2.1 Markov chains

Let $\mathcal {S}$ be a finite set and let M be a square matrix indexed by $\mathcal {S}\times \mathcal {S}$ with entries in $[0,1]$ . The matrix M is called stochastic if the sum of each row is 1, that is, for every $s\in \mathcal {S}$ ,

$$ \begin{align*} \sum_{s'\in \mathcal{S}} M(s,s')=1.\end{align*} $$

Given a stochastic matrix M and a probability measure $\mu $ on $\mathcal {S}$ , a discrete-time Markov chain with state space $\mathcal {S}$ , transition matrix M and initial distribution $\mu $ is a sequence of random variables $(X_n)_{n\in {\mathbb N}_0}$ on a probability space (with probability $\mathbb {P}$ ) such that $\mathbb {P}(X_0=s)=\mu (\{s\})=:\mu (s)$ , and

$$ \begin{align*}\mathbb{P}(X_{n+1}=s_{n+1}|X_n=s_n,\ldots,X_0=s_0)=\mathbb{P}(X_1=s_{n+1}|X_0=s_n)=M(s_n,s_{n+1}) \end{align*} $$

for every $n\in {\mathbb N}_0$ and $s_0,\ldots ,s_{n+1}\in \mathcal {S}$ . In this case,

$$ \begin{align*}\mathbb{P}(X_n=s')=\mu \cdot M^n(\cdot,s')=\sum_{s\in\mathcal{S}}\mu(s)M^n(s,s'). \end{align*} $$

For $s,s'\in \mathcal {S}$ , we say that $s'$ is accessible from s if $M^n(s,s')>0$ for some $n\in {\mathbb N}$ . The matrix M is called irreducible if for every $s,s'\in \mathcal {S}$ , there exists $n\in {\mathbb N}$ such that $M^n(s,s')>0$ (that is, all states are accessible from each other). We define the period of the state s as

$$ \begin{align*}\mathrm{Per}(s):=\gcd \{n\in {\mathbb N}: M^n(s,s)>0\}. \end{align*} $$

If every state has period 1, we say M is aperiodic. When M is irreducible, all states share the same period, so aperiodicity can be checked by finding one state with period 1.

The following classical result on the convergence of Markov chains will be a key tool in this paper.

Theorem 2.1. [Reference Levin and Peres35, Theorem 4.9]

Let X be a Markov chain with state space $\mathcal {S}$ (not necessarily finite), transition matrix M and initial distribution $\mu $ . Suppose M is irreducible and aperiodic, and there exists an invariant distribution $\unicode{x3bb} $ for M (that is, $\unicode{x3bb} =\unicode{x3bb} M$ ). Then, there exist $\rho \in (0,1)$ and $C>0$ such that

$$ \begin{align*}|\mathbb{P}(X_n=s)-\unicode{x3bb}(s)|\leq C\rho^n \end{align*} $$

for every $n\in {\mathbb N}$ and $s\in \mathcal {S}$ .

Notice that the invariant distribution for an irreducible and aperiodic transition matrix with a finite state space always exists (and it is unique) as a consequence of the Perron–Frobenius theorem. In addition, Theorem 2.1 is only dependent on the transition matrix, so that $\unicode{x3bb} $ , C and $\rho $ are independent of the initial distribution. A particularly interesting case occurs when the sum of each column of M is also 1, in which case, M is called doubly stochastic. In this case, the distribution $\unicode{x3bb} (s) = 1/|\mathcal {S}|$ is invariant for M. For further discussion on the convergence of Markov chains, see [Reference Norris44, §1.8] and [Reference Rosenthal45, §3].

2.2 Elements of symbolic dynamics

Let $\mathcal {D}$ be a finite set (in this paper, we assume $\mathcal {D}\subseteq \{0,\ldots ,g-1\}$ for some $g\in {\mathbb N}$ ) and consider the left-shift operator $\sigma $ over $\mathcal {D}^{{\mathbb N}_0}$ , defined by

$$ \begin{align*}\sigma: (w_n)_{n\in {\mathbb N}_0}\mapsto (w_{n+1})_{n\in {\mathbb N}_0}. \end{align*} $$

Endowing $\mathcal {D}$ with the discrete topology and considering the product topology on $\mathcal {D}^{{\mathbb N}_0}$ , $\sigma $ is a continuous function, and the topological system $(\mathcal {D}^{{\mathbb N}_0},\sigma )$ is called a full shift. If $\Sigma \subseteq \mathcal {D}^{{\mathbb N}_0}$ is closed and $\sigma (\Sigma )\subseteq \Sigma $ , the dynamical system $(\Sigma ,\sigma )$ is called a subshift. Although a subshift is a set of infinite sequences of symbols, it can be fully characterized by the finite sequences that appear in $\Sigma $ , referred to as words.

We define the language of $\Sigma $ , denoted by $\mathcal {L}(\Sigma )$ , as the set of all the words in $\Sigma $ ,

$$ \begin{align*} \mathcal{L}(\Sigma):= & \{w_0\cdots w_{n-1}:\ n\in {\mathbb N}_0, \\ & \quad \text{there exists } x\in \Sigma \text{ such that } x_0=w_0, \ x_1=w_1,\ldots,\ x_{n-1}=w_{n-1} \}.\end{align*} $$

When there is no confusion, we will simply denote the language by $\mathcal {L}$ . For any word $w=w_0\cdots w_{n-1}\in \mathcal {L}(\Sigma )$ , we say w has length n (and we denote the length by $|w|$ ). The set $\mathcal {L}^n(\Sigma )$ denotes all words in the language of length n. Also, we will commonly use the notions of concatenation of words and subwords. For $u=u_0\cdots u_{m-1}$ and ${v=v_0\cdots v_{n-1}}$ , the concatenation of u with v is written as $uv=u_0\cdots u_{m-1}v_0\cdots v_{n-1}$ and we say that $uv$ is an extension of u. For a given word u and $k\in {\mathbb N}$ , we also denote by $u^k=u\cdots u$ the concatenation k-times of u. We say w is a subword of x if the sequence of x contains w.

A subshift is a particular case of a topological dynamical system and some classical notions of topological dynamics have a special characterization in the context of symbolic systems.

However, the notion of topological entropy is a measure of the complexity of a dynamical system. In symbolic systems, it quantifies the complexity of the language.

Definition 2.3. The (topological) entropy of the subshift $\Sigma $ is defined as

$$ \begin{align*}h(\Sigma):=\lim_{N\to\infty}\dfrac{\log |\mathcal{L}^N(\Sigma)|}{N}.\end{align*} $$

See [Reference Lind and Marcus36, Ch. 4] for further discussion. Regarding the classification of subshifts, we will consider two important classes, shifts of finite type and sofic shifts.

Definition 2.4. Consider the full shift $X=\mathcal {D}^{{\mathbb N}_0}$ and let $\mathscr {F}\subseteq \mathcal {L}(X)$ be a finite set of forbidden words. We define the subshift

$$ \begin{align*}X_{\mathscr{F}}:=\{x\in X:\ x \text{ does not contain any } w\in \mathscr{F} \text{ as a subword} \}.\end{align*} $$

We say that $\Sigma \subseteq X$ is a shift of finite type (SFT for short) if there exists a finite set $\mathscr {F}$ such that $\Sigma =X_{\mathscr {F}}$ .

If the longest forbidden word has length $M+1$ , we say that $X_{\mathscr {F}}$ is an M-step shift of finite type. In a $1$ -step SFT, all forbidden words have length $2$ , so it can be described by a matrix indexed by $\mathcal {D}\times \mathcal {D}$ with values in $\{0,1\}$ . In this sense, $\Sigma $ is a $1$ -step SFT if and only if there exists a matrix T such that

$$ \begin{align*}\Sigma=\{x\in \mathcal{D}^{{\mathbb N}_0}:\ T(x_i,x_{i+1})=1 \text{ for all }i\}.\end{align*} $$

A more general way to construct subshifts is by reading infinite sequences in a labelled directed multi-graph. A labelled multi-graph is a triple $G=(V,E,\unicode{x3bb} )$ , where V is a finite set of nodes, E is a set of edges and $\unicode{x3bb} :E\to \mathcal {A}$ is a labelling function for some finite alphabet $\mathcal {A}$ . Each edge $e\in E$ has an associated starting node $s(e)\in V$ and a terminal node $t(e)\in V$ . A sequence of edges $\pi =e_0\cdots e_{n}$ is called a path in G if $t(e_i)=s(e_{i+1})$ for every $i\in \{0,\ldots ,n-1\}$ , and we say that $\pi $ has length $|\pi |:=n+1$ . Note that multiple edges may share the same starting and terminal nodes while having different labels.

