Counting Subwords Occurrences in Base-b Expansions

TL;DR

This paper investigates the occurrence of subwords in base-b expansions of non-negative integers, establishing recurrence relations and asymptotic behavior for a related counting sequence using a tree structure.

Contribution

It introduces a new approach using a tree structure to analyze subword counts in base-b expansions and proves the sequence's b-regularity.

Findings

Derived recurrence relations for the sequence

Proved the sequence's b-regularity

Established asymptotic behavior of the summatory function

Abstract

We count the number of distinct (scattered) subwords occurring in the base-b expansion of the non-negative integers. More precisely, we consider the sequence $(S_b(n))_{n\ge 0}$ counting the number of positive entries on each row of a generalization of the Pascal triangle to binomial coefficients of base-$b$ expansions. By using a convenient tree structure, we provide recurrence relations for $(S_b(n))_{n\ge 0}$ leading to the $b$-regularity of the latter sequence. Then we deduce the asymptotics of the summatory function of the sequence $(S_b(n))_{n\ge 0}$.

Tables

Table 1. Table 4: Values of c r , s subscript 𝑐 𝑟 𝑠 c_{r,s} for 0 ≤ r < b 2 0 𝑟 superscript 𝑏 2 0\leq r<b^{2} and 1 ≤ s ≤ b − 2 1 𝑠 𝑏 2 1\leq s\leq b-2 with x , y , z ∈ { 1 , … , b − 2 } 𝑥 𝑦 𝑧 1 … 𝑏 2 x,y,z\in\{1,\dots,b-2\} pairwise distinct.

${\mathrm{rep}}_b\left(r\right)$	$\epsilon$	$x$		$b-1$	$x0$		$\left(b-1\right)0$	$xx$
$s$	$z$	$x$	$z$	$z$	$x$	$z$	$z$	$x$	$z$
$c_{r,s}$	$0$	$1$	$0$	$-1$	$2$	$0$	$-2$	$2$	$0$
${\mathrm{rep}}_b\left(r\right)$	$\left(b-1\right)\left(b-1\right)$	$xy$			$x\left(b-1\right)$		$\left(b-1\right)x$
$s$	$z$	$x$	$y$	$z$	$x$	$z$	$x$	$z$
$c_{r,s}$	$-2$	$2$	$1$	$0$	$1$	$-1$	$-1$	$-2$

Table 2. Table 8: The values of a r , c r , 0 , c r , 1 subscript 𝑎 𝑟 subscript 𝑐 𝑟 0 subscript 𝑐 𝑟 1 a_{r},c_{r,0},c_{r,1} when b = 3 𝑏 3 b=3 and r ∈ { 0 , … , 8 } 𝑟 0 … 8 r\in\{0,\ldots,8\} .

Equations

$$L_b=\{1,\dots ,b-1\}\{0,\dots ,b-1{\}}^{\ast }\cup \{\epsilon \}$$

$$S_b(n):=\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{{\mathrm{rep}}_b(n)}{v}\right)>0\right\}.$$

$$A_b(n):=\sum _{j=0}^{n-1}{S}_b(j).$$

$$1,2,2,3,3,4,3,4,3,4,5,6,5,4,6,7,7,6,4,6,5,7,6,7,5,6,4,5,7,8,8,7,10,\dots$$

$$0,1,3,5,8,11,15,18,22,25,29,34,40,45,49,55,\dots$$

$$S_b(x{b}^{\ell }+r)$$

$$S_b(x{b}^{\ell }+x{b}^{\ell -1}+r)$$

$$S_b(x{b}^{\ell }+y{b}^{\ell -1}+r)$$

$$w=a_1^{n_1}\cdots a_M^{n_M},$$

$$\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{x00u}{v}\right)>0\right\}=2\cdot \#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{x0u}{v}\right)>0\right\}-\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{xu}{v}\right)>0\right\}.$$

$$2\cdot (2+\#T+2\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x\end{array}}\#R_i+\#R_{i_x})-(1+\#T+\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x\end{array}}\#R_i)=3+\#T+3\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x\end{array}}\#R_i+2\#R_{i_x}.$$

$$\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{xx0u}{v}\right)>0\right\}=\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{x0u}{v}\right)>0\right\}+\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{xu}{v}\right)>0\right\}.$$

$$\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{x0yu}{v}\right)>0\right\}=\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{xyu}{v}\right)>0\right\}+\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{yu}{v}\right)>0\right\}.$$

$$\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{xxyu}{v}\right)>0\right\}=2\cdot \#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{xyu}{v}\right)>0\right\}-\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{yu}{v}\right)>0\right\}.$$

$$\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{xyzu}{v}\right)>0\right\}=$$

$$-2\cdot \#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{zu}{v}\right)>0\right\}.$$

$$(2+2\#T+2\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_y,i_0\end{array}}\#R_i+\#R_{i_x}+2\#R_{i_y}+\#R_{i_0})$$

$$+2\cdot (2+2\#T+2\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_y,i_0\end{array}}\#R_i+2\#R_{i_x}+\#R_{i_y}+\#R_{i_0})$$

$$-2\cdot (1+\#T+\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_y,i_0\end{array}}\#R_i+\#R_{i_x}+\#R_{i_y})$$

$$=4+4\#T+4\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_y,i_0\end{array}}\#R_i+3\#R_{i_x}+2\#R_{i_y}+3\#R_{i_0}.$$

$$(2+2\#T+2\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_0\end{array}}\#R_i+\#R_{i_x}+\#R_{i_0})$$

$$+2\cdot (2+\#T+2\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_0\end{array}}\#R_i+2\#R_{i_x}+\#R_{i_0})$$

$$-2\cdot (1+\#T+\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_0\end{array}}\#R_i+\#R_{i_x})$$

$$=4+2\#T+4\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_0\end{array}}\#R_i+3\#R_{i_x}+3\#R_{i_0}.$$

$$(2+\#T+2\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_y,i_0\end{array}}\#R_i+2\#R_{i_y}+\#R_{i_0})$$

$$+2\cdot (2+2\#T+2\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_y,i_0\end{array}}\#R_i+\#R_{i_y}+\#R_{i_0})$$

$$-2\cdot (1+\#T+\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_y,i_0\end{array}}\#R_i+\#R_{i_y})$$

$$=4+3\#T+4\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_y,i_0\end{array}}\#R_i+2\#R_{i_y}+3\#R_{i_0}.$$

$$\#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{xy0u}{v}\right)>0\right\}=$$

$$-2\cdot \#\left\{v\in L_b\mid \left(\genfrac{}{}{0ex}{}{{\mathrm{rep}}_b({\mathrm{val}}_b(u))}{v}\right)>0\right\}.$$

$$(2+\#T+2\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_y\end{array}}\#R_i+\#R_{i_x}+2\#R_{i_y})$$

$$+2\cdot (2+\#T+2\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_y\end{array}}\#R_i+2\#R_{i_x}+\#R_{i_y})$$

