On the $k$-abelian complexity of the Cantor sequence

TL;DR

This paper proves that for all integers k ≥ 1, the k-abelian complexity of the Cantor sequence is a 3-regular sequence, revealing a structured regularity in its combinatorial complexity.

Contribution

It establishes that the k-abelian complexity function of the Cantor sequence is 3-regular for all k ≥ 1, a novel result in combinatorics on words.

Findings

k-abelian complexity is 3-regular for the Cantor sequence

The result applies to all integers k ≥ 1

Provides insight into the structure of the Cantor sequence's complexity

Abstract

In this paper, we prove that for every integer $k \geq 1$, the $k$-abelian complexity function of the Cantor sequence $\mathbf{c} = 101000101\cdots$ is a $3$-regular sequence.

Equations

$$\mathbf{c}:=c_0{c}_1{c}_2\cdots =101000101000000000101000101\cdots$$

$$c_{3n}=c_{3n+2}=c_n\text{ and }{c}_{3n+1}=0.$$

$${\mathrm{pref}}_{\ell }(u):=u_0{u}_1\cdots u_{\ell -1}\text{ and }{\mathrm{suff}}_{\ell }(u):=u_{n-\ell }\cdots u_{n-1};$$

$$\mathcal{P}^{(k)}_{\omega}(n)=\mathrm{Card}\big{(}\mathcal{F}_{\omega}(n)/\sim_{k}\big{)},$$

$${\mathcal{K}}_k(\mathbf{w})=\{(w_{k^{e}n+c}{)}_{n\ge 0}\mid e\ge 0,0\le c<k^e\}$$

$${\mathcal{R}\mathcal{S}}_{\mathbf{c}}(k)=\{0^k,{\mathrm{suff}}_k({\sigma }^i(010))\}\text{ and }{\mathcal{LS}}_{\mathbf{c}}(k)=\{0^k,{\mathrm{pref}}_k({\sigma }^i(010))\}.$$

$${\mathcal{P}}_{\omega }^{(1)}(n)=M_{\omega }(n)-m_{\omega }(n)+1,$$

$$M_{\omega }(n):=\max \left\{{\Sigma }_{j=i}^{i+n-1}{\omega }_j\mid i\ge 0\right\}\text{ and }{m}_{\omega }(n):=\min \left\{{\Sigma }_{j=i}^{i+n-1}{\omega }_j\mid i\ge 0\right\}.$$

$$\Sigma (i,n)\le \Sigma (0,n)\text{ for every integer }i\ge 0.$$

$$\left\{\begin{array}[]{ll}\Sigma(3i,3n)=2\Sigma(i,n),&\Sigma(3i+1,3n+2)=\Sigma(i,n+1)+\Sigma(i+1,n),\\ \Sigma(3i,3n+1)=\Sigma(i,n)+\Sigma(i,n+1),&\Sigma(3i+2,3n)=\Sigma(i,n)+\Sigma(i+1,n),\\ \Sigma(3i,3n+2)=\Sigma(i,n)+\Sigma(i,n+1),&\Sigma(3i+2,3n+1)=\Sigma(i,n+1)+\Sigma(i+1,n),\\ \Sigma(3i+1,3n)=\Sigma(i,n)+\Sigma(i+1,n),&\Sigma(3i+2,3n+2)=\Sigma(i,n+1)+\Sigma(i+1,n+1),\\ \Sigma(3i+1,3n+1)=\Sigma(i,n)+\Sigma(i+1,n).&\end{array}\right.$$

$$\Sigma (3i,3j+2)$$

$$\Sigma (3i+1,3j+2)$$

$$\Sigma (3i+2,3j+2)=\Sigma (i,j+1)+\Sigma (i+1,j+1)\le \Sigma (0,j)+\Sigma (0,j+1)=\Sigma (0,3j+2).$$

$$M_{\mathbf{c}}(3n)=2M_{\mathbf{c}}(n)\text{ and }{M}_{\mathbf{c}}(3n+1)=M_{\mathbf{c}}(3n+2)=M_{\mathbf{c}}(n)+M_{\mathbf{c}}(n+1).$$

$${\mathcal{P}}_{\mathbf{c}}^{(1)}(3n)=2{\mathcal{P}}_{\mathbf{c}}^{(1)}(n)-1\text{ and }{\mathcal{P}}_{\mathbf{c}}^{(1)}(3n+1)={\mathcal{P}}_{\mathbf{c}}^{(1)}(3n+2)={\mathcal{P}}_{\mathbf{c}}^{(1)}(n)+{\mathcal{P}}_{\mathbf{c}}^{(1)}(n+1)-1.$$

$$P(z,w):=\{\begin{array}{ll}1,& \text{if }z\text{ is a prefix of }w,\\ 0,& \text{otherwise,}\end{array}\text{ and }S(z,w):=\{\begin{array}{ll}1,& \text{if }z\text{ is a suffix of }w,\\ 0,& \text{otherwise.}\end{array}$$

$$|u{|}_z=\{\begin{array}{ll}|u{|}_{ay}-S(ay,u),& \text{if }\text{leavevmode}ay\notin R{S}_{\omega },\\ |u{|}_{yb}-P(yb,u),& \text{if }\text{leavevmode}yb\notin L{S}_{\omega },\\ |u{|}_{ay}-|u{|}_{ay(1-b)}-S(ay,u),& \text{if }\text{leavevmode}ay\in R{S}_{\omega },\\ |u{|}_{yb}-|u{|}_{(1-a)yb}-P(yb,u),& \text{if }\text{leavevmode}yb\in L{S}_{\omega }.\end{array}$$

$${\Delta }_i=\{\begin{array}{ll}|u{|}_{0^{3^{i}1}}+\frac{2}{3^i}|u{|}_{0^{3^i+1}}+1-P(0^{3^i}1,u)\text{leavevmode}(\mathrm{mod}\text{leavevmode}3),& \text{if }\text{leavevmode}P(0^{3^i+1},u)=S(0^{3^i+1},u)=0,\\ |u{|}_{0^{3^i}1}+1-S(0^{3^i+1},u)-P(0^{3^i}1,u)\text{leavevmode}(\mathrm{mod}\text{leavevmode}2),& \text{otherwise}.\end{array}$$

