$ with $a_i \in \{0,1,2\}$ for all $i$. Then:
\[
\begin{cases}
a_0 \equiv 2n_0 \bmod 3; \\
\angles{n_0} = \angles{n_1}a_0, \ \text{where $n_1= (2n_0-a_0)/3$}.
\end{cases}
\]
Moreover, we have $a_0=2n_0-3n_1$ and $a_0 \equiv n_1 \bmod 2$.
\end{proposition}

\begin{proof}
As $\angles{n_0}=a_k\cdots a_0$, we have
\[
\begin{split}
n_0	&amp; = \frac12\parens[\big]{a_k(3/2)^k+\dots+a_1(3/2)+a_0} \\
	&amp; = \parens[\Bigg]{\frac12\parens[\big]{a_k(3/2)^{k-1}+\dots+a_1}}\frac32+\frac{a_0}2. \\
\end{split}
\]
Hence $a_0\equiv 2n_0 \bmod 3$, and setting $n_1=(2n_0-a_0)/3$, we have $\angles{n_1}=a_k\cdots a_1$. The claims follow.
\end{proof}

\subsection{The odometer}\label{sec odometer}

The \emph{odometer} constructs the word $\angles{n+1}$ from the word $\angles{n}$. It is described as a finite automaton in~\cite{AFS}. Here is an equivalent description.

Let $\sigma=(2 \ 1 \ 0)$ be the cyclic permutation $2 \mapsto 1 \mapsto 0 \mapsto 2$. For a word $u \in \{0,1,2\}^*$, let $u^\sigma \in \{0,1,2\}^*$ denote the word obtained by applying $\sigma$ to each letter of $u$. For instance, $12012^\sigma=01201$, then identified to $1201$.

\begin{proposition}\label{prop odometer}
Let $n \in \N$. If\/ $\angles{n}=u0v \in \{0,1,2\}^*$ with $v \in \{1,2\}^*$, then $\angles{n+1}=u2v^\sigma$.
\end{proposition}

\begin{proof}
See~\cite[Section~3.2.3, pp.~66--67]{AFS}.
\end{proof}

The rule applies even if $u$ or $v$ is the empty word. In particular, if $\angles{n} = v \in \{1,2\}^*$, then identifying $v$ with $0v$, we get $\angles{n+1}=2v^\sigma$. Table~\ref{tab words}, listing $\angles{n}$ for $1 \le n \le 27$, was constructed by hand using this description of the odometer. The interested reader will effortlessly extend it much farther.

\begin{table}[h]
\caption{The words $\angles{n}$ for $1 \le n \le 27$.}\label{tab words}
$\begin{array}{ccccc}
\begin{array}{|r||r|}
\hline n &amp; \angles{n}\\
\hline 1 &amp; 2 \\
2 &amp; 21 \\
3 &amp; 210 \\
4 &amp; 212 \\
5 &amp; 2101 \\
6 &amp; 2120 \\
7 &amp; 2122 \\
8 &amp; 21011 \\
9 &amp; 21200 \\
\hline
\end{array} &amp; &amp;
\begin{array}{|r||r|}
\hline n &amp; \angles{n} \\
\hline 10 &amp; 21202 \\
11 &amp; 21221 \\
12 &amp; 210110 \\
13 &amp; 210112 \\
14 &amp; 212001 \\
15 &amp; 212020 \\
16 &amp; 212022 \\
17 &amp; 212211 \\
18 &amp; 2101100 \\
\hline
\end{array} &amp; &amp;
\begin{array}{|r||r|}
\hline n &amp; \angles{n}\\
\hline 19 &amp; 2101102 \\
20 &amp; 2101121 \\
21 &amp; 2120010 \\
22 &amp; 2120012 \\
23 &amp; 2120201 \\
24 &amp; 2120220 \\
25 &amp; 2120222 \\
26 &amp; 2122111 \\
27 &amp; 21011000 \\
\hline
\end{array}
\end{array}$
\end{table}

\subsection{The Collatz conjecture}\label{sec collatz}

Dating back to the 1930's, this conjecture deals with iterates $T^{(k)}$ of the Collatz function $T \colon \N \to \N$ defined in~\eqref{eq T}.

\begin{conjecture}\label{conj col1}
Let $n_0 \in \N$, $n_0 \ge 1$. Then there exists $k \ge 1$ such that $T^{(k)}(n_0)=1$.
\end{conjecture}