$$-2\cdot (1+\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_y\end{array}}\#R_i+\#R_{i_x}+\#R_{i_y})$$

$$=4+3\#T+4\sum _{\begin{array}{c}1\le i\le m\\ i\ne i_x,i_y\end{array}}\#R_i+3\#R_{i_x}+2\#R_{i_y}.$$

$$\begin{array}[]{rclr}S_{b}(xb^{\ell}+r)&=&2S_{b}(xb^{\ell-1}+r)-S_{b}(xb^{\ell-2}+r)&\text{(by Lemma~{}\ref{lem:arbrex00})}\\ &=&2(S_{b}(xb^{\ell-2}+r)+S_{b}(r))-S_{b}(xb^{\ell-2}+r)&\text{(by induction hypothesis)}\\ &=&S_{b}(xb^{\ell-2}+r)+S_{b}(r)+S_{b}(r)&\\ &=&S_{b}(xb^{\ell-1}+r)+S_{b}(r),&\text{(by induction hypothesis)}\\ \end{array}$$

$$\begin{array}[]{rclr}S_{b}(xb^{\ell}+xb^{\ell-1}+r)&=&S_{b}(xb^{\ell-1}+r)+S_{b}(xb^{\ell-2}+r)&\text{(by Lemma~{}\ref{lem:arbrexx0})}\\ &=&S_{b}(xb^{\ell-2}+r)+S_{b}(r))+S_{b}(xb^{\ell-2}+r)&\text{(using~{}\eqref{eq:rec_Sb_1})}\\ &=&2(S_{b}(xb^{\ell-2}+r)+S_{b}(r))-S_{b}(r)&\\ &=&2S_{b}(xb^{\ell-1}+r)-S_{b}(r),&\text{(using~{}\eqref{eq:rec_Sb_1})}\\ \end{array}$$

$$S_b(x{b}^{\ell }+y{b}^{\ell -1}+r)=S_b(x{b}^{\ell -1}+r)+S_b(y{b}^{\ell -1}+r)-2S_b(r),$$

$${\mathcal{K}}_b(s)=\{(s(b^{i}n+j){)}_{n\ge 0}|i\ge 0\text{ and }0\le j<b^i\}.$$

Full text

1 Introduction

A finite word is a finite sequence of letters belonging to a finite set called the alphabet. The binomial coefficient $\binom{u}{v}$ of two finite words $u$ and $v$ is the number of times $v$ occurs as a subsequence of $u$ (meaning as a “scattered” subword). All along the paper, we let $b$ denote an integer greater than $1$. We let $\operatorname{rep}_{b}(n)$ denote the (greedy) base-$b$ expansion of $n\in\mathbb{N}\setminus\{0\}$ starting with a non-zero digit. We set $\operatorname{rep}_{b}(0)$ to be the empty word denoted by $\varepsilon$. We let

[TABLE]

be the set of base-$b$ expansions of the non-negative integers. For all $w\in\{0,\ldots,b-1\}^{*}$, we also define $\operatorname{val}_{b}(w)$ to be the value of $w$ in base $b$, i.e., if $w=w_{n}\cdots w_{0}$ with $w_{i}\in\{0,\ldots,b-1\}$ for all $i$, then $\operatorname{val}_{b}(w)=\sum_{i=0}^{n}w_{i}b^{i}$.

Several generalizations and variations of the Pascal triangle exist and lead to interesting combinatorial, geometrical or dynamical properties [5, 6, 13, 14, 15]. Ordering the words of $L_{b}$ by increasing genealogical order, we introduced Pascal-like triangles $\mathrm{P}_{b}$ [15] where the entry $\mathrm{P}_{b}(m,n)$ is $\binom{\operatorname{rep}_{b}(m)}{\operatorname{rep}_{b}(n)}$. Clearly $\mathrm{P}_{b}$ contains $(b-1)$ copies of the usual Pascal triangle when only considering words of the form $a^{m}$ with $a\in\{1,\ldots,b-1\}$ and $m\geq 0$. In Figure 1, we depict the first few elements of $\mathrm{P}_{3}$ A284441 and its compressed version highlighting the number of positive elements on each line. The data provided by this compressed version is summed up in Definition 1.

Definition 1**.**

For $n\geq 0$, we define the sequence $(S_{b}(n))_{n\geq 0}$ by setting

[TABLE]

We also consider the summatory function $(A_{b}(n))_{n\geq 0}$ of the sequence $(S_{b}(n))_{n\geq 0}$ defined by $A_{b}(0)=0$ and for all $n\geq 1$,

[TABLE]

The quantity $A_{b}(n)$ can be thought of as the total number of base-$b$ expansions occurring as subwords in the base-$b$ expansion of integers less than $n$ (the same subword is counted $k$ times if it occurs in the base-$b$ expansion of $k$ distinct integers).

In some sense, the sequences $(S_{b}(n))_{n\geq 0}$ and $(A_{b}(n))_{n\geq 0}$ measure the sparseness of $\mathrm{P}_{b}$.

Example 2**.**

If $b=3$, then the first few terms of the sequence $(S_{3}(n))_{n\geq 0}$ A282715 are

[TABLE]

For instance, the subwords of the word $121$ are $\varepsilon,1,2,11,12,21,121$. Thus, $S_{3}(\operatorname{val}_{3}(121))=S_{3}(16)=7$. The first few terms of $(A_{3}(n))_{n\geq 0}$ A284442 are

[TABLE]

We studied [16] the triangle $\mathrm{P}_{2}$ A282714 and the sequence $(S_{2}(n))_{n\geq 0}$ A007306, which turns out to be the subsequence with odd indices of the Stern–Brocot sequence. The sequence $(S_{2}(n))_{n\geq 0}$ is $2$-regular in the sense of Allouche and Shallit [1]. We studied [17] the behavior of $(A_{2}(n))_{n\geq 0}$ A282720. To this aim, we exploited a particular decomposition of $A_{2}(2^{\ell}+r)$, for all $\ell\geq 1$ and all $0\leq r<2^{\ell}$, using powers of $3$.

1.1 Our contribution

We conjectured six recurrence relations for $(S_{3}(n))_{n\geq 0}$ depending on the position of $n$ between two consecutive powers of $3$; see [16]. Using the heuristic from [3] suggesting recurrence relations, the sequence $(S_{3}(n))_{n\geq 0}$ was expected to be $3$-regular. It was not obvious that we could derive general recurrence relations for $(S_{b}(n))_{n\geq 0}$ from the form of those satisfied by $(S_{2}(n))_{n\geq 0}$. We thought that $(b-1)b$ recurrence relations should be needed in the general case, leading to a cumbersome statement. Moreover it was computationally challenging to obtain many terms of $(S_{b}(n))_{n\geq 0}$ for large $b$ because the number of words of length $n$ in $L_{b}$ grows like $b^{n}$. Therefore we lack data to conjecture the $b$-regularity of $(S_{b}(n))_{n\geq 0}$.