$$|u{|}_{0^{3^i+1}}-|u{|}_{0^{3^i+2}}$$

$$=\sum _{\begin{array}{c}v\text{ is a block of zeros in }u\\ |v|\ge 3^i+1\end{array}}1=Z(3^i+1).$$

$${\Delta }_i=|u{|}_{0^{3^i}1}-2|u{|}_{0^{3^i+1}1}+1-S(0^{3^i+1},u)-P(0^{3^i}1,u).$$

$${\Delta }_i=|u{|}_{0^{3^i}1}-2Z(3^i+1)+1-P(0^{3^i}1,u).$$

$$|u{|}_{0^{3^i+1}}$$

$$-2Z(3^i+1)\equiv \frac{2}{3^i}|u{|}_{0^{3^i+1}}\text{leavevmode}(\mathrm{mod}\text{leavevmode}3).$$

$$E_1$$

$$E_2$$

$$|u{|}_z=|u{|}_{{\mathrm{pref}}_k(z)}-S({\mathrm{pref}}_k(z),u)=|v{|}_{{\mathrm{pref}}_k(z)}-S({\mathrm{pref}}_k(z),v)=|v{|}_z.$$

$$|u{|}_z=|u{|}_{{\mathrm{suff}}_k(z)}-P({\mathrm{suff}}_k(z),u)=|v{|}_{{\mathrm{suff}}_k(z)}-P({\mathrm{suff}}_k(z),v)=|v{|}_z.$$

$$|u{|}_{0^{k+1}}$$

$$\displaystyle=|u|_{0^{k}}-\big{(}|u|_{0^{k-1}1}-P(0^{k-1}1,u)\big{)}-S(0^{k},u)$$

$$\displaystyle=|v|_{0^{k}}-\big{(}|v|_{0^{k-1}1}-P(0^{k-1}1,v)\big{)}-S(0^{k},v)=|v|_{0^{k+1}}.$$

$$\{\begin{array}{rl}& |w{|}_{0^{k+1}}+|w{|}_{0^{k}1}=|w{|}_{0^k}-S(0^k,w),\\ & |w{|}_{0^{k+1}}+|w{|}_{10^k}=|w{|}_{0^k}-P(0^k,w),\\ & |w{|}_{10^k}+|w{|}_{10^{k-1}1}=|w{|}_{10^{k-1}}-S(10^{k-1},w),\\ & |w{|}_{0^{k+1}}+|w{|}_{10^{k-1}1}=|w{|}_{0^k}-1+{\Delta }_i,\end{array}$$

$$p_k(n,x,y):=\text{Card}\left({\mathcal{W}}_{n,x,y}/{\sim }_k\right),$$

Full text

This work was supported by NSFC (Nos. 11401188 and 11626110).

∗ Wen Wu is the corresponding author.

1. Introduction

This paper is devoted to the study of the $k$-abelian complexity of the Cantor sequence

[TABLE]

which satisfies $c_{0}=1$ and for all $n\geq 0$,

[TABLE]

The $k$-abelian complexity, which was introduced by Karhumäki in [8], is a measure of disorder of infinite words. It has been studied widely in [12, 13, 14, 15, 16]. Before we give its definition, we need some notations. Let $\mathcal{A}$ be a finite alphabet and $\mathcal{A}^{n}$ be the set of words of length $n$ for every positive integer $n$. Denote $\mathcal{A}^{*}$ the set of all finite words on $\mathcal{A}$. For two words $u,v\in\mathcal{A}^{*}$, $v$ is called a factor of $u$ if $u=wvw^{\prime}$ where $w,w^{\prime}\in\mathcal{A}^{*}$. For a word $u=u_{0}u_{1}\cdots u_{n-1}\in\mathcal{A}^{n}$, the prefix and suffix of length $\ell（\geq 1）$ are defined as

[TABLE]

while for $\ell\leq 0$, we define $\mathrm{pref}_{\ell}(u)=\varepsilon$ and $\mathrm{suff}_{\ell}(u)=\varepsilon$, where $\varepsilon$ is the empty word. Denote $|u|$ the length of a word $u$ and denote $|u|_{v}$ the number of occurrences of a word $v$ in $u$.

Definition 1** (see [17]).**

Let $k\geq 1$ be an integer. Two words $u,v\in\mathcal{A}^{*}$ are called $k$-abelian equivalent, written by $u\sim_{k}v$, if $\mathrm{pref}_{k-1}(u)=\mathrm{pref}_{k-1}(v)$, $\mathrm{suff}_{k-1}(u)=\mathrm{suff}_{k-1}(v)$ and $|u|_{w}=|v|_{w}$ for every $w\in\mathcal{A}^{k}$.

The above definition is one of the equivalent definitions of the $k$-abelian equivalence; see also [16]. The $k$-abelian equivalence is in fact an equivalence relation. The $k$-abelian complexity of an infinite word $\omega$ is the function $\mathcal{P}^{(k)}_{\omega}:\mathbb{N}\rightarrow\mathbb{N}$ and for every $n\geq 1$, $\mathcal{P}^{(k)}_{\omega}(n)$ is assigned to be the number of $k$-abelian equivalence classes of factors of $\omega$ of length $n$. Precisely, for every positive integer $n$,

[TABLE]

where $\mathcal{F}_{\omega}(n)$ is the set of all factors of length $n$ occurring in $\omega$.

In our first result, we reduce the $k$-abelian equivalence of any two factors of $\mathbf{c}$ to the abelian equivalence of such factors. In detail, we prove the following theorem.

Theorem 1**.**

Let $k\geq 1$ be an integer and let $u,v$ be two factors of $\mathbf{c}$ satisfying $|u|=|v|$. If $\mathrm{pref}_{k}(u)=\mathrm{pref}_{k}(v)$ and $\mathrm{suff}_{k}(u)=\mathrm{suff}_{k}(v)$, then $u\sim_{k+1}v$ if and only if $u\sim_{1}v$.

By using Theorem 1, we are able to study the $k$-abelian complexity of $\mathbf{c}$ for every $k\geq 1$, and we have the following result.

Theorem 2**.**

For every integer $k\geq 1$, the $k$-abelian complexity function of the Cantor sequence is a $3$-regular sequence.