Here is an equivalent version, as shown below.

\begin{conjecture}\label{conj col2}
Let $n_0 \in \N$, $n_0 \ge 1$. Then the proportion of even numbers in the sequence of iterates $T^{(k)}(n_0)$ tends to $1/2$ as $k$ tends to infinity.
\end{conjecture}

\begin{proposition}\label{prop_6}
Conjectures~\ref{conj col1} and~\ref{conj col2} are equivalent.
\end{proposition}

\begin{proof}
If Conjecture~\ref{conj col1} holds, then at some point the sequence of iterates $T^{(k)}(n_0)$ collapses to $1,2,1,2,1,2,\dots$, whence Conjecture~\ref{conj col2}.

Conversely, assume Conjecture~\ref{conj col2} holds. Given an integer $n \ge 1$, let $\Omega(n)=\parens[\big]{n,T(n),T^{(2)}(n),\dots}$ denote the trajectory of $n$ under iteration of $T$. Moreover, given $k \ge 1$, let
\[
\Omega_k(n)=\parens[\big]{n,T(n),T^{(2)}(n),\dots,T^{(k-1)}(n)},
\]
and let $k_1$ denote the number of odd integers in $\Omega_k(n)$. Assume for a contradiction that Conjecture~\ref{conj col1} fails. Then there is an integer $m \ge 10$ such that $\min \Omega(m)=m$. (Note that $\min \Omega(n)=1$ for $n \le 9$.) We claim that for all $k \ge 1$, we have
\begin{equation}\label{eq upper}
T^{(k)}(m) \le \frac{(3.1)^{k_1}}{2^k} m.
\end{equation}
The proof is straightforward, by induction on $k$. Indeed, \eqref{eq upper} holds for $k=1$, since $m$ is odd by minimality in $\Omega(m)$ and $T(m)=(3m+1)/2=(3+1/m)m/2 \le (3.1)m/2$ since $m \ge 10$. Let now $k \ge 1$ and assume that $T^{(k)}(m) \le (3.1)^{k_1}/2^k m$. Denote $n=T^{(k)}(m)$. If $n$ is even, then $T^{(k+1)}(m)= T(n) =n/2 \le (3.1)^{k_1}/2^{k+1} m$, as desired. If $n$ is odd, then
\[
T^{(k+1)}(m) = T(n) = (3+1/n)n/2 \le (3+1/m)n/2 \le (3.1)n/2
\]
since $n \ge m \ge 10$. Hence $T^{(k+1)}(m) \le (3.1)^{k_1+1}/2^{k+1}m$, as desired. The claim is proved.

Now $(3.1)^{k_1}/2^{k} \ge 1$ if and only if $k_1/k \ge \ln(2)/\ln(3.1)$, and $\ln(2)/\ln(3.1) &gt; 0.6$. By Conjecture~\ref{conj col2}, for $k$ large enough, the proportion $k_1/k$ will fall below $0.6$. Hence $T^{(k)}(m) \le (3.1)^{k_1}/2^k m &lt; m$, contradicting the minimality of $m$. This completes the proof of the proposition.
\end{proof}

At the time of writing, Conjecture~\ref{conj col1} has been verified up to $n_0 \le 2^{68}$. See~\cite{B}. With this bound, the results of~\cite{E} and its Table~2 on page 54 imply that, besides the trivial cycle $\{1,2\}$ under $T$, no other cycle, if any, can have length less than $\np{114208327604}$. See~\cite{L2} for a general reference on the conjecture, and~\cite{T} for recent strong results about it.

\section{Links with the Collatz function}\label{sec links}

We now address the two questions about $\LL=\{\angles{n} \mid n \in \N\}$ asked in the Introduction, and show that in both cases, the answer naturally involves the Collatz function $T$.