When studying $(A_{2}(n))_{n\geq 0}$, a possible extension seemed to emerge [17]. In particular, we prove that $A_{2}(2n)=3A_{2}(n)$ and, sustained by computer experiments, we conjectured that $A_{b}(nb)=(2b-1)A_{b}(n)$.

Surprisingly, for all $b\geq 2$, we show in Section 2 that the recurrence relations satisfied by $(S_{b}(n))_{n\geq 0}$ reduce to three forms; see Proposition 3. In particular, this proves the conjecture stated in [16]. Therefore, in Section 3, we deduce the $b$-regularity of $(S_{b}(n))_{n\geq 0}$; see Theorem 16. Moreover we obtain a linear representation of the sequence with $b\times b$ matrices. We also show that $(S_{b}(n))_{n\geq 0}$ is palindromic over $[(b-1)b^{\ell},b^{\ell+1}]$.

The key to study the asymptotics of $(A_{b}(n))_{n\geq 0}$ is to obtain specific recurrence relations for this sequence. In Proposition 26, we show that theses relations involve powers of $(2b-1)$. Therefore, we prove the conjecture about $A_{b}(nb)$. In Section 4, using the so-called $(2b-1)$-decompositions, we may apply the method introduced in [17].

We think that this paper motivates the quest for generalized Stern–Brocot sequences and analogues of the Farey tree [4, 7, 8, 11, 12, 18]. Namely can one reasonably define a tree structure, or some other combinatorial structure, in which the sequence $(S_{b}(n))_{n\geq 0}$ naturally appears?

Most of the results are proved by induction and the base case usually takes into account the values of $S_{b}(n)$ for $0\leq n<b^{2}$. These values are easily obtained from Definition 1 and summarized in Table 1.

2 General recurrence relations in base $b$

The aim of this section is to prove the following result exhibiting recurrence relations satisfied by the sequence $(S_{b}(n))_{n\geq 0}$. This result is useful to prove that the summatory function of the latter sequence also satisfies recurrence relations; see Section 4.

Proposition 3**.**

The sequence $(S_{b}(n))_{n\geq 0}$ satisfies $S_{b}(0)=1$, $S_{b}(1)=\cdots=S_{b}(b-1)=2$, and, for all $x,y\in\{1,\ldots,b-1\}$ with $x\neq y$, all $\ell\geq 1$ and all $r\in\{0,\ldots,b^{\ell-1}-1\}$,

[TABLE]

For the sake of completeness, we recall the definition of a particularly useful tool called the trie of subwords to prove Proposition 3. This tool is also useful to prove the $b$-regularity of the sequence $(S_{b}(n))_{n\geq 0}$; see Section 3.

Definition 4**.**

Let $w$ be a finite word over $\{0,\ldots,b-1\}$. The language of its subwords is factorial, i.e., if $xyz$ is a subword of $w$, then $y$ is also a subword of $w$. Thus we may associate with $w$, the trie111This tree is also called prefix tree or radix tree. All successors of a node have a common prefix and the root is the empty word. of its subwords. The root is $\varepsilon$ and if $u$ and $ua$ are two subwords of $w$ with $a\in\{0,\ldots,b-1\}$, then $ua$ is a child of $u$. We let $\mathcal{T}(w)$ denote the subtree in which we only consider the children $1,\ldots,b-1$ of the root $\varepsilon$ and their successors, if they exist.

Remark 5**.**

The number of nodes on level $\ell\geq 0$ in $\mathcal{T}(w)$ counts the number of subwords of length $\ell$ in $L_{b}$ occurring in $w$. In particular, the number of nodes of the trie $\mathcal{T}(\operatorname{rep}_{b}(n))$ is exactly $S_{b}(n)$ for all $n\geq 0$.

Definition 6**.**

For each non-empty word $w\in L_{b}$, we consider a factorization of $w$ into maximal blocks of consecutively distinct letters (i.e., $a_{i}\neq a_{i+1}$ for all $i$) of the form

[TABLE]

with $n_{\ell}\geq 1$ for all $\ell$. For each $\ell\in\{0,\dots,M-1\}$, we consider the subtree $T_{\ell}$ of $\mathcal{T}(w)$ whose root is the node $a_{1}^{n_{1}}\cdots a_{\ell}^{n_{\ell}}a_{\ell+1}$. For convenience, we set $T_{M}$ to be an empty tree with no node. Roughly speaking, we have a root of a new subtree $T_{\ell}$ for each new variation of digits in $w$. For each $\ell\in\{0,\dots,M-1\}$, we also let $\#T_{\ell}$ denote the number of nodes of the tree $T_{\ell}$.

Note that for $k-i\geq 2$, one could possibly have $a_{k}=a_{i}$. For each $\ell\in\{0,\dots,M-1\}$, we let $\mathrm{Alph}(\ell)$ denote the set of letters occurring in $a_{\ell+1}\cdots a_{M}$. Then for each letter $a\in\mathrm{Alph}(\ell)$, we let $j(a,\ell)$ denote the smallest index in $\{\ell+1,\ldots,M\}$ such that $a_{j(a,\ell)}=a$.

Example 7**.**

In this example, we set $b=3$ and $w=22000112\in L_{3}$. Using the previous notation, we have $M=4$, $a_{1}=2$, $a_{2}=0$, $a_{3}=1$ and $a_{4}=2$. For instance, $\mathrm{Alph}(0)=\{0,1,2\}$, $\mathrm{Alph}(2)=\{1,2\}$ and $j(0,0)=2$, $j(1,0)=3$, $j(2,0)=1$ and $j(2,1)=4$.

The following result describes the structure of the tree $\mathcal{T}(w)$. It directly follows from the definition.

Proposition 8** ([16, Proposition 27]).**

Let $w$ be a finite word in $L_{b}$. With the above notation about $M$ and the subtrees $T_{\ell}$, the tree $\mathcal{T}(w)$ has the following properties.

The node of label $\varepsilon$ has $\#(\mathrm{Alph}(0)\setminus\{0\})$ children that are $a$ for $a\in\mathrm{Alph}(0)\setminus\{0\}$. Each child $a$ is the root of a tree isomorphic $T_{j(a,0)-1}$. 2. 2.

For each $\ell\in\{0,\dots,M-1\}$ and each $i\in\{0,\dots,n_{\ell+1}-1\}$ with $(\ell,i)\neq(0,0)$, the node of label $x=a_{1}^{n_{1}}\cdots a_{\ell}^{n_{\ell}}a_{\ell+1}^{i}$ has $\#(\mathrm{Alph}(\ell))$ children that are $xa$ for $a\in\mathrm{Alph}(\ell)$. Each child $xa$ with $a\neq a_{\ell+1}$ is the root of a tree isomorphic to $T_{j(a,\ell)-1}$.