Definition 2.5. Let $G=(V,E,\unicode{x3bb} )$ be a labelled directed multi-graph. If $\mathcal {D}$ is the finite alphabet associated with the labelling function, we define the subshift $\Sigma _G\subset \mathcal {D}^{\mathbb {N}_0}$ as

$$ \begin{align*} \Sigma_G=\{(\unicode{x3bb}(e_j))_{j\in \mathbb{N}_0}\in \mathcal{D}^{\mathbb{N}_0}: \ s(e_{j+1})= t(e_j) \ \text{ for all } j\in \mathbb{N}_0 \}. \end{align*} $$

Thus, $\Sigma _G$ are all infinite sequences that we can read in some infinite path of G. It is well known that $\Sigma _G$ is a subshift.

Note that the cover of a sofic shift is not unique. Additionally, every shift of finite type is sofic.

To construct a cover for a sofic shift, we can rely on the notion of follower sets [Reference Lind and Marcus36, Ch. 3].

Definition 2.8. Let $\Sigma $ be a subshift of $\mathcal {D}^{{\mathbb N}_0}$ . For any $w\in \mathcal {L}(\Sigma )$ , the follower set of w, denoted $F_\Sigma (w)$ , is the set of all words that can extend w, that is,

$$ \begin{align*}F_\Sigma(w):=\{v\in \mathcal{L}(\Sigma):\ wv\in \mathcal{L}(\Sigma)\}.\end{align*} $$

We denote by $\mathcal {F}_\Sigma :=\{F_\Sigma (w): \ w\in \mathcal {L}(\Sigma )\}$ the collection of all follower sets. When the underlying subshift $\Sigma $ is clear from the context, we will simply write $F(w)$ and $\mathcal {F}$ .

While each word has its own follower set, a folklore result states that $\Sigma $ is sofic if and only if there are finitely many distinct follower sets [Reference Lind and Marcus36, Theorem 3.2.10], that is, ${|\mathcal {F}_\Sigma |<\infty} $ . If $d\in F(u)=F(v)$ , it follows that $F(ud)=F(vd)$ . Using these ideas, a cover can be constructed for $\Sigma $ .

The follower set graph is a cover of $\Sigma $ [Reference Lind and Marcus36, Proposition 3.2.9]. Notice that the follower set graph is right-resolving, meaning that for any node, all the edges going out have different labels. When the sofic shift is transitive, we would expect the cover to be an irreducible graph (that is, there is a path between each pair of vertices), but this is not always true for the follower set graph. To address this problem, a subgraph can be extracted, which remains a cover and is irreducible.

While not every subshift has synchronizing words, it is well known that transitive sofic shifts do. Moreover, every word can be extended to be synchronizing and any extension of a synchronizing word is also synchronizing. In the case of an M-step SFT, every word with length at least $M+1$ is synchronizing.

An important result due to Fischer [Reference Fischer21, Reference Fischer22] establishes the existence and uniqueness (up to isomorphism) of a minimal (in terms of the number of nodes), right-resolving and irreducible cover for a transitive sofic shift. This unique representation is known as the Fischer cover and plays an important role in this paper. The following outlines the construction of this cover.

2.3 Construction of the Fischer cover

Let $\Sigma \subseteq \mathcal {D}^{{\mathbb N}_0}$ be a transitive sofic shift. For each follower set $F\in \mathcal {F}_\Sigma $ , we create a node associated to F if there exists a synchronizing word w such that $F(w)=F$ ; in other words, we consider the set of nodes

$$ \begin{align*}\mathcal{V}_\Sigma:=\{F(w):\ w\in \mathcal{L}(\Sigma) \text{ is synchronizing}\}\subseteq \mathcal{F}_\Sigma.\end{align*} $$

Let $F,F'\in \mathcal {V}_\Sigma $ and $d\in \mathcal {D}$ . Then, there exists $w\in \mathcal {L}(\Sigma )$ such that $F=F(w)$ . We create an edge with starting node F, terminal node $F'$ and labelled by d if and only if ${F'=F(wd)}$ . The resulting graph, denoted by $\mathcal {G}_\Sigma =(\mathcal {V}_\Sigma ,\mathcal {E}_\Sigma ,\unicode{x3bb} _\Sigma )$ , is the Fischer cover for $\Sigma $ (see [Reference Lind and Marcus36, Ch. 3] and [Reference Spandl48, Lemma 6.11] for more details).

Thus, the Fischer cover is a subgraph of the follower set graph, constructed by considering only the nodes corresponding to follower sets of synchronizing words. Notice that for any synchronizing word $w \in \mathcal {L}(\Sigma )$ , every path in $\mathcal {G}_\Sigma $ labelled by w must end at the node associated with $F(w)$ . In the case where $\Sigma $ is a mixing sofic subshift, it is possible to find $n\in {\mathbb N}$ such that for every pair of nodes $F,F'\in \mathcal {V}$ and $m\geq n$ , there exists a path of length m from F to $F'$ . In such a case, for every pair of words u and v in the language, there exists $w\in \mathcal {L}^m(\Sigma )$ such that $uwv\in \mathcal {L}(\Sigma )$ .

2.4 Fractal sets of integers

In [Reference Glasscock, Moreira and Richter27], Glasscock, Moreira and Richter introduced the notion of multiplicatively invariant sets of integers (recall Definition 1.1), and studied them from a fractal perspective. Here, we focus on the notion of mass dimension, which is analogous to the box dimension in fractal geometry, but adapted to the setting of integers.

Definition 2.11. Let $A\subseteq {\mathbb N}_0$ be non-empty. We define the lower mass dimension and the upper mass dimension of A as

$$ \begin{align*} &\underline{\dim}_{\mathrm{{M}}}(A):=\liminf_{N\to \infty} \dfrac{\log |A\cap [0,N) |}{\log N}=\sup \bigg\{\gamma\geq 0: \ \liminf_{N\to \infty}\dfrac{|A\cap [0,N)|}{N^\gamma}>0 \bigg\},\\ &\overline{\dim}_{\mathrm{{M}}}(A):=\limsup_{N\to \infty} \dfrac{\log |A\cap [0,N) |}{\log N}=\sup \bigg\{\gamma\geq 0: \ \limsup_{N\to \infty}\dfrac{|A\cap [0,N)|}{N^\gamma}>0 \bigg\}. \end{align*} $$

If $\underline {\dim }_{\mathrm {{M}}}(A)=\overline {\dim }_{\mathrm {{M}}}(A)$ , we denote this value by $\dim _{\mathrm {{M}}}(A)$ and say that the mass dimension of A exists.

From [Reference Glasscock, Moreira and Richter27, Proposition 3.6], the mass dimension exists for any multiplicatively invariant set of integers. A natural way to generate $\times g$ -invariant sets is by considering subshifts of $X=\{0,\ldots ,g-1\}^{{\mathbb N}_0}$ .

Definition 2.12. The g-language set associated with a subshift $\Sigma \subseteq \{0,\ldots ,g-1\}^{{\mathbb N}_0}$ is the set $A_\Sigma \subseteq {\mathbb N}_0$ defined by

(2.1) $$ \begin{align} A_\Sigma:=\{(w)_g:=w_0+w_1g+\cdots+w_{n-1}g^{n-1}\ : \ w=w_0\cdots w_{n-1}\in \mathcal{L}(\Sigma) \}. \end{align} $$

In general, the expansion of a number in a certain base is read from left to right, where the leftmost digit represents the most significant digit. For a given word w, in this paper, we will consider the rightmost digit to be the most significant digit of the integer $(w)_g$ , this to be consistent with the theory of subshifts. However, when it comes to applications, this does not make a major distinction.

From the definition of $\times g$ -invariant sets and the shift-invariance of $\Sigma $ , it is evident that $A_\Sigma $ is $\times g$ -invariant. It is also possible to see that every multiplicatively invariant set can be written as $ A_\Sigma $ for some subshift $\Sigma $ and the mass dimension is related to the entropy of the subshift. This is summarized in the following proposition (see [Reference Glasscock, Moreira and Richter27, Proposition 3.10]).

Proposition 2.13. The g-language set $A_\Sigma \subseteq {\mathbb N}_0$ corresponding to any non-empty subshift $\Sigma \subseteq \{0,\ldots ,g-1\}^{{\mathbb N}_0}$ is a $\times g$ -invariant set and the mass dimension of $A_\Sigma $ is equal to the normalized topological entropy of the subshift $(\Sigma ,\sigma )$ , that is,

(2.2) $$ \begin{align} \dim_{\mathrm{{M}}}(A_\Sigma)=h(\Sigma)/\log(g). \end{align} $$

Moreover, for any $\times g$ -invariant set $A\subseteq {\mathbb N}_0$ , there exists a subshift $\Sigma \subseteq \{0,\ldots ,g-1\}^{{\mathbb N}_0}$ such that $A=A_\Sigma $ . In such a case, we say that A is represented by $\Sigma $ .

Corollary 2.14. Let $g\geq 2$ be an integer and let $\mathcal {D}\subseteq \{0,\ldots ,g-1\}$ . If we consider the full shift $\Sigma =\mathcal {D}^{{\mathbb N}_0}$ , $\mathcal {C}_{g,\mathcal {D}}=A_\Sigma $ . In addition,

$$ \begin{align*}\dim_{\mathrm{{M}}}(\mathcal{C}_{g,\mathcal{D}})=\log(|\mathcal{D}|)/\log(g).\end{align*} $$

Finally, the following result allows us to simplify the calculation of the dimension to simpler intervals.

