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\begin{center}
\vskip 1cm{\Large\bf On the Variation of the Sum of Digits in the\\ 
\vskip .02in
Zeckendorf Representation: An Algorithm to Compute\\
\vskip .1in
the Distribution and Mixing Properties}
\vskip 1cm
Yohan Hosten\\
Laboratoire Ami\'enois de Math\'ematique Fondamentale et Appliqu\'ee\\
Universit\'e de Picardie Jules Verne\\
33, rue Saint Leu\\ 
80000 Amiens\\
France\\
\href{mailto:yohan.hosten@math.cnrs.fr}{\tt yohan.hosten@math.cnrs.fr} \\
\end{center}

\vskip .2 in

\begin{abstract} We study probability measures defined by the variation of the sum of digits in the Zeckendorf representation. For $r\ge 0$ and $d\in\ZZ$, 
we consider $\mu^{(r)}(d)$, the density of integers $n\in\NN$ for which the sum of digits increases by $d$ when $r$ is added to $n$. We give a probabilistic interpretation of $\mu^{(r)}$ via the dynamical system provided by the odometer of Zeckendorf-adic integers and its unique invariant measure. We give an algorithm for computing $\mu^{(r)}$ and we prove the exponential decay of $\mu^{(r)}(d)$ as $d\to -\infty$, as well as the formula $\mu^{(F_{\ell})}=\mu^{(1)}$ where $F_{\ell}$ is a term of the Fibonacci sequence. Finally, we decompose the Zeckendorf representation of an integer $r$ into so-called ``blocks'' and show that when added to an adic Zeckendorf integer, the successive actions of these blocks can be seen as a sequence of mixing random variables.
\end{abstract}

\section{Introduction}
\subsection{Framework, notations and main results}
Throughout this article, we let $\NN:=\{0,1,2,\ldots\}$ denote the set of \emph{integers} and $\varphi:=\frac{1+\sqrt{5}}{2}$ represent the \emph{golden ratio}. We also define the well-known \emph{Fibonacci sequence} as follows: 

\begin{equation}\label{Fib_def}
    F_k:=\begin{cases}
        1, & \text{if $k=1$ or $k=2$;}\\
        F_{k-1}+F_{k-2}, & \text{if $k\ge 3$.}\\
    \end{cases}
\end{equation}

By Zeckendorf's theorem (proved by Lekkerkerker \cite[Theorem 1]{CGL} and Zeckendorf \cite[Theorem I.a, p.\ 179]{EZ}), every integer can be uniquely written as a sum of non-consecutive Fibonacci terms. That is, for every $n\in\NN$, there exists a unique sequence of \emph{digits} $(n_k)_{k\ge 2}\in\{0,1\}^{\infty}$ without two consecutive $1$'s, finitely many of them being equal to $1$ and such that
\begin{align}\label{Zeck_decomposition}
        n&=\sum_{k\ge 2}n_kF_k.
\end{align}

For $n\neq 0$ and $\ell:=\max\{k:~n_k\neq 0\}$, we introduce the notation $\left[n_{\ell}\cdots n_2\right]:=n$ that generalizes the usual way we write numbers in an integer base, and that we refer to as the \emph{(Zeckendorf) expansion} of $n$. This way to expand numbers is actually a particular case of an \emph{Ostrowski  numeration system} \cite{AO,VB,GBPL}. By convention, we set $\left[0\right]:=0$. Then we define the \emph{(Zeckendorf-)sum-of-digits} function as
\[s(n):=\sum_{k\ge 2} n_k.\]
A central object in our paper is the \emph{variation of the sum of digits} when we add a fixed integer $r$ to  $n$: for $r,n\in\NN$, we set
\begin{equation}\label{def delta zeck}
    \Delta^{(r)}(n):=s(n+r)-s(n).
\end{equation}
The analogous variation in an integer base has been studied extensively. The 
first appearance of the case of an integer base is in a paper from B\'esineau \cite{JB} in 1970. Using a statistical vocabulary, he showed the existence of the asymptotic density for the set of integers such that the variation is some integer $d\in\ZZ$. Given an integer $r$, these densities can be seen as a probability law. The variance of this law was studied, in the binary base, by Emme and Prikhod'ko \cite{JEAP} and Spiegelhofer and Wallner \cite{LSMW}. Emme and Hubert \cite{JEPH1} proved a central limit theorem in the binary case that was improved by Spiegelhofer and Wallner \cite{LSMW} (still in binary) and by the author in collaboration with Janvresse and de~la~Rue \cite{TREJYH} (in an arbitrary integer base). We also refer to \cite{JFMLS,LS,MDMKLS,LS2} for connected results. Some theorems were proved for the Zeckendorf expansion of an integer: for instance, Griffiths \cite{MG} about the digit proportions, and Labb\'e and Lep\v{s}ov\`a \cite{SBJL} about addition in this numeration system. Drmota, M\"ullner and Spiegelhofer \cite{MDCMLS} obtained results about the existence of prime numbers with a fixed Zeckendorf sum-of-digits. However, not much has been done about this variation in the Zeckendorf representation except the work from Dekking \cite{FMD} that characterizes the integers such that $\Delta^{(1)}$ is $>0$ (or $<0$ or $=0$) and from Spiegelhofer \cite[Lemma 1.30]{LS}, which adapted B\'esineau's techniques and proved that, for all $d\in\ZZ$, the following asymptotic density exists
\[\mu^{(r)}(d):=\lim\limits_{N\rightarrow +\infty}\frac{1}{N}\left\vert\left\{ n<N:~\Delta^{(r)}(n)=d\right\}\right\vert.\]
In the present paper, we adapt the ergodic theory point of view
introduced for the case of an integer base \cite{TREJYH} to the Zeckendorf
expansion, recovering Spiegelhofer's result and getting new results about
$\mu^{(r)}$ and $\Delta^{(r)}$. In particular, we provide an algorithm
that computes $\mu^{(r)}(d)$ and we represent $\Delta^{(r)}$ as the sum
of a stochastic mixing process. A perspective we have with this result
is to find a central limit theorem for $\Delta^{(r)}$.

To get our results, following the path initiated for the case of
an integer base \cite{TREJYH}, we study the variations of the sum-of-digits function in an appropriate probability space given by the compact set $\XX$ of \emph{(Zeckendorf-)adic numbers}. We consider the action of the odometer on $\XX$ (see Subsection \ref{ZIHTAI}), and endow $\XX$ with its unique invariant probability measure $\PP$.

We extend $\Delta^{(r)}$ almost everywhere on $\XX$ and show in Section~\ref{SOzeck} (Proposition~\ref{prop_moment Zeck}) that, for every $d\in\ZZ$
\[\mu^{(r)}(d)=\PP\left(\left\{x\in\XX~:~\Delta^{(r)}(x)=d\right\}\right). \]
Using the Rokhlin towers of the dynamical system (see Subsection \ref{RTEM}), we provide an algorithm to compute $\mu^{(r)}(d)$. This algorithm and its consequences can be adapted to an integer base. One implication is the following corollary on the behavior of the (negative) tail of the distribution:
\begin{corollary}\label{corollaire pour algo2}
    For $d$ small enough in $\ZZ$, we have the formula
    \[\mu^{(r)}(d-1)=\mu^{(r)}(d)\cdot \frac{1}{\varphi^{2}}.\]
\end{corollary}
\begin{remark}
One can show that the analogous result in base $b\ge 2$ is the same replacing the formula by $\mu^{(r)}(d-(b-1))=\mu^{(r)}(d)\cdot \frac{1}{b}$.
\end{remark}
Another implication is the next theorem about the measure $\mu^{(F_{\ell})}$. 
\begin{theorem}\label{theo muFk=mu1}
For $\ell\ge 3$
\begin{align}\label{muFk=mu1}
    \mu^{(F_{\ell})}=\mu^{(1)}.
\end{align}
\end{theorem}
The analogous result in an integer base $b$ replaces $F_{\ell}$ by $b^{\ell}$. Actually, it is a trivial result in base $b$. However, in the Zeckendorf decomposition, this result is much less obvious, due to the particular behavior of carry propagations that we describe in Subsection~\ref{HTAI}. The main difference with additions in base $b$ is, here, that carries propagate in {\it both\/}
directions.

Now to state our mixing result, we need to define the notion of \emph{blocks} in the expansion of an integer $r$ and to define a probabilistic notion of ($\alpha$-)\emph{mixing coefficients} (see the survey from Bradley \cite{RCB1} for others).
\begin{definition}\label{blocks Zeck}
       A \emph{block} in the expansion of an integer $r\in\NN$ is defined as a maximal sequence of the pattern $\left[10\right]$. (If $r_2=1$, we agree that a maximal sequence $\left[r_{2\ell}\cdots r_2\right]$ is a block if $r_{2k}=1$ for $k=1,\ldots, \ell$.) We define $\rho(r)$ as the number of blocks in the expansion of $r$.
\end{definition}
\begin{figure}[H]
    \centering
    \includegraphics[scale=0.35]{defblock.pdf}
    \caption{Two examples of decomposition into blocks.}
\end{figure}

\begin{definition}\label{coeff alpha melange Zeck}
      Let $(X_i)_{i\ge 1}$ be a (finite or infinite) sequence of random variables. The associated \emph{$\alpha$-mixing coefficients} $\alpha(k)$, $k\ge 1$, are defined by
   \[\alpha(k):=\sup_{p\ge 1} \hspace{1mm} \sup_{A,B} \hspace{2mm} \lvert \PP(A\cap B)-\PP(A)\PP(B) \rvert,\]
   where the second supremum is taken over all events $A$ and $B$ such that 
\begin{itemize}
    \item $A\in\sigma(X_i:1\le  i\le p)$ and
    \item $B\in\sigma(X_i:  i\ge k+p)$.
\end{itemize}
By convention, if $X_i$ is not defined when $i\ge k+p$ then the $\sigma$-algebra is trivial.
\end{definition}
For an integer $r$, we enumerate the blocks of $r$ from the units position and we define $X_i^{(r)}$ as the action of the $i^{\rm th}$ block once the previous blocks have already been taken into consideration (see Subsection~\ref{the process} for more details). These actions are constructed in order to have the equality
\begin{align}\label{decomposition delta zeck}
    \Delta^{(r)}=\sum_{i=1}^{\rho(r)}X_i^{(r)}.
\end{align}
We state the following theorem that gives an upper bound on the $\alpha$-mixing coefficients that is independent of $r$. 
\begin{theorem}\label{mixing action bloc zeck}
 The $\alpha$-mixing coefficients of $(X_i^{(r)})_{i=1,\ldots,\rho(r)}$ satisfy
\[\forall k\ge 1, \hspace{5mm} \alpha(k)\le 12\left(1-\frac{1}{\varphi^8}\right)^{\frac{k}{6}}+\frac{1}{\varphi^{2k}}.\]
\end{theorem}
The author et al.\ \cite{TREJYH} proved a central limit theorem for $\mu^{(r)}$ in the case of an integer base using a similar result together with some estimate of the variance depending on the number of blocks. Such an estimate is, here, more difficult than in base $b$ since we do not have inductive relations on $\mu^{(r)}$.

\subsection{Roadmap}

Section~\ref{HTA} is devoted to understanding the effect of adding ``$1$'' and ``$F_k$'' on the digits of Zeckendorf-adic integers. We highlight stopping patterns in the carry propagation, which play an important role in the analysis.

In Section~\ref{UEO}, we focus on the unique ergodic measure $\PP$ of the odometer. We show that $\PP$ satisfies some renewal properties (Proposition~\ref{renouvellement}) and we estimate the $\phi$-mixing coefficients for the coordinates of a Zeckendorf-adic integer (Proposition~\ref{Melange_coord}).

Then in Section~\ref{SOzeck}, we place the study of the measures $\mu^{(r)}$ in the context of the odometer on $\XX$. We extend $\Delta^{(r)}$ almost everywhere on $\XX$ and we show that the convergence
\[\lim\limits_{N\to\infty}\frac{1}{N}\sum_{n<N}f(\Delta^{(r)}(n))=\int_{\XX} f(\Delta^{(r)}(x)) \mathrm{d}\PP(x)\]
is satisfied for functions $f:\ZZ\rightarrow \CC$ of polynomial growth (Proposition \ref{prop_moment Zeck}) and, more generally, for functions $f$ such that $f\circ\Delta^{(r)}$ is integrable (Proposition \ref{prop_gene Zeck}). We deduce from Proposition \ref{prop_moment Zeck} that $\mu^{(r)}(d)=\PP(\{x\in \XX: \Delta^{(r)}(x)=d \})$ and that $\mu^{(r)}$ has finite moments. In particular, we show that $\mu^{(r)}$ is of zero-mean.

In Section~\ref{HTCM}, we construct an algorithm that computes $\mu^{(r)}$. Pseudocode is given in Subsection~\ref{PC}. In Subsection~\ref{ROTA}, we prove Corollary~\ref{corollaire pour algo2} that gives an estimate on the tail of $\mu^{(r)}$.

Section~\ref{HTPMFM1} is devoted to the proof of Theorem~\ref{theo muFk=mu1}. The proof consists of applying the algorithm.

In the last section, we build a finite sequence of random variables associated with addition of an integer $r$ that gives the decomposition of $\Delta^{(r)}$ mentioned in \eqref{decomposition delta zeck}. Using that sequence, we prove Theorem~\ref{mixing action bloc zeck} on the estimation of the $\alpha$-mixing coefficients for this sequence.

\section{How to do additions}\label{HTA}

    \subsection{How to add integers}\label{HTAI}
    
Here, we describe the algorithm for addition using the Zeckendorf way to represent numbers. We start with the addition of $1$. There are two cases.
\begin{enumerate}
    \item Either there exists $\ell\ge 0$ such that the Zeckendorf decomposition of $n$ is 
    \[n=\sum_{k=1}^{\ell}F_{2k+1}+\sum_{k\ge 2\ell+4}n_kF_k.\] 
    Then with the relations \eqref{Fib_def}, we get 
    \[\begin{array}{ccccccc}
         (n)    &       & \cdots    & n_{2\ell+4}   & 0 & 0 & (10)^{\ell} \\
                &+      &           &               &   &   & 1 \\ \hline
         (n+1)  &=      & \cdots    & n_{2\ell+4}   & 0 & 1 & (00)^{\ell}   
    \end{array}\]
    Indeed, $F_3+\cdots + F_{2\ell+1}+F_2=F_{2\ell+2}$.
    \item Or there exists $\ell\ge 0$ such that 
    \[n=\sum_{k=1}^{\ell+1}F_{2k}+\sum_{k\ge 2\ell+5}n_kF_k.\]
    Then for the same reasons 
    \[\begin{array}{ccccccccc}
         (n)    &       & \cdots    & n_{2\ell+5}   & 0 & 0 & 1 & (01)^{\ell} \\
                &+      &           &                &   &   &    & 1 \\ \hline
         (n+1)  &=      & \cdots    & n_{2\ell+5}   & 0 & 1 & 0 & (00)^{\ell}   
    \end{array}\]
\end{enumerate}
We observe that adding $1$ to the rightmost digit of a block as defined
in \eqref{blocks Zeck} modifies that block into a chain of $0$ digits of
the same length and put a $1$ in the first left position of the block.

Of course, in order to compute the sum of two integers, adding $1$ as many times as needed is enough. However, we want to show the main difference with 
addition in an integer base. For instance, in an integer base, adding $1$ at some position $k\ge 2$ to an integer $n$ may change the digits of the expansion of $n$ of higher indices, due to carry propagation. Here it can also change the digits of lower indices. 
We consider the addition $n+F_k$ where $k\ge 3$. We observe that a consequence of \eqref{Fib_def} is 
\begin{equation}\label{Fib_relat2}
    2F_k=\begin{cases}
        F_2+F_4, & \text{if $k=3$;}\\
        F_{k+1}+F_{k-2}, & \text{otherwise.}
    \end{cases}    
\end{equation}

Many cases appear. For simplicity, the digit in \textcolor{webred}{\textbf{color}} represents the digit at position $k$ and we do not represent digits that remain the same in the expansion of $n$ and $n+F_k$. We start with the cases that change only digits of indices $\ge k$.
\begin{enumerate}
    \item If $n=\left[\cdots 0\textcolor{webred}{\textbf{0}}0\cdots\right]$ then 
        \[\begin{array}{cccccccc}
         (n)      &       & \cdots    & 0 & \textcolor{webred}{\textbf{0}} & 0& \cdots\\
         (+F_k)   &+      &           &   & \textcolor{webred}{\textbf{1}} &  &  \\ \hline
         (n+F_k)  &=      & \cdots    & 0 & \textcolor{webred}{\textbf{1}} & 0& \cdots
    \end{array}\]
    \item If there exists $\ell\ge 0$ such that $n=\left[\cdots 001(01)^{\ell}\textcolor{webred}{\textbf{0}}\cdots\right]$ then 
    \[\begin{array}{cccccccccc}
         (n)        &       & \cdots    & 0 & 0 & 1 & (01)^{\ell} &\textcolor{webred}{\textbf{0}} &  \cdots\\
         (+F_k)     &+      &           &   &   &   &             &\textcolor{webred}{\textbf{1}} &   \\ \hline
         (n+F_k)    &=      & \cdots    & 0 & 1 & 0 &(00)^{\ell}  &\textcolor{webred}{\textbf{0}} &  \cdots
    \end{array}\]
\end{enumerate}
We continue with the first case where the digit of index $k-1$ is changed.

\begin{enumerate}[resume]
     \item If there exists $\ell\ge 0$ such that $n=\left[\cdots 00(10)^{\ell}\textcolor{webred}{\textbf{0}}1\cdots\right]$ then \[\begin{array}{ccccccccc}
         (n)      &       & \cdots    & 0 & 0 & (10)^{\ell} &\textcolor{webred}{\textbf{0}} & 1 & \cdots\\
         (+F_k)   &+      &           &   &   &             &\textcolor{webred}{\textbf{1}} &   &  \\ \hline
         (n+F_k)  &=      & \cdots    & 0 & 1 & (00)^{\ell} &\textcolor{webred}{\textbf{0}} & 0 & \cdots
    \end{array}\]
\end{enumerate}