The $k$-regular sequence was introduced by Allouch and Shallit [2] as an extension of the $k$-automatic sequence. The definitions of the $k$-automatic sequences and the $k$-regular sequences are stated below; see also [1, 6].

Definition 2**.**

For an integer $k\geq 1$, a sequence $\mathbf{w}=(w_{n})_{n\geq 0}$ is a $k$-automatic sequence if its $k$-kernel

[TABLE]

is a finite set. The sequence $\bf{w}$ is called a $k$-regular sequence if the $\mathbb{Z}$-module generated by its $k$-kernel is finitely generated.

Karhumäki, Saarela and Zamboni [14] studied the $k$-abelian complexity of the Thue-Morse sequence, which is a $2$-automatic sequence. Vandomme, Parreau and Rigo [17] conjectured that the $2$-abelian complexity of the Thue-Morse sequence is a $2$-regular sequence. This has been proved independently in [12] by Greinecker and in [13] by Parreau, Rigo, Rowland and Vandomme.

Our result (Theorem 2) supports the following more general conjecture, which has been posed in [13].

Conjecture 1**.**

The $k$-abelian complexity of any $\ell$-automatic sequence is an $\ell$-regular sequence.

This paper is organized as follows. In Section 2, we give the recurrence relations for the abelian complexity function of the sequence $\mathbf{c}$. As a consequence, the abelian complexity function of the Cantor sequence is a $3$-regular sequence. In Section 3, we prove Theorem 1. In the last section, we give the proof of Theorem 2.

2. Abelian complexity

The abelian complexity of an infinite word $\omega$ is in fact the $1$-abelian complexity of $\omega$. For more details of the abelian complexity, see [4, 5, 8, 9, 10, 11] and references therein. In this section, we shall investigate the abelian complexity of $\mathbf{c}.$

First we introduce a useful result which characterizes the left and right special factors of $\mathbf{c}.$ Recall that a factor $v$ of $w$ is called right special (resp. left special) if both $va$ and $vb$ (resp. $av$ and $bv$) are factors of $w$ for distinct letters $a,b\in\mathcal{A}$. We denote $\mathcal{RS}_{w}(n)$ (resp. $\mathcal{LS}_{w}(n)$) the set of all right special (resp. left special) factors of $w$ of length $n$.

Lemma 1**.**

For every $i\geq 0$ and $3^{i}<k\leq 3^{i+1}$,

[TABLE]

Proof.

The result follows from [7, Theorem 1] and the fact that every left special factor in $\mathbf{c}$ is the reversal of some right special factor in $\mathbf{c}$. ∎

Let $\omega=\omega_{0}\omega_{1}\omega_{2}\cdots$ be an infinite sequence on $\{0,1\}$. It is proved in [3, Proposition 2.2] that the abelian complexity of $\omega$ is related to its digit sums in the following way: for every $n\geq 1$,

[TABLE]

where

[TABLE]

For the digit sums of the Cantor sequence $\mathbf{c}$, we have the following lemma.

Lemma 2**.**

For every integer $n\geq 1$, $M_{\mathbf{c}}(n)=\sum_{i=0}^{n-1}c_{i}$ and $m_{\mathbf{c}}(n)=0$.

Proof.

Since $0^{n}$ is always a factor of $\mathbf{c}$ for every $n\geq 1$, we have $m_{\mathbf{c}}(n)=0$ for every $n\geq 1$.

For every $i\geq 0$ and $n\geq 1$, let $\Sigma(i,n):=\sum_{j=i}^{i+n-1}c_{j}$. We only need to show that $M_{\mathbf{c}}(n)\leq\Sigma(0,n)$ for every $n\geq 1$, since the inverse inequality always holds by definition. For this purpose, we shall prove that for every $n\geq 1$,

[TABLE]

Since ‘$1$’ occurs in $\mathbf{c}$ and ‘$11$’ does not occur in $\mathbf{c}$, we have $\Sigma(i,1)\leq 1=\Sigma(0,1)$ and $\Sigma(i,2)\leq 1=\Sigma(0,2)$. Now suppose (2.2) holds for $n<m$. We first deal with the case: $m=3j+2$. By (1.1), we have the following nine recurrence relations:

[TABLE]

By the above equations and the inductive assumption, for every $i\geq 0$,

[TABLE]

Note that at least one of $c_{i}$ and $c_{i+1}$ must be zero. So

[TABLE]

Therefore, (2.2) holds in the case $m=3j+2$. Following the same way, we can verify (2.2) when $m=3j,\,3j+1$. ∎

Corollary 1**.**

$M_{\mathbf{c}}(1)=1$, $M_{\mathbf{c}}(2)=1$ and for every $n\geq 1$,

[TABLE]

Moreover, $\{M_{\mathbf{c}}(n)\}_{n\geq 1}$ is a $3$-regular sequence.

Proposition 1**.**

$\mathcal{P}^{(1)}_{\mathbf{c}}(1)=2$, $\mathcal{P}^{(1)}_{\mathbf{c}}(2)=2$ and for every $n\geq 1$,

[TABLE]

Moreover, $\{\mathcal{P}^{(1)}_{\mathbf{c}}(n)\}_{n\geq 1}$ is a $3$-regular sequence.

Proof.

It follows from Lemma 2, Corollary 1 and (2.1). ∎

3. From $k$-abelian equivalence to $1$-abelian equivalence

In this section, we give a key theorem, which implies that under certain condition, $k$-abelian equivalence can be reduced to $1$-abelian equivalence. Using this theorem, we deduce the regularity of the $k$-abelian complexity of $\mathbf{c}$ from that of the abelian complexity of $\mathbf{c}$. Before stating the result, we give two auxiliary lemmas. For $z,w\in\mathcal{A}^{*}$, we define

[TABLE]

Lemma 3**.**

Let $\mathcal{\omega}\in\{0,1\}^{\mathbb{N}}$ and $u,z\in\mathcal{F}_{\mathcal{\omega}}$ with $|u|\geq|z|$. Suppose $z=ayb,$ where $a,b\in\{0,1\}$. We have

[TABLE]

Proof.