Lemma 2.15. [Reference Glasscock, Moreira and Richter27, Lemma 3.2. (IV)].

For $r\in \mathbb {N}$ , $r\geq 2$ ,

(2.3) $$ \begin{align} \overline{\dim}_{\mathrm{{M}}}(A)=\limsup_{N\to \infty}\dfrac{\log |A\cap [0,r^N) |}{N\log r}.\end{align} $$

Analogously, we can compute $\underline {\dim }_{\mathrm {{M}}}(A)$ by using $\liminf $ instead of $\limsup $ in (2.3).

3.1 Transition matrix

To define the transition matrix, we will group the elements of $\mathcal {L}(\Sigma )$ according to how they extend other words. Recalling that $(w_0\cdots w_n)_g:=w_0+w_1g+\cdots +w_ng^n$ , we introduce the following subsets of $\mathcal {L}^p(\Sigma )$ .

Definition 3.4. Let $i\in {\mathbb N}$ . For any $\mathbf {b}\in {\mathbb Z}_{\mathbf {a}}={\mathbb Z}_{a_1}\times \cdots \times {\mathbb Z}_{a_r}$ and $F,F'\in \mathcal {V}$ , we define the set

$$ \begin{align*}E_{i}(\mathbf{b},F,F'):=\{w\in \mathcal{L}^p(\Sigma): \mathbf{f}(g^i(w)_g)\equiv \mathbf{b}\ \mathrm{mod}\ \mathbf{a},\ F\overset{w}{\to}F'\}.\end{align*} $$

Notice that $E_i$ depends on g, $\mathbf {f}$ , $\mathbf {a}$ , $\Sigma $ and p, but as in Remark 3.3, we set these elements, but we will not make the dependency explicit to lighten the notation. The following lemma clarifies the motivation for defining these sets.

Proof. Since $F(x)\overset {w}{\to }F'$ , $xw\in \mathcal {L}(\Sigma )$ and $F(xw)=F'$ . Also,

$$ \begin{align*}\mathbf{f}((xw)_g)=\mathbf{f}\bigg(\sum_{j=0}^{i-1}x_jg^j+\sum_{j=i}^{i+p-1}w_jg^j\bigg)=\mathbf{f}((x)_g)+\mathbf{f}(g^i(w)_g)\equiv \mathbf{f}((x)_g)+\mathbf{b}\ \mathrm{mod}\ \mathbf{a}.\end{align*} $$

If $i\geq \ell $ , $\mathbf {f}(g^i(y)_g)=\mathbf {f}(g^{i+np}(y)_g)$ for every $n\in {\mathbb N}$ and $y\in \mathcal {L}^p(\Sigma )$ , concluding that ${E_i=E_{i+np}.}$

Despite that the collection of sets $E_{i}$ is not a partition of $\mathcal {L}^p$ (since a word might be read along different paths in $\mathcal {G}$ ), we still can understand it as a partition of the paths of length p in $\mathcal {G}$ .

Proposition 3.6. Let $i\in {\mathbb N}$ . For every $F \in \mathcal {V}$ ,

(3.2) $$ \begin{align} \sum_{(\mathbf{b},F')\in {\mathbb Z}_{\mathbf{a}}\times \mathcal{V}}|E_{i}(\mathbf{b},F,F')|=|\{\pi=(\pi_0\cdots\pi_{p-1}) \text{ is a path of } \mathcal{G} \text{ and } s(\pi)=F\}|.\end{align} $$

Similarly, for every $F'\in \mathcal {V}$ ,

(3.3) $$ \begin{align} \sum_{(\mathbf{b},F)\in {\mathbb Z}_{\mathbf{a}}\times \mathcal{V}}|E_{i}(\mathbf{b},F,F')|=|\{\pi=(\pi_0\cdots\pi_{p-1}) \text{ is a path of } \mathcal{G} \text{ and } t(\pi)=F'\}|.\end{align} $$

Proof. Let $F,F'\in \mathcal {V}$ . Notice that the sets $E_{i}(\mathbf {b},F,F')$ and $E_{i}(\mathbf {b}',F,F')$ are disjoint when $\mathbf {b}\neq \mathbf {b}'\in {\mathbb Z}_{\mathbf {a}}$ . For any $\mathbf {b}\in {\mathbb Z}_{\mathbf {a}}$ , if $w\in E_{i}(\mathbf {b},F,F')$ , there exists a path $\pi $ in $\mathcal {G}$ of length p such that $\unicode{x3bb} (\pi )=w$ from F to $F'$ . However, any path $\pi $ of length p from F to $F'$ represents a word $x:=\unicode{x3bb} (\pi )$ of $E_{i}(\mathbf {b},F,F')$ , where $\mathbf {b}:=\mathbf {f}(g^i(x)_g)$ . Since F and $F'$ are fixed, each of these paths represents a different word. Then,

$$ \begin{align*} |\{\pi=(\pi_0\cdots\pi_{p-1}) \text{ is a path in } \mathcal{G} \text{ from } F \text{ to } F'|&=\bigg|\bigcup_{\mathbf{b}\in {\mathbb Z}_{\mathbf{a}}}E_{i}(\mathbf{b},F,F')\bigg|\\ &=\sum_{\mathbf{b}\in {\mathbb Z}_{\mathbf{a}}} |E_{i}(\mathbf{b},F,F')|. \end{align*} $$

By summing over $F'$ , the equality (3.2) is obtained and similarly, (3.3) follows by summing over F.

Usually, (3.2) and (3.3) do not agree. For this work, we require an extra regularity of $\mathcal {G}$ to ensure that the number of paths of length p starting at some $F\in \mathcal {V}$ is equal to the number of paths of length p with terminal node $F'\in \mathcal {V}$ . From graph theory, a directed multi-graph is said to be k-regular if each vertex has k edges going in and k edges going out (two edges are different if they differ in the starting node, terminal node or label). We can pose this definition in subshifts.

If a graph is $1$ -regular, it must represent a full shift in one symbol. Thus, we are interested in the case $k\geq 2$ , so we include it directly in the definition. Trivially, any full shift satisfies this condition, with k being the number of symbols. When $\Sigma $ is k-regular and $F \in \mathcal {V}$ , there are $k^p$ paths of length p starting from F and $k^p$ paths of length p ending at F. From this idea, we can define a transition matrix for the Markov chains.

Definition 3.8. In the framework of Remark 3.3 and assuming $\Sigma $ is k-regular, we define the state space $\mathcal {S}:={\mathbb Z}_{\mathbf {a}}\times \mathcal {V}$ and the matrix $M_{i}$ indexed by $\mathcal {S}\times \mathcal {S}$ as

$$ \begin{align*}M_{i}((\mathbf{b},F),(\mathbf{b}',F')):=k^{-p}\cdot |E_{i}(\mathbf{b}'-\mathbf{b},F,F')|\end{align*} $$

for any $(\mathbf {b},F), (\mathbf {b}',F')\in \mathcal {S}.$

Clearly, the matrix $M_i$ depends on several elements, including the choice of p, but we will not make the dependence explicit. The regularity condition allows us to show that this matrix $M_i$ is suitable for defining a Markov chain.

Proof. Let $(\mathbf {b},F)\in \mathcal {S}$ . From (3.2),

$$ \begin{align*} \sum_{s\in \mathcal{S}} M_{i}((\mathbf{b},F),s)&=k^{-p}\sum_{(\mathbf{b}',F')\in \mathcal{S}}|E_{i}(\mathbf{b}'-\mathbf{b},F,F')| \\ &=k^{-p} \sum_{(\hat{\mathbf{b}},F')\in \mathcal{S}} |E_{i}(\hat{\mathbf{b}},F,F')|\\ &=k^{-p}|\{\pi=(\pi_0\pi_1\cdots\pi_{p-1}) \text{ is a path in } \mathcal{G} \text{ and } s(\pi)=F\}|.\end{align*} $$

We conclude that $ \sum _{s\in \mathcal {S}} M_{i}((\mathbf {b},F),s)=1$ from k-regularity. Analogously, from (3.3), we obtain $\sum _{s\in \mathcal {S}} M_{i}(s,(\mathbf {b},F))=1.$

Note that the regularity is important to ensure that the normalization factor to make the matrix stochastic is the same for all rows. Without this, while a stochastic matrix could still be defined, the link between $M_i$ and the distribution of integers becomes unclear, and our method no longer seems to work.

3.2 Initial distribution

In our attempt to study $A_\Sigma $ , the idea is to extend all the $\ell $ -long words that have a follower set in $\mathcal {V}$ , as we can fully describe those extensions using the Fischer cover. However, for some $w\in \mathcal {L}^{\ell }$ , it might be that $F(w)\notin \mathcal {V}$ . This leads to the following definition.