Now we consider the cases where many digits of indices $<k$ are changed. It is due to \eqref{Fib_relat2}.
\begin{enumerate}[resume]
    \item\label{item 4} If $k\ge 5$ and there exist $\ell,\ell'\ge 0$ such that $n=[\cdots 00(10)^{\ell}\textcolor{webred}{\textbf{1}}(01)^{\ell'}000\cdots]$ then
        \[\begin{array}{cccccccccccc}
         (n)      &       & \cdots    & 0 & 0 & (10)^{\ell} &\textcolor{webred}{\textbf{1}} & (01)^{\ell'} & 0 & 0 & 0 & \cdots\\
        (+F_k)    &+      &           &   &   &             &\textcolor{webred}{\textbf{1}} &              &   &   &   & \\ \hline
         (n+F_k)  &=      & \cdots    & 0 & 1 & (00)^{\ell} &\textcolor{webred}{\textbf{0}} & (10)^{\ell'} & 0 & 1 & 0 & \cdots
    \end{array}\]
    \item\label{item 5} If $k\ge 6$ and there exist $\ell,\ell'\ge 0$ such that $n=[\cdots 00(10)^{\ell}\textcolor{webred}{\textbf{1}}(01)^{\ell'}0010\cdots]$ then
        \[\begin{array}{ccccccccccccc}
         (n)      &       & \cdots    & 0 & 0 & (10)^{\ell} &\textcolor{webred}{\textbf{1}} & (01)^{\ell'} & 0 & 0 & 1 & 0 &\cdots\\
         (+F_k)   &+      &           &   &   &             &\textcolor{webred}{\textbf{1}} &              &   &   &   &   & \\ \hline
         (n+F_k)  &=      & \cdots    & 0 & 1 & (00)^{\ell} &\textcolor{webred}{\textbf{0}} & (10)^{\ell'} & 1 & 0 & 0 & 0 &\cdots
    \end{array}\]
\end{enumerate}
Finally, we consider the ``boundary'' cases where all the digits of indices $<k$ are changed.
\begin{enumerate}[resume]
    \item If $k\ge 4$, $k$ is even and there exists $\ell\ge 0$ such that $n=[\cdots 00(10)^{\ell}\textcolor{webred}{\textbf{1}}(01)^{\frac{k-2}{2}}]$ then
        \[\begin{array}{cccccccc}
         (n)      &       & \cdots    & 0 & 0 & (10)^{\ell} &\textcolor{webred}{\textbf{1}} & (01)^{\frac{k-2}{2}}\\
        (+F_k)    &+      &           &   &   &             &\textcolor{webred}{\textbf{1}} & \\ \hline
         (n+F_k)  &=      & \cdots    & 0 & 1 & (00)^{\ell} &\textcolor{webred}{\textbf{0}} & (10)^{\frac{k-2}{2}}
    \end{array}\]
    \item If $k\ge 5$, $k$ is odd and there exists $\ell\ge 0$ such that $n=[\cdots 00(10)^{\ell}\textcolor{webred}{\textbf{1}}(01)^{\frac{k-3}{2}}0]$ then
        \[\begin{array}{ccccccccc}
         (n)      &       & \cdots    & 0 & 0 & (10)^{\ell} &\textcolor{webred}{\textbf{1}} & (01)^{\frac{k-3}{2}} & 0 \\
        (+F_k)    &+      &           &   &   &             &\textcolor{webred}{\textbf{1}} &                      &\\ \hline
         (n+F_k)  &=      & \cdots    & 0 & 1 & (00)^{\ell} &\textcolor{webred}{\textbf{0}} & (10)^{\frac{k-3}{2}} & 1  
    \end{array}\]
\end{enumerate}

    \subsection{The Zeckendorf-adic integers and how to add one of them to an integer}\label{ZIHTAI}

We define the \textit{Zeckendorf-adic integers} (or \textit{Z-adic integers} for simplicity) as elements of \[\XX:=\big\{x\in\{0;1\}^{\NN_{\ge 2}}: \forall k\ge 2~x_kx_{k+1}=0\big\}.\]
Also, coordinates of a Z-adic integer $x\in \XX$ are interpreted as digits in the Zeckendorf representation: elements of $\XX$ can be viewed as ``generalized integers having possibly infinitely many nonzero digits in their Zeckendorf representation''. An element $x=(x_k)_{k\ge 2}\in \XX$ is represented as a left-infinite sequence $(\ldots, x_3, x_2)$, with $x_2$ being the unit digit. We endow $\XX$ with the product topology that turns it into a compact metrizable space. The set $\NN$ can be identified with the subset of sequences with finite support. More precisely, using the inclusion function 
\begin{align*}
    i:n=[n_{\ell}\cdots n_2] \in\NN \longmapsto (\ldots, 0,n_{\ell},\ldots, n_2) \in \XX
\end{align*}
we identify $\NN$ and $i(\NN)$. We can also identify the notation 
\[(\ldots, x_3, x_2)=[\cdots x_3 x_2].\]
We take the opportunity to define $\XX_f$ as the set of finite sequences of $0$'s and $1$'s without two consecutive $1$'s. For instance, the Zeckendorf expansion of a given integer is composed using a sequence in $\XX_f$. Let us define, for $\ell\ge 2$ and $(n_k)_{k\ge 2}\in\XX_f$, the \textit{cylinder} $C_{n_{\ell}\cdots n_2}$ as the set of sequences $x\in\XX$ such that $x_i=n_i$ for $ i=2,\ldots, \ell$. We observe that $n_{\ell}\cdots n_2$ is not necessarily the Zeckendorf expansion of a given integer: the leftmost digit(s) can be $0$('s).

We want to extend the transformation $n\mapsto n+1$, defined on $\NN$, on $\XX$. From the description given in Subsection \ref{HTAI}, it is convenient to consider the transformation $T$ defined on $\XX$ by the following formula, where $\ell\in\NN$
\begin{displaymath}
    T(x):=\begin{cases}
        [\cdots x_{2\ell+4}00(10)^{\ell}],& \text{if $x=[\cdots x_{2\ell+4}01(00)^{\ell}]$;}\\
        [\cdots x_{2\ell'+3}001(01)^{\ell}],& \text{if $x=[\cdots x_{2\ell'+3}010(00)^{\ell}]$;}\\
        [(10)^{\infty}]& \text{if $x=[0^{\infty}]$;}\\
        [(01)^{\infty}]& \text{if $x=[0^{\infty}]$.}
    \end{cases}
\end{displaymath}
Indeed, by Subsection \ref{HTAI} we observe that $T_{|\NN}(n)=n+1$. Now if we take $x\in\XX$ whose expansion contains two consecutive $0$'s, we observe that the sequence $(\left[x_{\ell}\cdots x_2\right]+1)_{\ell\ge 2}$ converges to $T(x)$ because the digits are not changed eventually so we can define $x+1:=T(x)$ in that case. Otherwise, if $x\in\XX$ does not have two consecutive $0$'s in its expansion, there are two cases: $\left[(10)^{\infty}\right]$ and $\left[(01)^{\infty}\right]$. Adding $1$ to the truncated sequence
$([(10)^{\ell}])_{\ell\in\NN}$ converges to $[(10)^{\infty}]+1:=[0^{\infty}]$. It is the same for $[(01)^{\infty}]$. Thus, the transformation $T$ can be described in a simpler way as 
\begin{displaymath}
    T:\begin{cases}
        \XX &\longrightarrow  \XX \\
        x   &\longmapsto x+1\text{.}
    \end{cases}
\end{displaymath}
Due to the two pre-images of $0^{\infty}$, the transformation $T$ is not a homeomorphism on $\XX$, but remains continuous and surjective on $\XX$. Thus, $(\XX,T)$ is a topological dynamical system that we call the \emph{odometer}.

We know how to add $1$ to a Z-adic integer $x$. Repeating this operation enables us to add an integer $r$ to $x$. For later purposes, we need to specify how to add $F_k$ ($\ge 3$) directly. In Subsection~\ref{HTAI}, we have to compute $x+F_k$ for the $x\in\XX$ with two consecutive $0$'s at indices $>k$. Thus, we only need to focus on what happens if $x$ does not have two consecutive $0$'s at indices $>k$. There are only finitely many cases to consider, which we detail below. Again for simplicity, we write 
the digits at position $k$ 
in \textcolor{webred}{\textbf{color}},
and we do not represent digits that are not modified. We start with the cases where the only digits that change are those of indices $\ge k-1$.
\begin{enumerate}
    \item If $x=[(10)^{\infty}\textcolor{webred}{\textbf{0}}1\cdots]$ then 
    \[\begin{array}{cccccc}
        (x)    &   & (10)^{\infty} & \textcolor{webred}{\textbf{0}} & 1 & \cdots  \\
        (F_k)  & + &               & \textcolor{webred}{\textbf{1}} &   &       \\ \hline
        (x+F_k)& = & (00)^{\infty} & \textcolor{webred}{\textbf{0}} & 0 & \cdots
    \end{array}\]
    \item If $x=[(01)^{\infty}\textcolor{webred}{\textbf{0}}\cdots]$ then 
    \[\begin{array}{cccccc}
        (x)    &   & (01)^{\infty} & \textcolor{webred}{\textbf{0}} &  \cdots  \\
        (F_k)  & + &               & \textcolor{webred}{\textbf{1}} &        \\ \hline
        (x+F_k)& = & (00)^{\infty} & \textcolor{webred}{\textbf{0}} &  \cdots
    \end{array}\]
\end{enumerate}
Now we consider cases where some
digits of small indices are changed, but not all them.
\begin{enumerate}[resume]
    \item If $k\ge 5$ and there exists $\ell'\ge0$ such that $x=[(10)^{\infty}\textcolor{webred}{\textbf{1}}(01)^{\ell'}000\cdots]$ then 
    \[\begin{array}{ccccccccc}
        (x)    &   & (10)^{\infty} & \textcolor{webred}{\textbf{1}} & (01)^{\ell'} & 0 & 0 & 0 & \cdots  \\
        (F_k)  & + &               & \textcolor{webred}{\textbf{1}} &              &   &   &   & \\ \hline
        (x+F_k)& = & (00)^{\infty} & \textcolor{webred}{\textbf{0}} & (10)^{\ell'} & 0 & 1 & 0 &  \cdots
    \end{array}\]
    \item If $k\ge 6$ and there exists $\ell'\ge0$ such that $x=[(10)^{\infty}\textcolor{webred}{\textbf{1}}(01)^{\ell'}0010\cdots]$ then 
    \[\begin{array}{cccccccccc}
        (x)    &   & (10)^{\infty} & \textcolor{webred}{\textbf{1}} & (01)^{\ell'} & 0 & 0 & 1 & 0 & \cdots  \\
        (F_k)  & + &               & \textcolor{webred}{\textbf{1}} &              &   &   &   &   & \\ \hline
        (x+F_k)& = & (00)^{\infty} & \textcolor{webred}{\textbf{0}} & (10)^{\ell'} & 1 & 0 & 0 & 0 & \cdots
    \end{array}\]
\end{enumerate}
Finally, we consider cases where the whole prefix of $x$ is modified.
\begin{enumerate}[resume]
    \item If $k\ge 4$ and is even and $x=[(10)^{\infty}\textcolor{webred}{\textbf{1}}(01)^{\frac{k-2}{2}}]$ then 
    \[\begin{array}{ccccc}
        (x)     &   & (10)^{\infty} & \textcolor{webred}{\textbf{1}} & (01)^{\frac{k-2}{2}} \\
        (F_k)   & + &               & \textcolor{webred}{\textbf{1}} & \\ \hline
        (x+F_k) & = & (00)^{\infty} & \textcolor{webred}{\textbf{0}} & (10)^{\frac{k-2}{2}}
    \end{array}\]
    \item If $k\ge 3$ and is odd and $x=[(10)^{\infty}\textcolor{webred}{\textbf{1}}(01)^{\frac{k-3}{2}}0]$ then 
    \[\begin{array}{cccccc}
        (x)     &   & (10)^{\infty} & \textcolor{webred}{\textbf{1}} & (01)^{\frac{k-3}{2}} & 0 \\
        (F_k)   & + &               & \textcolor{webred}{\textbf{1}} &                      & \\ \hline
        (x+F_k) & = & (00)^{\infty} & \textcolor{webred}{\textbf{0}} & (10)^{\frac{k-3}{2}} & 1
    \end{array}\]
\end{enumerate}
We can sum up all these cases in the following proposition.

\begin{proposition}\label{T^{F_n}}
The table below summarizes the action of the addition of $F_k$ ($k\ge 2)$ on the digits of $x\in\XX$. Here, $\ell,\ell'\in\NN$ and we represent in \textcolor{webred}{\textbf{color}}, the digits at position $k$. The digits that are not explicitly written remain untouched by the operation.
\begin{align}
    T^{F_k}:    \XX &\longrightarrow  \XX \notag \\
                [\cdots 0\textcolor{webred}{\textbf{\emph{0}}}0\cdots]& \longmapsto  [\cdots 0\textcolor{webred}{\textbf{\emph{1}}}0\cdots] \\
                [\cdots 001(01)^{\ell}\textcolor{webred}{\textbf{\emph{0}}}\cdots] & \longmapsto  [\cdots 010(00)^{\ell}\textcolor{webred}{\textbf{\emph{0}}}\cdots] \label{avancee1}\\
                [\cdots00(10)^{\ell}\textcolor{webred}{\textbf{\emph{0}}}1\cdots]&\longmapsto   [\cdots 01(00)^{\ell}\textcolor{webred}{\textbf{\emph{0}}}0\cdots] \label{avancee2}\\
                [\cdots00(10)^{\ell}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\ell'}000\cdots]&\longmapsto   [\cdots 01(00)^{\ell}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\ell'}010\cdots] \label{recul1} \\
                [\cdots 00(10)^{\ell}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\ell'}0010\cdots]&\longmapsto   [\cdots 01(00)^{\ell}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\ell'}1000\cdots] \label{recul2}\\
                [\cdots 00(10)^{\ell}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\frac{k-5}{2}}001]&\longmapsto   [\cdots 01(00)^{\ell}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\frac{k-5}{2}}100] \label{11}\\
                [\cdots 00(10)^{\ell}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\frac{k-4}{2}}00]&\longmapsto   [\cdots 01(00)^{\ell}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\frac{k-4}{2}}01]\label{12}\\
                [\cdots 00(10)^{\ell}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\frac{k-2}{2}}]&\longmapsto   [\cdots 01(00)^{\ell}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\frac{k-2}{2}}] \label{recul max} \\
                [\cdots 00(10)^{\ell}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\frac{k-3}{2}}0]&\longmapsto   [\cdots 01(00)^{\ell}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\frac{k-3}{2}}1]\label{17} \\
                [(01)^{\infty}\textcolor{webred}{\textbf{\emph{0}}}\cdots]&\longmapsto   [0^{\infty}\textcolor{webred}{\textbf{\emph{0}}}\cdots] \label{avancee infinie}\\
                 [(10)^{\infty}\textcolor{webred}{\textbf{\emph{0}}}1\cdots]&\longmapsto   [0^{\infty}\textcolor{webred}{\textbf{\emph{0}}}0\cdots] \\
                [(10)^{\infty}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\ell'}000\cdots]&\longmapsto   [0^{\infty}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\ell'}010\cdots] \label{recul3}\\
                [(10)^{\infty}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\ell'}0010\cdots]&\longmapsto   [0^{\infty}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\ell'}1000\cdots]\label{recul4}\\
                [(10)^{\infty}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\frac{k-5}{2}}001]&\longmapsto   [0^{\infty}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\frac{k-5}{2}}100]\label{14.bis} \\
                [(10)^{\infty}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\frac{k-4}{2}}00]&\longmapsto   [ 0^{\infty}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\frac{k-4}{2}}01]\label{15.bis} \\                
                [(10)^{\infty}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\frac{k-2}{2}}]&\longmapsto   [0^{\infty}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\frac{k-2}{2}}] \label{16.bis}\\
                [(10)^{\infty}\textcolor{webred}{\textbf{\emph{1}}}(01)^{\frac{k-3}{2}}0]&\longmapsto   [0^{\infty}\textcolor{webred}{\textbf{\emph{0}}}(10)^{\frac{k-3}{2}}1]\label{17.bis}.
\end{align}
(We specify that $k$ is even in cases \eqref{12}, \eqref{recul max}, \eqref{15.bis} and \eqref{16.bis}, whereas $k$ is odd in cases \eqref{11}, \eqref{17}, \eqref{14.bis} and \eqref{17.bis}. Also, in some cases, $k$ is assumed to be large enough for the operation to be possible.)
\end{proposition}
    
    \subsection{Stopping conditions when adding an integer to an adic number}\label{subsection Focus}

    Through the cases described in Proposition~\ref{T^{F_n}}, we observe that if there is a $1$ at position $k$ in the expansion of $x$, 
the addition of $F_k$ yields a carry propagation in both directions:
\begin{itemize}
    \item to the left, modifying digits of higher indices (as in an integer base),
    \item to the right, modifying digits of lower indices.
\end{itemize}
The propagation (in both directions) happens through a maximal sequence of alternative $0$'s and $1$'s and is stopped at the first occurrence of two consecutive $0$'s. But the modifications depend on the propagation direction.
\begin{itemize}
    \item In propagation to the left, the maximal subword of alternative $1$'s and $0$'s is transformed into a subword of $0$'s of the same length (case \eqref{recul1} for instance), and the stopping pattern $00$ is transformed into $01$. Note that this propagation also happens if $x_k=0$ (case \eqref{avancee1} for instance).
    \item In propagation to the right, which only happens if $x_k=1$, the maximal subword of alternative $1$'s and $0$'s is transformed into a symmetrical subword where the $1$'s become $0$'s and vice-versa (cases \eqref{recul1} and \eqref{recul2} for instance). Then the first occurrence of $00$ (in the sense the largest index $\le k$ such that the digits of $x$ are $00$) can either be part of the pattern $w_0:=\left[01000\right]$ or $w_1:=\left[10010\right]$. (We call $w_0$ and $w_1$ the \emph{right-stopping pattern}.) 
\end{itemize}
Depending on the right-stopping pattern ($\left[01000\right]$ or $\left[10010\right]$), the modifications of digits at these indices are given by the next scheme.
\begin{figure}[H]
    \centering
    \includegraphics[scale=0.35]{stopping_pattern_behavior.pdf}
    \caption{Modifications at the position of the right-stopping pattern.}
    \label{Scheme}
\end{figure}
Note that in both cases, we get a new right-stopping pattern in the same position as before the addition of $F_k$. This addition of $F_k$ to $x\in\XX$ modifies some digits at positions $\le k-2$ only if $x_k=1$, and, in this case, the modification of the digits takes place up to the first occurrence of one of the right-blocking patterns. 

