Avoiding squares over words with lists of size three amongst four symbols

TL;DR

This paper proves the existence of infinite square-free words with certain list constraints and provides lower bounds on their counts, advancing understanding of combinatorial word structures with restricted alphabet lists.

Contribution

It establishes new lower bounds for the number of square-free words over lists of size four and constructs such words over subsets of size three within a four-letter alphabet.

Findings

Number of square-free words over lists of size four is at least 2.45^n.

Constructs square-free words over subsets of size three within a four-letter alphabet.

Provides computational evidence supporting the conjecture.

Abstract

In 2007, Grytczuk conjecture that for any sequence $(\ell_i)_{i\ge1}$ of alphabets of size $3$ there exists a square-free infinite word $w$ such that for all $i$, the $i$-th letter of $w$ belongs to $\ell_i$. The result of Thue of 1906 implies that there is an infinite square-free word if all the $\ell_i$ are identical. On the other, hand Grytczuk, Przyby{\l}o and Zhu showed in 2011 that it also holds if the $\ell_i$ are of size $4$ instead of $3$. In this article, we first show that if the lists are of size $4$, the number of square-free words is at least $2.45^n$ (the previous similar bound was $2^n$). We then show our main result: we can construct such a square-free word if the lists are subsets of size $3$ of the same alphabet of size $4$. Our proof also implies that there are at least $1.25^n$ square-free words of length $n$ for any such list assignment. This proof relies on the existence of a set of coefficients verified with a computer. We suspect that the full conjecture could be resolved by this method with a much more powerful computer (but we might need to wait a few decades for such a computer to be available).