\subsection{Extending admissible words}\label{sec extend}

Here is the first question. Given $w \in \LL$, which extensions among $w0,w1,w2$ still belong to $\LL$?

\begin{proposition}\label{prop link 1}
Let $w \in \LL$, say $w = \angles{n}$ for some $n \in \N$. Then:
\[
\begin{cases}
w1 \in \LL \iff n \equiv 1 \bmod 2, \ \text{and then} \ w1=\angles{(3n+1)/2} =\angles{T(n)}; \\
w0 \in \LL \iff n \equiv 0 \bmod 2 \iff w2 \in \LL, \ \text{and then} \ w0=\angles{3n/2} \ \text{and} \ w2=\angles{(3n+2)/2}.
\end{cases}
\]
\end{proposition}

\begin{proof}
If $\angles{n}1\in \LL$, Proposition~\ref{prop n1} implies $1 \equiv n \bmod 2$ and $w1=\angles{(3n+1)/2}$. Similarly, if $\angles{n}0 \in \LL$ then $0 \equiv n \bmod 2$ and $w0=\angles{3n/2}$. Finally, if $\angles{n}2 \in \LL$ then $2 \equiv n \bmod 2$ and $w2=\angles{(3n+2)/2}$. The stated equivalences follow.
\end{proof}

\begin{corollary}\label{cor n odd}
Let $n \in \N$ be odd. Then $\langle T(n) \rangle=\langle n \rangle1$.
\end{corollary}

\begin{proof}
Immediate from the above proposition.
\end{proof}

\begin{remark}
The map $n \mapsto n/2$ for $n$ even is straightforward in the base~$2$ representation of $n$: just remove the last digit, $0$. Symmetrically, Corollary~\ref{cor n odd} shows that the map $n \mapsto (3n+1)/2$ for $n$ odd is straightforward in the base~$3/2$ representation of $n$: just add $1$ as rightmost digit. By any chance, could it happen that considering the map $n \mapsto T(n)$ simultaneously in base~$2$ and in rational base~$3/2$ may lead to new insights on the Collatz conjecture?
\end{remark}

\subsection{Saturated words}\label{sec saturated}

Given a number system $\mathcal{B}$ over a finite set of digits, we say that a positive integer $n \ge 1$ is \emph{saturated} if its representation $\angles{n}_\mathcal{B}$ in base $\mathcal{B}$ is of length strictly less than that of $\angles{m}_\mathcal{B}$ for all $m \ge n+1$. Accordingly, we say that $\angles{n}_\mathcal{B}$ is a \emph{saturated word}. For instance in base~$10$, the saturated integers and corresponding words are the $n=10^\ell-1=9\cdots9$.

The digits $1,2$ of the saturated words in rational base~$3/2$ turn out to be unpredictable in some sense and closely related to the $3x+1$ problem. Recall that $\abs{w}$ denotes the length of the finite word $w$.

\begin{notation}
Given $k \ge 1$, we denote by $\sat(k)$ the largest positive integer whose representation in base~$3/2$ is of length $k$. That is,
\[
\sat(k) = \max \braces[\big]{n \in \N, \ \abs{\angles{n}} = k}.
\]
\end{notation}

\begin{proposition}\label{prop sat}
For all $k \ge 1$, $\angles{\sat(k)}$ is the unique admissible word of length $k$ with digits in $\{1,2\}$.
\end{proposition}

\begin{proof}
By the odometer as described in Proposition~\ref{prop odometer}, if $\angles{n}$ is of length $k$ and contains a $0$ digit, then $\angles{n+1}$ is of the same length $k$. This shows that the inequality $\abs{\angles{n+1}} &gt; \abs{\angles{n}}$ happens exactly when $\angles{n}$ contains no $0$ digit except possibly as an irrelevant prefix.
\end{proof}

For instance, a look at $\angles{n}$ for $1 \le n \le 27$ in Table~\ref{tab words} yields $\sat(k)$ and the corresponding word for $k \le 7$ in Table~\ref{tab sat}. Observe in Table~\ref{tab sat} that $\angles{\sat(k-1)}$ is a prefix of $\angles{\sat(k)}$. Indeed, this is true for all $k \ge 1$.