Formally: let $x\in\XX$ and $k\ge 2$.
\begin{itemize}
    \item If $x_k=0$ then $(x+F_k)_n=x_n$ for all $n\le k-2$.
    \item If $x_k=1$ and if there exists $j\le k+1$ such that $x_jx_{j-1}\cdots x_{j-4}$ is a right-stopping patterns $w_i$ ($i\in\{0,1\})$---we let $j'$ denote the largest index with this property---then 
    \begin{itemize}
        \item $(x+F_k)_n=x_n$ for all $n\le j'-4$,
        \item $w_{1-i}$ appears in $(x+F_k)$ in the same position $j'$, unless $w_i=w_0$ and $j'=k+1$, in which case we might have $00010$ instead of $w_1$ at position $j'$ in $x+F_k$ (case~\eqref{recul1} with $\ell=\ell'=0$).
    \end{itemize}
\end{itemize}
A straightforward consequence is the following result.
\begin{proposition}
Assume that the right-stopping pattern $w_i$ ($i=0,1$) appears in $x\in\XX$ at position $j\ge 5$, that is, assume $x_jx_{j-1}\cdots x_{j-4}=w_i$. Let $k\ge j-1$. Then
\begin{itemize}
    \item for all $n\le j-4$, $(x+F_k)_n=x_n$, and
    \item $w_0$, $w_1$ or $00010$ appears in $x+F_k$ at some position $j'$ with $k+1\ge j'\ge j$.
\end{itemize}
\end{proposition}
We now state the following corollary that enhances that property when we add not only a Fibonacci term but an integer whose expansion involves Fibonacci numbers of high indices.
\begin{corollary}\label{blocage recul}
    Let $x\in\XX$. Assume that, for some $\ell\ge 2$, $x_{\ell}=x_{\ell+1}=0$. Let $r\in\NN$ be such that $r_j=0$ for $j=2,\ldots,\ell+1$.
    Then for each $n\le\ell-2$, we have $(x+r)_n=x_n$.
\end{corollary}
\begin{proof}
    We are considering the following addition:
    \[
    \begin{array}{ccccccccccc}
        (x)     &       & \cdots    & x_{\ell+3}    & x_{\ell+2}    & 0     & 0     & x_{\ell-1}    & x_{\ell-2}    & x_{\ell-3}    & \cdots   \\
        (r)     & +     & \cdots    & r_{\ell+3}    & r_{\ell+2}    & 0     & 0     & 0             & 0             & 0             & \cdots  \\ \hline
        (x+r)   & =     & \cdots    & \star         & \star         & \star & \star & \star         & x_{\ell-2}    & x_{\ell-3}    & \cdots 
    \end{array}
    \]
    Let $r=F_{k_{s(r)}}+\cdots +F_{k_1}$ be the Zeckendorf decomposition of $r$ with $k_{s(r)}>\cdots>k_2>k_1\ge \ell+2$.
    We first consider the addition of $F_{k_1}$ to $x$:
    \begin{itemize}
        \item if, for all $j$ such that $k_1\ge j\ge \ell$, we have $x_j=0$, then for every $n\le\ell+1$ we have $(x+F_{k_1})_n=x_n$ (in particular $00$ appears in the same place in $x+F_{k_1}$);
        \item otherwise, there exists a largest integer $j'$ with $k_1+1\ge j'\ge \ell+2$, such that one of the right-stopping pattern appears in $x$ at position $j'$. Then we can apply the above proposition, which proves that 
        \begin{itemize}
            \item either $w_0$ or $w_1$ appears at position $j'$ in $x+F_{k_1}$,
            \item or $00$ appears at position $j'$ in $x+F_{k_1}$ and $j'\ge \ell+3$.
        \end{itemize}
        In each case, we still have $(x+F_{k_1})_n=x_n$ for $n\le\ell-2$.
    \end{itemize}
    Then we prove by induction on $t$ such that for each $t$, $1\le t\le s(r)$, the above is true for $x+F_{k_1}+\cdots+F_{k_t}$.
\end{proof}

The following lemma ensures that the pattern $00$ is a left-stopping condition: it stops the propagation of a carry coming from the right.
\begin{lemma}\label{blocage_avancee}
    Let $x\in\XX$, $r\in\NN$. Let $\ell\ge 2$ be such that $r<F_{\ell+1}$ and assume $x_{\ell+2}=x_{\ell+3}=0$. Then we have $(x+r)_k=x_k$ for all $k\ge \ell+3$.
\end{lemma}
\begin{proof}
The assumption $r<F_{\ell+1}$ implies $r=\left[r_{\ell}\cdots r_2\right]$ with $(r_{\ell},\cdots,r_2)\in\XX_f$. Since $\left[x_{\ell+1}\cdots x_2\right]+r<F_{\ell+2}+F_{\ell+1}=F_{\ell+3}$, a carry cannot propagate on digits of indices $\ge \ell+3$: we have the addition 
\[\begin{array}{cccccccccc}
    (x)  &   & \cdots & x_{\ell+4} & 0 & 0              & x_{\ell+1}   & x_{\ell}    & \cdots & x_2 \\
    (r)  & + &        &            &   &                &              & r_{\ell}    & \cdots & r_2 \\ \hline
    (x+r)& = & \cdots & x_{\ell+4} & 0 & (x+r)_{\ell+2} &(x+r)_{\ell+1}& (x+r)_{\ell}& \cdots & (x+r)_2
\end{array}\]
\end{proof}
The next lemma enhances the previous one: it shows that, given $x$ with some restrictions, right-stopping patterns can appear in the expansion $x+F_k$ when $k$ is a small integer. 
\begin{lemma}\label{blocage avancee + motifs}
Let $x\in\XX$ and $r\in\NN$. Let $\ell\ge 2$ be such that $x_{\ell+1}=x_{\ell+2}=x_{\ell+3}=x_{\ell+4}=0$ and $r_{\ell+2}=1$. Assume $r<F_{\ell+3}$. Then for $k=1,2,3,4$ we have
\[(x+r)_{\ell+k}=r_{\ell+k},\]
or
\[(x+r)_{\ell+1}=0 \textrm{ and } (x+r)_{\ell+3}=1.\]
\end{lemma}
\begin{proof}
    We are actually considering the following addition:
\[    \begin{array}{ccccccccccc}
        (x) &   & \cdots & x_{\ell+5}   & 0  & 0  & 0   & 0 & x_{\ell} &  \cdots    & x_2 \\
         (r)& + & \cdots &              &    &    & 1   & 0 & r_{\ell} & \cdots     & r_2
    \end{array}
 \]
We want to show that the expansion of $x+r$ is either
   \begin{align}\label{case 1}
      \left[\cdots x_{\ell+5}0010(x+r)_{\ell}\cdots(x+r)_2\right], \tag{C1}
  \end{align}
 or
  \begin{align}\label{case 2}
      \left[\cdots x_{\ell+5}0100(x+r)_{\ell}\cdots(x+r)_2\right]. \tag{C2}
 \end{align}
 Thanks to Lemma~\ref{blocage_avancee}, we have $(x+r)_k=x_k$ for every $k\ge\ell+4$, which means that the digits of $x+r$ of indices $\le \ell+3$ are given by the addition $r+\left[x_{\ell}\cdots x_2\right]$.
 Now we claim
 \[(x+r)_{\ell+3} \textrm{ or } (x+r)_{\ell+2} \textrm{ is }1.\]
 Indeed, if it is not the case then $\left[x_{\ell}\cdots x_2\right]+r<F_{\ell+2}$ while $r\ge F_{\ell+2}$ (since $r_{\ell+2}=1$).
 We consider now both possibilities.
 \begin{itemize}
     \item If $(x+r)_{\ell+2}=1$ then we obtain \eqref{case 1}.
     \item If $(x+r)_{\ell+3}=1$ then we must have $(x+r)_{\ell+1}=0$ since we have
     \begin{align*}
         \left[x_{\ell}\cdots x_2\right]+r-F_{\ell+3}&=\left[x_{\ell}\cdots x_2\right]+\left[r_{\ell}\cdots r_2\right]+F_{\ell+2}-F_{\ell+3}\\
         &< F_{\ell+1}.
     \end{align*}
     Thus, the expansion of $x+r$ is \eqref{case 2}.
 \end{itemize}
\end{proof}

We deduce the next corollary.
\begin{corollary}\label{stopping condition bloc 1}
Let $x\in\XX$ and $r\in\NN$ such that it exists $\ell\ge 2$ with $x_{\ell}=\cdots=x_{\ell+5}=0$, $r_{\ell+1}=r_{\ell+5}=0$ and $r_{\ell+3}=1$. Let $\widetilde{r}:=\left[r_{\ell}\cdots r_2\right]$. Then for all $k\ge \ell+4$ we have
    \begin{align}\label{ccl 1 pour bloc de taille 1}
      (x+\widetilde{r}+F_{\ell+3})_k=(x+\widetilde{r})_k=x_k, 
    \end{align}
    and, for all $k\le \ell+2$
    \begin{align}\label{ccl2 pour bloc de taille 1}
      (x+r)_k=(x+\widetilde{r}+F_{\ell+3})_k=(x+\widetilde{r})_k.  
    \end{align}
\end{corollary}
\begin{remark}\label{rem stop cdts 1}
The hypothesis means that we are considering the following addition:
\[\begin{array}{ccccccccccccc}
    (x) &  & \cdots & x_{\ell+6} & 0 & 0 & 0 & 0 & 0 & 0      & x_{\ell-1}  & \cdots    & x_2  \\
    (r) & +& \cdots & r_{\ell+6} & 0 & 0 & 1 & 0 & 0 & r_\ell     & \cdots  & \cdots    & r_2 \\
\end{array}\]
and the conclusion ensures that the digits of indices $\le \ell+2$ 
(i.e., those on the right-hand side of the pattern we have imposed) of $x+r$ are the same if we compute this previous addition as if we compute 
\[\begin{array}{ccccccccccccc}
    (x)                     &  & \cdots & x_{\ell+6}   & 0 & 0 & 0 & 0 & 0 & 0      & x_{\ell-1}  & \cdots    & x_2  \\
    (\widetilde{r}+F_{\ell+3}) & +&        &           &   &   & 1 & 0 & 0 & r_\ell     & \cdots  & \cdots    & r_2 \\
\end{array}\]
or if we compute
\[\begin{array}{ccccccccccccc}
    (x)            &  & \cdots & x_{\ell+6}    & 0 & 0 & 0 & 0 & 0 & 0      & x_{\ell-1}  & \cdots    & x_2  \\
    (\widetilde{r})& +&        &            &   &   &   &   &   & r_\ell     & \cdots  & \cdots    & r_2 \\
\end{array}\]

In other words, the corollary states that the conditions we put on $x$ and $r$ stop the propagation of carries in both directions. 
\end{remark}
\begin{proof}
    First, we have the following addition
    \[\begin{array}{ccccccccccccc}
    (x)                &  & \cdots & x_{\ell+6}    & 0   & 0 & 0 & 0 & 0                         & 0     & x_{\ell-1}   & \cdots    & x_2  \\
    (\widetilde{r})    & +&       &             &     &   &   &   &                           & r_\ell   & \cdots    & \cdots    & r_2 \\ \hline
    (x+\widetilde{r})  & =& \cdots&  x_{\ell+6}    & 0  & 0 & 0 & 0 & (x+\widetilde{r})_{\ell+1}   & \cdots& \cdots    & \cdots    & (x+\widetilde{r})_2 
\end{array}\]
    Indeed, due to Lemma~\ref{blocage_avancee}, we have that $(x+\widetilde{r})_j=x_j$ for every $j\ge \ell+2$. When we add $F_{\ell+3}$, it gives $x+\widetilde{r}+F_{\ell+3}$, whose Zeckendorf expansion is
    \[(x+\widetilde{r}+F_{\ell+3})=\left[\cdots x_{\ell+6}0010(x+\widetilde{r})_{\ell+1}\cdots(x+\widetilde{r})_2\right].\]
    We thus get the relation~\eqref{ccl 1 pour bloc de taille 1}.
    
    We now consider addition (where $\widetilde{x}:=x+\widetilde{r}+F_{\ell+3}$)
    \[
    \begin{array}{ccccccccccc}
    (\widetilde{x}) &   & \cdots    & x_{\ell+6}    & 0 & 0 & 1 & 0 & (x+\widetilde{r})_{\ell+1}    & \cdots    & (x+\widetilde{r})_2       \\
                    & + & \cdots    & r_{\ell+6}    & 0 & 0 & 0 & 0 &   0                           & \cdots    & 0 
    \end{array}
    \]
    Now using Corollary~\ref{blocage recul}, we obtain \eqref{ccl2 pour bloc de taille 1}.
\end{proof}
The statement of Corollary~\ref{stopping condition bloc 1} means that, given a block of length $1$ (in the sense that it has one pattern $10$) in $r$, we are able to control the propagation of carries so that the left part of the addition does not change the expansion on the right part and vice versa. We now want to have the same kind of control for larger blocks. We could assume that $x$ has many $0$'s facing the block in $r$ that we want to control. However, this condition would be more and more ``expensive'' (in the sense that the probability for $x$ to satisfy it would decrease to $0$) as the length of the block increases. To avoid this, we are looking for conditions on $x$ that affect only a bounded number of digits, regardless of the length of the block. In other words, we want our conditions to appear in ``most'' of $x\in\XX$. We obtain the following corollary where the length of the block is $m+2$ and where we fixed $8$ digits in $x$.
\begin{corollary}\label{stopping cdt bloc >1}
Let $x\in\XX$, $r\in\NN$ and $m\ge0$. Assume that there exists $\ell\ge 2$ such that 
    \begin{itemize}
        \item $x_i=0$ where $i\in\{\ell,\ell+1,\ell+2,\ell+3,\ell+2m+4,\ell+2m+5,\ell+2m+6,\ell+2m+7\}$,
        \item $r_{\ell+1}=r_{\ell+2m+7}=0$ and $r_{\ell+2i+3}=1$ for $i=0,\ldots,m+1$.
    \end{itemize}
    Let $\widetilde{r}:=\left[r_{\ell}\cdots r_2\right]$. Then we have 
    \begin{align}\label{ccl 1 lemme stopping cdt}
        (x+\widetilde{r}+\sum_{i=0}^{m+1}F_{\ell+2i+3})_k=(x+\widetilde{r})_k=x_k,
    \end{align}
    for all $k\ge\ell+2m+7$ and
    \begin{align}\label{ccl 2 lemme stopping cdt}
      (x+r)_k=(x+\widetilde{r}+\sum_{i=0}^{m+1}F_{\ell+2i+3})_k,  
    \end{align}
     for all $k\le \ell+2m+2$.
\end{corollary}
\begin{proof}
For simplicity, we write $B$ as the block $\sum_{i=0}^{m+1}F_{\ell+2i+3}$. We decompose the sum $x+r$ in several steps. First we add $\widetilde{r}$. With the hypothesis on $x$, we actually consider the following addition:
\[
\begin{array}{cccccccccccccccccc}
     (x)                &   & \cdots    & x_{\ell+8+2m} & 0 & 0 & 0 & 0 & x_{\ell+2m+3} & \cdots    & x_{\ell+4}    & 0 & 0 & 0 & 0         & x_{\ell-1}    & \cdots    & x_2  \\
     (\widetilde{r})    & + &           &               &   &   &   &   &               &           &               &   &   &   & r_{\ell}  & r_{\ell-1}    & \cdots    & r_2
\end{array}
\]
Thanks to Lemma~\ref{blocage_avancee}, we know that this addition can only modify digits of $x$ of indices $\le \ell+1$. We continue with the addition of $B$. For simplicity, we write $\widetilde{x}:=x+\widetilde{r}$:
\[
\begin{array}{cccccccccccccccccc}
    (\widetilde{x}) &   & \cdots    & x_{\ell+8+2m} & 0 & 0 & 0 & 0 & x_{\ell+2m+3} & \cdots    & x_{\ell+4}    & 0 & 0 & \widetilde{x}_{\ell+1}  & \widetilde{x}_{\ell}    & \widetilde{x}_{\ell-1}  & \cdots    & \widetilde{x}_2 \\
    (B)             & + &           &               &   &   & 1 & 0 & 1             & \cdots    & 0             & 1 & 0 & 0                         & 0                         & 0                         & \cdots    & 0
\end{array}
\]
Now Lemma~\ref{blocage avancee + motifs} ensures that the expansion of $\widetilde{x}$ is not modified for digits of indices $\ge \ell+2m+7$. We thus prove \eqref{ccl 1 lemme stopping cdt}. But more precisely, Lemma~\ref{blocage avancee + motifs} concludes that the expansion of $\widetilde{x}+B$ is either
\[\left[\cdots x_{\ell+8+2m}0100(\widetilde{x}+B)_{\ell+2m+3}\cdots(\widetilde{x}+B)_2\right],\]
or
\[\left[\cdots x_{\ell+8+2m}0010(\widetilde{x}+B)_{\ell+2m+3}\cdots(\widetilde{x}+B)_2\right].\]
In both case, a pattern $00$ appears between the indices $\ell+2m+4$ and $\ell+2m+7$. Thus, we can apply Corollary~\ref{blocage recul}: the final step of the addition, which consists of adding what remains in $r$, does not modify the digits of indices $\le \ell+2m+2$. We obtain the conclusion \eqref{ccl 2 lemme stopping cdt}.
\end{proof}

\section{Unique ergodicity of the odometer}\label{UEO}

    \subsection{Rokhlin towers and the ergodic measure}\label{RTEM}
    
In this subsection, we focus on the Rokhlin towers of the odometer $(\XX,T)$. A \emph{Rokhlin tower} is a family of disjoint subsets $(A_0, A_1, \cdots, A_{k-1})$ such that $T(A_i)=A_{i+1}$ for $i<k-1$. We say that the subset $A_i$ is a \emph{level} of the Rokhlin tower and that $k$ is the \emph{height} of the tower. This family is usually represented by a tower (see Figure~\ref{RT4_Zeck}). In our case, we show that we can construct a partition of $\XX$ using two Rokhlin towers.

More precisely, we focus on the action of $T$ on cylinders. For each $k\ge 1$, we consider the partition of $\XX$ into $F_{k+2}$ cylinders corresponding to all possible blocks formed by the rightmost $k$ digits of a Z-adic integers (we call them \emph{cylinders of order $k$}).
For example, at order $1$, we partition $\XX$ into $C_0$ and $C_1$. At order $2$, we get $\XX=C_{00}\sqcup C_{01}\sqcup C_{10}$. At order $3$
\[\XX=C_{000}\sqcup C_{001} \sqcup C_{010} \sqcup C_{100} \sqcup C_{101}.\]
In general, cylinders of order $k$ are ordered lexicographically:
\begin{itemize}
    \item first, those whose name has a $0$ in the leftmost position (there are $F_{k+1}$ of them),
    \item then those whose name has a $1$ in the leftmost position (there are $F_{k}$ of them).
\end{itemize}
We observe that each cylinders of order $k$ with a $0$ in the leftmost position, except the last one, is mapped by $T$ onto the next one, giving rise to a Rokhlin tower of height $F_{k+1}$: we call it the \emph{large} tower of order $k$. Similarly, each cylinders or order $k$ with a $1$ in the leftmost position, except the last one, is mapped by $T$ onto the next one, giving rise to another Rokhlin tower of height $F_k$: we call it the \emph{small} tower of order $k$. Thus, for each $k\ge 1$, we get a partition of $\XX$ into two Rokhlin towers whose levels are all cylinders of order $k$. For instance, for $k=4$, we get  the
partition depicted in Figure~\ref{RT4_Zeck}.
It remains to describe the transition from the Rokhlin towers of order $k$ to those of order $k+1$. First, note that all levels of the small tower of order $k$ are also levels of the large tower of order $k+1$, since a cylinder of order $k$ with a $1$ in the leftmost position coincides with the cylinders of order $k+1$ with an additional $0$ concatenated at the left of its name (e.g., $C_{101000}=C_{0101000}$).