Definition 3.10. Let $\Sigma $ be a transitive sofic shift. We define the $\ell $ -restricted language of $\Sigma $ as all words of $\mathcal {L}(\Sigma )$ that are extensions of a word of length $\ell $ with follower set in $\mathcal {V}$ ,

$$ \begin{align*}\mathcal{L}_{\mathcal{V},\ell}(\Sigma):=\{w\in \mathcal{L}(\Sigma): |w|\geq \ell \text{ and } F(w_0\cdots w_{\ell-1})\in \mathcal{V}\}.\end{align*} $$

For every $i\geq \ell $ , we also define $\mathcal {L}_{\mathcal {V},\ell }^i(\Sigma ):=\{w\in \mathcal {L}_{\mathcal {V},\ell }(\Sigma ):\ |w|=i\}$ .

The reason for considering $\ell $ large enough in Remark 3.3 is to ensure that this set is not empty. If $\Sigma $ is an M-step shift of finite type, any word w such that $|w|\geq M+1$ is synchronizing; therefore, $\mathcal {L}_{\mathcal {V},\ell }^{i}(\Sigma )=\mathcal {L}^i(\Sigma )$ for all $i\geq M+1$ . Under the condition of regularity, it is easy to describe the cardinality of $\mathcal {L}_{\mathcal {V},\ell }^i$ . For the upcoming results, if $w = w_0 \cdots w_n$ is a word, we will identify it with a pair of words $(x, y)$ such that $w = xy$ .

Proposition 3.11. Let $\Sigma $ be a transitive sofic shift and k-regular. Then,

$$ \begin{align*}|\mathcal{L}^i_{\mathcal{V},\ell}(\Sigma)|=|\mathcal{L}^{\ell}_{\mathcal{V},\ell}(\Sigma)|\cdot k^{i-\ell}\end{align*} $$

for every $i\geq \ell $ .

Proof. Notice $x\in \mathcal {L}^i_{\mathcal {V},\ell }$ if and only if $F:=F(x_0\cdots x_{\ell -1})\in \mathcal {V}$ and there is a path $\pi $ in the Fischer cover of length $i-\ell $ from the vertex associated to F, and $\pi $ is labelled by $x_\ell \cdots x_{i-1}$ . Using the identification,

(3.4) $$ \begin{align} \mathcal{L}^i_{\mathcal{V},\ell}=\bigcup_{F\in \mathcal{V}}\{w\in \mathcal{L}^\ell_{\mathcal{V},\ell}: F(w)=F\}\times \{\unicode{x3bb}(\pi):\ \pi=\pi_0\cdots \pi_{i-\ell-1}\in \mathcal{G},\ s(\pi)=F \}. \end{align} $$

Since the Fischer cover is right-resolving, all paths in $\{ \pi =\pi _0\cdots \pi _{i-\ell -1}\in \mathcal {G}, s(\pi )=F \}$ have different labels. Then, the conclusion follows by using the k-regularity in (3.4).

We introduce a probability measure on $\mathcal {S}$ .

Definition 3.12. For every $i\geq \ell $ , the distribution $\mu _{i}$ on $\mathcal {S}={\mathbb Z}_{\mathbf {a}}\times \mathcal {V}$ is defined as

$$ \begin{align*}\mu_{i}(\{(\mathbf{b},F)\}):=\dfrac{|\{w\in \mathcal{L}_{\mathcal{V},\ell}^{i}(\Sigma):\ \mathbf{f}((w)_g)\equiv \mathbf{b}\ \mathrm{mod}\ \mathbf{a},\ F(w)=F\}|}{|\mathcal{L}_{\mathcal{V},\ell}^{i}(\Sigma)|}\end{align*} $$

for any $(\mathbf {b},F)\in \mathcal {S}$ .

We will denote $\mu _{i}(\mathbf {b},F):=\mu _{i}(\{(\mathbf {b},F)\})$ . By understanding the powers of the matrix $M_i$ , we can describe the evolution of the measure $\mu _{i+np}$ with respect to $n\in {\mathbb N}$ .

Proposition 3.13. For any $i,n\in {\mathbb N}$ and $s=(\mathbf {b}, F),s'=(\mathbf {b}',F')\in \mathcal {S}$ ,

(3.5) $$ \begin{align} M_{i}^n(s,s')=k^{-pn}|\{w\in \mathcal{L}^{pn}(\Sigma):\ \mathbf{f}(g^{i}(w)_g)\equiv \textbf{b}'-\textbf{b}\ \mathrm{mod}\ \textbf{a}, \ F\overset{w}{\to} F'\}|. \end{align} $$

Moreover, for every $i\geq \ell $ and $n\in {\mathbb N}$ ,

(3.6) $$ \begin{align} \mu_{i+np}=\mu_{i} M_{i}^n. \end{align} $$

Proof. We proceed by induction to show (3.5). Let $s=(\mathbf {b}, F), s'=(\mathbf {b}',F')\in \mathcal {S}$ . The case $n=1$ holds by definition. Suppose that the result holds for some $n\in {\mathbb N}$ , so we have to prove the result for $n+1$ . Since

$$ \begin{align*}M_i^{n+1}(s,s')=\sum_{\tilde{s}\in \mathcal{S}}M_i^{n}(s,\tilde{s})M_i(\tilde{s},s'),\end{align*} $$

from the induction hypotheses, it will be sufficient to prove that

(3.7) $$ \begin{align} \begin{aligned} &\{w\in \mathcal{L}^{p(n+1)}(\Sigma):\ \mathbf{f}(g^{i}(w)_g)\equiv \textbf{b}'-\textbf{b}\ \mathrm{mod}\ \textbf{a}, \ F\overset{w}{\to} F'\}\\& \quad =\bigcup_{(\tilde{\mathbf{b}},\tilde{F})\in \mathcal{S}} ( \{x\in \mathcal{L}^{pn}(\Sigma):\ \mathbf{f}(g^{i}(x)_g)\equiv \tilde{\textbf{b}}-\textbf{b}\ \mathrm{mod}\ \textbf{a}, \ F\overset{x}{\to} \tilde{F}\}\\ &\hspace{2cm}\times\{y\in \mathcal{L}^{p}(\Sigma):\ \mathbf{f}(g^{i}(w)_g)\equiv \textbf{b}'-\tilde{\textbf{b}}\ \mathrm{mod}\ \textbf{a}, \ \tilde{F}\overset{y}{\to} F'\}). \end{aligned} \end{align} $$

Let $w\in \mathcal {L}^{p(n+1)}$ in the left-hand side of (3.7) and consider the decomposition $w=(x,y)$ , where $x=w_0\cdots w_{pn-1}$ and $y=w_{pn}\cdots w_{n(p+1)-1}$ . Since $\mathbf {f}(g^i(w)_g)=\mathbf {f}(g^i(x)_g)+\mathbf {f}(g^{i+pn}(y)_g),$ the eventual periodicity leads to

$$ \begin{align*} \mathbf{f}(g^i(w)_g)\equiv_{\mathbf{a}}\mathbf{b}'-\mathbf{b} \text{ if and only if }\mathbf{f}(g^i(x)_g)\equiv_{\mathbf{a}}\tilde{\mathbf{b}}-\mathbf{b}\text{ and }\mathbf{f}(g^i(y)_g)\equiv_{\mathbf{a}}\mathbf{b}'-\tilde{\mathbf{b}}, \end{align*} $$

where $\tilde {\mathbf {b}}:=\mathbf {f}(g^i(x)_g)+\mathbf {b}\ \mathrm {mod}\ \mathbf {a}$ . In addition, $F\overset {w}{\to }F'$ is equivalent to the existence of a path $\pi $ in $\mathcal {G}$ of length $(n+1)p$ from F to $F'$ labelled by w. If we denote by $\tilde {F}$ the terminal node for the subpath $\pi _0\cdots \pi _{np-1}$ , $F\overset {w}{\to }F' \text { if and only if }F\overset {x}{\to }\tilde {F}\text { and }\tilde {F}\overset {y}{\to }F'$ . Thus, we conclude (3.7). However,

$$ \begin{align*}(\mu_iM^n_i)(s)=\sum_{s'\in \mathcal{S}}\mu_i(s')M_i^n(s',s).\end{align*} $$

By Proposition 3.11, $k^{pn}|\mathcal {L}^i_{\mathcal {V},\ell }(\Sigma )|=|\mathcal {L}^{i+np}_{\mathcal {V},\ell }(\Sigma )|$ . Hence, using (3.5), we can conclude (3.6) by showing that

(3.8) $$ \begin{align} \begin{aligned} &\{w\in \mathcal{L}_{\mathcal{V},\ell}^{i+(n+1)p}(\Sigma):\ \mathbf{f}((w)_g)\equiv \mathbf{b}\ \mathrm{mod}\ \mathbf{a},\ F(w)=F\}\\& \quad =\bigcup_{s'\in \mathcal{S}}(\{x\in \mathcal{L}_{\mathcal{V},\ell}^{i}(\Sigma):\ \mathbf{f}((w)_g)\equiv \mathbf{b}'\ \mathrm{mod}\ \mathbf{a},\ F(w)=F'\}\\ &\hspace{1cm}\times \{y\in \mathcal{L}^{pn}(\Sigma):\ \mathbf{f}(g^{i}(y)_g)\equiv \textbf{b}-\textbf{b}'\ \mathrm{mod}\ \textbf{a}, \ F'\overset{y}{\to} F\}). \end{aligned} \end{align} $$

If $w\in \mathcal {L}_{\mathcal {V},\ell }^{i+(n+1)p}(\Sigma )$ , it is straightforward that $w_0\cdots w_{i-1}\in \mathcal {L}^i_{\mathcal {V},\ell }(\Sigma )$ . The rest of the proof follows by replicating the argument used to prove (3.7).