Example 9**.**

Let us continue Example 7. The tree $\mathcal{T}(22000112)$ is depicted in Figure 2. We use three different colors to represent the letters $0,1,2$.

The tree $T_{0}$ (resp., $T_{1}$; resp., $T_{2}$; resp., $T_{3}$) is the subtree of $\mathcal{T}(w)$ with root $2$ (resp., $2^{2}0$; resp., $2^{2}0^{3}1$; resp., $2^{2}0^{3}1^{2}2$). These subtrees are represented in Figure 2 using dashed lines. The tree $T_{3}$ is limited to a single node since the number of nodes of $T_{M-1}$ is $n_{M}$, which is equal to $1$ in this example.

Using tries of subwords, we prove the following five lemmas. Their proofs are essentially the same, so we only prove two of them.

Lemma 10**.**

For each letter $x\in\{1,\ldots,b-1\}$ and each word $u\in\{0,\ldots,b-1\}^{*}$, we have

[TABLE]

Proof.

Recall that from Remark 5, we need to prove that $\#\mathcal{T}(x00u)=2\#\mathcal{T}(x0u)-\#\mathcal{T}(xu)$.

Assume first that $u$ is of the form $u=0^{n}$, $n\geq 0$. The tree $\mathcal{T}(xu)$ is linear and has $n+2$ nodes, $\mathcal{T}(x0u)$ has $n+3$ nodes and $\mathcal{T}(x00u)$ has $n+4$ nodes. The formula holds.

Now suppose that $u$ contains other letters than [math]. We let $a_{1},\ldots,a_{m}$ denote all the pairwise distinct letters of $u$ different from [math]. They are implicitly ordered with respect to their first appearance in $u$. If $x\in\{a_{1},\ldots,a_{m}\}$, we let $i_{x}\in\{1,\ldots,m\}$ denote the index such that $a_{i_{x}}=x$. For all $i\in\{1,\ldots,m\}$, we let $u_{i}a_{i}$ denote the prefix of $u$ that ends with the first occurrence of the letter $a_{i}$ in $u$, and we let $R_{i}$ denote the subtree of $\mathcal{T}(xu)$ with root $xu_{i}a_{i}$.

First, observe that the subtree $T$ of $\mathcal{T}(xu)$ with root $x$ is equal to the subtree of $\mathcal{T}(x0u)$ with root $x0$ and also to the subtree of $\mathcal{T}(x00u)$ with root $x00$.

Secondly, for all $i\in\{1,\ldots,m\}$, the subtree of $\mathcal{T}(x0u)$ with root $xa_{i}$ is $R_{i}$. Similarly, $\mathcal{T}(x00u)$ contains two copies of $R_{i}$: the subtrees of root $xa_{i}$ and $x0a_{i}$.

Finally, for all $i\in\{1,\ldots,m\}$ with $i\neq i_{x}$, the subtree of $\mathcal{T}(x0u)$ with root $a_{i}$ is $R_{i}$ and the subtree of $\mathcal{T}(x00u)$ with root $a_{i}$ is $R_{i}$.

The situation is depicted in Figure 3 where we put a unique edge for several indices when necessary, e.g., the edge labeled by $a_{i}$ stands for $m$ edges labeled by $a_{1},\ldots,a_{m}$. The claimed formula holds since

[TABLE]

∎

Lemma 11**.**

For each letter $x\in\{1,\ldots,b-1\}$ and each word $u\in\{0,\ldots,b-1\}^{*}$, we have

[TABLE]

Proof.

The proof is similar to the proof of Lemma 10. ∎

Lemma 12**.**

For all letters $x,y\in\{1,\ldots,b-1\}$ and each word $u\in\{0,\ldots,b-1\}^{*}$, we have

[TABLE]

Proof.

The proof is similar to the proof of Lemma 10. Observe that one needs to divide the proof into two cases according to whether $x$ is equal to $y$ or not. As a first case, also consider $u=y^{n}$ with $n\geq 0$ instead of $u=0^{n}$ with $n\geq 0$. ∎

Lemma 13**.**

For all letters $x,y\in\{1,\ldots,b-1\}$ and each word $u\in\{0,\ldots,b-1\}^{*}$, we have

[TABLE]

Proof.

The proof is similar to the proof of Lemma 12. ∎

The next lemma having a slightly more technical proof, we present it.

Lemma 14**.**

For all letters $x,y\in\{1,\ldots,b-1\}$ with $x\neq y$, $z\in\{0,\ldots,b-1\}$ and each word $u\in\{0,\ldots,b-1\}^{*}$, we have

[TABLE]

Proof.

Let $x,y\in\{1,\ldots,b-1\}$ with $x\neq y$, $z\in\{0,\ldots,b-1\}$, and let $u\in\{0,\ldots,b-1\}^{*}$. Our reasoning is again based on the structure of the associated trees. The proof is divided into two cases depending on the fact that $z=0$ or not.

$\bullet$ As a first case, suppose that $z\neq 0$. Now assume that $u$ is of the form $u=z^{n}$, $n\geq 0$. If $x\neq z$ and $y\neq z$, the tree $\mathcal{T}(zu)$ is linear and has $n+2$ nodes, $\mathcal{T}(xzu)$ and $\mathcal{T}(yzu)$ have $2(n+2)$ nodes and $\mathcal{T}(xyzu)$ has $4(n+2)$ nodes and the claimed formula holds. If $x\neq z$ and $y=z$, the tree $\mathcal{T}(zu)$ is linear and has $n+2$ nodes, $\mathcal{T}(xzu)$ has $2(n+2)$ nodes, $\mathcal{T}(yzu)$ has $n+3$ nodes and $\mathcal{T}(xyzu)$ has $2(n+3)$ nodes and the claimed formula holds. If $x=z$ and $y\neq z$, the tree $\mathcal{T}(zu)$ is linear and has $n+2$ nodes, $\mathcal{T}(xzu)$ has $n+3$ nodes, $\mathcal{T}(yzu)$ has $2(n+2)$ nodes and $\mathcal{T}(xyzu)$ has $3(n+2)+1$ nodes and the claimed formula holds.

Now suppose that $u$ contains other letters than $z$. We let $a_{1},\ldots,a_{m}$ denote all the pairwise distinct letters of $u$ different from $z$. They are implicitly ordered with respect to their first appearance in $u$. If $x,y,0\in\{a_{1},\ldots,a_{m}\}$, we let $i_{x},i_{y},i_{0}\in\{1,\ldots,m\}$ respectively denote the indices such that $a_{i_{x}}=x$, $a_{i_{y}}=y$ and $a_{i_{0}}=0$. For all $i\in\{1,\ldots,m\}$, we let $u_{i}a_{i}$ denote the prefix of $u$ that ends with the first occurrence of the letter $a_{i}$ in $u$, and we let $R_{i}$ denote the subtree of $\mathcal{T}(zu)$ with root $zu_{i}a_{i}$.