Each level of the large tower of order $k$ is partitioned into two cylinders of order $k+1$: one obtained by concatenating a $0$ at the left of its name and the other obtained by concatenating a $1$. Thus, the large Rokhlin tower of order $k$ is cut into two subtowers:
\begin{itemize}
    \item the first one is the bottom part of the large Rokhlin tower of order $k+1$ (the top part being nothing but the small Rokhlin tower of order $k$);
    \item the second one is the small Rokhlin tower of order $k+1$.
\end{itemize}
This transition is referred to as the \emph{cutting-and-stacking process}.
See Figure~\ref{C_S_Zeck}.

\begin{figure}[htb]
    \centering
    \includegraphics[scale=0.35]{tour_ordre_4.pdf}
    \caption{Rokhlin towers of order $4$ (the action of $T$ is represented by the arrows).}
    \label{RT4_Zeck}
\end{figure}

\begin{figure}[htb]
    \centering
    \includegraphics[scale=0.26]{cut_and_stack_Zeck.pdf}
    \caption{Visual description of the  cutting-and-stacking process (for $k=4$).}
    \label{C_S_Zeck}
\end{figure}

The analysis of the action of $T$ on cylinders, yielding to the construction of the Rokhlin towers, enables us to describe the unique $T$-invariant probability measure.
\begin{proposition}\label{prop caract mesure zeck}
$(\XX,T)$ is uniquely ergodic and the unique $T$-invariant 
measure $\PP$ satisfies 
\begin{equation}\label{caract mesure zeck}
    \PP(C_{r_{\ell}\cdots r_2})=\begin{cases}
        \dfrac{1}{\varphi^{\ell-1}},& \text{if $r_{\ell}=0$;}\\
        \dfrac{1}{\varphi^{\ell}},& \text{otherwise.}
    \end{cases}
\end{equation}
$\forall \ell\ge 2$ and $(r_{\ell},\ldots,r_2)\in\XX_f$.
\end{proposition}

\begin{proof}
Let $\PP$ be a $T$-invariant measure. For each order $k$ we observe that, by $T$-invariance, all levels in the large tower have the same probability (and similarly in the small tower).
We claim that 
\begin{align}\label{mesure C_1}
    \PP(C_1)=\frac{1}{\varphi^2}.
\end{align} 
Indeed, let $u_k$ (resp.\ $v_k$) denote the number of levels included in $C_1$ in the large (resp.\ small) tower of order $k$. From the cutting-and-stacking process, for $k\ge 1$ we get the following inductive equations
\begin{displaymath}
\begin{cases}
    u_{k+1}=u_k+v_k,   \\
     v_{k+1}=u_k,
\end{cases}
\end{displaymath}
with initial conditions $u_1=0$ and $v_1=1$. We deduce that $u_k=F_{k-1}$ and $v_k=F_{k-2}$ for $k\ge 2$. We also have the identity
\begin{align*}
    \PP(C_1)&= u_k~\PP~(x\in \text{one fixed level in the large tower of order $k$})\\
        & \hspace{5mm} + v_k~\PP~(x\in\text{one fixed level in the small tower of order $k$}) \\
        &= \frac{u_k}{F_{k+1}}~\PP~(x\in \text{the large tower of order $k$}) \\
        & \hspace{5mm} + \frac{v_k}{F_k}~\PP~(x\in\text{the small tower of order $k$}) \\
        &=\frac{v_k}{F_k} + (\frac{u_k}{F_{k+1}}-\frac{v_k}{F_k})~\PP~(x\in \text{the large tower of order $k$}).
\end{align*}
Taking the limit as $k\to\infty$ gives \eqref{mesure C_1}. As $\XX=C_0\sqcup C_1$, we deduce that $\PP(C_0)=\frac{1}{\varphi}$.

Then assume that, at some order $k\ge 1$, each level in the large (resp.\ small) tower has measure $\frac{1}{\varphi^k}$ (resp.\ $\frac{1}{\varphi^{k+1}}$). Since a level in the small tower of order $k$ is also a level in the large tower of order $k+1$, we deduce that each level in the large tower of order $k+1$ also has measure $\frac{1}{\varphi^{k+1}}$. Let $p\in [0,1]$ denote the common measure of all levels in the small tower of order $k+1$. Since there are $F_{k+2}$ (resp.\ $F_{k+1}$) levels in the large (resp.\ small) tower of order $k+1$, we have the equation
\begin{align*}
    \frac{F_{k+2}}{\varphi^{k+1}} + p F_{k+1} &=1,
\end{align*}
which is equivalent to the relation
\begin{align*}
    \varphi F_{k+2} + p\varphi^{k+2} F_{k+1} &= \varphi^{k+2}.
\end{align*}
Combining with the classical identity $\varphi F_{k+2}+F_{k+1}=\varphi^{k+2}$, we deduce that $p=\frac{1}{\varphi^{k+2}}$. By induction, we prove \eqref{caract mesure zeck}.
\end{proof}

    \subsection{Probabilistic interpretation of the measure}\label{PIM}
    
In the case of the $b$-adic odometer \cite[p.\ 7]{TREJYH} ($b\ge 2)$, the $T$-invariant measure can be interpreted as an independent choice of each digit according to the uniform law on the set of possible digits. For the Z-adic odometer, it is not as easy to describe. First, the digits do not follow the same law. 
Indeed, for all $k\ge 2$, we have $\PP(x_k=1)=\frac{F_{k-1}}{\varphi^{k}}$, which depends on $k$.
(This provides the classical asymptotic formula on the frequency of $1$'s in the Zeckendorf expansion: $\PP(x_k=1)\tend{k}{+\infty}{}{\frac{1}{\varphi^2+1}}$; see~\cite{CGL,MG} for other proofs.) Therefore, if $\sigma$ is the shift on $\XX$, the law of $x$ is not the same as the law of $\sigma^k(x)$ for $k\ge 1$, which means $\PP$ is not stationary. Furthermore, the choice of a digit is not independent of the other digits because a $1$ must be followed by a $0$. However, the lack of stationarity and
independence of $\PP$ is compensated by the following renewal property.
\begin{proposition}\label{renouvellement}
   Let $C$ be a cylinder, $k\ge 2$, and $(r_k,\ldots,r_2)\in\XX_f$. Then
\begin{enumerate}
    \item $\PP\left(\sigma^{k}x\in C~\lvert~x\in C_{0r_{k}\cdots r_2}\right)=\PP(C)$,
\end{enumerate}
and
\begin{enumerate}[resume]
    \item $\PP\left(\sigma^{k+1}x\in C~\lvert~x\in C_{1r_{k}\cdots r_2}\right)=\PP(C)$ (if $r_{k}=0$).
\end{enumerate}
\end{proposition}
\begin{proof}
Without loss of generality, we can assume that $C$ is a cylinder of order $k_0$ with a $0$ at the left side of its name for some $k_0\ge 1$, so $\PP(C)=\frac{1}{\varphi^{k_0}}$. Then
\begin{align*}
    \PP\left(\sigma^{k+2}x\in C~\lvert~x\in C_{0r_{k}\cdots r_2}\right) &=\frac{\PP\left(x\in C_{0r_{k}\cdots r_2}\cap\sigma^{k+2}x\in C\right)}{\PP(x\in C_{0r_{k}\cdots r_2})}.
\end{align*}
We observe that the set $\{x\in C_{0r_{k}\cdots r_2}\cap\sigma^{k+2}x\in C\}$ is actually a cylinder of order $k+k_0$ with a leftmost $0$ in its name. Its measure is therefore $\frac{1}{\varphi^{k+k_0}}$.
\begin{align*}
    \PP\left(\sigma^{k+2}x\in C~\lvert~x\in C_{0r_{k}\cdots r_2}\right) &=\frac{\varphi^{k}}{\varphi^{k+k_0}}=\frac{1}{\varphi^{k_0}}=\PP(C).
\end{align*}
We get the first point of the proposition. Then we observe that, since $C_{1r_k\cdots r_2}=C_{01r_k\cdots r_2}$, the second is a particular case of the first one (with $k+1$ instead of $k$).
\end{proof}
Once we know the value of $x_k$, the conditional law of $x_{k+1}$ depends neither on $k$ or $x_{k-1},\ldots,x_2$. In particular, we get that
\begin{displaymath}
\PP(x_{k+2}=1~\lvert~x_{k+1},\ldots,x_2)=\begin{cases}
    1/\varphi^{2},& \textrm{ if  $x_{k+1}=0$;} \\
    0, & \text{otherwise,}
\end{cases}
\end{displaymath}
and
\begin{displaymath}
\PP(x_{k+2}=0~\lvert~x_{k+1},\ldots,x_2)=\begin{cases}
    1/\varphi,& \textrm{ if  $x_{k+1}=0$;} \\
    1, & \text{otherwise.}
\end{cases}
\end{displaymath}
Therefore, under $\PP$, the digits $x_2$, $x_3$, $\ldots$ form a Markov chain with transition probabilities given on Figure~\ref{Markov chain}, starting with the initial law 
\[\PP(x_2=1)=\frac{1}{\varphi^2} \textrm{ and } \PP(x_2=0)=\frac{1}{\varphi}.\]

\begin{figure}[H]
    \centering
    \begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto] 
        \node[state] (q_0)   {$0$}; 
        \node[state](q_3) [right=of q_0] {$1$};
        \path[->] 
        (q_0) edge [bend left] node [] {$\frac{1}{\varphi^2}$} (q_3)
        (q_3) edge [bend left] node [] {1} (q_0)
        (q_0) edge [loop above] node [] {$\frac{1}{\varphi}$} (q_0);
    \end{tikzpicture}
    \caption{Transition probabilities of the Markov chain.}
    \label{Markov chain}
\end{figure}
    
    \subsection{Reminders on some notions of mixing coefficients}\label{RNMC}
    
In Definition~\ref{coeff alpha melange Zeck}, we introduced the notion of $\alpha$-mixing coefficients. It happens that the distribution of the digits in a $Z$-adic integer satisfies a good inequality for another (better) notion of mixing coefficients: the $\phi$-mixing coefficients. They are defined as follows.
\begin{definition}\label{coeff phi melange Zeck}
   Let $(X_j)_{j\ge 1}$ be a (finite or infinite) sequence of random variables. The associated \emph{$\phi$-mixing coefficients} $\phi(k)$, $k\ge 1$, are defined by
   \[\phi(k):=\sup_{p\ge 1} \hspace{1mm} \sup_{A,B} \hspace{2mm} \lvert \PP_A(B)-\PP(B) \rvert,\]
   where the second supremum is taken over all events $A$ and $B$ such that 
\begin{itemize}
    \item $A\in\sigma(X_j:1\le  j\le p)$,
    \item $\PP(A)>0$ and,
    \item $B\in\sigma(X_j:  j\ge k+p)$.
\end{itemize}
By convention, if $X_j$ is not defined when $j\ge k+p$, then the $\sigma$-algebra is trivial.
\end{definition}
In the case of a finite sequence $(X_1,\ldots,X_n)$, the convention implies that $\phi(k)=0$ for $k\ge n$.

Both $\alpha$ and $\phi$-mixing coefficients are linked together and, as written above, $\phi$-mixing coefficients are ``better'' than $\alpha$-mixing coefficients. Indeed, we have the following property from Bradley \cite{RCB1}:
\begin{proposition}[Bradley]\label{Lien coeff phi et alpha Zeck}
For every $k\ge 1$
\[\alpha(k)\le \frac{1}{2}\phi(k).\]
\end{proposition}

    \subsection{\texorpdfstring{$\phi$}{0}-mixing property for Z-adic digits}\label{MPZD}

As mentioned in the previous subsection, there exists a good upper bound on the $\phi$-mixing coefficients for the coordinates of $x\in\XX$.
\begin{proposition}\label{Melange_coord}
For $x=(x_j)_{j\ge 2}\in\XX$ randomly chosen with law $\PP$ and $k\ge 1$ we have 
\[\phi(k)\le \frac{2}{\varphi^{2k}}.\]
\end{proposition}
Our way to prove this result needs to introduce a parametrization of the Markov chain. So, we define the function on $\{0,1\}\times [0,1]$ as follows:
\begin{displaymath}
\psi(\varepsilon,t) := \begin{cases}
1, & \text{if $\varepsilon=0$ and $t<\frac{1}{\varphi^2}$;}\\
0, & \text{if $\varepsilon=0$ and $t\ge\frac{1}{\varphi^2}$;}\\
0, & \text{otherwise.}
\end{cases}
\end{displaymath}

Now let $(U_j)_{j\ge 2}$ be independent and identically distributed random variables following a uniform law on the real unit interval and define the sequence $(y_j)_{j\ge 1}$ as follows:
\begin{equation}\label{genere_coord}
y_j := \begin{cases}
0, & \text{if $j=1$;}\\
\psi(y_{j-1},U_j), & \text{otherwise.}\\
\end{cases}
\end{equation}
\begin{lemma}\label{Automate=Loi}
    The law of $(y_j)_{j\ge 2}$ is $\PP$.
\end{lemma}
\begin{proof}[Proof of Lemma~\ref{Automate=Loi}]
The transition matrix is the same as for $\PP$ and so is the initial law: the probability that $y_2$ is $0$ is the same as the event $U_2\ge \frac{1}{\varphi^2}$ that is $\frac{1}{\varphi}=\PP(C_0)$.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{Melange_coord}]
Let $k\ge 1$ and $p\ge 2$. From Lemma~\ref{Automate=Loi}, choosing $x$ with law $\PP$ is equivalent to constructing a sequence $y=(y_j)_{j\ge 2}$ using the process $\eqref{genere_coord}$. Thus, we have a sequence of independent and identically distributed random variables $(U_j)_{j\ge 2}$ with a uniform law on the real unit interval. We consider the event 
\[C:=\left\{\exists j\in]p,p+k]: U_j\ge \frac{1}{\varphi^2}\right\},\]
the probability of which is
\[\PP(C)=1-\frac{1}{\varphi^{2k}}>0.\]
If $y\in C$ then this implies that $y_j=0$ because of $\eqref{genere_coord}$. Then we take $A\in\sigma(x_j:2\le  j\le p)$ and $B\in\sigma(x_j:  j\ge k+p)$. We observe that, due to \eqref{genere_coord}, $A\in\sigma(U_j: 2\le j \le p)$ while $C\in\sigma(U_j: p<j\le k+p)$ thus $A$ and $C$ are independent. Also, due to Proposition~\ref{renouvellement}, we observe $B\cap C\in\sigma(U_j:j>p)$. So $A$ and $B\cap C$ are independent. 
We also have
\begin{equation*}
        \lvert\PP_A(B)-\PP(B)\rvert  \le\lvert\PP_A(B)-\PP_{A\cap C}(B)\rvert +\lvert\PP_{A\cap C}(B)-\PP_C(B)\rvert
          + \lvert\PP_C(B)-\PP(B)\vert.        
\end{equation*}
But the independence between $A$ and $C$ and $A$ and $B\cap C$ gives that 
\begin{equation}\label{ineg triang Zeck}
    \begin{split}
        \lvert\PP_A(B)-\PP(B)\rvert & \le\lvert\PP_A(B)-\PP_{A\cap C}(B)\rvert + \lvert\PP_C(B)-\PP(B)\vert.         
    \end{split}
\end{equation}
As shown in the case of an integer base \cite[Lemma 4.3]{TREJYH}, if we let $\overline{C}$ denote the complement of $C$, then for every event $D$ we have the general inequality 
\begin{align}\label{ineg generale conditionnement Zeck}
    \lvert\PP_C(D)-\PP(D)\rvert &\le \PP(\overline{C}),
\end{align}
which implies in \eqref{ineg triang Zeck} that
\[\lvert\PP_A(B)-\PP(B)\rvert\le 2\PP(\overline{C}).\]
\end{proof}
Another property about the measure is that if $\PP(x_k=1)$ indeed 
depends on $k$, then it is actually bounded between two positive values. We generalize this fact with the next proposition.
\begin{proposition}\label{borne proba Zeck}
Let $(k_0,\ldots,k_{\ell})$ be a collection of integers such that 
\[2\le k_0<k_1<\cdots<k_{\ell}.\]
Let $A$ be a union of cylinders of order $k_0-1$ such that $x_{k_0}=0$ if $x\in A$ and such that $\PP(A)>0$. Then
\[\frac{1}{\varphi^{\ell}}\le \PP_A~(\forall 1\le i\le \ell~:~x_{k_i}=0)\le \left(\frac{2}{\varphi^2}\right)^{\ell}.\]
\end{proposition}
\begin{proof}
Let $I\subset \NN$. Since there is a finite number of cylinders of order $k_0-1$, we can write $A$ as a disjoint union $\sqcup_{i\in I} C^{(i)}$ where $C^{(i)}$ is a cylinder of order $k_0-1$. We have the identity
\begin{align}\label{etap prop borne proba}
    \PP_A~(\forall 1\le i\le \ell~:~x_{k_i}=0)&=\prod_{i=1}^{\ell}\PP~(x_{k_i}=0~\lvert~A\cap \bigcap_{j=1}^{i-1}(x_{k_j}=0)),
\end{align}
with the convention $\cap_{j=1}^0 (x_{k_j}=0)=\XX$.
But we have 
\begin{align*}
    \PP_A(x_{k_1}=0)&=\frac{\PP\left(\left(x_{k_1}=0\right)\cap \left(x\in\sqcup_{i\in I} C^{(i)}\right)\right)}{\PP(A)}\\
        &=\frac{1}{\PP(A)}\sum_{i\in I} \PP\left((x_{k_1}=0)\cap x\in C^{(i)}\right).
\end{align*}
We recall that $\PP(A)=\frac{\lvert I\rvert}{\varphi^{k_0-1}}$, and  that the summand is the probability of a disjoint union of $F_{k_1-k_0+1}$ cylinders of order $k_1-1$ such that $x_{k_1}=0$, so its measure is $\frac{F_{k_1-k_0+1}}{\varphi^{k_1-1}}$. Thus, we obtain
\[\PP_A(x_{k_1}=0)=\PP(x_{k_1-k_0+1}=0).\]
We proceed similarly for the other terms in \eqref{etap prop borne proba} and get
\begin{align}
    \PP_A~(\forall 1\le i\le \ell~:~x_{k_i}=0)&=\prod_{i=1}^{\ell}\PP~(x_{k_i-k_{i-1}+1}=0).
\end{align}
The last equality is given using the renewal of $\PP$. Now we claim that for all $k\ge 2$ we have
\[\frac{1}{\varphi}\le \PP(x_k=0)\le \frac{2}{\varphi^2}.\]
Indeed, we recall that $\PP(x_k=0)=\frac{F_k}{\varphi^{k-1}}$, and we observe that the subsequences $\left(\frac{F_{2k}}{\varphi^{2k-1}}\right)$ and $\left(\frac{F_{2k+1}}{\varphi^{2k}}\right)$ are adjacent sequences.
\end{proof}

    \subsection{Ergodic convergence}

For $x$ in $\XX$, we define the sequence of empirical probability measures along the (beginning of the) orbit of $x$: for every $N\ge 1$, we set
\[\epsilon_N(x):=\frac{1}{N}\sum_{0\le n<N}\delta_{T^nx}\],
where $\delta_y$ is the Dirac measure on $y\in\XX$.

