On some problems of Harju concerning squarefree arithmetic progressions in infinite words

TL;DR

This paper addresses open problems posed by Harju about the existence and properties of square-free arithmetic progressions in infinite words, providing solutions to two of these problems.

Contribution

It solves two of Harju's open problems related to square-free arithmetic progressions in infinite words.

Findings

Solved two of Harju's open problems.

Established new results on square-free arithmetic progressions.

Contributed to the theoretical understanding of infinite words.

Abstract

In a recent paper, Harju posed three open problems concerning square-free arithmetic progressions in infinite words. In this note we solve two of them.

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1 Introduction

The study of infinite words avoiding squares is a classical problem in combinatorics on words. A square is a word of the form $xx$, such as tartar. One of the most well-studied squarefree words [10, 6] is the word $\mbox{\hskip 0.72229pt{\tt vtm}}=012021012102012\cdots$ obtained by iterating the map $0\mapsto 012$, $1\mapsto 02$, $2\mapsto 1$.

Harju [7] studied the following question and showed that it has a positive solution for all $k\geq 3$:

Given $k$, does there exist an infinite squarefree sequence $(w_{n})_{n\geq 0}$ over a ternary alphabet such that the subsequence $(w_{kn})_{n\geq 0}$ is squarefree?

Carpi [3], Currie and Simpson [5], and Kao et al. [8] also studied similar problems. Harju ended his paper with three open problems:

Does there exist a squarefree sequence $(w_{n})_{n\geq 0}$ over a ternary alphabet such that for every $k\geq 3$, the subsequence $(w_{kn})_{n\geq 0}$ contains a square? 2. 2.

Do there exist pairs $(k,\ell)$ of relatively prime integers such that there exists a squarefree sequence $(w_{n})_{n\geq 0}$ over a ternary alphabet for which both $(w_{kn})_{n\geq 0}$ and $(w_{\ell n})_{n\geq 0}$ are squarefree? 3. 3.

It is true that for all squarefree words $(w_{n})_{n\geq 0}$ over a ternary alphabet, there exists a word $(v_{n})_{n\geq 0}$ and an integer $k\geq 3$ such that $(v_{kn})_{n\geq 0}=(w_{n})_{n\geq 0}$?

In this note we show that the word  vtm gives a positive answer to the first problem. We also show that there a positive answer to the third problem for every $k\geq 23$.

2 The main results

We recall that $\mbox{\hskip 0.72229pt{\tt vtm}}=(v_{n})_{n\geq 0}$ is the fixed point (starting with [math]) of the morphism that maps $0\mapsto 012$, $1\mapsto 02$, $2\mapsto 1$.

Theorem 1**.**

For each $k\geq 2$ the sequence $(v_{kn})_{n\geq 0}$ contains either the square $00$ or the square $22$.

Proof.

The first part of the proof relies on the fact that  vtm is a $2$-automatic sequence. Berstel [1] studied several different ways to generate the sequence  vtm; in particular, he showed that  vtm is generated by the $2$-DFAO (deterministic finite automaton with output) in Figure 1. The automaton takes the binary representation of $n$ as input, and if the computation ends in a state labeled $a$, the automaton outputs $a$, indicating that $v_{n}=a$.

Since  vtm is an automatic sequence, we can use Walnut [9] to verify that it has certain combinatorial properties. We verify with Walnut that for every $k\geq 2$, the sequence  vtm contains a length $k+1$ factor of the form $0u0$ or $2u2$. The Walnut command to do this is:

eval same_first_last ‘‘Ei (VTM[i]=@0 & VTM[i+k]=@0)|(VTM[i]=@2 & VTM[i+k]=@2)’’;

The Walnut output for this command is the automaton in Figure 2, which shows that the given predicate holds for all $k\geq 2$.

To complete the proof, it suffices to show that  vtm contains an occurrence of the length $k+1$ factor $0u0$ or $2u2$ at a position congruent to [math] modulo $k$. Blanchet-Sadri et al. [2, Theorem 3] showed that for each odd $k$ and each factor $w$ of  vtm, the set of positions at which $w$ occurs in  vtm contains all congruence classes modulo $k$, including, in particular, [math] modulo $k$. This establishes the claim for all odd $k\geq 2$.

If $k$ is even, write $k=2^{a}k^{\prime}$, where $k^{\prime}$ is odd. Suppose that $k^{\prime}\geq 3$. We have already seen that  vtm contains an occurrence of a length $k^{\prime}+1$ factor $0u0$ or $2u2$ at a position $i\equiv 0\pmod{k^{\prime}}$. From the automaton generating  vtm, we see that if $v_{i}=0$ (resp. $v_{i}=2$), then $v_{2^{a}i}=0$ (resp. $v_{2^{a}i}=2$), which establishes the claim for $k^{\prime}\geq 3$.

Finally, suppose that $k$ is a power of $2$. Then the binary representations of $k$ and $2k$ have the form $10^{\ell}$ and $10^{\ell+1}$ respectively, for some $\ell\geq 1$. From the automaton generating  vtm, we see that $v_{k}=v_{2k}=2$, as required. This completes the proof. ∎

This resolves Harju’s first problem in the affirmative. For the third, we use the result of Currie [4] (also referenced in [7]), who showed that for $k\geq 23$ there exists a cyclic squarefree $k$-uniform morphism on the ternary alphabet. Let $h_{k}:\{0,1,2\}^{*}\to\{0,1,2\}^{*}$ denote such a morphism. By cyclic we mean that if $\sigma$ denotes the morphism defined by the cyclic permutation of the alphabet $0\mapsto 1\mapsto 2\mapsto 0$, then for $i\in\{0,1,2\}$ we have $h_{k}(i)=\sigma^{i}(h_{k}(0))$. By $k$-uniform we mean that for $i\in\{0,1,2\}$ the images $h_{k}(i)$ all have length $k$. Finally, we say that the morphism $h_{k}$ is squarefree if it maps squarefree words to squarefree words. Without loss of generality, suppose that $h_{k}(i)$ begins with the letter $i$. Now if $w=(w_{n})_{n\geq 0}$ is a squarefree word, then $v=h_{k}(w)$ is also squarefree and moreover $(v_{kn})_{n\geq 0}=(w_{n})_{n\geq 0}$, as required. This resolves Harju’s third problem, and furthermore, shows that solutions exist for every $k\geq 23$.