From the previous results, we can characterize the distribution using Markov chains.

5.1 Uniform distribution of integers with missing digits

Let $g\geq 2$ be an integer and let $\mathcal {D}\subseteq \{0,\ldots ,g-1\}$ be a subset of digits with at least two elements. We will write $\mathcal {D}=\{d_1,\ldots ,d_t\}$ , where $d_1<d_2<\cdots <d_t$ . Recalling that $\mathcal {C}_{g,\mathcal {D}}=A_\Sigma $ for $\Sigma =\mathcal {D}^{{\mathbb N}_0}$ , it is easy to see that the Fischer cover for $\Sigma $ is the labelled graph with a unique node F and for each $d\in \mathcal {D}$ , there is a path from F to F labelled by d. Thus, $\Sigma $ is mixing, sofic and regular, so we can apply Theorem 4.9. Moreover, the existence of a single follower set allows us to obviate this notion in the following discussion.

Definition 5.1. Let $A\subseteq {\mathbb N}_0$ be a $\times g$ -invariant set and $a,a'\in {\mathbb N}$ . We say A is jointly uniformly distributed $\ \mathrm {mod}\ (a,a')$ if

$$ \begin{align*}\lim_{N\to\infty}\dfrac{|\{n\in A:\ n\equiv b\ \mathrm{mod}\ a,\ S_g(n)\equiv b'\ \mathrm{mod}\ a'\}\cap[0,N)|}{|A\cap [0,N)|}=\dfrac{1}{aa'}\end{align*} $$

for every $b\in {\mathbb Z}_a,b'\in {\mathbb Z}_{a'}$ .

From Proposition 3.2, we can find an eventual period of the form $(p,\ell )$ for $\mathbf {f}=(\operatorname {\mathrm {id}},S_g)$ . Thus, checking that $M_i$ is irreducible reduces to show that for all $(b,b')\in {\mathbb Z}_a\times {\mathbb Z}_{a'}$ ,

(5.3) $$ \begin{align} \text{there is } n\in {\mathbb N} \text{ and } w\in \mathcal{D}^{np} \text{ such that }g^i(w)_g\equiv b\ \mathrm{mod}\ a \text{ and }S_g(w)\equiv b'\ \mathrm{mod}\ a'. \end{align} $$

Similarly, $M_i$ is aperiodic if

(5.4) $$ \begin{align} \gcd \{n\kern1.4pt{\in}\kern1.4pt {\mathbb N}: \text{there exists } w\in \mathcal{D}^{pn} \text{ such that } (w)_g\equiv 0\ \mathrm{mod}\ a \text{ and } S_g(w)\equiv 0\ \mathrm{mod}\ a'\}\kern1.4pt{=}\kern1.4pt1. \end{align} $$

Aperiodicity in the case $\gcd (g,a)=1$ follows immediately from the following result.

Proof. Let $n\in {\mathbb N}$ . Notice that $\gcd (g,a(g-1))=1$ , so the Euler’s theorem implies that $g^p\equiv 1\ \mathrm {mod}\ a(g-1)$ . If $w=d\cdots d\in \mathcal {D}^{np}$ ,

(5.5) $$ \begin{align} (w)_g=d+dg+\cdots+dg^{np-1}=d\dfrac{g^{np}-1}{g-1}. \end{align} $$

Since $g^{np}-1$ is divisible by $a(g-1)$ , $({g^{np}-1})/({g-1})\equiv 0\ \mathrm {mod}\ a$ , concluding that $(w)_g\equiv 0\ \mathrm {mod}\ a$ . However, $S_g(w)=npd=a'(n\phi (a(g-1))d)\equiv 0\ \mathrm {mod}\ a'$ .

The following application of Bézout’s identity is fundamental to check the irreducibility of $M_i$ .

Lemma 5.3. Let $g\geq 2$ be an integer and let $\mathcal {D}=\{d_1,\ldots ,d_t\}$ be a set of digits, where $d_1$ is the smallest digit. Consider $a,a'\in {\mathbb N}$ such that $\gcd (g,a)=1$ and define $p:=a'\phi (a(g-1))$ , where $\phi $ is the Euler’s totient function. Then, there exists a word $w\in \mathcal {L}(\mathcal {D}^{{\mathbb N}_0})$ such that

$$ \begin{align*}|w|\equiv 0\ \mathrm{mod}\ p, \quad (w)_g\equiv \delta\ \mathrm{mod}\ a\quad \text{and}\quad S_g(w)\equiv \delta\ \mathrm{mod}\ a',\end{align*} $$

where $\delta :=\gcd (aa',d_2-d_1,\ldots ,d_t-d_1).$

Proof. From Bézout’s identity, there exist $n_1,n_2,\ldots ,n_t\in {\mathbb Z}$ such that

$$ \begin{align*}\delta=n_1(aa')+n_2(d_2-d_1)+\cdots+n_t(d_t-d_1).\end{align*} $$

For all $j\in \{2,\ldots ,t\}$ , choose $k_j\in {\mathbb N}$ large enough such that $m_j:=k_j(aa')+n_j>0$ . Then,

(5.6) $$ \begin{align} m_2(d_2-d_1)+\cdots+m_t(d_t-d_1)\equiv \delta\ \mathrm{mod}\ aa'. \end{align} $$

Let us define the word $w(j):=d_jd_1\cdots d_1\in \mathcal {D}^p$ for $j\in \{2,\ldots ,t\}$ . From Lemma 5.2,

$$ \begin{align*}(w(j))_g=(d_j-d_1)+d_1+d_1g+\cdots+d_1g^{p-1}\equiv d_j-d_1\ \mathrm{mod}\ a.\end{align*} $$

Similarly, $S_g(w(j))=pd_1+(d_j-d_1)\equiv d_j-d_1 \ \mathrm {mod}\ a'$ . Let us define

$$ \begin{align*}w:=\underbrace{w(2)\cdots w(2)}_{m_2 \text{ times}}\cdots \underbrace{w(t)\cdots w(t)}_{m_t \text{ times}}.\end{align*} $$

Notice that $|w|=p(m_2+\cdots +m_t)$ . Also, the periodicity $g^p\equiv 1 \ \mathrm {mod}\ a$ and (5.6) imply that

$$ \begin{align*}(w)_g\equiv \sum_{j=2}^t(w(j))_gm_j\equiv \sum_{j=2}^t(d_j-d_1)m_j\equiv \delta\ \mathrm{mod}\ a.\end{align*} $$

Using (5.6) again, we can conclude

$$ \begin{align*}S_g(w)=\sum_{j=2}^t S_g(w(j))m_j\equiv \sum_{j=2}^t (d_j-d_1)m_j\equiv \delta\ \mathrm{mod}\ a'.\\[-47pt] \end{align*} $$

Leveraging Theorem 4.9 and Lemma 5.3, we can establish a complete description of the phenomenon of joint distribution (Definition 5.1) for missing digits sets when $\gcd (g,a)=\gcd (a,a')=1$ . In particular, this shows Theorem A.

Theorem 5.4. Let $g\geq 2$ be an integer and let $\mathcal {D}=\{d_1,\ldots ,d_t\}$ be a set of digits, where $d_1$ is the smallest digit. Suppose that $\gcd (g,a)=\gcd (a,a')=1$ and recall the missing digits set $\mathcal {C}_{g,\mathcal {D}}$ defined in (1.1). Defining $\delta :=\gcd (aa',d_2-d_1,\ldots ,d_t-d_1)$ , we get the following.

1. (i) $\mathcal {C}_{g,\mathcal {D}}$ is jointly uniformly distributed $\ \mathrm {mod}\ (a,a')$ if and only if $\delta =1$ .

More generally, concerning the limit

(5.7) $$ \begin{align}L(b,b'):=\lim_{N\to\infty}\dfrac{|\{n\in \mathcal{C}_{g,\mathcal{D}}:\ n\equiv b\ \mathrm{mod}\ a,\ S_g(n)\equiv b'\ \mathrm{mod}\ a'\}\cap[0,N)|}{|\mathcal{C}_{g,\mathcal{D}}\cap [0,N)|},\end{align} $$

we have the following.