First, observe that the subtree $T$ of $\mathcal{T}(zu)$ with root $z$ is equal to the subtree of $\mathcal{T}(xzu)$ with root $xz$, to the subtree of $\mathcal{T}(yzu)$ with root $yz$ and also to the subtree of $\mathcal{T}(xyzu)$ with root $xyz$.

Suppose that $x\neq z$ and $y\neq z$.

Using the same reasoning as in the proof of Lemma 10, the situation is depicted in Figure 4. The claimed formula holds since

[TABLE]

Suppose that $x\neq z$ and $y=z$.

The situation is depicted in Figure 5. The claimed formula holds since

[TABLE]

Suppose that $x=z$ and $y\neq z$.

The situation is depicted in Figure 6. The claimed formula holds since

[TABLE]

$\bullet$ As a second case, suppose that $z=0$. Then, by convention, leading zeroes are not allowed in base-$b$ expansions and we must prove that the following formula holds

[TABLE]

It is useful to note that $\operatorname{rep}_{b}(\operatorname{val}_{b}(\cdot)):\{0,\ldots,b-1\}^{*}\mapsto L_{b}$ plays a normalization role. It removes leading zeroes.

If $u=0^{n}$, with $n\geq 0$, then $\operatorname{rep}_{b}(\operatorname{val}_{b}(u))=\varepsilon$ and the tree $\mathcal{T}(\operatorname{rep}_{b}(\operatorname{val}_{b}(u)))$ has only one node. The trees $\mathcal{T}(x0u)$ and $\mathcal{T}(y0u)$ both have $n+3$ nodes and the tree $\mathcal{T}(xy0u)$ has $3(n+2)+1$ nodes and the claimed formula holds.

Now suppose that $u$ contains other letters than [math]. We let $a_{1},\ldots,a_{m}$ denote all the pairwise distinct letters of $u$ different from [math]. They are implicitly ordered with respect to their first appearance in $u$. If $x,y\in\{a_{1},\ldots,a_{m}\}$, we let $i_{x},i_{y}\in\{1,\ldots,m\}$ respectively denote the indices such that $a_{i_{x}}=x$ and $a_{i_{y}}=y$. For all $i\in\{1,\ldots,m\}$, we let $u^{\prime}_{i}a_{i}$ denote the prefix of $\operatorname{rep}_{b}(\operatorname{val}_{b}(u))$ that ends with the first occurrence of the letter $a_{i}$ in $\operatorname{rep}_{b}(\operatorname{val}_{b}(u))$, and we let $R_{i}$ denote the subtree of $\mathcal{T}(\operatorname{rep}_{b}(\operatorname{val}_{b}(u)))$ with root $u^{\prime}_{i}a_{i}$.

The situation is depicted in Figure 7. Observe that the subtree $T$ of $\mathcal{T}(y0u)$ with root $y0$ is equal to the subtree of $\mathcal{T}(x0u)$ with root $x0$ and to the subtree of $\mathcal{T}(xy0u)$ with root $xy0$. The claimed formula holds since

[TABLE]

∎

Those five lemmas can be translated into recurrence relations satisfied by the sequence $(S_{b}(n))_{n\geq 0}$ using Definition 1.

Proof of Proposition 3.

The first part is clear using Table 1. Let $x,y\in\{1,\ldots,b-1\}$ with $x\neq y$. Proceed by induction on $\ell\geq 1$.

Let us first prove (2). If $\ell=1$, then $r=0$ and (2) follows from Table 1. Now suppose that $\ell\geq 2$ and assume that (2) holds for all $\ell^{\prime}<\ell$. Let $r\in\{0,\ldots,b^{\ell-1}-1\}$, and let $u$ be a word in $\{0,\ldots,b-1\}^{*}$ such that $|u|\geq 1$ and $\operatorname{rep}_{b}(xb^{\ell}+r)=x0u$. The proof is divided into two parts according to the first letter of $u$. If $u=0u^{\prime}$ with $u^{\prime}\in\{0,\ldots,b-1\}^{*}$, then

[TABLE]

which proves (2). Now if $u=zu^{\prime}$ with $z\in\{1,\ldots,b-1\}$ and $u^{\prime}\in\{0,\ldots,b-1\}^{*}$, then (2) directly follows from Definition 1 and Lemma 12.

Let us prove (3). If $\ell=1$, then $r=0$ and (2) follows from Table 1. Now suppose that $\ell\geq 2$ and assume that (3) holds for all $\ell^{\prime}<\ell$. Let $r\in\{0,\ldots,b^{\ell-1}-1\}$, and let $u$ be a word in $\{0,\ldots,b-1\}^{*}$ such that $|u|\geq 1$ and $\operatorname{rep}_{b}(xb^{\ell}+xb^{\ell-1}+r)=xxu$. The proof is divided into two parts according to the first letter of $u$. If $u=0u^{\prime}$ with $u^{\prime}\in\{0,\ldots,b-1\}^{*}$, then

[TABLE]

which proves (3). Now if $u=zu^{\prime}$ with $z\in\{1,\ldots,b-1\}$ and $u^{\prime}\in\{0,\ldots,b-1\}^{*}$, then (3) directly follows from Definition 1 and Lemma 13.

Let us finally prove (4). If $\ell=1$, then $r=0$ and (2) follows from Table 1. Now suppose that $\ell\geq 2$ and assume that (4) holds for all $\ell^{\prime}<\ell$. Let $r\in\{0,\ldots,b^{\ell-1}-1\}$, let $z$ be a letter of $\{1,\ldots,b-1\}$ and let $u$ be a word in $\{0,\ldots,b-1\}^{*}$ such that $\operatorname{rep}_{b}(xb^{\ell}+yb^{\ell-1}+r)=xyzu$. Using Definition 1 and Lemma 14, we directly have that

[TABLE]

which proves (4). ∎

3 Regularity of the sequence $(S_{b}(n))_{n\geq 0}$

The sequence $(S_{2}(n))_{n\geq 0}$ is shown to be $2$-regular; see [16]. We recall that the $b$-kernel of a sequence $s=(s(n))_{n\geq 0}$ is the set

[TABLE]

A sequence $s=(s(n))_{n\geq 0}\in\mathbb{Z}^{\mathbb{N}}$ is $b$-regular if there exists a finite number of sequences $(t_{1}(n))_{n\geq 0}$, …, $(t_{\ell}(n))_{n\geq 0}$ such that every sequence in the $\mathbb{Z}$-module $\langle\mathcal{K}_{b}(s)\rangle$ generated by the $b$-kernel $\mathcal{K}_{b}(s)$ is a $\mathbb{Z}$-linear combination of the $t_{r}$’s. In this section, we prove that the sequence $(S_{b}(n))_{n\geq 0}$ is $b$-regular. As a consequence, one can get matrices to compute $S_{b}(n)$ in a number of matrix multiplications proportional to $\log_{b}(n)$. To prove the $b$-regularity of the sequence $(S_{b}(n))_{n\geq 0}$ for any base $b$, we first need a lemma involving some matrix manipulations.