Note that $|u|_{ay}-S(ay,u)$ is the number of occurrences of a right extendable $ay$ in $u$. When $ay$ is not right special, every right extension of a right extendable $ay$ must be $z$. So, $|u|_{ay}-S(ay,u)=|u|_{z}$. When $ay$ is right special, its right extensions are either $z$ or $ay(1-b)$. So, $|u|_{ay}-S(ay,u)=|u|_{z}+|u|_{ay(1-b)}$. The rest cases can be verified in the same way. ∎

Lemma 4**.**

For every $i\geq 0$, $u\in\mathcal{F}_{\mathbf{c}}$, let $\Delta_{i}:=|u|_{0^{3^{i}+2}}+|u|_{10^{3^{i}}1}-|u|_{0^{3^{i}+1}}+1$. Then $\Delta_{i}\in\{0,1,2\}$ and

[TABLE]

Proof.

Let $Z(\ell)$ $(\ell\geq 1)$ be the number of blocks of zeros (in $u$) of length not less than $\ell$. For example, when $u=0010100$, then $Z(1)=3$ and $Z(2)=2$. Note that, for every $\ell\geq 3^{i}+1$, $|0^{\ell}|_{0^{3^{i}+1}}-|0^{\ell}|_{0^{3^{i}+2}}=1$. So,

[TABLE]

On the other hand, $10^{3^{i}}1$ only occurs in $\sigma^{i+1}(1)$. Thus, there is a block of zeros of length $3^{i+\ell}$ (for some $\ell\geq 1$) between two consecutive $10^{3^{i}}1$. Since the block of zeros could also be the prefix or suffix of $u$, we have $|u|_{10^{3^{i}}1}-1\leq Z(3^{i}+1)\leq|u|_{10^{3^{i}}1}+1$, which implies $\Delta_{i}\in\{0,1,2\}$.

When $P(0^{3^{i}+1},u)=1$ or $S(0^{3^{i}+1},u)=1$, there is at least one block of zeros of length not less than $3^{i}+1,$ which is not located between two consecutive $10^{3^{i}}1$. This implies that $|u|_{10^{3^{i}}1}\leq Z(3^{i}+1)\leq|u|_{10^{3^{i}}1}+1$. So, in this case, $\Delta_{i}\in\{0,1\}$. Applying Lemma 3 to $|u|_{0^{3^{i}+2}}$ and $|u|_{10^{3^{i}}1}$, we have

[TABLE]

Since $\Delta_{i}\in\{0,1\}$, by (3.1), $\Delta_{i}=|u|_{0^{3^{i}}1}+1-S(0^{3^{i}+1},u)-P(0^{3^{i}}1,u)\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 2).$

Now, suppose $P(0^{3^{i}+1},u)=S(0^{3^{i}+1},u)=0$. Applying Lemma 3 to $|u|_{0^{3^{i}+1}1}$, by (3.1), we have

[TABLE]

Let $\sum_{v}$ denote the sum over all blocks of zeros $v$ of $u$ of length not less than $3^{i}+1$. Then

[TABLE]

Note that, in this case, all blocks of zeros of $u$ are of length $3^{i+\ell}$ for some $\ell\geq 1$. So,

[TABLE]

The result of this case follows from (3.3) and (3.2). ∎

Now, we prove Theorem 1.

Proof of Theorem 1.

Let $u,v\in\mathcal{F}_{\mathbf{c}}$ satisfying $|u|=|v|$, $\mathrm{pref}_{k}(u)=\mathrm{pref}_{k}(v)$ and $\mathrm{suff}_{k}(u)=\mathrm{suff}_{k}(v)$. When $k\geq|u|$, the assumption gives $u=v$. In this case, the result is trivial. In the following, we alway assume that $k<|u|$.

The ‘only if’ part follows directly from the definition of $k$-abelian equivalence. For the ‘if’ part, we only need to show that $u\sim_{k}v$ implies that for every $z\in\mathcal{F}_{\mathbf{c}}(k+1)$, $|u|_{z}=|v|_{z}$. For this purpose, we separate $\mathcal{F}_{\mathbf{c}}(k+1)$ into two disjoint parts, i.e., $\mathcal{F}_{\mathbf{c}}(k+1)=E_{1}\cup E_{2}$, where

[TABLE]

Suppose $z\in E_{1}$. If $\mathrm{pref}_{k}(z)\notin\mathcal{RS}_{\mathbf{c}}(k)$, then by Lemma 3,

[TABLE]

If $\mathrm{suff}_{k}(z)\notin\mathcal{LS}_{\mathbf{c}}(k)$, then by Lemma 3,

[TABLE]

So, for every $z\in E_{1}$, $|u|_{z}=|v|_{z}$.

Now, let $z\in E_{2}$. Suppose $3^{i}<k\leq 3^{i+1}$ for some $i\geq 0$. When $k\neq 3^{i}+1$, by Lemma 1, $E_{2}=\{0^{k+1}\}$. By Lemma 3 and the assumptions of this result,

[TABLE]

When $k=3^{i}+1$, by Lemma 1, $E_{2}=\{0^{k+1},0^{k}1,10^{k},10^{k-1}1\}$. For every $w\in\mathcal{F}_{\mathbf{c}}$, by Lemma 3 and 4, we have the following linear system:

[TABLE]

which determines $(|w|_{z})_{z\in E_{2}}$ uniquely. If $u\sim_{k}v$, then the linear systems (3.4) for $u$ and $v$ turn out to be the same one. So, $u\sim_{k}v$ implies $|u|_{z}=|v|_{z}$ for every factor $z\in E_{2}$. ∎

We may now apply Theorem 1 repeatedly to reduce the $k$-abelian equivalence to the $1$-abelian equivalence under the condition of Theorem 1.

Corollary 2**.**

Let $k\geq 1$ and $u,v\in\mathcal{F}_{\mathbf{c}}$ satisfying $|u|=|v|$. If $\mathrm{pref}_{k}(u)=\mathrm{pref}_{k}(v)$ and $\mathrm{suff}_{k}(u)=\mathrm{suff}_{k}(v)$, then $u\sim_{k+1}v\leavevmode\nobreak\ \text{if and only if}\leavevmode\nobreak\ u\sim_{1}v$.