1. (ii) Let $\delta _a:=\gcd (\delta ,a)$ and $\delta _{a'}:=\gcd (\delta ,a')$ . If $d_1\equiv 0\ \mathrm {mod}\ \delta _a$ and $d_1\equiv 0\ \mathrm {mod}\ \delta _{a'}$ , then $L(b,b')=\delta /aa'$ for every $(b,b')$ in the subgroup $\langle \delta _{a}\rangle \times \langle \delta _{a'}\rangle \subseteq {\mathbb Z}_a\times {\mathbb Z}_{a'}$ . However, if $(b,b')\notin \langle \delta _{a}\rangle \times \langle \delta _{a'}\rangle $ ,

   $$ \begin{align*}\{n\in \mathcal{C}_{g,\mathcal{D}}:\ n\equiv b\ \mathrm{mod}\ a, \ S_g(n)\equiv b'\ \mathrm{mod}\ a'\}=\emptyset.\end{align*} $$

2. (iii) In any other case, for every $(b,b')\in {\mathbb Z}_a\times {\mathbb Z}_{a'}$ , the limit $L(b,b')$ either equals zero or does not exist.

Proof. Let us define $p=a'\phi (a(g-1))$ , where $\phi $ is the Euler’s totient function, and note that $(p,1)$ is an eventual period for $\mathbf {f}=(\operatorname {\mathrm {id}},S_g)$ with respect to $\mathbf {a}=(a,a')$ . For $i\in {\mathbb N}$ , let $X^i$ be the Markov chain associated to $g, \mathbf {a}, \mathbf {f}, p, \ell $ and the subshift $\Sigma =\mathcal {D}^{{\mathbb N}_0}$ , as defined in Definition 3.14. We denote by $M_i$ its transition matrix and $\mu _i$ its initial distribution.

The aperiodicity of $M_i$ follows from Lemma 5.2 and (5.4).

Let us suppose $\delta =1$ . From Lemma 5.3, there exists a word w using symbols of $\mathcal {D}$ such that $|w|\equiv 0\ \mathrm {mod}\ p$ , $(w)_g\equiv 1\ \mathrm {mod}\ a$ and $S_g(w)\equiv 1\ \mathrm {mod}\ a'$ . Since $\gcd (a,a')=1$ , the element $(1,1)$ is a generator for ${\mathbb Z}_a\times {\mathbb Z}_{a'}$ . Then, for every $(b,b')\in {\mathbb Z}_a\times {\mathbb Z}_{a'}$ , there exists $m\in {\mathbb N}$ such that $m\equiv (g^i)^{-1}b\ \mathrm {mod}\ a$ and $m\equiv b'\ \mathrm {mod}\ a'$ . Let $y\in \mathcal {L}(\mathcal {D}^{{\mathbb N}_0})$ be constructed by concatenating w with itself m times. The periodicity $g^p\equiv 1\ \mathrm {mod}\ a$ implies that $g^i(y)_g\equiv g^i m(w)_g\equiv b\ \mathrm {mod}\ a$ . However, $S_g(y)\equiv mS_g(w)\equiv b'\ \mathrm {mod}\ a'$ . Therefore, $M_i$ is irreducible for every $i\in {\mathbb N}$ . This concludes that the Markov condition holds, then Theorem 4.9 implies that $\mathcal {C}_{g,\mathcal {D}}$ is jointly uniformly distributed $\ \mathrm {mod}\ (a,a').$

Now, let us suppose that $\delta \neq 1$ . Since $\delta $ divides $aa'$ and $\gcd (a,a')=1$ , we can write $\delta =\delta _{a}\cdot \delta _{a'}$ , where $\delta _{a}:=\gcd (\delta ,a)$ and $\delta _{a'}:=\gcd (\delta ,a')$ . In particular, $\gcd (\delta _a,\delta _{a'})=1.$

From the definition of $\delta $ , note that $d_j\equiv d_1\ \mathrm {mod}\ \delta $ for all $j\in \{1,\ldots ,t\}$ . Therefore, for any $m\in {\mathbb N}$ and $x=x_0\cdots x_{m-1}\in \mathcal {D}^m$ ,

(5.8) $$ \begin{align} (x)_g=\sum_{j=0}^{m-1}(x_j-d_1)g^j+d_1g^j\equiv d_1\sum_{j=0}^{m-1}g^j\ \mathrm{mod}\ \delta_a \quad \text{and}\quad S_g(x)\equiv md_1\ \mathrm{mod}\ \delta_{a'}. \end{align} $$

In particular, for every $x\in \mathcal {L}(\mathcal {D}^{{\mathbb N}_0})$ such that $|x|\equiv 0\ \mathrm {mod}\ p$ , Lemma 5.2 implies that $(x)_g\equiv 0\ \mathrm {mod}\ \delta _a$ and $S_g(x)\equiv 0\ \mathrm {mod}\ \delta _{a'}$ . However, from Lemma 5.3, we can find a word y such that $|y|\equiv 0\ \mathrm {mod}\ p$ , $(y)_g\equiv \delta \ \mathrm {mod}\ a$ and $S_g(y)\equiv \delta \ \mathrm {mod}\ a'.$ By the Chinese remainder theorem, $(\delta ,\delta )$ is a generator for the subgroup $\mathcal {S}(\delta _a,\delta _{a'}):=\langle \delta _a\rangle \times \langle \delta _{a'}\rangle \subseteq {\mathbb Z}_a\times {\mathbb Z}_{a'}$ . Therefore, recalling that $\gcd (g,\delta _a)=1$ ,

(5.9) $$ \begin{align} \bigcup_{n\in{\mathbb N}}\{w\in \mathcal{D}^{np}:\ g^i(w)_g\equiv b\ \mathrm{mod}\ a,\ S_g(w)\equiv b'\ \mathrm{mod}\ a'\}\neq \emptyset \end{align} $$

if and only if $(b,b')\in \mathcal {S}(\delta _a,\delta _{a'})$ . In particular, this already ensures that $M_i$ is not irreducible on ${\mathbb Z}_a\times {\mathbb Z}_{a'}$ , so the Markov condition does not hold. However, we can show that $X^i$ is an irreducible and aperiodic Markov chain when restricting the state space. Recall the initial distribution of $X^i$ given by

$$ \begin{align*} \mu_{i}(b,b')=\dfrac{|\{w\in \mathcal{D}^i: \ (w)_g\equiv b\ \mathrm{mod}\ a,\ S_g(w)\equiv b'\ \mathrm{mod}\ a'\}|}{|\mathcal{D}|^i}. \end{align*} $$

Defining $\delta _a(i):= d_1\sum _{j=0}^{i-1}g^{i-1}\ \mathrm {mod}\ \delta _a$ and $\delta _{a'}(i):= id_1\ \mod {\delta _{a'}}$ , (5.8) implies that $(x)_g\equiv \delta _a(i)\ \mathrm {mod}\ \delta _a$ and $S_g(x)\equiv \delta _{a'}(i)\ \mathrm {mod}\ \delta _{a'}$ for every $x\in \mathcal {D}^i$ . Therefore, ${\mu _i(b,b')\neq 0}$ implies $b\equiv \delta _a(i) \ \mathrm {mod}\ \delta _a$ and $b'\equiv \delta _{a'}(i)\ \mathrm {mod}\ \delta _{a'}$ . If we define

$$ \begin{align*} \mathcal{S}(i):=(\delta_a(i),\delta_{a'}(i))+\mathcal{S}(\delta_a,\delta_{a'}), \end{align*} $$

the equivalence in (5.9) indicates that $X^i$ never visits the states in ${\mathbb Z}_a\times {\mathbb Z}_{a'}\setminus \mathcal {S}(i)$ , and also $X^i$ is irreducible and aperiodic when restricted to the state space $\mathcal {S}(i)$ . Note that $M^i$ restricted to $\mathcal {S}(i)$ is still doubly stochastic, since we only ignore zero-value entries. Therefore, as $|\mathcal {S}(i)|=|\mathcal {S}(\delta _a,\delta _{a'})|=(a/\delta _a)\cdot (a'/\delta _{a'})$ ,

(5.10) $$ \begin{align} \lim_{N\to \infty}\mu_{i+Np}(b,b')=\begin{cases} \delta/aa' & \text{if }(b,b')\in \mathcal{S}(i),\\ 0& \text{if }(b,b')\notin \mathcal{S}(i). \end{cases} \end{align} $$

If $d_1\equiv 0\ \mathrm {mod}\ \delta _a$ and $d_1\equiv 0\ \mathrm {mod}\ \delta _{a'}$ , $\mathcal {S}(i)=\mathcal {S}(\delta _a,\delta _{a'})$ for every $i\in {\mathbb N}$ . This shows that each Markov chain $X^i$ is irreducible and aperiodic in a common state space $\mathcal {S}(\delta _a,\delta _{a'})$ . Note that the Markov condition means that every $X^i$ is irreducible and aperiodic over the state space ${\mathbb Z}_a\times {\mathbb Z}_{a'}$ , but it is easy to see that the procedure of §4 is identical if we assume that each $X^i$ is irreducible and aperiodic over a subgroup of ${\mathbb Z}_a\times {\mathbb Z}_{a'}$ (and the Markov chain never visits the other states). Thus, by replicating the argument of §4, we can conclude

$$ \begin{align*} \lim_{N\to \infty}\dfrac{|\{n\in \mathcal{C}_{g,\mathcal{D}}:\ n\equiv_a b,\ S_g(n)\equiv_{a'} b'\}\cap [0,N)|}{|\mathcal{C}_{g,\mathcal{D}}\cap [0,N)|}=\begin{cases} \delta/aa' & \text{if } (b,b')\in \mathcal{S}(\delta_a,\delta_{a'}), \\ 0& \text{if } (b,b')\notin \mathcal{S}(\delta_a,\delta_{a'}). \end{cases} \end{align*} $$