\begin{proposition}\label{prop sat k-1}
For all $k \ge 2$, we have $\angles{\sat(k)}=\angles{\sat(k-1)}a_0$ where $a_0 \in \{1,2\}$ and $a_0 \equiv \sat(k-1) \bmod 2$.
\end{proposition}

\begin{proof}
Since $\LL$ is prefix-closed, it follows from Proposition~\ref{prop sat} that the length $k-1$ prefix of $\angles{\sat(k)}$ equals $\angles{\sat(k-1)}$. Thus $\angles{\sat(k)}=\angles{\sat(k-1)}a_0$ where $a_0 \in \{1,2\}$, and $a_0 \equiv \sat(k-1) \bmod 2$ by Proposition~\ref{prop n1}.
\end{proof}

\begin{table}[h]
\caption{The first seven saturated numbers and words.}\label{tab sat}
$\begin{array}{|r||c|l|}
\hline k &amp; \sat(k) &amp; \angles{\sat(k)} \\
\hline 1 &amp; 1 &amp; 2 \\
2 &amp; 2 &amp; 21 \\
3 &amp; 4 &amp; 212 \\
4 &amp; 7 &amp; 2122 \\
5 &amp; 11 &amp; 21221 \\
6 &amp; 17 &amp; 212211 \\
7 &amp; 26 &amp; 2122111 \\
\hline
\end{array}$
\end{table}

\subsection{The function \texorpdfstring{$U$}{U}}\label{sec U}

It turns out that the sequence $\parens[\big]{\sat(k)}_{k \ge 1}$ coincides with the sequence of iterates $U^{(k)}(0)$ of a certain function $U$ closely related to the Collatz function $T$. See Proposition~\ref{prop sat rec} below.

\begin{notation}
Let $U \colon \N \to \N$ be defined as
\[
U(n) =
\begin{cases}
(3n+2)/2=T(n)+n+1 		&amp; \text{if $n \equiv 0 \bmod 2$},\\
(3n+1)/2=T(n) 			&amp; \text{if $n \equiv 1 \bmod 2$}.
\end{cases}
\]
\end{notation}

\begin{remark}\label{rem U and T}
We have $U(n) \equiv T(n)+n+1 \bmod 2$ for all $n \in \N$.
\end{remark}

Starting from any $n_0 \in \N$, let us consider the sequence of iterates $U^{(k)}(n_0)$ for $k \ge 1$. The following proposition shows that, for the sequence of corresponding words $\angles{U^{(k)}(n_0)}$ in rational base~$3/2$, \emph{each term is a prefix of the next one and the added digit is either $1$ or\/ $2$.}

\begin{proposition}\label{prop u 2}
Let $n \in \N$. Then:
\[
\angles{U(n)}=
\begin{cases}
\angles{n}1 		&amp; \text{if $n \equiv 1 \bmod 2$}, \\
\angles{n}2 		&amp; \text{if $n \equiv 0 \bmod 2$}.
\end{cases}
\]
\end{proposition}

\begin{proof}
Straightforward from Proposition~\ref{prop link 1}.
\end{proof}

A particularly interesting sequence of iterates $U^{(k)}(n_0)$ is for $n_0=0$.

\begin{proposition}\label{prop sat rec}
For all $k \ge 1$, we have $\sat(k)=U^{(k)}(0)$.
\end{proposition}

\begin{proof}
For $k=1$, we have $\sat(1)=2$ and $U(1)=T(1)=2$. For $k \ge 2$, we only need show $\sat(k)=U\parens[\big]{\sat(k-1)}$. Indeed, it follows from Proposition~\ref{prop sat k-1} that
\[
\sat(k)= \frac{3\sat(k-1)+a_0}2
\]
where $a_0 \in \{1,2\}$ and $a_0 \equiv \sat(k-1) \bmod 2$. Hence $\sat(k)=U\parens[\big]{\sat(k-1)}$ by definition of $U$, as desired.
\end{proof}

Consequently, the growth factor of the sequence $\sat(k)$ tends to $3/2$.

\section{Open questions}\label{sec questions}

We conclude this note with two questions and two conjectures.

\begin{question}\label{ques 1}
Is the Collatz conjecture equivalent to some behavior of the iteration of the function~$U$? For instance, starting with any $n_0 \in \N$, is it true that the proportion of $1$'s in $\angles{U^{(k)}(n_0)}$ tends to $1/2$ as $k$ goes to infinity, as supported by numerical evidence? Note that by Proposition~\ref{prop u 2}, the number of digits ``0'' remains constant along the trajectory of $n_0$.
\end{question}