Since the space of probability measures on $\XX$ is compact for the weak-$*$ topology, we can always extract a convergent subsequence. Moreover, every subsequential limit of $(\epsilon_N(x))$ is a $T$-invariant probability measure. By the uniqueness of the $T$-invariant probability measure, for every $x\in \XX$ we have $\epsilon_N(x)\rightarrow \PP$. In other words, we have the convergence 
\begin{equation}\label{uniq_ergo+birk1 Zeck}
        \forall x\in \XX, \hspace{2mm} \forall f\in\mathcal{C}(\XX), \hspace{5mm} \frac{1}{N}\sum_{0\le n< N}f(T^nx) \tend{N}{+\infty}{}{\int_{\XX} f \mathrm{d}\PP}.
\end{equation}
Here, we are interested in the special case $x=0$,
because $\NN=\{T^n0:n\in\NN\}$. Then \eqref{uniq_ergo+birk1 Zeck} becomes
\begin{equation}\label{uniq_ergo+birk Zeck} 
        \forall f\in\mathcal{C}(\XX), \hspace{5mm} \frac{1}{N}\sum_{0\le n< N}f(n) \tend{N}{+\infty}{}{\int_{\XX} f \mathrm{d}\PP}.
\end{equation}
Equation \eqref{uniq_ergo+birk Zeck} shows that, for a continuous function $f$, averaging $f$ over $\NN$ (for the natural density) amounts to averaging over $\XX$ (for $\PP$). The next section shows how this convergence can be extended to some non-continuous functions related to the sum-of-digits function.

\section{Sum of digits on the odometer}\label{SOzeck}

In this section, we adapt the techniques used for the case of an integer base \cite{TREJYH}. The main difference is the need of the new Lemma~\ref{On Delta1 Zeck}. For every integer $k\ge 2$,  we define the continuous map $s_k:\XX\rightarrow\ZZ$ as the sum of the digits of indices $\le k$, that is to say 
\begin{align*}
    s_k(x):=x_k+\cdots +x_2.
\end{align*}
Let $r\in\NN$. We define the functions $\Delta^{(r)}_k:\XX\rightarrow\ZZ$ by 
\begin{align*}
    \Delta_k^{(r)}(x):=s_k(x+r)-s_k(x).
\end{align*}
The functions $\Delta_k^{(r)}$ are well-defined, continuous (and bounded) on $\XX$. By \eqref{uniq_ergo+birk Zeck}, we have 
\begin{align}\label{birkhoff_result_delta_k Zeck}
         \frac{1}{N}\sum_{n< N} \Delta_k^{(r)}(n)&=\frac{1}{N}\sum_{n< N} \Delta_k^{(r)}(T^n0)\tend{N}{+\infty}{}{\int_{\XX} \Delta_k^{(r)} \mathrm{d}\PP}.
\end{align}
Although the sum-of-digits function $s$ is not well defined on $\XX$, we can extend the function $\Delta^{(r)}$ defined by \eqref{def delta zeck} on the set of $x\in\XX$ for which the number of different digits between $x$ and $x+r$ is finite. This subset contains the $Z$-adic integers $x$ such that there exists an index $k\ge 2+\max(\{\ell: r_{\ell}\neq 0\})$ such that $x_{k}=x_{k+1}=0$ (see Lemma \ref{blocage_avancee}). So, except for a finite number of $Z$-adic integers, we can define 
\[\Delta^{(r)}(x):=\lim\limits_{k\to\infty}\Delta^{(r)}_k(x).\]
\begin{remark}
Let $t,u$ be two integers. For every integer $k$ we have the decomposition formula 
\begin{align}\label{decomp_sumk Zeck}
    \Delta_k^{(t+u)}&=\Delta_k^{(t)}+\Delta_k^{(u)}\circ T^t.
\end{align}
So, taking $\PP$-almost everywhere the limit as $k\to\infty$, we get
\begin{align}\label{decomp_sum Zeck}
    \Delta^{(t+u)}&=\Delta^{(t)}+\Delta^{(u)}\circ T^t \hspace{2mm}  (\PP\textrm{-a-s.}).
\end{align}
Then by induction on $t$, we deduce
\begin{align}\label{decomp_gene Zeck}
    \Delta^{(t)}&=\Delta^{(1)}+\Delta^{(1)}\circ T+\cdots+\Delta^{(1)}\circ T^{t-1} \hspace{2mm}  (\PP\textrm{-a-s.}).
\end{align}
\end{remark}
The function
$\Delta^{(r)}$ is not bounded on $\XX$, and therefore it is not continuous. So
Eq.~\eqref{uniq_ergo+birk Zeck} is not applicable for $f\circ \Delta^{(r)}$ with $f$ a continuous map on $\XX$. However, it is possible to get the same convergence as in \eqref{uniq_ergo+birk Zeck} with weaker assumptions on $f$ than continuity.
\begin{proposition}\label{prop_moment Zeck}
Let $r\ge 1$ and $f:\ZZ\rightarrow\CC$. Assume that there exist  $\alpha\ge 1$ and $C$ in $\RR^*_+$  such that for every $n\in\ZZ$  we have
\begin{align}\label{hyp_prop_moment2 Zeck}
    \lvert f(n) \rvert &\le C\lvert n \rvert^{\alpha} +\lvert f(0)\rvert.
\end{align}
Then $f\circ \Delta^{(r)}\in L^1\left(\PP\right)$ and we have the convergence
\begin{align*}
    \lim\limits_{N\to\infty}\frac{1}{N}\sum_{n<N}f(\Delta^{(r)}(n))&=\int_{\XX} f(\Delta^{(r)}(x)) \mathrm{d}\PP(x)\\
            &= \lim\limits_{k\to\infty} \int_{\XX} f(\Delta^{(r)}_k(x)) \mathrm{d}\PP(x).
\end{align*}
\end{proposition}

\begin{corollary}\label{corollaire_prop_moment Zeck}
 For every $d\in\ZZ$ we have
\begin{align}\label{lien_mu^r_et_mu Zeck}
        \mu^{(r)}(d)&:=\lim\limits_{N\to\infty}\frac{1}{N} \left|\left\{n<N: \Delta^{(r)}(n)=d\right\}\right| \notag \\
        &=\PP\left(\left\{x\in \XX: \Delta^{(r)}(x)=d\right\}\right).
\end{align}
Moreover, $\Delta^{(r)}$ has zero-mean and has finite moments. 
\end{corollary}
The proof of this corollary is exactly the same as for the
case of an integer base \cite{TREJYH}. In particular, we just prove the existence of the asymptotic densities of the sets $\{n\in\NN: \Delta^{(r)}(n)=d\}$, where $d\in\ZZ$. In the case of an integer base, this is easy to prove (see~\cite{JB,TREJYH}). Here, it is harder to follow B\'esineau's proof because it uses the arithmetic properties of the sum-of-digits  (in an integer base) function that we do not have any longer. 
\begin{remark}\label{et pour 0 ? Zeck}
    Using trivial arguments, Proposition~\ref{prop_moment Zeck} and Corollary~\ref{corollaire_prop_moment Zeck} are also true when $r=0$. We observe that $\mu^{(0)}=\delta_0$.
\end{remark}
Before proving this proposition and its corollary, we need the following lemma that looks like Lemma~1.29 from Spiegelhofer \cite{LS}.
\begin{lemma}\label{lemme_technique Zeck}
    Let $r\ge 1$. For $N\ge 1$, $k\ge 2$ and $d,d'\in\ZZ$, we have the inequality
    \begin{equation}\label{ineg_techni2 Zeck}
        \begin{split}
             \frac{1}{N}\left| \{n<N: (\Delta^{(r)}(n),\Delta^{(r)}_k(n))=(d,d')\} \right| & \\
             & \hspace{-3 cm}\le r\varphi^3  \hspace{1mm}\PP \hspace{-1mm} \left( \{x\in \XX: (\Delta^{(r)}(x),\Delta^{(r)}_k(x))=(d,d')\}\right).
        \end{split}
    \end{equation}
    In particular, we have
    \begin{align}\label{ineg_techni Zeck}
        \frac{1}{N}\left| \{n<N: \Delta^{(r)}(n)=d\} \right| \le r\varphi^3 \hspace{1mm} \PP \hspace{-1mm} \left( \{x\in \XX: \Delta^{(r)}(x)=d\}\right).
    \end{align}
\end{lemma}
\begin{proof}[Proof of Lemma~\ref{lemme_technique Zeck}]
    We adapt the proof of \cite[Lemma~2.3]{TREJYH}
    for the case of an integer base. First, \eqref{ineg_techni2 Zeck} implies \eqref{ineg_techni Zeck} so we just prove \eqref{ineg_techni2 Zeck}. We fix $k\ge 2$. For $\ell\ge 1$, let $V_{\ell}$ be the set of the values reached by the pair $(\Delta^{(r)},\Delta^{(r)}_k)$ on the $F_{\ell+1}-r$ (resp.\ $F_{\ell}-r$) first levels of the large (resp.\ small) tower of order $\ell$. Of course, if $F_{\ell+1}-r\le 0$ then $V_{\ell}:=\emptyset$. In particular, for every $r\ge 1$, $V_1=\emptyset$. Also, if $F_{\ell}\le r<F_{\ell+1}$ then $V_{\ell}$ is defined considering only the large tower of order $\ell$. For every $\ell$, we observe that $V_{\ell}$ is a finite set. On each level that is not in the $r$ top levels of the large or small tower, the first $\ell$ digits of both $x$ and $x+r$ are constant, and digits of higher order are the same. Therefore, $\Delta^{(r)}$ and $\Delta^{(r)}_k$ are constant on such a level. We observe that the sequence $(V_{\ell})_{\ell\ge 1}$ is increasing for the inclusion.
    
    Now for $d,d'\in\ZZ$, there are two cases.
    \begin{enumerate}
        \item If $(d,d')\notin\cup_{\ell\ge 2} V_{\ell}$, then for each $n\in\NN$ we have $(\Delta^{(r)}(n),\Delta^{(r)}_k(n))\neq(d,d')$. Indeed, for each $n\in\NN$, there exists a smallest integer $\ell\ge 2$ such that $n$ is in the first $F_{\ell+1}-r$ levels of the large tower of order $\ell$, hence $(\Delta^{(r)}(n),\Delta^{(r)}_k(n)) \in V_{\ell}$. In this case, \eqref{ineg_techni2 Zeck} is trivial.
        \item If $(d,d')\in\cup_{\ell\ge 1} V_{\ell}$ then there exists a unique $\ell\ge 2$ such that $(d,d')\in V_{\ell}\backslash V_{\ell-1}$. We observe that, due to the  cutting-and-stacking process, the value $(d,d')$ must appear firstly in the large tower of order $\ell$. Since $(\Delta^{(r)},\Delta^{(r)}_k)$ is constant on each of the first $F_{\ell+1}-r$ levels of the large tower, it takes the value $(d,d')$ on at least one whole such level, which is of measure $\frac{1}{\varphi^{\ell}}$. So, we have  
        \begin{align}\label{etap1 lemme tech zeck}
            \PP\hspace{-1mm} \left( \{x\in \XX: (\Delta^{(r)}(x),\Delta^{(r)}_k(x))=(d,d')\}\right)\ge \frac{1}{\varphi^{\ell}}.
        \end{align}
    Also, since the pair $(d,d')\notin V_{\ell-1}$, we claim that for every $N\ge 1$
    \[\frac{1}{N}\left| \{n<N: (\Delta^{(r)}(n),\Delta^{(r)}_k(n))=(d,d')\} \right|\le \frac{r}{F_{\ell-1}}.\]
    Indeed, 
    \begin{enumerate}
        \item If $r\ge F_{\ell-1}$ then the inequality is trivial.
        \item If $r< F_{\ell-1}$ then since $(d,d')$ is not in $V_{\ell-1}$,
	the pair $(d,d')$ can only appear inside a part of the  $r$ highest levels of the big or small towers of order $\ell-1$. Let $C$ denote the union of these $r$ highest levels. Since $0$ lies in the bottom level of the large tower of order $\ell-1$, the set $S$ of integers $n\ge 0$ such that $T^n0\in C$ has the following properties:
        \begin{itemize}
            \item $\{0,\ldots,F_{\ell}-r-1\}\cap S=\emptyset$ and
            \item $S$ is the union of subsets formed by $r$ consecutive integers separated by gaps of length $F_{\ell}-r$ or $F_{\ell-1}-r$.
        \end{itemize}
        So, we have
        \begin{align*}
            \frac{1}{N}\left| \{0\le n<N: (\Delta^{(r)}(n),\Delta^{(r)}_k(n))=(d,d')\} \right|&\\
            &\hspace{-1.38cm}\le \frac{1}{N}\left| \{0\le n<N: T^n0\in C \}\right|\\
            &\hspace{-1.38cm}\le\frac{r}{F_{\ell-1}}.
        \end{align*}
    \end{enumerate}
    Since $F_{\ell-1}\ge \varphi^{\ell-3}$ (by double induction), we get
    \begin{align}\label{etap2 lemme tech zeck}
        \frac{1}{N}\left| \{n<N: (\Delta^{(r)}(n),\Delta^{(r)}_k(n))=(d,d')\} \right|\le \frac{r}{\varphi^{\ell-3}}.
    \end{align}
    Combining inequalities \eqref{etap1 lemme tech zeck} and \eqref{etap2 lemme tech zeck} gives \eqref{ineg_techni2 Zeck}.
    \end{enumerate}
    \begin{figure}[H]
        \centering
        \includegraphics[scale=0.33]{V_lambda_dans_tour.pdf}
        \caption{Visual description of $V_{\ell-1}$, $V_{\ell}$ and $V_{\ell}\backslash{V_{\ell-1}}$.}
        \label{Vl Zeck}
    \end{figure}
\end{proof}
As shown in \eqref{decomp_gene Zeck}, the understanding of $\Delta^{(1)}$ (and so $\Delta_k^{(1)}$ for $k\ge 2$) is fundamental to understand $\Delta^{(r)}$,
so we also state the following lemma.
\begin{lemma}\label{On Delta1 Zeck}
    The function $\Delta^{(1)}$ is well-defined for $x\in\XX$ if and only if $x\in C_{00(10)^{d}}\cup C_{001(01)^d}$ for some $d\ge 0$. 
    
    Furthermore, if $x\in C_{00(10)^{d}}$ then $\Delta^{(1)}(x)=1-d$ and 
    \begin{displaymath}
     \Delta^{(1)}_k(x) = \begin{cases}
        \frac{2-k}{2}, & \text{if $k \equiv \modd{0} {2}$ and $k\le2d$;}\\
        \frac{1-k}{2}, & \text{if $k \equiv \modd{1} {2}$ and $k\le2d+1$;}\\
        1-d, & \text{if $k\ge 2d+2$.}
        \end{cases}
    \end{displaymath}    
    Also, if $x\in C_{001(01)^{d}}$ then $\Delta^{(1)}(x)=-d$ and 
    \begin{displaymath}
     \Delta^{(1)}_k(x) = \begin{cases}
        \frac{-k}{2}, & \text{if $k \equiv \modd{0} {2}$ and $k\le2d+2$;}\\
        \frac{1-k}{2}, & \text{if $k \equiv \modd{1} {2}$ and $k\le2d+1$;}\\
        \frac{3-k}{2}, & \text{if $k=2d+3$;}\\
        -d, & \text{if $k\ge 2d+4$.}
        \end{cases}
    \end{displaymath}
\end{lemma}
\begin{proof}[Proof of Lemma \ref{On Delta1 Zeck}]
    If $x\in C_{00(10)^{d}}$ for some $d\ge 0$, the definition of $T$ gives that $x+1\in C_{01(00)^{d}}$ with the other digits unchanged. Thus the sequence $(\Delta^{(1)}_k(x))$ is stationary, $\Delta^{(1)}(x)$ is well defined, and $\Delta^{(1)}(x)=1-d$. Also, if $x\in C_{001(01)^{d}}$ for some $d\ge 0$, then $x+1\in C_{010(00)^{d}}$, $\Delta^{(1)}(x)$ is well defined and $\Delta^{(1)}(x)=-d$. If $x$ does not belong to those cylinders then $x=(01)^{\infty}$ or  $x=(10)^{\infty}$. In that case, $x+1=0$ and the sequence $(\Delta^{(1)}_k(x))$ diverges to $-\infty$. This proves equivalence. The values of $\Delta_k^{(1)}(x)$ can be easily found by looking at the addition we write in Subsection \ref{HTAI}.
\end{proof}
\begin{proof}[Proof of Proposition \ref{prop_moment Zeck}]
We adapt the proof of \cite[Prop.~2.1]{TREJYH} for the case of
an integer base. Let $\varepsilon>0$. For every integer $k\ge 2$, we have
\begin{align*}
    \left\vert\frac{1}{N}\sum_{n<N} f(\Delta^{(r)}(n))-\int_{\XX} f(\Delta^{(r)}(x)) \mathrm{d}\PP(x)\right\vert 
        &\le A_1+A_2+A_3,
\end{align*}
where
\begin{align*}
    A_1 &:= \displaystyle\frac{1}{N}\sum_{n<N} \left\vert f(\Delta^{(r)}(n))-f(\Delta^{(r)}_k(n))\right\vert, \\ 
    A_2 &:= \displaystyle\left\vert\frac{1}{N}\sum_{n<N}f(\Delta^{(r)}_k(n))-\int_{\XX} f(\Delta^{(r)}_k(x)) \mathrm{d}\PP(x)\right\vert, \\
    A_3 &:=\displaystyle\int_{\XX} \left\vert f(\Delta^{(r)}(x))-f(\Delta^{(r)}_k(x))\right\vert \mathrm{d}\PP(x).
\end{align*}
However, we have
\begin{align*}
    A_1&=\frac{1}{N}\sum_{n<N} \sum_{j,j'\in\ZZ}\Big\vert f(j)-f(j')\Big\vert\mathbbm{1}_{(j,j')}\left(\Delta^{(r)}(n),\Delta^{(r)}_k(n)\right)\\
    &= \sum_{j,j'\in\ZZ}\Big\vert f(j)-f(j')\Big\vert\frac{1}{N}\sum_{n<N}\mathbbm{1}_{(j,j')}\left(\Delta^{(r)}(n),\Delta^{(r)}_k(n)\right)\\
    &\le r \varphi^3\sum_{j,j'\in\ZZ}\Big\vert f(j)-f(j')\Big\vert \hspace{1mm}\PP \hspace{-1mm} \left( \Delta^{(r)}(n) = j , \Delta^{(r)}_k(n)=j'\right)\\
    &=r \varphi^3\int_{\XX} \Big\vert f(\Delta^{(r)}(x))-f(\Delta^{(r)}_k(x))\Big\vert\mathrm{d}\PP(x)=r \varphi^3 A_3.
\end{align*}
Then
\begin{align*}
    \left\vert\frac{1}{N}\sum_{n<N} f(\Delta^{(r)}(n))-\int_{\XX} f(\Delta^{(r)}(x)) \mathrm{d}\PP(x)\right\vert 
        &\le A_2 + (1+r \varphi^3)A_3.
\end{align*}
We want to apply the dominated convergence theorem to deal with $A_3$ (observe that we have the convergence $f(\Delta^{(r)}_k(x))\tend{k}{\infty}{}{f(\Delta^{(r)}(x))}$ $\PP$-almost-surely). For this, we need to find a good dominant function. We define $g_i:=\sup_{k\ge 2}|\Delta^{(1)}_k\circ T^i(x)|$ for $i=0,\ldots,r-1$ and by \eqref{hyp_prop_moment2 Zeck} we get the inequalities
\[\left\lvert f\circ\Delta^{(r)}_k(x)\right\rvert\le C \sum_{j_0+\cdots+j_{r-1}=\alpha}\sum_{i=0}^{r-1}\frac{\binom{\alpha}{j_0,\ldots,j_{r-1}}}{r}g_i(x)^{rj_i}+\Big\lvert f(0)\Big\rvert\]
and
\[\left\lvert f\circ\Delta^{(r)}(x)\right\rvert\le C \sum_{j_0+\cdots+j_{r-1}=\alpha}\sum_{i=0}^{r-1}\frac{\binom{\alpha}{j_0,\ldots,j_{r-1}}}{r}g_i(x)^{rj_i}+\Big\lvert f(0)\Big\rvert.\]
We need to prove that $g_i^{rj_i}$ is integrable for the measure $\PP$. It is equivalent to show that $\sum_m\PP(\{x\in\XX: g_i^{rj_i}(x)>m\})$ is a convergent series. We have 
\begin{align*}
    g_i(x)> m^{\frac{1}{rj_i}} &\Leftrightarrow \sup_{k\ge 2}|\Delta^{(1)}_k\circ T^i(x)|> m^{\frac{1}{rj_i}}\\&\Leftrightarrow \exists k\ge 2, \hspace{2mm} \lvert\Delta_k^{(1)}(T^ix) \rvert > m^{\frac{1}{rj_i}}. \\
\end{align*}
From Lemma~\ref{On Delta1 Zeck}
\begin{align*}
    \exists k\ge 2, \hspace{2mm} \lvert\Delta_k^{(1)}(T^ix) \rvert > m^{\frac{1}{rj_i}} &\Leftrightarrow T^ix\in C_{00(10)^{\lfloor m^{\frac{1}{rj_i}}\rfloor}}\cup C_{0(01)^{\lfloor m^{\frac{1}{rj_i}}\rfloor}}.
\end{align*}
It follows, by $T$-invariance, that
\[\PP(\{x\in\XX: g_i^{rj_i}(x)>m\})\le \left(\frac{1}{\varphi}\right)^{2+2\lfloor m^{\frac{1}{rj_i}}\rfloor}+\left(\frac{1}{\varphi}\right)^{1+2\lfloor m^{\frac{1}{rj_i}}\rfloor}=\left(\frac{1}{\varphi}\right)^{2\lfloor m^{\frac{1}{rj_i}}\rfloor}.\]
The quantity on the RHS is the general term of a convergent series, which shows that $g_i^{rj_i}$ is integrable for the measure $\PP$. The dominated convergence theorem can be applied and, for $k$ large enough, we have $(1+r \varphi^3)A_3\le \frac{\varepsilon}{2}$ for every $N\ge 1$. Now once we have fixed such a $k$, for $N$ large enough, $A_2$ is bounded by $\frac{\varepsilon}{2}$ because of \eqref{uniq_ergo+birk Zeck} and the continuity of $\Delta_k^{(r)}$ and $f$. The convergence in the statement is thus proved.
 