Remark 1**.**

A similar result for Sturmian words is obtained by Karhumäki, Saarela and Zamboni [16, Corollary 3.1]. We would like to ask that in general, what kind of infinite words share a property similar to Corollay 2?

4. $k$-abelian complexity

In this section, we first give the regularity of the $2$-abelian complexity of $\mathbf{c}$. Then, by using Theorem 1 properly, we deduce the regularity of the $k$-abelian complexity of $\mathbf{c}$. We start by classifying the $k$-abelian equivalent classes of $\mathcal{F}_{\mathbf{c}}(n)$ by their prefixes and suffixes of length $k-1$.

For every $k\geq 2$, $x,y\in\mathcal{F}_{\mathbf{c}}(k-1)$ and every $n\geq 1$, let

[TABLE]

where

[TABLE]

Here $p_{k}(n,x,y)$ denotes the number of $k$-abelian equivalent classes with the prefix $x$ and the suffix $y.$ Then, for every $n\geq 1$,

[TABLE]

By Theorem 1,

[TABLE]

4.1. Regularity of the $2$-abelian complexity of $\mathbf{c}$

Recall that the Cantor sequence $\mathbf{c}$ is the fixed point of the morphism $\sigma:0\mapsto 000,1\mapsto 101$ starting by $1,$ i.e., $\mathbf{c}=\sigma^{\infty}(1).$

Lemma 5**.**

For all $i,j\geq 1$, let $d_{j}$ be the number of ‘[math]’ between the $j$-th ‘$1$’ and the $(j+1)$-th ‘$1$’ in $\mathbf{c}$, and let $f(i,j)=j+\sum_{\ell=i}^{i+j-1}d_{\ell}$. Then, for every $j\geq 1$,

[TABLE]

Moreover, for all $i,j\geq 1$,

[TABLE]

Proof.

While applying $\sigma$ to ‘$1$’ or a block of ‘[math]’s, we obtain only one block of ‘[math]’s in both cases. Note that in $\mathbf{c}$, every ‘$1$’ is followed by a block of ‘[math]’s. Before the $i$-th ‘$1$’, the number of occurrences of ‘$1$’ is $(i-1)$ and there are $(i-1)$ blocks of ‘[math]’s in $\mathbf{c}$. So, while applying $\sigma$ to $\mathbf{c}$, the $i$-th ‘$1$’ will generate the $(2i-1)$-th block of ‘[math]’s, which implies $d_{2i-1}=1$. For the same reason, the $i$-th block of ‘[math]’s will generate the $2i$-th block of ‘[math]’s. So, $d_{2i}=3d_{i}.$ This proves (4.3).

The recurrence relations (4.4) follows directly from (4.3). We verify the first one as an example:

[TABLE]

∎

Proposition 2**.**

$p_{2}(1,0,0)=p_{2}(1,1,1)=1$, $p_{2}(1,0,1)=p_{2}(1,1,0)=0$ and for every $n\geq 2$,

[TABLE]

Proof.

The initial values can be showed by enumerating all the factors of length $1$ and $2$. Now, let $n\geq 2$ and suppose $n<3^{i}$ for some $i\geq 1$.

Clearly, for every $w\in\mathcal{W}_{n,0,0}$, $|w|_{1}\leq M_{\mathbf{c}}(n-2)$. So, $p_{2}(n,0,0)\leq M_{\mathbf{c}}(n-2)+1$. We prove the inverse inequality in the following. For every $0\leq\ell\leq n-1$, let $W_{\ell}=0^{n-\ell}\mathrm{pref}_{\ell}(\sigma^{i}(1))$ that is a factor of $\sigma^{i}(01)$ and hence, a factor of $\mathbf{c}$. Note that $|W_{0}|_{1}=0$ and $|W_{n-2}|_{1}=M_{\mathbf{c}}(n-2)$. Since $|W_{\ell}|_{1}\leq|W_{\ell+1}|_{1}\leq|W_{\ell}|_{1}+1$, we know that $|W_{\ell}|_{1}$ changes continuously from [math] to $M_{\mathbf{c}}(n-2)$ while $\ell$ takes values from [math] to $n-2$. Therefore, for every $0\leq s\leq M_{\mathbf{c}}(n-2)$, there exists $0\leq\ell\leq n-2$ such that $|W_{\ell}|_{1}=s$. If the last letter of $W_{\ell}$ is [math], then $W_{\ell}\in\mathcal{W}_{n,0,0}$. Otherwise, $|W_{\ell+1}|_{1}=|W_{\ell}|_{1}=s$ since $11$ is not a factor of $\mathbf{c}$. So, $W_{\ell+1}\in\mathcal{W}_{n,0,0}$. This implies that $p_{2}(n,0,0)\geq M_{\mathbf{c}}(n-2)+1$ which proves (4.5a).

Since for every factor $w$ of $\mathbf{c}$, its reversal $\bar{w}$ is also a factor of $\mathbf{c}$, we have $p_{2}(n,1,0)=p_{2}(n,0,1)$. Then, applying a similar argument on the words $W_{\ell}^{\prime}=\mathrm{suff}_{\ell}(\sigma^{i}(1))0^{n-\ell}$ where $1\leq\ell\leq n-1$, we obtain (4.5b).

(In the rest of the proof, the symbol ‘$\equiv$’, otherwise stated, means equality modulo $2$.)

Now, we prove (4.5c) for the case $n\equiv 0$. We first observe that for every $w\in\mathcal{W}_{n,1,1}$, $|w|\equiv 1$. Since the number of [math] between two successive $1$ must be $3^{j}$ for some $j\geq 0$ and $3^{j}\equiv 1$, we have $|w|_{0}\equiv|w|_{1}-1$ for every $w\in\mathcal{W}_{n,1,1}$. Therefore, $|w|=|w|_{0}+|w|_{1}\equiv 1$. Hence, $\mathcal{W}_{n,1,1}=\emptyset$ when $n$ is an even number, which implies $p_{2}(n,1,1)=0$ when $n\equiv 0$.