Lemma 15**.**

Let $I$ and [math] respectively be the identity matrix of size $b^{2}\times b^{2}$ and the zero matrix of size $b^{2}\times b^{2}$. Let $M_{b}$ be the block-matrix of size $b^{3}\times b^{3}$

[TABLE]

This matrix is invertible and its inverse is given by

[TABLE]

For the proof of the previous lemma, simply proceed to the multiplication of the two matrices. Using this lemma, we prove that the sequence $(S_{b}(n))_{n\geq 0}$ is $b$-regular.

Theorem 16**.**

For all $r\in\{0,\ldots,b^{2}-1\}$, we have

[TABLE]

where the coefficients $a_{r}$ and $c_{r,s}$ are unambiguously determined by the first few values $S_{b}(0)$, $S_{b}(1)$,…, $S_{b}(b^{3}-1)$ and given in Table 2, Table 3 and Table 4. In particular, the sequence $(S_{b}(n))_{n\geq 0}$ is $b$-regular. Moreover, a choice of generators for $\langle\mathcal{K}_{b}(s)\rangle$ is given by the $b$ sequences $(S_{b}(n))_{n\geq 0}$, $(S_{b}(bn))_{n\geq 0}$, $(S_{b}(bn+1))_{n\geq 0}$, …, $(S_{b}(bn+b-2))_{n\geq 0}$.

Proof.

We proceed by induction on $n\geq 0$. For the base case $n\in\{0,1,\dots,b^{2}-1\}$, we first compute the coefficients $a_{r}$ and $c_{r,s}$ using the values of $S_{b}(nb^{2}+r)$ for $n\in\{0,\ldots,b-1\}$ and $r\in\{0,\dots,b^{2}-1\}$. Then we show that (5) also holds with these coefficients for $n\in\{b,\ldots,b^{2}-1\}$.

Base case. Let $I$ denote the identity matrix of size $b^{2}\times b^{2}$. The system of $b^{3}$ equations (5) when $n\in\{0,\ldots,b-1\}$ and $r\in\{0,\dots,b^{2}-1\}$ can be written as $MX=V$ where the matrix $M\in\mathbb{Z}^{b^{3}}_{b^{3}}$ is equal to

[TABLE]

and the vectors $X,V\in\mathbb{Z}^{b^{3}}$ are respectively given by

[TABLE]

Observe that in the vector $X$, the coefficients $c_{r,s}$ are first sorted by $s$ then by $r$. Using Table 1, the matrix $M$ is equal to the matrix $M_{b}$ of Lemma 15. By this lemma, the previous system has a unique solution given by $X=M^{-1}_{b}V$. Consequently, using Lemma 15, we have, for all $r\in\{0,\ldots,b^{2}-1\}$ and all $s\in\{1,\ldots,b-2\}$,

[TABLE]

The values of the coefficients can then be computed using Table 1 and are stored in Table 2, Table 3 and Table 4.

For $n\in\{b,\dots,b^{2}-1\}$, the values of $S_{b}(nb^{2}+r)$ are given in Table 5, Table 6 and Table 7 according to whether $\operatorname{rep}_{b}(n)$ is of the form $x0$, $xx$ or $xy$ with $x\neq y$. The proof that (5) holds for each $n\in\{b,\ldots,b^{2}-1\}$ only requires easy computations that are left to the reader.

Inductive step. Consider $n\geq b^{2}$ and suppose that the relation (5) holds for all $m<n$. Then $|\operatorname{rep}_{b}(n)|\geq 3$. Like for the base case, we need to consider several cases according to the form of the base-$b$ expansion of $n$. More precisely, we need to consider the following five forms, where $u\in\{0,\dots,b-1\}^{*}$, $x,y,z\in\{1,\dots,b-1\}$, $x\neq z$, and $t\in\{0,\dots,b-1\}$:

[TABLE]

Let us focus on the first form of $\operatorname{rep}_{b}(n)$ since the same reasoning can be applied for the other ones. Assume that $\operatorname{rep}_{b}(n)=x00u$ where $x\in\{1,\ldots,b-1\}$ and $u\in\{0,\ldots,b-1\}^{*}$. For all $r\in\{0,\ldots,b^{2}-1\}$, there exist $r_{1},r_{2}\in\{0,\ldots,b-1\}$ such that $\operatorname{val}_{b}(r_{1}r_{2})=r$. We have

[TABLE]

which proves (5).

$b$-regularity. From the first part of the proof, we directly deduce that the $\mathbb{Z}$-module $\left\langle\mathcal{K}_{b}(S_{b})\right\rangle$ is generated by the $(b+1)$ sequences

[TABLE]

We now show that we can reduce the number of generators. To that aim, we prove that

[TABLE]

We proceed by induction on $n\geq 0$. As a base case, the proof that (6) holds for each $n\in\{b,\ldots,b^{2}-1\}$ only requires easy computations that are left to the reader (using Table 1). Now consider $n\geq b^{2}$ and suppose that the relation (6) holds for all $m<n$. Then $|\operatorname{rep}_{b}(n)|\geq 3$. Mimicking the first induction step of this proof, we need to consider several cases according to the form of the base-$b$ expansion of $n$. More precisely, we need to consider the following five forms, where $u\in\{0,\dots,b-1\}^{*}$, $x,y,z\in\{1,\dots,b-1\}$, $x\neq z$, and $t\in\{0,\dots,b-1\}$:

[TABLE]

Let us focus on the first form of $\operatorname{rep}_{b}(n)$ since the same reasoning can be applied for the other ones. Assume that $\operatorname{rep}_{b}(n)=x00u$ where $x\in\{1,\ldots,b-1\}$ and $u\in\{0,\ldots,b-1\}^{*}$. We have

[TABLE]

which proves (5).

The $\mathbb{Z}$-module $\left\langle\mathcal{K}_{b}(S_{b})\right\rangle$ is thus generated by the $b$ sequences

[TABLE]

∎

Example 17**.**

Let $b=2$. Using Table 2, Table 3 and Table 4, we find that $a_{0}=-1$, $a_{1}=1$, $a_{2}=4$, $a_{3}=5$, $c_{0,0}=2$, $c_{1,0}=1$, $c_{2,0}=-1$ and $c_{3,0}=-2$. In this case, there are no $c_{r,s}$ with $s>0$. Applying Theorem 16 and from (6), we get

[TABLE]

for all $n\geq 0$. This result is a rewriting of [16, Theorem 21]. Observe that the third and the fifth identities are redundant: they follow from the other ones.

Example 18**.**

Let $b=3$. Using Table 2, Table 3 and Table 4, the values of the coefficients $a_{r}$, $c_{r,0}$ and $c_{r,1}$ can be found in Table 8.