 Note that the argument of the dominated convergence theorem also proves that $f\circ\Delta^{(r)}\in L^1(\PP)$ and $\int_{\XX} f\circ \Delta^{(r)}\mathrm{d}\PP=\lim\limits_{k\rightarrow \infty}\int_{\XX} f\circ\Delta^{(r)}_k\mathrm{d}\PP$.
\end{proof}

More generally, we have the following convergence.
\begin{proposition}\label{prop_gene Zeck}
Let $r\ge 1$ and $f:\ZZ\rightarrow\CC$ be such that $f\circ\Delta^{(r)}\in L^1(\PP)$. Then 
    \[\lim\limits_{N\to\infty}\frac{1}{N}\sum_{n<N}f(\Delta^{(r)}(n))=\int_{\XX} f(\Delta^{(r)}(x)) \mathrm{d}\PP(x).\]
\end{proposition}
The proof is exactly the same as Proposition 2.4 in \cite{TREJYH}. We have shown that the random variable $\Delta^{(r)}$ satisfies some good properties such as the finiteness of moments of every order of $\mu^{(r)}$, the law of $\Delta^{(r)}$. The next section focuses on another natural question: how to compute the law $\mu^{(r)}$?

\section{How to compute \texorpdfstring{$\mu^{(r)}$}{yr}}\label{HTCM}

In this section, we present an algorithm that exactly computes the measure $\mu^{(r)}$ for $r$ in $\NN$ and its consequences.

    \subsection{Description of the algorithm}\label{DOTA}
    
This algorithm is completely different from the classical way 
to compute $\mu^{(r)}$ in an integer base, as
developed by B\'esineau \cite[p.\ 14]{JB}.
In an integer base $b\ge 2$, the computation of $\mu^{(r)}$ relies on inductive relations for the expansion of $r$, which are easy to prove in that case. For instance, one of the relation states that if $r$ is a multiple of $b$ then the variation when adding $r$ to $x$ is the same as the variation when adding $\frac{r}{b}$ to $\sigma x$. So the laws of the variation when adding $r$ or $\frac{r}{b}$ are the same. In the Zeckendorf representation, this trivial argument is false because the carries can modify the digits on both sides (unlike the case of an
integer base, where only the left side can be affected). However, we find an algorithm to compute $\mu^{(r)}$, in this Zeckendorf system that relies on Rokhlin towers. This algorithm can be adapted to an integer base.

First, let $r\ge 1$ ($r=0$ is irrelevant) and let $\ell\ge 2$ be the unique integer such that $F_{\ell}\le r< F_{\ell+1}$. At a given order $k\ge 1$, we define $\CST{k}$ as the union of the levels of the large and small towers of order $k$, except the $r$ highest levels. We observe $\CST{1}=\emptyset$. On each level of $\CST{k}$, $\Delta^{(r)}$ is constant (see the proof of Lemma \ref{lemme_technique Zeck}). But, due to the  cutting-and-stacking process, the value of $\Delta^{(r)}$ on these ``constant'' levels of $\CST{k}$ may be deduced from those taken in $\CST{k-1}$. That is why we also introduce, for $k\ge 2$, the \emph{new information zone of the Rokhlin tower of order $k$} as the set 
\[\NIZ{k}:=\CST{k}\backslash{\CST{k-1}}.\]

Of course, everything depends on $r$ but we do not emphasize on that in the notation of $\ell$, $\NIZ{k}$ and $\CST{k}$ for simplicity. Let us define clearly this set $\NIZ{k}$.
\begin{proposition}\label{Def NIZ_k}
    In a large tower, we enumerate, starting by $1$, the levels from the base of the tower to the top of it. We have the following description of $\NIZ{k}$.
    \begin{itemize}
        \item If $2\le k\le \ell-1$, $\NIZ{k}=\emptyset$.
        \item $\NIZ{\ell}$ is the union of the first $F_{\ell+1}-r$ levels of the large tower of order $\ell$.
        \item $\NIZ{\ell+1}$ is the union of the $F_{\ell}$ levels between the $F_{\ell+1}-r+1^{\rm th}$ and the $F_{\ell+2}-r^{\rm th}$ levels of the large tower of order $\ell+1$.
        \item If $k>\ell+1$, $\NIZ{k}$ is the union of the $r$ levels between the $F_{k}-r+1^{\rm th}$ and the $F_k^{\rm th}$ levels of the large tower of order $k$.
    \end{itemize}
\end{proposition}
\begin{remark}
    By definition, for every $k\ge 1$, the set $\NIZ{k}$ is either empty or a disjoint union of cylinders of order $k$ (exactly $F_{\ell+1}-r$ if $k=\ell$, $F_{\ell}$ if $k=\ell+1$ or $r$ if $k\ge\ell+2$). Moreover, we observe that $(\NIZ{k})_{k\ge \ell}$ is a partition of $\XX\backslash{\{(01)^{\infty}, (10)^{\infty}\}}$.
\end{remark}

\begin{example}\label{Example NIZ pour r=4} Figure~\ref{fig7} illustrates
an example with $r=4=\left[101\right]$.
\begin{figure}[H]
    \centering
    \includegraphics[scale=0.34]{NIZ_pour4.pdf}
    \caption{Visualization of $(\NIZ{k})_{k\ge 1}$ when $r=4=\left[101\right]$.}
    \label{fig7}
\end{figure}
Figure~\ref{NIZ general} is a more general figure to aid understanding.
\begin{figure}[H]
    \centering
    \includegraphics[scale=0.35]{NIZ_general.pdf}
    \caption{Visualization of $\NIZ{k}$, $k\ge \ell+2$.}
    \label{NIZ general}
\end{figure}
\end{example}
Through the following lemmas, we state some observations on this sequence $(\NIZ{k})_{k\ge 1}$. We start by a description of the cylinders of order $k$ that appear in $\NIZ{k}$ for $k\ge \ell+2$.
\begin{lemma}\label{Contenu NIZ_k}
    If $k\ge 2$, the set $\NIZ{k}$ is composed of cylinders of order $k$ whose name has a pattern $00$ in the leftmost positions.
\end{lemma}
\begin{proof}
    Let $k\ge \ell+2$, consider a cylinder $C_{n_{k+1}\cdots n_2}$ of order $k$ that belongs in $\NIZ{k}$ and where $(n_{k+1},\ldots,n_2)\in\XX_f$. Since this cylinder is part of the large Rokhlin tower of order $k$, $n_{k+1}=0$. Moreover, this cylinder is not in the $F_{k-1}$ highest cylinders of the same tower,
    so $n_k=0$, too.
\end{proof}
Now we show a relation between cylinders of order $k$ in $\NIZ{k}$ and those of order $k+2$ in $\NIZ{k+2}$ for $k\ge \ell$.
\begin{lemma}\label{relation cylindres NIZ}
    Let $k\ge 2$ and let $(n_{k-1},\ldots,n_2)\in\XX_f$. We have the implication
    \begin{align}
        C_{00n_{k-1}\cdots n_2}\subset \NIZ{k} \implies C_{0010n_{k-1}\cdots n_2}\subset \NIZ{k+2} 
    \end{align}
\end{lemma}
\begin{remark}
If $k=2$, we agree that the word $n_{k-1}\cdots n_2$ is the empty word.
\end{remark}
\begin{proof}
    For $r=1$, we let the reader check that $C_{00}\subset \NIZ{2}$ and $C_{0010}\subset \NIZ{4}$. For $r\ge 2$, we have $\NIZ{2}=\emptyset$. So, we can assume $k\ge 3$. Let $C_{00n_{k-1}\cdots n_2}\subset \NIZ{k}$ and $n:=\sum_{i=2}^{k-1} n_iF_i$. Then $n\in \NIZ{k}\cap [0,F_k[$ where the interval is an integer interval and considered as being part of $\XX$.
    \begin{enumerate}
        \item If $k\ge \ell+2$, we observe that $\NIZ{k}\cap [0,F_k[=[F_k-r,F_k-1[$. It follows that $F_{k+1}+n$, whose Zeckendorf expansion is $\left[10n_{k-1}\cdots n_2\right]$, belongs to $[F_{k+2}-r,F_{k+2}-1]\subset \NIZ{k+2}$. Furthermore $F_{k+1}+n\in C_{0010n_{k-1}\cdots n_2}$, which is a level of the large tower of order $k+2$. So  $C_{0010n_{k-1}\cdots n_2}\subset \NIZ{k+2}$.
        \item For $k=\ell$, we observe that $\NIZ{\ell}\cap [0,F_{\ell}[=[0,F_{\ell+1}-r-1[$. It follows that $n+F_{\ell+1}\in[F_{\ell+1},2F_{\ell+1}-r-1]$. We claim that
        \begin{align}\label{etape relation cylindres NIZ} 
            F_{\ell+2}-r \le F_{\ell+1}\le 2F_{\ell+1}-r-1\le  F_{\ell+2}-1.
        \end{align}
        Indeed, since $F_{\ell}\le r < F_{\ell+1}$;
        \begin{itemize}
            \item  $F_{\ell+2}-r\le F_{\ell+2}-F_{\ell}=F_{\ell+1}$,
            \item  $F_{\ell+1}-r-1\ge 0$ and we deduce $2F_{\ell+1}-r-1\ge F_{\ell+1}$,
            \item $2F_{\ell+1}-r-1\le F_{\ell+1}+F_{\ell-1}-1\le F_{\ell+2}-1$.
        \end{itemize}
        From \eqref{etape relation cylindres NIZ}, we get $n+F_{\ell+1}\in [F_{\ell+2}-r,F_{\ell+2}-1]$. Since $[F_{\ell+2}-r,F_{\ell+2}-1]\subset \NIZ{\ell+2}$, we conclude as in the first point.
        \item If $k=\ell+1$, we get $\NIZ{k}\cap [0,F_{\ell}[=[F_{\ell+1}-r,F_{\ell+2}-r-1]$. It follows that $n+F_{\ell+2}\in[F_{\ell+3}-r,2F_{\ell+2}-r-1]$. Since $r\ge F_{\ell}$, $2F_{\ell+2}-r-1\le F_{\ell+3}-1$, and so $n+F_{\ell+2}\in[F_{\ell+3}-r,F_{\ell+3}-1]\subset \NIZ{\ell+2}$. We conclude as in the first point.
    \end{enumerate}
\end{proof}
\begin{lemma}\label{relation par Delta cylindres NIZ}
    Let $k\ge\ell$ and a collection $(n_{k-1},\ldots, n_2)\in\XX_f$ such that $C_{00n_{k-1}\cdots n_2}$ is a cylinder of $\NIZ{k}$. Let $d\in\ZZ$ such that 
    \[\Delta^{(r)}_{\lvert C_{00n_{k-1}\cdots n_2}}=d,\]
    then
    \[\Delta^{(r)}_{\lvert C_{0010n_{k-1}\cdots n_2}}=d-1.\]
\end{lemma}

These lemmas imply the following corollary on which our algorithm is based.
\begin{corollary}\label{corollaire pour algo}
    Let $k\ge \ell+4$, the values taken by $\Delta^{(r)}$ on the cylinders of order $k+2$ that compose $\NIZ{k+2}$ are exactly the values taken on cylinders of order $k$ in $\NIZ{k}$ shifted by $-1$.
\end{corollary}
\begin{proof}[Proof of Lemma~\ref{relation par Delta cylindres NIZ}] 
Let $n:=\sum_{i=2}^{k-1} n_iF_i$ such that $C_{00n_{k-1}\cdots n_2}\subset\NIZ{k}$. We have $C_{00n_{k-1}\cdots n_2}\subset C_{0n_{k-1}\cdots n_2}$, which is, due to the  cutting-and-stacking process, one of the $r$ top levels of the large tower of order $k-1$. Also, due to the  cutting-and-stacking process, $n+r$ must lies in the small tower of order $k-1$ (see Figure~\ref{main argument}) so $(n+r)_k$ must be $1$.
\begin{figure}[H]
    \centering
    \includegraphics[scale=0.30]{dessin_preuve_cylindre.pdf}
    \caption{Main argument to justify $(n+r)_k=1$.}
    \label{main argument}
\end{figure}
So, we have the following addition 
    \[\begin{array}{ccccccccc}
        (n)   &     & 0 & 0 & n_{k-1}   &\cdots & n_{\ell}      & \cdots    & n_2 \\
        (r)   & +   &   &   &           &       & r_{\ell}      & \cdots    & r_2\\ \hline
        (n+r) & =   & 0 & 1 & 0         &\cdots & (n+r)_{\ell}  & \cdots    &(n+r)_2
    \end{array}\]
    with, by hypothesis, $\Delta^{(r)}(n)=d$.
    Then $n+F_{k+1}+r$ is calculated as follows
    \[\begin{array}{ccccccccccc}
        (n+F_{k+1})     & 0 & 0 & 1 & 0 & n_{k-1}   &\cdots & n_{\ell}      & \cdots    & n_2 \\
        (r)             & + &   &   &   &   &           &       & r_{\ell}      & \cdots    & r_2\\ \hline
        (n+F_{k+1}+r)   & =   & 0 & 0 & 1 & 1 & 0         &\cdots & (n+r)_{\ell}  & \cdots    &(n+r)_2\\
        (n+F_{k+1}+r)   & =   & 0 & 1 & 0 & 0 & 0         &\cdots & (n+r)_{\ell}  & \cdots    &(n+r)_2
    \end{array}\]
    The sum of digits of $n+F_{k+1}+r$ is the same as those of $n+r$ shifted by $-1$ due to the correction of the expansion made. Hence, we get $\Delta^{(r)}(n+F_{k+1})=\Delta^{(r)}(n)-1$, i.e.,
    \[\Delta^{(r)}_{\lvert C_{0010n_{k-1}\cdots n_2}}=\Delta^{(r)}_{\lvert C_{00n_{k-1}\cdots n_2}}-1.\]
\end{proof}
\begin{proof}[Proof of Corollary~\ref{corollaire pour algo}]
Definition~\ref{Def NIZ_k} ensures that the same number of cylinders of order $k+2$ in $\NIZ{k+2}$ is equal to
the number of cylinders of order $k$ in $\NIZ{k}$: there are $r$ cylinders of each order. Lemma~\ref{relation cylindres NIZ} and Lemma~\ref{relation par Delta cylindres NIZ} enable us to conclude.
\end{proof}
These results enable us to give an algorithm that computes the value of $\mu^{(r)}(d)$ for every $d\in\ZZ$. This algorithm has a low complexity (polynomial). Indeed, Corollary~\ref{corollaire pour algo} implies that we can guess the values that appear in the new information zone of order $k$. So we just need to compute $\Delta^{(r)}$ on a finite number of integers to compute exactly $\mu^{(r)}(d)$ ($d\in\ZZ$). We explain this in details further in this section. Before, we give pseudocode and provide an example of computation.
    