In the following, we will prove (4.5c) when $n\equiv 1$. For every $w\in\mathcal{W}_{n,1,1}$,

[TABLE]

for some $i\geq 1$. (Since if a word occurs in $\mathbf{c}$, then it will occur infinitely many times in $\mathbf{c}$, we can assume $i\geq 3$.) Therefore, we only need to prove that for every $m\geq 1$, there is only one integer $t_{m}\geq 2$ satisfying

[TABLE]

for some $i\geq 1$. We reason by induction. Since $\mathcal{W}_{3,1,1}=\{101\}$ and $\mathcal{W}_{5,1,1}=\{10001\}$, it follows that (4.6) holds for $m=1$ and $2$. Assuming that (4.6) holds for every $\ell\leq m$, we prove it for $m+1$. We only give the proof for the case $m=3m^{\prime}$; the other cases follow in a similar way. In this case, by inductive assumptions and (4.4),

[TABLE]

which implies that there is a solution of (4.6) for $m+1$. Now, we prove the uniqueness. Let $t\geq 2$ be a solution of (4.6) for $m+1$. Then,

[TABLE]

which implies $f(i,t)\equiv 2\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 3)$. According to (4.4), this happens only if $(i,t)\equiv(1,1)$. Write $i=2i^{\prime}+1$ and $t=2t^{\prime}+1$. Then, by (4.4) and (4.7),

[TABLE]

By the inductive assumption, we know that $t^{\prime}$ is the unique solution of (4.6) for $m^{\prime}$. So, the only solution of (4.6) for $m+1$ is $2t_{m^{\prime}}+1$. ∎

By Proposition 2, for every $n\geq 2$, we have

[TABLE]

4.2. Regularity of the $k$-abelian complexity of $\mathbf{c}$

In this part, we prove the regularity of the $k$-abelian complexity of the Cantor sequence for every $k\geq 3$.

Let $\mathcal{F}_{\mathbf{c}}$ denote the set of all factors of $\mathbf{c}.$ For every $u\in\mathcal{F}_{\mathbf{c}}$ and $\ell\geq 1,$ we define

[TABLE]

The elements in $\mathrm{Type}(\ell,u)$ are called types of $u$ (with respect to $\ell$). Clearly, for every $\ell$ and $u\in\mathcal{F}_{\mathbf{c}}$, $\mathrm{Card}(\mathrm{Type}(\ell,u))\geq 1$.

Every type of $u$ gives a decomposition of $u$ in the following sense. For every ${j}\in\mathrm{Type}(\ell,u)$, there is an integer $n\geq 0$ such that

[TABLE]

where $h=\lfloor\frac{|u|+{j}}{3^{\ell}}\rfloor$, $j_{0}=3^{\ell}-{j}$ and $j_{1}=j+|u|-3^{\ell}h$. The following lemma shows that every non-zero factor of $\mathbf{c}$, which is long enough, occurs in a (relatively) fixed position, i.e., has only one type. By a non-zero factor we mean a factor that contains at least one letter ‘$1$’.

Lemma 6**.**

For every integer $\ell\geq 1$ and every non-zero factor $u\in\mathcal{F}_{\mathbf{c}}$ with $|u|>3^{\ell}$,

[TABLE]

Proof.

We prove by induction on $\ell$. We first prove the result for $\ell=1$. Now, we show that $\mathrm{Card}(\mathrm{Type}(1,u))=1$ for $u\in\mathcal{F}_{\mathbf{c}}(4)$ with $|u|_{1}>0$. We only verify the case $u=0001$ as an example; the rest can be verified in the same way. Suppose $0001=c_{n}c_{n+1}c_{n+2}c_{n+3}$. Since $c_{n+3}=1$, by (1.1), we have $n\not\equiv 1\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 3)$. If $n\equiv 2\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 3)$, then by (1.1), $0=c_{n+1}=c_{n+3}=1$, which is a contradiction. Thus, $\mathrm{Type}(1,0001)=\{0\}$.

For every non-zero factor $u\in\mathcal{F}_{\mathbf{c}}$ with $|u|>4$, let $u=xvy$ where $v$ is the the first non-zero factor of length $4$ of $u$. Since $\mathrm{Type}(1,v)+|x|\equiv\mathrm{Type}(1,u)\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 3)$, we have

[TABLE]

Suppose the result holds for $\ell$. We prove it for $\ell+1$. Let $u\in\mathcal{F}_{\mathbf{c}}$ with $|u|>3^{\ell+1}$ and $i_{0}\in\mathrm{Type}(\ell,u)$. Then,

[TABLE]

for some $n\geq 0$. By (4.8), $u$ uniquely determines $i_{0}$, $|u|$ and $c_{n}c_{n+1}\cdots c_{n+h}$ where $h=\lfloor\frac{|u|+i_{0}}{3^{\ell}}\rfloor$. Since $h\geq 3$, $n\equiv i_{1}\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 3)$ where $i_{1}\in\mathrm{Type}(1,c_{n}\cdots c_{n+h})$. Therefore,

[TABLE]

By the inductive assumptions, $\mathrm{Card}(\mathrm{Type}(1,c_{n}\cdots c_{n+h}))=1$ and $\mathrm{Card}(\mathrm{Type}(\ell,u))=1$. So, by (4.9), we have

[TABLE]

∎

Lemma 7**.**

For every integer $\ell\geq 1$ and every non-zero factor $u\in\mathcal{F}_{\mathbf{c}}$ with $3^{\ell}<|u|\leq 3^{\ell+1}$,

[TABLE]

Proof.

Let $u\in\mathcal{F}_{\mathbf{c}}$ with $3^{\ell}<|u|\leq 3^{\ell+1}$ and $i_{0}\in\mathrm{Type}(\ell,u)$. Then, $u=c_{3^{\ell}n+i_{0}}\cdots c_{3^{\ell}n+i_{0}+|u|-1}$ for some $n\geq 0$. By (4.8), $u$ uniquely determines $i_{0}$, $|u|$ and $c_{n}c_{n+1}\cdots c_{n+h}=:v,$ where $h=\lfloor\frac{|u|+i_{0}}{3^{\ell}}\rfloor$. Note that $v$ is a non-zero factor. Write $q(v):=\max\{j\mid 0^{j}\text{ is a prefix of }v\}$. Then $c_{n+q(v)}=1$, which implies $n+q(v)\not\equiv 1\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 3)$ by (1.1). So,

[TABLE]

The result follows from Lemma 6 and the above formula. ∎

In the rest of this section, let $i$ be the integer satisfying

[TABLE]

To study the regularity of $\{p_{k}(n,x,y)\}_{n\geq 1}$ for $x,y\in\mathcal{F}_{\mathbf{c}}(k-1)$, our idea is the following. We first give the upper bound of $p_{k}(n,\cdot,\cdot)$ by using $M_{\mathbf{c}}(\cdot)$, which is a $3$-regular sequence according to Corollary 1. Then, by constructing sufficiently many words which belong to different $k$-abelian equivalence classes, we show that the upper bound can be reached. Therefore, the regularity of $\{p_{k}(n,x,y)\}_{n\geq 1}$ follows from the regularity of $\{M_{\mathbf{c}}(n)\}_{n\geq 1}$.