Applying Theorem 16 and from (6), we get

[TABLE]

for all $n\geq 0$. This result is a proof of [16, Conjecture 26]. Observe that the fourth, the seventh and the tenth identities are redundant.

Remark 19**.**

Combining (5) and (6) yield $b^{2}+1$ identities to generate the $\mathbb{Z}$-module $\left\langle\mathcal{K}_{b}(S_{b})\right\rangle$. However, as illustrated in Example 17 and Example 18, only $b^{2}-b+1$ identities are useful: the relations established for the sequences $(S_{b}(b^{2}n+br+b-1))_{n\geq 0}$, with $r\in\{0,\ldots,b-1\}$, can be deduced from the other identities.

Remark 20**.**

Using Theorem 16 and (6) and the set of $b$ generators of the $\mathbb{Z}$-module $\left\langle\mathcal{K}_{b}(S_{b})\right\rangle$ being

[TABLE]

we get matrices to compute $S_{b}(n)$ in a number of steps proportional to $\log_{b}(n)$. For all $n\geq 0$, let

[TABLE]

Consider the matrix-valued morphism $\mu_{b}:\{0,1,\ldots,b-1\}^{*}\to\mathbb{Z}^{b}_{b}$ defined, for all $s\in\{0,\ldots,b-2\}$, by

[TABLE]

and

[TABLE]

Observe that the number of generators explains the size of the matrices above. For each $s\in\{0,\ldots,b-2\}$, exactly $b-1$ identities from Theorem 16 are used to define the matrix $\mu_{b}(s)$. If $s,s^{\prime}\in\{0,\ldots,b-2\}$ are such that $s\neq s^{\prime}$, then the relations used to define the matrices $\mu_{b}(s)$ and $\mu_{b}(s^{\prime})$ are pairwise distinct. Finally, the first row of the matrix $\mu_{b}(b-1)$ is (6) and the other rows are $b-1$ identities from Theorem 16, which are distinct from the previous relations. Consequently, $(b-1)(b-1)+b$ identities are used, which corroborates Remark 19.

Using the definition of the morphism $\mu$, we can show that $V_{b}(bn+s)=\mu_{b}(s)V_{b}(n)$ for all $s\in\{0,\ldots,b-1\}$ and $n\geq 0$. Consequently, if $\operatorname{rep}_{b}(n)=n_{k}\cdots n_{0}$, then

[TABLE]

For example, when $b=2$, the matrices $\mu_{2}(0)$ and $\mu_{2}(1)$ are those given in [16, Corollary 22]. When $b=3$, we get

[TABLE]

The class of $b$-synchronized sequences is intermediate between the classes of $b$-automatic sequences and $b$-regular sequences. These sequences were first introduced in [9].

Proposition 21**.**

The sequence $(S_{b}(n))_{n\geq 0}$ is not $b$-synchronized.

Proof.

The proof is exactly the same as [16, Proposition 24]. ∎

To conclude this section, the following result proves that the sequence $(S_{b}(n))_{n\geq 0}$ has a partial palindromic structure as the sequence $(S_{2}(n))_{n\geq 0}$; see [16]. For instance, the sequence $(S_{3}(n))_{n\geq 0}$ is depicted in Figure 8 inside the interval $[2\cdot 3^{4},3^{5}]$.

Proposition 22**.**

Let $u$ be a word in $\{0,1,\ldots,b-1\}^{*}$. Define $\bar{u}$ by replacing in $u$ every letter $a\in\{0,1,\ldots,b-1\}$ by the letter $(b-1)-a\in\{0,1,\ldots,b-1\}$. Then

[TABLE]

In particular, there exists a palindromic substructure inside of the sequence $(S_{b}(n))_{n\geq 0}$, i.e., for all $\ell\geq 1$ and $0\leq r<b^{\ell}$,

[TABLE]

Proof.

The trees $\mathcal{T}((b-1)u)$ and $\mathcal{T}((b-1)\bar{u})$ are isomorphic. Indeed, on the one hand, each node of the form $(b-1)x$ in the first tree corresponds to the node $(b-1)\bar{x}$ in the second one and conversely. On the other hand, if there exist letters $a\in\{1,\ldots,b-2\}$ in the word $(b-1)u$, the position of the first letter $a$ in the word $(b-1)u$ is equal to the position of the first letter $(b-1)-a$ in the word $(b-1)\bar{u}$ and conversely. Consequently, the node of the form $ax$ in the first tree corresponds to the node of the form $((b-1)-a)\bar{x}$ in the second tree and conversely.

For the special case, note that for every word $z$ of length $\ell$, there exists $r\in\{0,\ldots,b^{\ell}-1\}$ such that $\operatorname{rep}_{b}((b-1)\cdot b^{\ell}+r)=(b-1)z$ and

[TABLE]

Hence, $(b-1)\bar{z}=\operatorname{rep}_{b}((b-1)\cdot b^{\ell}+b^{\ell}-1-r)$. Using (1), we obtain the desired result. ∎

4 Asymptotics of the summatory function $(A_{b}(n))_{n\geq 0}$

In this section, we consider the summatory function $(A_{b}(n))_{n\geq 0}$ of the sequence $(S_{b}(n))_{n\geq 0}$; see Definition 1. The aim of this section is to apply the method introduced in [17] to obtain the asymptotic behavior of $(A_{b}(n))_{n\geq 0}$. As an easy consequence of the $b$-regularity of $(S_{b}(n))_{n\geq 0}$, we have the following result.

Proposition 23**.**

For all $b\geq 2$, the sequence $(A_{b}(n))_{n\geq 0}$ is $b$-regular.

Proof.

This is a direct consequence of Theorem 16 and of the fact that the summatory function of a $b$-regular sequence is also $b$-regular; see [2, Theorem 16.4.1]. ∎

From a linear representation with matrices of size $d\times d$ associated with a $b$-regular sequence, one can derive a linear representation with matrices of size $2d\times 2d$ associated with its summatory function; see [10, Lemma 1]. Consequently, using Remark 20, one can obtain a linear representation with matrices of size $2b\times 2b$ for the summatory function $(A_{b}(n))_{n\geq 0}$. The goal is to decompose $(A_{b}(n))_{n\geq 0}$ into linear combinations of powers of $(2b-1)$. We need the following two lemmas.

Lemma 24**.**

For all $\ell\geq 0$ and all $x\in\{1,\ldots,b-1\}$, we have

[TABLE]

Proof.

We proceed by induction on $\ell\geq 0$. If $\ell=0$ and $x\in\{1,\ldots,b-1\}$, then using Table 1, we have

[TABLE]

If $\ell=1$ and $x\in\{1,\ldots,b-1\}$, then we have

[TABLE]

Using Table 1, we get $A_{b}(xb)=(2x-1)(2b-1)$.