    \subsection{Pseudocode}\label{PC}

\begin{algorithm}[H]
\caption{Compute $\mu^{(r)}(d)$}
\begin{algorithmic} 
\REQUIRE $r \geq 1$ and  $d\in\ZZ$
\ENSURE $\mu^{(r)}(d)$
\STATE \textbf{STEP 0}: Find the unique $\ell\ge 1$ such that $F_{\ell}\le r < F_{\ell+1}$
\STATE \textbf{STEP 1}: Passage in $\NIZ{\ell}$
\STATE Define $\App{\ell}\leftarrow 0$ (number of apparition of $d$ during the for-loop)
\FOR{$n=0,\ldots,F_{\ell+1}-r-1$}
    \IF{$\Delta^{(r)}(n)=d$}
        \STATE $\App{\ell}\leftarrow \App{\ell}+1$
    \ENDIF
\ENDFOR
\STATE \textbf{STEP 2}: Passage in $\NIZ{\ell+1}$
\STATE Define $\App{\ell+1}\leftarrow 0$
\FOR{$n=F_{\ell+1}-r\ldots,F_{\ell+2}-r-1$}
    \IF{$\Delta^{(r)}(n)=d$}
        \STATE $\App{\ell+1}\leftarrow \App{\ell+1}+1$
    \ENDIF
\ENDFOR
\STATE \textbf{STEP 3}: Passage in $\NIZ{\ell+2}$
\STATE Define $\App{\ell+2}\leftarrow 0$
\FOR{$n=F_{\ell+2}-r,\ldots, F_{\ell+2}-1$}
    \STATE Store $\Delta^{(r)}(n)$ in an array \textbf{ARR1} (with multiplicity)
    \IF{$\Delta^{(r)}(n)=d$}
        \STATE $\App{\ell+2}\leftarrow \App{\ell+2}+1$
    \ENDIF
\ENDFOR
\end{algorithmic}
\end{algorithm}

\newpage

\begin{algorithm}[H]
\begin{algorithmic}
\STATE \textbf{STEP 4}: Passage in $\NIZ{\ell+3}$
\STATE Define $\App{\ell+3}\leftarrow 0$
\FOR{$n=F_{\ell+3}-r,\ldots, F_{\ell+3}-1$}
    \STATE Store $\Delta^{(r)}(n)$ in an array \textbf{ARR2} (with multiplicity)
    \IF{$\Delta^{(r)}(n)=d$}
        \STATE $\App{\ell+3}\leftarrow \App{\ell+3}+1$
    \ENDIF
\ENDFOR
\STATE \textbf{STEP 5}: Search for $d$ in higher orders
\FOR{$i\in \textbf{ARR1}$}
    \IF{$i>d$}
        \STATE $d$ appears at the order $\ell+2+2(i-d)$ so
        \IF{$\App{\ell+2+2(i-d)}$ is defined}
            \STATE $\App{\ell+2+2(i-d)}\leftarrow \App{\ell+2+2(i-d)}+1$
            \ELSE 
            \STATE $\App{\ell+2+2(i-d)}\leftarrow 1$
        \ENDIF     
    \ENDIF
\ENDFOR
\FOR{$i\in \textbf{ARR2}$}
    \IF{$i>d$}
        \STATE $d$ appears at the order $\ell+3+2(i-d)$ so
        \IF{$\App{\ell+3+2(i-d)}$ is defined}
            \STATE $\App{\ell+3+2(i-d)}\leftarrow \App{\ell+3+2(i-d)}+1$
            \ELSE 
            \STATE $\App{\ell+3+2(i-d)}\leftarrow 1$
        \ENDIF     
    \ENDIF
\ENDFOR
\STATE \textbf{STEP 6}: Compute $\mu^{(r)}(d)$
\STATE $\mu^{(r)}(d)\leftarrow \displaystyle\sum_{k\ge \ell} \dfrac{\App{k}}{\varphi^k}$
\end{algorithmic}
\end{algorithm}

    \subsection{An example: computation of \texorpdfstring{$\mu^{(4)}$}{u4}}

Let us compute the measure $\mu^{(r)}$ for $r=4$. We have $\ell=4$. Looking
at the Rokhlin towers (see Example~\ref{Example NIZ pour r=4}) gives that 
\begin{itemize}
    \item $\NIZ{k}=\emptyset$ if $k=1,2,3$,
    \item $\NIZ{4}=C_{0000}$,
    \item $\NIZ{5}=C_{00001}\sqcup C_{00010} \sqcup C_{00100}$,
    \item $\NIZ{6}=C_{000101} \sqcup C_{001000} \sqcup C_{001001} \sqcup C_{001010}$,
    \item $\NIZ{7}=C_{0010000} \sqcup C_{0010001} \sqcup C_{0010010} \sqcup C_{0010100}$.
\end{itemize}
The next terms of $\NIZ{k}$ are found using Lemma~\ref{relation cylindres NIZ}. Using Corollary~\ref{corollaire pour algo}, one can construct the table
in Figure{Value of Delta r} that gives the values taken by $\Delta^{(r)}$ (with multiplicity) on each cylinder of $\NIZ{k}$.
\begin{figure}[H]
    \centering
    \includegraphics[scale=0.35]{tableau_algo.pdf}
    \caption{Table of the values of $\Delta^{(4)}$ on $\NIZ{k}$.}
    \label{Value of Delta r}
\end{figure}
Thus, it is possible to know how many times some integer $d\in\ZZ$ appear in each column, i.e., at each order. Since we know the measure of a cylinder at each order, then it is possible to deduce the value of $\mu^{(r)}(d)$. For instance
\begin{itemize}
    \item The value $2$ appears once at order $4$, so $\mu^{(4)}(2)=\frac{1}{\varphi^4}$.
    \item The value $1$ appears twice at order $5$, once at order $6$, so $\mu^{(4)}(1)=\frac{2}{\varphi^5} + \frac{1}{\varphi^6}$.
    \item The value $0$ appears once at order $5$ and $8$. It also appears twice at order $6$ and $7$, we deduce that $\mu^{(4)}(0)=\frac{1}{\varphi^5} + \frac{2}{\varphi^6} + \frac{2}{\varphi^7} + \frac{1}{\varphi^8}$.
    \item The value $-1$ appears once at order $6$ and $10$. It appears twice at order $7$, $8$ and $9$. We deduce $\mu^{(4)}(-1)=\frac{1}{\varphi^6} + \frac{2}{\varphi^7} + \frac{2}{\varphi^8} + \frac{2}{\varphi^9} + \frac{1}{\varphi^{10}}$.
    \item Since $-1$ is a value at order $\ell+2=6$ that does not appear in the previous order, we deduce by Corollary~\ref{corollaire pour algo} that the value $-2$ appears as many times as $-1$ appears, but at orders incremented by $2$. In other words, the value $-2$ appears once at order $8$ and $12$. It appears twice at order $9$, $10$ and $11$. Thus $\mu^{(4)}(-2)=\mu^{(4)}(-1)\cdot \frac{1}{\varphi^2}$. We observe that $-2$ does not appear in the lower orders. This observation
    is explained in Proposition~\ref{conseq algo 1} and Corollary~\ref{corollaire pour algo2}.
\end{itemize}
So, after simplification, we get
\begin{displaymath}
    \mu^{(4)}(d) = \begin{cases}
    0, & \text{if  $d>2$;}  \\
    \frac{1}{\varphi^4}, & \text{if $d=2$;}\\
    \frac{1}{\varphi^3}, & \text{if $d=1$;}\\
    \frac{2}{\varphi^4}, & \text{if $d=0$;} \\
    \mu^{(4)}(-1)\cdot \varphi^{2d+2}& \text{if $d<0$},\\
    \end{cases}
\end{displaymath}
(where $\mu^{(4)}(-1)=\frac{1}{\varphi^4}+\frac{1}{\varphi^{6}}$).
\begin{figure}[H]
    \centering
    \includegraphics[scale=0.24]{bar_chart_mu_4.pdf}
    \caption{Bar chart of $\mu^{(4)}$.}
\end{figure}
As an exercise, the reader can check that $\mu^{(1)}=\mu^{(F_3)}=\mu^{(F_4)}$. In the next section, we are going to prove that, for $k\ge 2$ we have
\[\mu^{(F_k)}=\mu^{(1)}.\]
    
    \subsection{Remarks on the algorithm}\label{ROTA}
    
Once again, the algorithm described in Subsections~\ref{DOTA} and \ref{PC} can be adapted to an integer base,
so the following observations can also be adapted to an integer base. Also, we write an algorithm that returns, for a given $r\ge 1$ and $d\in\ZZ$, the value $\mu^{(r)}(d)$. Of course, the algorithm can also be adapted to return $\mu^{(r)}(d)$ for every $d\in\ZZ$ or, at least to be executed on a computer, for a finite number of integers $d$ without more computations (the adaptation only consists of manipulations of arrays). 

In this subsection, let $r\ge 1$. Let also $\ell\ge 1$ be the unique integer such that $F_{\ell}\le r< F_{\ell+1}$. The first important observation due to the algorithm is that we only need to compute $\Delta^{(r)}(n)$ for a finite number of $n$'s to know exactly the quantity $\mu^{(r)}$ . More precisely, we need to compute $\Delta^{(r)}$ for 
\begin{itemize}
    \item $F_{\ell+1}-r$ integers during \textbf{STEP 1},
    \item $F_{\ell}$ integers during \textbf{STEP 2},
    \item $r$ integers during \textbf{STEP 3} and
    \item $r$ integers during \textbf{STEP 4}.
\end{itemize}
So, we need to compute the image of $F_{\ell+2}+r$ different integers to compute $\mu^{(r)}$ exactly. One can prove that, when $r=F_{\ell}$, it is possible to adapt the algorithm to reduce that number to $F_{\ell+2}$. The adaptation consists of forgetting \textbf{STEP 4} and store the $F_{\ell}$ values in \textbf{STEP 2}.

The algorithm implies the following properties.
\begin{proposition}\label{conseq algo 1}
If $d$ is small enough in $\ZZ$ then there exist exactly $2r$ cylinders of (non-necessarily distinct) orders $k_1,\ldots,k_{2r}$, respectively in $\NIZ{k_1},\ldots,\NIZ{k_{2r}}$ on which $\Delta^{(r)}$ takes the value $d$.
\end{proposition}
\begin{remark}
    The ``small enough''  assumption is specified in the proof.
\end{remark}
It leads to Corollary~\ref{corollaire pour algo2} we state again here.
\begin{corollaire*}
    For $d$ small enough in $\ZZ$, we have the formula
    \[\mu^{(r)}(d-1)=\mu^{(r)}(d)\cdot \frac{1}{\varphi^{2}}.\]
\end{corollaire*}
\begin{proof}[Proof of Proposition~\ref{conseq algo 1}]
    For $k\ge \ell +2$, there are $r$ cylinders of order $k$ that compose $\NIZ{k}$ so $\Delta^{(r)}$ takes, at most, $r$ different values on those cylinders. For $k\ge \ell+2$, let $(d_i^{(k)})_{1\le i\le r}$ be the collection of values taken by $\Delta^{(r)}$ on the cylinders of order $k$ in $\NIZ{k}$, repeated with multiplicity. Let \[m:=\min\left\{d_i^{(k)}~|~i=1,\ldots,r~\textrm{and}~k=\ell+2,\ell+3\right\}.\]
    Now let $d\le m$. We observe that due to Lemma~\ref{Contenu NIZ_k} and Lemma~\ref{relation par Delta cylindres NIZ}, $d$ is actually smaller than every value taken by $\Delta^{(r)}$ on $\NIZ{\ell}$ or $\NIZ{\ell+1}$. Furthermore, for $i\in [1,r]$ and $k\in\{\ell+2,\ell+3\}$, there exists a unique $j_i^{(k)}\in\NN$ such that $d=d_i^{(k)}-j_i^{(k)}$. Due to Corollary~\ref{corollaire pour algo}, $d$ is a value reached by $\Delta^{(r)}$ on the corresponding cylinder of order $k+2j_i^{(k)}$ (by repeated use of Lemma~\ref{etape relation cylindres NIZ} and \ref{relation par Delta cylindres NIZ}). There are $2r$ cylinders at order $\ell+2$ and $\ell+3$ so $d$ appears exactly on $2r$ cylinders.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{corollaire pour algo2}]
    With the same notation as in the proof of Proposition~\ref{conseq algo 1}, let $d\le m$. Then $d$ is a value taken by exactly $2r$ different cylinders of some orders. But, due to Proposition~\ref{conseq algo 1} and Corollary~\ref{corollaire pour algo}, $\Delta^{(r)}$ takes the value $d-1$ on the same number of cylinders but with the orders of those cylinders shifted by $+2$ so their measures are divided by $\varphi^{2}$. 
\end{proof}

\section{How to prove \texorpdfstring{$\mu^{(F_{\ell})}=\mu^{(1)}$}{uFk=u1}}\label{HTPMFM1}

There is an analogous relation in the integer base $b$ case. The law of adding $b^{\ell}$ is the same as the one of adding $1$. It is trivial in base $b$ since the addition $x+b^{\ell}$ does not change the first few digits of $x$, the addition really consists of adding $1$ from a certain position. In the Zeckendorf representation case, it is not trivial since carries propagate in both directions. Here, the proof consists of looking at our algorithm described in Section~\ref{HTCM} in the special case $r=F_{\ell}$. However, before starting, we need to compute $\mu^{(1)}$.
\begin{proposition}\label{mu1}
For $d\in\ZZ$, we have
\begin{displaymath}
    \mu^{(1)}(d) = \begin{cases}
    0, & \text{if $d\ge 2$;}\\
    \frac{1}{\varphi^2}, & \text{if $d=1$;}\\
    \frac{1}{\varphi^{2-2d}}, & \text{otherwise.}
    \end{cases}
\end{displaymath}
\end{proposition}
\begin{proof}
    We consider the following partition:
    \begin{align}\label{partition pour Delta 1}
        \XX\backslash \{(01)^{\infty},(10)^{\infty}\}=C_{00}\bigsqcup_{d\le 0} \left(C_{0(01)^{1-d}}\bigsqcup C_{00(10)^{1-d}}\right).
    \end{align}
    We observe that $\Delta^{(1)}(C_{00})=1$ and, for every $d\le 0$, 
    \[\Delta^{(1)}(C_{0(01)^{1-d}})=\Delta^{(1)}(C_{00(10)^{1-d}})=d.\] 
    Using Proposition~\ref{prop caract mesure zeck}, we conclude the proof.
\end{proof}

Now let $\ell\ge 3$. We follow the path given by the algorithm. We can rewrite Proposition~\ref{Def NIZ_k} in our special case $r=F_{\ell}$:
\begin{enumerate}
    \item if $2\le k\le \ell-1$, $\NIZ{k}=\emptyset$,
    \item $\NIZ{\ell}$ is the union of the first $F_{\ell-1}$ levels of the large tower of order $\ell$ and
    \item if $k\ge \ell+1$, $\NIZ{k}$ is the union of the $F_{\ell}$ levels between the $F_k-F_{\ell}^{\rm th}$ and the $F_k^{\rm th}$ levels of the large tower of order $k$.
\end{enumerate}
Executing the algorithm, we want to compute $\Delta^{(F_{\ell})}$ on the levels of $\NIZ{\ell}$. We obtain the following lemma.
\begin{lemma}\label{DeltaFl_orderl}
    \[\Delta^{(F_{\ell})}(\NIZ{\ell})=1.\]
\end{lemma}
\begin{proof}
Indeed, consider a cylinder of order $\ell$ in $\NIZ{\ell}$. Its name has for leftmost digits $000$ since the cylinders contains one element of the integer interval $[0,F_{\ell-1}-1]$. So the addition with $F_{\ell}$ considered in these cylinders are
\[\begin{array}{ccccccc}
     (x)            &   &\cdots & 0 & 0 & 0 & \cdots  \\
     (F_{\ell})     & + &       &   & 1 &   & \\ \hline
     (x+F_{\ell})   & = &\cdots & 0 & 1 & 0 & \cdots
\end{array}\]
Thus the variation of the sum of digits is $1$.
\end{proof}
We now look at $\NIZ{\ell+1}$, which is composed of $F_{\ell}$ cylinders of order $\ell+1$. We have the following proposition.
\begin{proposition}\label{DeltaFl_orderl+1}
In $\NIZ{\ell+1}$, there are 
\begin{itemize}
    \item $F_{\ell-1}$ levels on which $\Delta^{(F_{\ell})}=0$ and
    \item $F_{\ell-2}$ levels on which $\Delta^{(F_{\ell})}=1$.
\end{itemize}
\end{proposition}
For technical reasons, we need to separate the proof in two parts: one when $\ell$ is odd and the other if $\ell$ is even. However, since both are treated the same way (the differences are only technicalities), we only consider the case where $\ell$ is odd. We are going to prove that the values taken by $\Delta^{(F_{\ell})}$ are the first $F_{\ell}$ terms of the sequence 
\begin{itemize}
    \item $1,0,1,0,0,1,1,1,0,0,0,0,0,1,1,1,1,1,1,1,1, \ldots$ (if $\ell$ is odd) or
    \item  $0,1,0,1,1,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,\ldots$ (if $\ell$ is even),
\end{itemize}
where both sequences start with $0$ or $1$ depending on the parity and then the construction consists of concatenating alternatively $0$'s or $1$'s a Fibonacci number of times.
\begin{proof}[Proof of Proposition~\ref{DeltaFl_orderl+1} if $\ell$ is odd.]
    Let $\ell'\ge 1$ and $\ell:=2\ell'+1\ge 3$ . The highest level in $\NIZ{\ell+1}$ is the cylinder $C_{00(10)^{\ell'}}$ since it contains $F_{\ell+1}-1$; the lowest is the cylinder $C_{00010^{\ell-3}}$ since it contains $F_{\ell-1}$.