This shows (5.4). If either $d_1\not \equiv 0\ \mathrm {mod}\ \delta _a$ or $d_1\not \equiv 0\ \mathrm {mod}\ \delta _{a'}$ , notice that $\mathcal {S}(1)\cap \mathcal {S}(2)=\emptyset $ . Then, for every $(b,b')\in {\mathbb Z}_a\times {\mathbb Z}_{a'}$ , there exists $j\in {\mathbb N}$ such that the Markov chain $X^j$ does not visit the state $(b,b')$ . From this fact, we want to show that (5.7) either equals zero or the limit does not exist. Let $(b,b')\in {\mathbb Z}_a\times {\mathbb Z}_{a'}$ and let $j\in {\mathbb N}$ such that $X^j$ never visits the state $(b,b')$ . By contradiction, suppose that $L(b,b')>0$ . Since $X^j$ does not visit the state $(b,b')$ ,

$$ \begin{align*}\{n\in \mathcal{C}_{g,\mathcal{D}}:\ n\equiv b\ \mathrm{mod}\ a,\ S_g(n)\equiv b'\ \mathrm{mod}\ a'\}\cap [g^{j+mp-1},g^{j+mp})=\emptyset\end{align*} $$

for every $m\in {\mathbb N}$ . The contradiction follows by noting that

$$ \begin{align*} L(b,b') & = \lim_{m\to \infty}\dfrac{|\{n\in \mathcal{C}_{g,\mathcal{D}}:\ n\equiv b\ \mathrm{mod}\ a,\ S_g(n)\equiv b'\ \mathrm{mod}\ a'\}\cap [0,g^{j+mp-1})|}{|\mathcal{C}_{g,\mathcal{D}}\cap [0,g^{j+mp-1})|}\\ &>\liminf_{m\to \infty}\dfrac{|\{n\in \mathcal{C}_{g,\mathcal{D}}:\ n\equiv b\ \mathrm{mod}\ a,\ S_g(n)\equiv b'\ \mathrm{mod}\ a'\}\cap [0,g^{j+mp})|}{|\mathcal{C}_{g,\mathcal{D}}\cap [0,g^{j+mp})|}. \end{align*} $$

This concludes case (5.4).

Note that in the cases (5.4) and (5.4), we can conclude from Proposition 4.6 instead of using Theorem 4.9, as $\Sigma $ is a full shift. Hence, the order of convergence $O(N^{-\gamma })$ for some $\gamma>0$ can be provided for the limit (5.7). An immediate consequence follows by taking $a'=1$ , which allows us to fully describe the phenomenon of uniform distribution $\ \mathrm {mod}\ a$ when $\gcd (g,a)=1$ . This shows Corollary A.

Note that the condition $\gcd (g,a)= 1$ is commonly assumed in results of this kind. In the framework of Corollary 5.5, when $\gcd (g,a)\neq 1$ , the matrices $M_i$ are no longer irreducible over ${\mathbb Z}_a$ . However, the Markov chains constructed can still capture the behaviour of the distribution by breaking $X^i$ into pieces determined by the communication classes of the Markov chain (see [Reference Norris44, §1.2] for details about this notion). The disadvantage of this situation is that the convergence of $X^i$ depends on the initial distribution.

The following simple example shows that when $\gcd (a,g)\neq 1$ , the limiting behaviour depends heavily on the initial distribution.

Example 5.6. Let $a=12$ , $g=6$ and $\mathcal {D}=\{1,2,4\}$ . Notice that a divides $g^i$ for every $i\geq 2$ , then $(w_0\cdots w_n)_g\equiv w_0+w_1g\ \mathrm {mod}\ a$ . Therefore,

$$ \begin{align*}\lim_{N\to \infty}\dfrac{|\{n\in \mathcal{C}_{6,\mathcal{D}}:\ n\equiv b\ \mathrm{mod}\ 12\}\cap [0,N)|}{|\mathcal{C}_{6,\mathcal{D}}\cap [0,N)|}=\begin{cases} 2/9 & \text{if } b\in \{1,2,4\},\\ 1/9 & \text{if } b\in\{7,8,10\},\\ 0 & \text{if } b\in\{0,3,5,6,9,11\}. \end{cases}\end{align*} $$

Notice that this distribution agrees with the distribution when considering $[0,g^2)$ . Also, we can see that the limit is different and non-zero for some classes, behaviour which is not observed under the assumption $\gcd (g,a)=1$ .

Finally, we will present an equivalent condition for uniform distribution $\ \mathrm {mod}\ (a,a')$ without the assumption $\gcd (a,a')=1$ , however, at the cost of having a more complicated condition to check.

Proof. If $\mathcal {C}_{g,\mathcal {D}}$ is jointly uniformly distributed $\ \mathrm {mod}\ (a,a')$ , in particular, it is uniformly distributed $\ \mathrm {mod}\ a$ . From Corollary 5.5, it follows that $\gcd (a,d_2-d_1,\ldots ,d_t-d_1)=1$ . In addition, there exist infinitely many words w such that $(w)_g\equiv 0\ \mathrm {mod}\ a$ , $S_g((w)_g)\equiv 1\ \mathrm {mod}\ a'$ and $|w|\equiv 0 \ \mathrm {mod}\ p$ . Otherwise, for every $m\in {\mathbb N}$ ,

$$ \begin{align*} \{n\in \mathcal{C}_{g,\mathcal{D}}:\ n\equiv 0\ \mathrm{mod}\ a,\ S_g(n)\equiv 1\ \mathrm{mod}\ 1\}\cap [g^{mp-1},g^{mp})=\emptyset, \end{align*} $$

contradicting that

$$ \begin{align*} \lim_{N\to\infty}\dfrac{|\{n\in \mathcal{C}_{g,\mathcal{D}}:\ n\equiv 0\ \mathrm{mod}\ a,\ S_g(n)\equiv 1\ \mathrm{mod}\ a'\}\cap[0,N)|}{|\mathcal{C}_{g,\mathcal{D}}\cap [0,N)|}=\dfrac{1}{aa'}>0. \end{align*} $$

However, let us suppose that $\gcd (a,d_2-d_1,\ldots ,d_t-d_1)=1$ and that there exists $w\in \mathcal {L}(\mathcal {D}^{{\mathbb N}_0})$ such that $(w)_g\equiv 0\ \mathrm {mod}\ a$ , $S_g(w)\equiv 1\ \mathrm {mod}\ a'$ and $|w|\equiv 0 \ \mathrm {mod}\ p$ . From Lemma 5.3, there exists $y\in \mathcal {L}(\mathcal {D}^{{\mathbb N}_0})$ such that $(y)_g\equiv 1\ \mathrm {mod}\ a$ and $|y|\equiv 0\ \mathrm {mod}\ p$ . To prove that $M_i$ is irreducible, we will check (5.3). Let $b\in {\mathbb Z}_a$ and $b'\in {\mathbb Z}_{a'}$ . If $\tilde {y}$ be the concatenation $((g^{i})^{-1}b\ \mathrm {mod}\ a)$ -times of the word y (in the case where $b=0$ , we concatenate a times), the periodicity $g^p\equiv 1\ \mathrm {mod}\ a$ implies that $g^i(\tilde {y})_g\equiv g^i(g^{i})^{-1}b\equiv b \ \mathrm {mod}\ a$ . If $s:=S_g(\tilde {y})$ , we denote by $\tilde {w}$ the concatenation $(b'-s\ \mathrm {mod}\ a')$ -times of w. Thus, we can conclude that $g^i(\tilde {y}\tilde {w})_g\equiv b\ \mathrm {mod}\ a$ and $S_g(\tilde {y}\tilde {w})\equiv b'\ \mathrm {mod}\ a'$ . Clearly, $|\tilde {y}\tilde {w}|\equiv 0\ \mathrm {mod}\ p$ . Therefore, $M_i$ is irreducible. The matrix $M_i$ is also aperiodic as a consequence of Lemma 5.2. Thus, the Markov condition holds and the result follows by Theorem 4.9.

5.2 Revisiting Gelfond’s theorem: an equivalent condition

Let $g\geq 2$ be an integer and $a,a'\in {\mathbb N}$ . As presented in Corollary 1.2, Gelfond [Reference Gelfond25] studied the long-term behaviour of

(5.11) $$ \begin{align} |\{n\leq N:\ n\equiv b\ \mathrm{mod}\ a,\ S_g(n)\equiv b'\ \mathrm{mod}\ a'\}|, \end{align} $$

showing that, under the assumption $\gcd (g-1,a')=1$ , the set $A={\mathbb N}$ is jointly uniformly distributed $\ \mathrm {mod}\ (a,a')$ . In this section, we present an extension of Corollary 1.2, obtaining an equivalent condition to uniform distribution $\ \mathrm {mod}\ (a,a')$ .