The following lemma contributes to the construction of words that belong to different $k$-abelian equivalence classes.

Lemma 8**.**

Let $\alpha\in\{0,1\}$. For every $\ell\geq 1$ and every $h=1,2,\cdots,M_{\mathbf{c}}(\ell)$, there is a word $W_{h}\in\mathcal{F}_{\mathbf{c}}(\ell+3)$ such that $|W_{h}|_{1}=h$ and $W_{h}=00U_{h}\alpha$, where $U_{h}\in\mathcal{F}_{\mathbf{c}}(\ell)$.

Proof.

For all $j=0,1,\cdots,\ell+1$, let

[TABLE]

where $s\in\mathbb{N}$ satisfying $3^{s}>\ell+1$. Since $|W_{j}|_{1}\leq|W_{j+1}|_{1}\leq|W_{j}|_{1}+1$ and $|W_{\ell}|_{1}=M_{\mathbf{c}}(\ell)$, we know that $|W_{j}|_{1}$ changes from 0 to $M_{\mathbf{c}}(\ell)$ continuously while $j$ takes values from [math] to $\ell$. So, for every $h=1,\cdots,M_{\mathbf{c}}(\ell)$, there is a $j_{h}(\leq\ell)$ such that $|W_{j_{h}}|_{1}=h$. Moreover, we can require that the last letter of $W_{j_{h}}$ is [math]. Otherwise, $1$ is the last letter of $W_{j_{h}}$. Then, $W_{j_{h}+1}$ ends with [math] and $|W_{j_{h}+1}|_{1}=|W_{j_{h}}|_{1}$.

There also is a $j^{\prime}_{h}$ such that $|W_{j^{\prime}_{h}}|_{1}=h$, of which the last letter is $1$. Otherwise, [math] is the last letter of $W_{j^{\prime}_{h}}$. Let $m_{h}:=\max\{q\mid 0^{q}\leavevmode\nobreak\ \text{is a suffix of}\leavevmode\nobreak\ W_{j^{\prime}_{h}}\}$. Since $|W_{j^{\prime}_{h}}|_{1}=h\geq 1$, we always have $m_{h}<j^{\prime}_{h}$. Then, $W_{j^{\prime}_{h}-m_{h}}$ ends with $1$ and $|W_{j^{\prime}_{h}-m_{h}}|_{1}=|W_{j^{\prime}_{h}}|_{1}$. If $m_{h}>j^{\prime}_{h}$, ∎

Now, we shall show the regularity of $\{p_{k}(n,x,y)\}_{n\geq 1}$ for all $x,y\in\mathcal{F}_{\mathbf{c}}(k-1)$.

Lemma 9**.**

$\{p_{k}(n,0^{k-1},0^{k-1})\}_{n\geq 1}$* is a $3$-regular sequence.*

Proof.

Without loss of generality, we can assume that $n\geq 2\cdot 3^{i+1}+2k-2$, since changing finite terms of a sequence does not change its regularity. Noticing that $3^{i}<k-1\leq 3^{i+1}$, the occurrence of each $w\in\mathcal{W}_{n,0^{k-1},0^{k-1}}$ in $\mathbf{c}$ must be one of the four forms in Figure 1.

In all the four forms, we have $|w|_{1}=2^{i+1}|u|_{1}$ and $|u|=\ell$ or $\ell-1$, which implies

[TABLE]

where $\ell=\left\lfloor\frac{n-2k+2}{3^{i+1}}\right\rfloor$. Next, we prove the inverse of (4.11). That is

[TABLE]

Applying Lemma 8 for the above $\ell$ and $\alpha=0$, we have

[TABLE]

for all $h=1,2,\cdots,M_{\mathbf{c}}(\ell)$. Set $t:=n-3^{i+1}\ell-k+1$. Then, $k-1\leq t<k-1+3^{i+1}$. Therefore,

[TABLE]

for every $h=1,\cdots,M_{\mathbf{c}}(\ell)$. Noting also that $0^{n}\in\mathcal{W}_{n,0^{k-1},0^{k-1}}$, the inequality (4.12) holds. The result then follows from (4.11), (4.12) and Corollary 1. ∎

For every non-zero factor $v\in\mathcal{F}_{\mathbf{c}}(k-1)$, let $z_{v}:=\max\{p\mid 0^{p}\text{ is a suffix of }v\}$ and

[TABLE]

where $c_{q}$ is the last $1$ in $v$. Then, it follows from Lemma 7 that $1\leq\text{Card}(\tilde{L}_{v})\leq 2$. Moreover, if $\tilde{L}_{v}=\{q_{1},q_{2}\}$, where $0\leq q_{1}<q_{2}\leq 3^{i+1}-1$, then by (4.10), we have $q_{2}=q_{1}+2\cdot 3^{i}$.

For a word $w=w_{0}w_{1}\cdots w_{n-1}\in\mathcal{A}^{n}$, the reversal of $w$ is defined to be $\bar{w}=w_{n-1}\cdots w_{1}w_{0}$. When $w=uv$, we write $wv^{-1}:=u$ and $u^{-1}w:=v$ by convention.

Lemma 10**.**

For all non-zero factors $x,y\in\mathcal{F}_{\mathbf{c}}(k-1)$, two sequences $\{p_{k}(n,0^{k-1},y)\}_{n\geq 1}$ and $\{p_{k}(n,x,0^{k-1})\}_{n\geq 1}$ are both $3$-regular sequences.