More specifically, for $n_0=0$, we have $U^{(k)}(0)=\sat(k)$ for all $k \ge 1$ by Proposition~\ref{prop sat rec}. Having computed $\sat(k)$ up to $k = 2^{20}= \np{1048576}$, we find that the word $w=\angles{\sat(2^{20})}$, of length $k=2^{20}$, counts exactly $\np{525067}$ digits ``1'' and $\np{523509}$ digits ``2''. Moreover, for all $1 \le \ell &lt; 17$, the word $w$ contains all $2^\ell$ words in $\{1,2\}^\ell$ as subwords. See Table~\ref{tab 200} for the word $\angles{\sat(200)}$, with exactly 100 digits ``1''. This numerical evidence leads us to the following conjecture.

\begin{conjecture}\label{conj our}
The proportion of digits ``1'' in the word $\angles{\sat(k)}$ tends to $1/2$ as $k$ tends to infinity.
\end{conjecture}

Is Conjecture~\ref{conj our} equivalent to the Collatz conjecture? The question arises from the close relationship between the functions $T$ and $U$, Remark~\ref{rem U and T} and the analogy with Conjecture~\ref{conj col2}. At any rate, even if not equivalent to Conjecture~\ref{conj col2}, Conjecture~\ref{conj our} might well be of the same order of difficulty.

In fact, the numerical evidence below Question~\ref{ques 1} leads to an even stronger conjecture. Since $\angles{\sat(k-1)}$ is a prefix of $\angles{\sat(k)}$ for all $k \ge 2$, we may define the limit
\[
\angles{\sat(\infty)} = \lim_{k \to \infty} \angles{\sat(k)},
\]
an infinite word over $\{1,2\}$. Informally, recall that an infinite word $w$ over a finite alphabet $\mathcal{B}$ of cardinality $\abs{\mathcal{B}}=b$ is said to be \emph{normal} if for all $\ell \ge 1$, all $b^\ell$ words in $\mathcal{B}^\ell$ occur as subwords in $w[n]$, the length $n$ prefix of $w$, with frequency tending to $1/b^\ell$ as $n \to \infty$. Our stronger conjecture runs as follows.

\begin{conjecture}\label{conj our 2}
The word $\angles{\sat(\infty)}$ is normal.
\end{conjecture}

\begin{table}[h]
\caption{The word $\angles{\sat(200)}=\angles{U^{200}(0)}$.}\label{tab 200}
\begin{tabular}{ll}
&amp; 2122111221211221212112212112121111122121111111212222211221212212121\\
\noindent &amp; 1211222212212222221211111221221212211211222211122111222122221211212\\
\noindent &amp; 122122212122212121211111111212122222111112211121212222122111112112
\end{tabular}
\end{table}

Our second question is about fully understanding the Collatz map $T$ in rational base~$3/2$.

\begin{question}\label{ques n even}
For $n$ odd, we have seen that $\angles{T(n)}=\angles{n}1$. For $n$ even, how does one determine $\angles{T(n)}$ from $\angles{n}$? Is any answer susceptible to lead to new insights on the Collatz conjecture?
\end{question}

Note that determining the parity of $n \in \N$ from the digits $a_i$ of the word $\angles{n}=a_k\cdots a_0 \in \LL$ is more cumbersome than in base~$2$ or $10$. Indeed, the following criterion is straightforward but not quite practical for calculations by hand.

\begin{proposition}
Let $n \in \N \setminus \{0\}$ and $\angles{n}=a_k\cdots a_0 \in \LL$. Then
\[
n \equiv 0 \bmod 2 \ \iff \ \sum_{i=0}^k 2^{k-i}3^ia_i \equiv 0 \bmod 2^{k+2}.
\]
\end{proposition}

\begin{proof}
We have $n=\frac12\parens[\big]{\sum_{i=0}^k (3/2)^ia_i}=2^{-(k+1)}\parens[\big]{\sum_{i=0}^k 2^{k-i}3^ia_i}$. Hence $n \equiv 0 \bmod 2 \iff 2^{k+1}n \equiv 0 \bmod 2^{k+2} \iff \sum_{i=0}^k 2^{k-i}3^ia_i \equiv 0 \bmod 2^{k+2}$.
\end{proof}

\section*{Acknowledgments}

We thank the anonymous referee for their careful reading and constructive comments, in particular for insisting to formally state Proposition~\ref{prop_6} and carefully prove it, as opposed to just a remark with informal arguments as in the original version of this article.

\section*{Declaration of interests}

The authors do not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and have declared no affiliations other than their research organizations.

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