Now suppose that $\ell\geq 1$ and assume that the result holds for all $\ell^{\prime}\leq\ell$. To prove the result, we again proceed by induction on $x\in\{1,\ldots,b-1\}$. When $x=1$, we must show that $A_{b}(b^{\ell+1})=(2b-1)^{\ell+1}$. We have

[TABLE]

By decomposing the sum into three parts accordingly to Proposition 3, we get

[TABLE]

and, using Proposition 3,

[TABLE]

By observing that for all $y$,

[TABLE]

and that

[TABLE]

we obtain

[TABLE]

and finally

[TABLE]

Using the induction hypothesis, we obtain

[TABLE]

which ends the case where $x=1$.

Now suppose that $x\in\{2,\ldots,b-1\}$ and assume that the result holds for all $x^{\prime}<x$. The proof follows the same lines as in the case $x=1$ with the difference that we decompose the sum into

[TABLE]

Applying Proposition 3 and using (10) and (11) leads to the equality

[TABLE]

The induction hypothesis ends the computation. ∎

Lemma 25**.**

For all $\ell\geq 1$ and all $x,y\in\{1,\ldots,b-1\}$, we have

[TABLE]

Proof.

The proof of this lemma is similar to the proof of Lemma 24 so we only proof the formula for $A_{b}(xb^{\ell}+xb^{\ell-1})$, the other being similarly handled. We proceed by induction on $\ell\geq 1$. If $\ell=1$, the result follows from Table 1. Assume that $\ell\geq 2$ and that the formulas hold for all $\ell^{\prime}<\ell$. We have

[TABLE]

Applying Proposition 3 and using (10) and (11) leads to the equality

[TABLE]

Using Lemma 24 completes the computation. ∎

Lemma 24 and Lemma 25 give rise to recurrence relations satisfied by the summatory function $(A_{b}(n))_{n\geq 0}$ as stated below. This is a key result that permits us to introduce $(2b-1)$-decompositions (Definition 28 below) of the summatory function $(A_{b}(n))_{n\geq 0}$ and allows us to easily deduce Theorem 30; see [17] for similar results in base $2$.

Proposition 26**.**

For all $x,y\in\{1,\ldots,b-1\}$ with $x\neq y$, all $\ell\geq 1$ and all $r\in\{0,\ldots,b^{\ell-1}\}$,

[TABLE]

Proof.

We first prove (12). Let $x\in\{1,\ldots,b-1\}$, $\ell\geq 1$ and $r\in\{0,\ldots,b^{\ell-1}\}$. If $r=0$, then (12) holds using Lemma 24. Now suppose that $r\in\{1,\ldots,b^{\ell-1}\}$. Applying successively Proposition 3 and Lemma 24, we have

[TABLE]

which proves (12).

The proof of (13) and (14) are similar, thus we only prove (13). Let $x\in\{1,\ldots,b-1\}$, $\ell\geq 1$ and $r\in\{0,\ldots,b^{\ell-1}\}$. If $r=0$, then (13) holds using Lemma 25. Now suppose that $r\in\{1,\ldots,b^{\ell-1}\}$. Applying Proposition 3, we have

[TABLE]

Using Lemma 24 and Lemma 25, we get

[TABLE]

which proves (13). ∎

The following corollary was conjectured in [17].

Corollary 27**.**

For all $n\geq 0$, we have $A_{b}(nb)=(2b-1)A_{b}(n)$.

Proof.

Let us proceed by induction on $n\geq 0$. It is easy to check by hand that the result holds for $n\in\{0,\ldots,b-1\}$. Thus consider $n\geq b$ and suppose that the result holds for all $n^{\prime}<n$. The reasoning is divided into three cases according to the form of the base-$b$ expansion of $n$. As a first case, we write $n=xb^{\ell}+r$ with $x\in\{1,\ldots,b-1\}$, $\ell\geq 1$ and $0\leq r<b^{\ell-1}$. By Proposition 26, we have

[TABLE]

We conclude this case by using the induction hypothesis. The other cases can be handled using the same technique. ∎

Using Proposition 26, we can define $(2b-1)$-decompositions as follows.

Definition 28**.**

Let $n\geq b$. Applying iteratively Proposition 26 provides a unique decomposition of the form

[TABLE]

where $d_{i}(n)$ are integers, $d_{0}(n)\neq 0$ and $\ell_{b}(n)$ stands for $\lfloor\log_{b}n\rfloor-1$. We say that the word

[TABLE]

is the $(2b-1)$-decomposition of $A_{b}(n)$. For the sake of clarity, we also write $(d_{0}(n),\ldots,d_{\ell_{b}(n)}(n))$. Also notice that the notion of $(2b-1)$-decomposition is only valid for integers in the sequence $(A_{b}(n))_{n\geq 0}$.

Example 29**.**

Let $b=3$. Let us compute the $5$-decomposition of $A_{3}(150)=1665$. We have $\operatorname{rep}_{3}(150)=12120$ and $\ell_{3}(150)=3$. Applying once Proposition 26 leads to

[TABLE]

Applying again Proposition 26, we get

[TABLE]

Using Proposition 26, we find

[TABLE]

Using Lemma 24, we have $A_{3}(3^{1})=5^{1}$ and $A_{3}(2\cdot 3^{1})=3\cdot 5^{1}$. Plugging all those values together in (15), we finally have

[TABLE]

The $5$-decomposition of $A_{3}(150)$ is thus $(4,32,82,-45)$.

The proof of the next result follows the same lines as the proof of [17, Theorem 1]. Therefore we only sketch it.

Theorem 30**.**

There exists a continuous and periodic function $\mathcal{H}_{b}$ of period 1 such that, for all large enough $n$,

[TABLE]

As an example, when $b=3$, the function $\mathcal{H}_{3}$ is depicted in Figure 9 over one period.

Sketch of the proof of Theorem 30.

Let us start by defining the function $\mathcal{H}_{b}$. Given any integer $n\geq 1$, we let $\phi_{n}$ denote the function

[TABLE]

where $e_{n}(\alpha)=b^{n+1}+b\lfloor b^{n}\alpha\rfloor+1$. With a proof analogous to the one of [17, Proposition 20], the sequence of functions $(\phi_{n})_{n\geq 1}$ uniformly converges to a function $\Phi_{b}$. As in [17, Theorem 5], this function is continuous on $[0,1]$ and such that $\Phi_{b}(0)=\Phi_{b}(1)=1$. Furthermore, it satisfies

[TABLE]

see [17, Lemma 24]. Using Corollary 27, we get that, for all $n=b^{j}(b^{k}+r)$, $j,k\geq 0$ and $r\in\{0,\ldots,b^{k}-1\}$,

[TABLE]

The function $\mathcal{H}_{b}$ is defined by $\mathcal{H}_{b}(x)=\Phi_{b}(b^{\{x\}}-1)$ for all real $x$ ($\{\cdot\}$ stands for the fractional part). ∎