Proof.

For every $x\in\mathcal{F}_{\mathbf{c}}$, its reversal $\bar{x}\in\mathcal{F}_{\mathbf{c}}$, since $x$ is a factor of $\sigma^{{m}}(1)$ for some ${m}\geq 1$ and $\overline{\sigma^{{m}}(1)}=\sigma^{{m}}(1)$. So, $p_{k}(n,x,0^{k-1})=p_{k}(n,0^{k-1},\bar{x})$ for every $n\geq 1$. Thus, we only need to verify the regularity of $\{p_{k}(n,0^{k-1},y)\}_{n\geq 1}$ for every non-zero factor $y\in\mathcal{F}_{\mathbf{c}}(k-1)$.

Since changing finite terms of a sequence does not change its regularity, we can assume that $n\geq 2\cdot 3^{i+1}+2k-2$. Recall that $3^{i}<k-1\leq 3^{i+1}$. Each occurrence of every $w\in\mathcal{W}_{n,0^{k-1},y}$ in $\mathbf{c}$ must be one of the six forms in Figure 2. In all the six forms, for every $o_{y}\in\tilde{L}_{y}$, we have

[TABLE]

and $|\tilde{u}|=\ell(o_{y})$ or $\ell(o_{y})+1$, where

[TABLE]

When $\text{Card}(\tilde{L}_{y})=1$, write $\tilde{L}_{y}=\{o_{y}\}$. By (4.13), we have

[TABLE]

On the other hand, applying Lemma 8 for $\ell(o_{y})$ and $\alpha=1$, we have

[TABLE]

for all $h=1,\cdots,M_{\mathbf{c}}(\ell+1)$. Set $t:=n-3^{i+1}\ell-o_{y}-z_{y}-1$; so $k-1\leq t<k-1+3^{i+1}$. Therefore,

[TABLE]

and

[TABLE]

for all $h=1,\cdots,M_{\mathbf{c}}(\ell+1)$. This implies that $p_{k}(n,0^{k-1},y)\geq M_{\mathbf{c}}\left(\ell(o_{y})+1\right)$. The previous inequality, (4.14) and Corollary 1 give the result in the case $\text{Card}(\tilde{L}_{y})=1$.

Now suppose $\text{Card}(\tilde{L}_{y})=2$ and set $\tilde{L}_{y}=\{o_{y},o_{y}^{\prime}:=o_{y}+2\cdot 3^{i}\}$ with $0\leq o_{y}\leq 3^{i}-1$. From (4.13), we know that $n_{o_{y}^{\prime}}\equiv n_{o_{y}}+2^{i}\mod 2^{i+1}$. Therefore,

[TABLE]

For every $q\in\tilde{L}_{y}$, applying Lemma 8 for $\ell(q)$ and $\alpha=1$, we have

[TABLE]

for every $h=1,\cdots,M_{\mathbf{c}}(\ell_{1}+1)$. Set $t(q):=n-3^{i+1}\ell(q)-q-z_{y}-1$; so $k-1\leq t(q)<k-1+3^{i+1}$. Therefore, for every $q\in\tilde{L}_{y}$,

[TABLE]

and

[TABLE]

for all $h=1,\cdots,M_{\mathbf{c}}(\ell_{1}+1)$. Since $|V_{o_{y}}|_{1}\equiv|V_{o_{y}^{\prime}}|_{1}-2^{i}$ (mod $2^{i+1}$), $V_{o_{y}}$ and $V_{o_{y}^{\prime}}$ belongs to different $k$-abelian equivalence classes. Therefore,

[TABLE]

Combining (4.15), (4.16) and Corollary 1, the result follows. ∎

Lemma 11**.**

For two non-zero factors $x,y\in\mathcal{F}_{\mathbf{c}}(k-1)$, $\{p_{k}(n,x,y)\}_{n\geq 1}$ is ultimately periodic.

Proof.

Without loss of generality, we can assume that $n\geq 2\cdot 3^{i+1}+2k-2$ since changing finite terms of a sequence does not change its regularity. Noticing that $3^{i}<k-1\leq 3^{i+1}$, for every pair of factors $x,y$ of length $k-1,$ the occurrence of each $w\in\mathcal{W}_{n,x,y}$ in $\mathbf{c}$ must be one of the nine forms in Figure 3.

For every fixed pair of $o_{x}\in\tilde{L}_{x}$ and $o_{y}\in\tilde{L}_{y}$, in all the nine forms, we have

[TABLE]

and

[TABLE]

where $\ell(o_{x},o_{y}):=(k-1-o_{x}-z_{x}+o_{y}+z_{y})<2\cdot 3^{i+1}$ and

[TABLE]

Further, according to (4.5c), $\tilde{u}$ in (4.19) must satisfy $|\tilde{u}|\equiv 1\mod 2$. This fact and (4.17) yield that $\mathcal{W}_{n,x,y}=\emptyset$ when $n\not\equiv\ell(o_{x},o_{y})\mod 2\cdot 3^{i+1}$.

Now we deal with the case $n\equiv\ell(o_{x},o_{y})\mod 2\cdot 3^{i+1}$. Note that by (4.5c), we have $p(2j+1,1,1)=1$ for all $j\geq 1$. This fact and (4.18) imply that for all $n=2\cdot 3^{i+1}j+\ell(o_{x},o_{y})$,

[TABLE]

In conclusion, let $\mathcal{I}_{x,y}=\{2\cdot 3^{i+1}j+\ell(o_{x},o_{y})\mid j\geq 1,o_{x}\in\tilde{L}_{x},o_{y}\in\tilde{L}_{y}\}$. We have

[TABLE]

Therefore, $\{p_{k}(n,x,y)\}_{n\geq 1}$ is ultimately periodic with a period $2\cdot 3^{i+1}$. ∎

Proposition 3**.**

$\{\mathcal{P}_{\mathbf{c}}^{(k)}(n)\}_{n\geq 1}$* is a $3$-regular sequence for every $k\geq 3.$*

Proof.

It follows directly from Lemmas 9, 10 and 11 and (4.1). ∎

Theorem 2 follows from Propositions 1, 2 and 3.