Notice that the obstruction in applying our method arises from the lack of the condition $\gcd (g,a)=1$ , which implies that the Markov chains associated with this problem are not irreducible. However, we can still break each Markov chain $X^i$ into pieces, generating irreducible and aperiodic Markov chains. In that way, the behaviour of (5.11) can be understood using our method.

Theorem 5.8. Let $g\geq 2$ be an integer and $a,a'\in {\mathbb N}$ . Then,

(5.12) $$ \begin{align}\lim_{N\to\infty}\dfrac{ |\{n\leq N:\ n\equiv b\ \mathrm{mod}\ a,\ S_g(n)\equiv b'\ \mathrm{mod}\ a'\}|}{N}=\dfrac{1}{aa'} \end{align} $$

for every $(b,b')\in {\mathbb Z}_a\times {\mathbb Z}_{a'}$ if and only if there exists $n\in {\mathbb N}$ such that $n\equiv 0\ \mathrm {mod}\ a$ and $S_g(n)\equiv 1\ \mathrm {mod}\ a'.$

Proof. If (5.12) holds, there are infinitely many $n\in {\mathbb N}$ such that $n\equiv 0\ \mathrm {mod}\ a$ and $S_g(n)\equiv 1\ \mathrm {mod}\ a'$ .

For the other direction, let $n\in {\mathbb N}$ be such that $n\equiv 0\ \mathrm {mod}\ a$ and $S_g(n)\equiv 1\ \mathrm {mod}\ a'$ . For $\mathbf {f}=(\operatorname {\mathrm {id}},S_g)$ , let $p\in {\mathbb N}$ be given by Proposition 3.2. Then, $(p,a-1)$ is an eventual period for $\mathbf {f}$ with respect to $\mathbf {(a,a')}$ and notice that $\Sigma =\{0,\ldots ,g-1\}^{{\mathbb N}_0}$ is such that ${\mathbb N}=A_\Sigma $ . For $i\geq a-1$ , let $X^i$ be the Markov chain of Definition 3.14, with transition matrix $M_i$ and initial distribution $\mu _i$ .

Let $w=10\cdots 0$ such that $|w|=p$ and clearly $g^i(w)_g\equiv g^i\ \mathrm {mod}\ a$ . If y is the base-g expansion of n (by adding $0$ s, we can suppose that y has length divisible by p), then $(y)_g\equiv 0\ \mathrm {mod}\ a$ and $S_g(y)\equiv 1\ \mathrm {mod}\ a'$ . Hence, for any $(b,b')\in {\mathbb Z}_a\times {\mathbb Z}_{a'}$ , there exists $k\in {\mathbb N}$ such that $(w^{b}y^{k})_g\equiv b\ \mathrm {mod}\ a$ and $S_g(w^{b}y^{k})_g\equiv b'\ \mathrm {mod}\ a'$ (where $z^j$ denotes the concatenation of $z j$ -times). Therefore,

(5.13) $$ \begin{align} \bigcup_{m\in{\mathbb N}}\{w\in \mathcal{D}^{mp}:\ g^i(w)_g\equiv b\ \mathrm{mod}\ a,\ S_g(w)\equiv b'\ \mathrm{mod}\ a'\}\neq \emptyset \end{align} $$

if and only if $(b,b')\in \langle g^i \rangle \times {\mathbb Z}_{a'}\subseteq {\mathbb Z}_a\times {\mathbb Z}_{a'}.$ Denoting $\delta _i:=\gcd (g^i,a)$ , it follows that $\sum _{m\in {\mathbb N}}M_i^m((b_0,b_1),(b_0',b_1'))>0$ if and only if $b_0\equiv b_0' \ \mathrm {mod}\ \delta _i$ . Thus, if the Markov chain $X^i$ starts at $X_0^i=(b,b')$ , that trajectory will only visit classes in $ (b+\langle \delta _i \rangle )\times {\mathbb Z}_{a'}\subseteq {\mathbb Z}_a\times {\mathbb Z}_{a'}.$ In this way, we can split $X^i$ into pieces depending on the starting point.

For each $j\in \{0,\ldots ,\delta _i-1\}$ , we define $\mathcal {S}(i,j):=j+\langle \delta _i \rangle $ and

$$ \begin{align*}\mu_{i,j}(b,b'):=\begin{cases} \dfrac{\mu_i(b,b')}{\bar{\mu}_{i,j}}& \text{if } b\equiv j\ \mathrm{mod}\ \delta_i, \\ 0& \text{otherwise,} \end{cases}\end{align*} $$

where $\mu _i$ is the initial distribution of $X^i$ and

$$ \begin{align*}\bar{\mu}_{i,j}:=\sum_{b'\in {\mathbb Z}_{a'}}\sum_{b\in \mathcal{S}(j)}\mu_i(b,b')=\dfrac{|w\in \{0,\ldots,g-1\}^i:\ (w)_g\equiv j\ \mathrm{mod}\ \delta_i|}{g^i}=\dfrac{1}{\delta_i}.\end{align*} $$

Let $X^{i,j}$ be a Markov chain on the state space $\mathcal {S}(i,j)$ , with initial distribution $\mu _{i,j}$ and transition matrix $M_i$ , both restricted to indexes of $\mathcal {S}(i,j)$ . The aperiodicity follows directly as $0\cdots 0$ is in the language of $\{0,\ldots ,g-1\}^{{\mathbb N}_0}$ , and $M_{i}$ restricted to $\mathcal {S}(i,j)$ is irreducible by (5.13). Therefore, the Markov chain $X^{i,j}$ converges to its invariant distribution given by $1/|\mathcal {S}(i,j)|=\delta _i/aa'.$ It is not difficult to see that

(5.14) $$ \begin{align} \mu_{i+np}=\sum_{j=0}^{\delta_i-1}\bar{\mu}_{i,j}(\mu_{i,j}M^n_i)=\dfrac{1}{\delta_i}\sum_{j=0}^{\delta_i-1}\mu_{i,j}M^n_i. \end{align} $$

From Theorem 2.1, there exist $C_{i,j}>0$ and $\rho _{i,j}\in (0,1)$ such that

(5.15) $$ \begin{align} \bigg|(\mu_{i,j}M_i^n)(b,b')-\dfrac{\delta_i}{aa'}\bigg|\leq C_{i,j}\rho_{i,j}^n \end{align} $$

for every $(b,b')\in \mathcal {S}(i,j).$ Clearly, $(\mu _{i,j}M_i^n)(b,b')=0$ if $(b,b')\notin \mathcal {S}(i,j)$ . Since ${\mathbb Z}_a\times {\mathbb Z}_{a'}=\bigsqcup _{j\in \{0,\ldots ,\delta _i-1\}}\mathcal {S}(i,j)$ , from (5.14) and (5.15), we can obtain that, for every $(b,b')\in {\mathbb Z}_a\times {\mathbb Z}_{a'}$ ,

(5.16) $$ \begin{align} \bigg|\mu_{i+np}(b,b')-\dfrac{1}{aa'}\bigg|\leq C_i\rho_i^n, \end{align} $$

where $ C_i:=\max _{j=0,\ldots ,\delta _i-1}C_{i,j}$ and $\rho _i:=\max _{j=0,\ldots ,\delta _i-1}\rho _{i,j}$ . Equivalently,

(5.17) $$ \begin{align} \bigg|\mathbb{P}(X^i_n=(b,b'))-\dfrac{1}{aa'}\bigg|\leq C_i\rho_i^n. \end{align} $$

Despite each $X^i$ being not irreducible, we still have convergence to the uniform measure $1/aa'$ . Note that (5.17) is equivalent to the estimation (4.3) in the case of $\Sigma $ being a full shift, and essentially, this is what is required to retrieve the results of §4 and, in particular, Theorem 4.9. Therefore, we can conclude that ${\mathbb N}$ is jointly uniformly distributed $\ \mathrm {mod}\ (a,a')$ by replicating the arguments of §4.

5.3 Forbidding combinations of digits

So far, we have focused on applying Theorem 4.9 in restricted digit sets, where restrictions come from forbidding some digits (missing digits sets). Given the flexibility in choosing digits and restrictions in the general case, it does not seem possible to establish general conditions like those in §5.1 within the context of general $\times g$ -invariant sets. In such a case, Theorem 4.9 is still a useful tool on a case-by-case basis. Seeking greater generality, we construct a class of sets with more flexible digit restrictions that exhibit joint uniform distribution. While we focus on restrictions of length 2, some of these ideas can be extended to longer lengths.

Let $g\geq 2$ be an integer and $a,a'\in {\mathbb N}$ . Let $\mathcal {D}\subseteq \{0,\ldots ,g-1\}$ be a set of digits and let T be a matrix on $\mathcal {D}$ such that $T(d,d')\in \{0,1\}$ for every $d,d'\in \mathcal {D}$ .