    The cylinder of order $\ell+1$ just below $C_{00(10)^{\ell'}}$ is $C_{00(10)^{\ell'-1}01}$. We observe that, if $\ell=3$ (or, equivalently, $\ell'=1$ or $F_{\ell}=2$), this last cylinder $C_{00(10)^{\ell'-1}01}$ is actually the same as $C_{00010^{\ell-3}}$. In general, there are two possibilities:
    \begin{itemize}
        \item Either there are no other cylinders of order $\ell+1$, which means $\ell=3$ (or, equivalently, $\ell'=1$ or $F_{\ell}=2$). So far, we have mentioned $2$ cylinders of order $4$ in $\NIZ{4}$ and the values of $\Delta^{(F_3)}$ on these cylinders are $0$ and $1$ (with an easy computation): it is the conclusion of the proposition if $\ell=3$ (if we agree $F_0:=1$).
        \item Or there are other cylinders to find in $\NIZ{\ell+1}$, which means  $F_{\ell}>2$ (or, equivalently, $\ell\ge 5$ or $F_{\ell}\ge 5$) and implies that there are $F_{\ell}-F_2\ge 3$ levels to identify.
    \end{itemize}
    To continue, we assume $\ell\ge 5$ (or, equivalently, $\ell'\ge2$) and identify the three cylinders that are below $C_{00(10)^{\ell'-1}01}$: they are
    \begin{itemize}
        \item $C_{00(10)^{\ell'-1}00}$ on which $\Delta^{(F_{\ell})}$ equals 1,
        \item $C_{00(10)^{\ell'-2}0101}$ on which $\Delta^{(F_{\ell})}$ equals 0 and
        \item $C_{00(10)^{\ell'-2}0100}$ on which $\Delta^{(F_{\ell})}$ equals 0.
    \end{itemize}
    Once again, we observe that, if $\ell=5$ (or, equivalently, $\ell'=2$), $C_{00(10)^{\ell'-2}0100}=C_{00010^{\ell-3}}$. There are again two possibilities.
    \begin{itemize}
        \item Either there is no other cylinders of order $\ell+1$, which means $\ell=5$ (or, equivalently, $\ell'=2$ or $F_{\ell}=5$). So far again, we have mentioned $5$ cylinders of order $6$ in $\NIZ{6}$ and $\Delta^{(F_5)}$ takes thrice the value $0$ and twice the value $1$ on these cylinders: it is the conclusion of the proposition if $\ell=5$.
        \item Or there are other cylinders to find in $\NIZ{\ell+1}$, which means  $F_{\ell}>5$ (or, equivalently, $\ell\ge 7$ or $F_{\ell}\ge 13$) and implies that there are $F_{\ell}-F_5\ge 8$ levels to identify.
    \end{itemize}
    We continue the same way. If we assume there are, for some $k\in[1,\ell-1]$, $F_{\ell}-F_{2k-1}\ge F_{2k}$ cylinders. There are two kind of cylinder.
    \begin{itemize}
        \item Those whose names are $C_{00(10)^{\ell'-k+1}00n_{2k-3}\cdots n_2}$ for all $(n_{2k-3}\cdots n_2)\in\XX_f$: there are $F_{2k-2}$ such cylinders and one can compute that $\Delta^{(F_{\ell})}$ is $1$.
        \item Those whose names are $C_{00(10)^{\ell'-k}010n_{2k-2}\cdots n_2}$ for all $(n_{2k-2}\cdots n_2)\in\XX_f$: there are $F_{2k-1}$ such cylinders and one can compute that $\Delta^{(F_{\ell})}$ is $0$.
    \end{itemize}
    We observe that if $\ell=2k+1$ (or, equivalently, $\ell'=k$) the last cylinder identified is $C_{00(10)^{\ell'-k}010^{2k-2}}$ and is actually the same as $C_{00010^{\ell-3}}$. So far, we have identified $1+F_1+(F_2+F_3)+(F_4+F_5)+\cdots +(F_{2k-2}+F_{2k-1})=F_{2k+1}$ cylinders and 
    \begin{itemize}
        \item on $1+F_2+F_4+F_{2k-2}= F_{2k-1}$ of them, $\Delta^{(F_{\ell})}$ is $1$;
        \item on $F_1+F_3+F_5+\cdots +F_{2k-1}=F_{2k}$ of them, $\Delta^{(F_{\ell})}$ is $0$.
    \end{itemize}
    It is exactly the conclusion when $\ell=2k+1$.
\end{proof}



Finally, we look at $\NIZ{\ell+2}$, which also contains $F_{\ell}$ cylinders of order $\ell+2$. 
\begin{proposition}\label{DeltaF1_orderl+2}
 \[\Delta^{(F_{\ell})}(\NIZ{\ell+2})=0.\]
\end{proposition}
\begin{proof}
    Consider a cylinder of order $\ell+2$ in $\NIZ{\ell+2}$. Since it contains exactly one integer of the interval $[F_{\ell+1},F_{\ell+2}-1]$, its name is $C_{0010n_{\ell-1}\cdots n_2}$ for every $(n_{\ell-1}\cdots n_2)\in\XX_f$. We compute that, on such a cylinder, $\Delta^{(F_{\ell})}$ is null.
\end{proof}
We do not need to prove anything for $\NIZ{\ell+3}$, the algorithm teaches us that $\Delta^{(F_{\ell})}$ is going to take the same values, but shifted by $-1$, on the same number of levels as in $\NIZ{\ell+1}$. Now it is time to prove our theorem.
\begin{theorem*}
For every $\ell\ge 2$, $\mu^{(F_{\ell})}=\mu^{(1)}$.
\end{theorem*}
\begin{proof}
    We recall 
    \begin{displaymath}
    \mu^{(1)}(d) = \begin{cases}
    0, & \text{if $d\ge 2$;}\\
    \frac{1}{\varphi^2}, & \text{if $d=1$;}\\
    \frac{1}{\varphi^{2-2d}}, & \text{otherwise.}
    \end{cases}
\end{displaymath}
    For $\ell=2$, this is trivial. Let $\ell\ge 3$ and $d\in\ZZ$. It is also trivial when $d\ge 2$ since $\Delta^{(F_{\ell})}$ does not take this value. Assume that $d=1$, this value appears in $F_{\ell-1}$ levels in $\NIZ{\ell}$, $F_{\ell-2}$ levels in $\NIZ{\ell+1}$ and nowhere else. So
    \[\mu^{(F_{\ell})}(1)=\frac{F_{\ell-1}}{\varphi^{\ell}}+\frac{F_{\ell-2}}{\varphi^{\ell+1}}=\frac{\varphi^{\ell-1}}{\varphi^{\ell+1}}=\frac{1}{\varphi^{2}}=\mu^{(1)}(1).\]
    If $d\le 0$, the value $d$ appears, due to Corollary~\ref{corollaire pour algo}, on $F_{\ell-1}$ levels at order $\ell+1-2d$, $F_{\ell}$ levels at order $\ell+2-2d$, $F_{\ell-2}$ levels at order $\ell+3-2d$ and nowhere else. Thus
    \begin{align*}
        \mu^{(F_{\ell})}(d)&=\frac{F_{\ell-1}}{\varphi^{\ell+1-2d}}+\frac{F_{\ell}}{\varphi^{\ell+2-2d}}+\frac{F_{\ell-2}}{\varphi^{\ell+3-2d}}\\
            &=\frac{\varphi^{\ell+1}}{\varphi^{\ell+3-2d}}\\
            &=\frac{1}{\varphi^{2-2d}}\\
            &=\mu^{(1)}(d).
    \end{align*}
\end{proof}

\section{\texorpdfstring{$\Delta^{(r)}$}{Dr} as a mixing process}\label{DMP Zeck}

We work on the probability space $(\XX,\mathscr{B}(\XX),\PP)$. For a given integer $r$, $\Delta^{(r)}$ is viewed as a random variable with law $\mu^{(r)}$ by Corollary~\ref{corollaire_prop_moment Zeck} (the randomness comes from the argument $x$ of $\Delta^{(r)}$, considered as a random outcome in $\XX$ with law $\PP$). In this section, we decompose $\Delta^{(r)}$ as a sum composed of a finite number of random variables satisfying a universal inequality on their mixing coefficients.

    \subsection{The process}\label{the process}

The case $r=0$ is irrelevant so we assume $r\ge 1$. For $1\le i\le \rho(r)$, we write $B_i$ as the $i^{\rm th}$ block present in the expansion of $r$, starting from the unit digit. We define $r[i]$ as the integer whose expansion is given by the first $i$ blocks of the expansion of $r$ (see Figure below). We observe that $r[\rho(r)]=r$.
\begin{figure}[H]
    \centering
    \includegraphics[scale=0.38]{block.pdf}
    \caption{Example of the construction of $(r[i])_{i=0,\ldots,\rho(r)}$.}
    \label{construction r[i] Zeck}
\end{figure} 
With the convention $r[0]:=0$, we observe the trivial equality 
\[r=\sum_{i=1}^{\rho(r)}(r[i]-r[i-1]).\]
For $1\le i\le \rho(r)$, we define almost everywhere on $\XX$ (see Subsection~\ref{SOzeck})
\[X_i^{(r)}:=\Delta^{(r[i]-r[i-1])}\circ T^{r[i-1]}.\] 
Since $r[i]-r[i-1]=\left[B_i 0\cdots 0\right]$, the function  $X_i^{(r)}$ is a random variable corresponding to the action of the $i^{\rm th}$ block $B_i$ once the previous blocks have already been taken into consideration.
From \eqref{decomp_sum Zeck}, we get
\[\Delta^{(r)}=\sum_{i=1}^{\rho(r)}X_i^{(r)}.\]
In particular, if $x\in\XX$ is randomly chosen with law $\PP$, then $\displaystyle\sum_{i=1}^{\rho(r)}X_i^{(r)}(x)$ follows the law $\mu^{(r)}$.

    \subsection{The \texorpdfstring{$\alpha$}{a}-mixing coefficients on the actions of blocks}

This part is devoted to the proof of one of the main theorem stated in the introduction: we are going to show that the $\alpha$-mixing coefficients for the process $(X_i^{(r)})_{i=1,\ldots,\rho(r)}$ satisfy a universal upper bound that is independent of $r$. In particular, these coefficients exponentially decrease to $0$.
\begin{theorem*}
 The $\alpha$-mixing coefficients of $(X_i^{(r)})_{i=1,\ldots,\rho(r)}$ satisfy
\[\forall k\ge 1, \hspace{5mm} \alpha(k)\le 12\left(1-\frac{1}{\varphi^8}\right)^{\frac{k}{6}}+\frac{1}{\varphi^{2k}}.\]
\end{theorem*}

\begin{proof}
    Let $k,p\ge 1$. Let $A\in\sigma(X_i^{(r)}:1\le  i\le p)$ and $B\in\sigma(X_i^{(r)}:  i\ge k+p)$. Let $C\in\sigma(X_i^{(r)}:  p< i <k+p)$ such that $\PP(C)>0$. Then we have the following inequality:
     \begin{align}\label{ineg triang 2 Zeck}
        & \begin{array}{lll}
        \lvert\PP(A\cap B)-\PP(A)\PP(B)\rvert & \le \lvert\PP(A\cap B)-\PP_C(A\cap B)\rvert \\
        & \hspace{1 cm}+ \lvert\PP_C(A\cap B)-\PP_C(A)\PP_C(B)\rvert \\
        & \hspace{1 cm} + \lvert\PP_C(A)\PP_C(B)-\PP(A)\PP(B)\rvert.
        \end{array}
    \end{align}
    Using \eqref{ineg generale conditionnement Zeck}, we get
    \begin{align}\label{etape 1 mixing block}
        \lvert\PP(A\cap B)-\PP(A)\PP(B)\rvert & \le 3\PP(\overline{C}) + \lvert\PP_C(A\cap B)-\PP_C(A)\PP_C(B)\rvert.
    \end{align}
    Following the idea used in the case of an integer base \cite{TREJYH}, we will consider an event $C$ that has a probability close to $1$ and such that, conditionally to $C$, the events $A$ and $B$ are ``almost'' independent. A difference with the integer-base case is that, as the digits of a random Zeckendorf-adic numbers are not independent, the last term on the right-hand side of \eqref{etape 1 mixing block} does not totally vanish and, thus, some extra work is needed. However, we can anticipate that this problem of ``almost'' independence will be solved using the inequality on the $\phi$-mixing coefficients (on the law of coordinates) stated in Proposition~\ref{Melange_coord}.
    
    To define the event $C$, we use the ideas developed in Subsection~\ref{subsection Focus} (especially Corollaries~\ref{stopping condition bloc 1} and ~\ref{stopping cdt bloc >1}). Since we are working with blocks, we introduce $\ell_i$ as the number of patterns $10$ in $B_i$, the $i^{\rm th}$ block of $r$. We also introduce $n_i$ as the minimal index of digits in $B_i$. Then for $i=1,\ldots, \rho(r)$ we define the set Adm$(i)$ as the set of indices corresponding to the digits of $x$ involved in the assumptions of Corollaries~\ref{stopping condition bloc 1} and \ref{stopping cdt bloc >1} if we want to apply them in the area of the block $B_i$ in $r$. More precisely, if $\ell_i=1$, then 
    \[\textrm{Adm}(i):=[n_i-2, n_i+3].\]
    And, if $\ell_i\ge 2$, then
    \[\textrm{Adm}(i):=[n_i-2, n_i+1] \sqcup [n_i-2+2\ell_i, n_i+1+2\ell_i].\]
    Observe that $\lvert \textrm{Adm}(i)\rvert\le 8$.
    \begin{figure}[H]
        \centering
        \includegraphics[scale=0.4]{adm.pdf}
        \caption{Indices of Adm$(i)$ and conditions put on $x$.}
        \label{Adm(i,l(i))}
    \end{figure}
    Then for $i=1,\ldots,\rho(r)$ we define 
    \[C^{(i)}:=\{x\in\XX~: ~\forall j\in \textrm{Adm}(i), ~ x_j=0\}.\]
    We also define the events
    \[ C_1:=\bigcup_{\substack{i=p+2 \\ \textrm{even} }}^{p+\frac{k}{3}}C^{(i)} \hspace{5mm} \textrm{ and } \hspace{5mm} C_2:=\bigcup_{\substack{i=p+\frac{2k}{3} \\ \textrm{even} }}^{p+k-2}C^{(i)}\]
    and, finally 
    \[C:=C_1\cap C_2.\]
    \begin{figure}[H]
        \centering
        \includegraphics[scale=0.35]{c1c2.pdf}
        \caption{Location of the events $C_1$ and $C_2$ (dots represent blocks or possible zeros).}
        \label{visual}
    \end{figure}
    Thus, $x$ belongs to $C$ if and only if the expansion of $x$ satisfies the hypotheses of Corollary~\ref{stopping condition bloc 1} or \ref{stopping cdt bloc >1} for, at least, two blocks placed on the leftmost third of the window (for at least one block) and the rightmost third of the window (for at least one block).
    
    We prove now that $C$ is an event that has a high probability to happen. We claim that
    \begin{align}\label{majoration C block}
        \PP(C)\ge 1 -2(1-\frac{1}{\varphi^8})^{\frac{k}{6}}.
    \end{align}
    Indeed, if we let $\overline{C_1}$ (resp.\ $\overline{C_2}$) denote the complement of $C_1$ (resp.\ $C_2$) in $\XX$, then we have        \[\PP(C)=1-\PP(\overline{C_1})-\PP(\overline{C_2})+\PP(\overline{C_1}\cap\overline{C_2})\ge\PP(C_1)+\PP(C_2)-1.\]
    Then we obtain from Lemma~\ref{borne proba Zeck}
    \begin{align*}
        \PP(C_1)&=1-\PP\left(\bigcap_{\substack{i=p+2 \\ \textrm{even} }}^{p+\frac{k}{3}} \overline{C^{(i)}}\right) \\
                &=1-\prod_{\substack{i=p+2 \\ \textrm{even} }}^{p+\frac{k}{3}} \PP\left(\overline{C^{(i)}}~\bigg\lvert~\bigcap_{\substack{j=p+2 \\ \textrm{even} }}^{i-2}\overline{C^{(j)}}\right)\\
                & \ge 1-(1-\frac{1}{\varphi^8})^{\frac{k}{6}},
    \end{align*}
    because the product contains $\lfloor\frac{p+\frac{k}{3}-p+2-2}{2}\rfloor=\lfloor\frac{k}{6}\rfloor$ terms. We show that $\PP(C_2$) satisfies the same inequality. We thus obtain \eqref{majoration C block}. Observe that, if $k\ge 194$, $\PP(C)>0$.
    
    Now we want to estimate the term $\lvert\PP_C(A\cap B)-\PP_C(A)\PP_C(B)\rvert$ that appears in \eqref{etape 1 mixing block}. We want to use Proposition~\ref{Melange_coord}, and for that we have to clarify which digits of $x$ the events $A$ and $B$ depends on.
    
    But, conditionally to $C$, from Corollaries~\ref{stopping condition bloc 1} and \ref{stopping cdt bloc >1} we know that the actions of the first $p$ blocks only modify digits of indices $\le N_1+2$ where $N_1$ is the index of digit of the leftmost $1$ of the block $B_{\lfloor p+\frac{k}{3}\rfloor}$ of $r$. Thus, there exists $A'\in\sigma(x_i~:~i\le N_1+2)$ such that $A\cap C=A'\cap C$. Likewise, the actions of the $\rho(r)-k-p+1$ last blocks only modify digits of indices $\ge N_2$ where  $N_2$ the index of the rightmost $0$ of the block $B_{\lfloor p+\frac{2k}{3}\rfloor}$. Thus, there also exists $B'\in\sigma(x_i~:~i\ge N_2)$ such that  $B'\cap C=B\cap C$. 
    \begin{figure}[H]
        \centering
        \includegraphics[scale=0.4]{N1N2.pdf}
        \caption{Location of indices $N_1$ and $N_2$ in the expansion of $r$.}
        \label{visual2}
    \end{figure}
  Then 
     \begin{align}\label{etape 2 mixing block}
        \vert\PP_C(A\cap B)-\PP_C(A)\PP_C(B)\rvert&=\vert\PP_C(A'\cap B')-\PP_C(A')\PP_C(B')\rvert.
    \end{align}
    Then we get 
    \begin{align*}
        \vert\PP_C(A'\cap B')-\PP_C(A')\PP_C(B')\rvert &\le \vert\PP_C(A'\cap B')-\PP(A'\cap B')\rvert \\
                &\hspace{1cm} +\vert\PP(A'\cap B')-\PP(A')\PP(B')\rvert \\
                &\hspace{1cm} + \vert\PP(A')\PP(B')-\PP_C(A')\PP_C(B')\rvert.
    \end{align*}
    The term $\vert\PP(A'\cap B')-\PP(A')\PP(B')\rvert$ can be estimated using the $\phi$-mixing coefficient on the law of coordinates. Indeed, observe that $N_2-N_1$ is at least about order $k$. Then using \eqref{ineg generale conditionnement Zeck} and Propositions~\ref{Lien coeff phi et alpha Zeck} and \ref{Melange_coord}, we obtain
    \begin{align}\label{etape 3 mixing block}
        \vert\PP_C(A'\cap B')-\PP_C(A')\PP_C(B')\rvert&\le 3\PP(\overline{C})+\frac{1}{2}\phi(k).
    \end{align}
    Finally, combining \eqref{etape 1 mixing block}, \eqref{etape 2 mixing block} and \eqref{etape 3 mixing block}, we obtain \begin{align*}
         \lvert\PP(A\cap B)-\PP(A)\PP(B)\rvert &\le 6\PP(\overline{C})+\frac{1}{2}\phi(k).
    \end{align*}
    Taking the supremum on $A$ and $B$, we conclude the proof.
\end{proof}

\section{Acknowledgments}
The author wishes to thank \'Elise Janvresse and Thierry de la Rue who introduced him to this subject. They gave the author a lot of valuable advice.

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\end{thebibliography}


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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11A63; Secondary 11B39, 11K55, 37A25.

\noindent \emph{Keywords: } Zeckendorf expansion, sum-of-digits, generalized odometer, mixing process.

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\vspace*{+.1in}
\noindent
Received October 13 2023; 
revised versions received October 18 2023; January 19 2024; February 23 2024.
Published in {\it Journal of Integer Sequences}, February 24 2024.

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