There exist infinite cube-free words over any sequence of binary alphabets

TL;DR

This paper proves the existence of infinite cube-free words over any sequence of binary alphabets and shows that such words can be constructed with computational efficiency if the alphabet sequence is computable.

Contribution

It establishes the existence of infinite cube-free words over arbitrary binary alphabet sequences and demonstrates their computability under certain conditions.

Findings

Existence of infinite cube-free words over any binary alphabet sequence.

At least 1.35^n cube-free words of length n in the product of the first n alphabets.

Computability of such words when the alphabet sequence is computable.

Abstract

We prove that for any sequence of binary alphabets $\mathcal{A}_1,\mathcal{A}_2,\dots$, there exists a cube-free word $c_1c_2\dots$ so that $c_1\in\mathcal{A}_1,c_2\in\mathcal{A}_2,\dots$. In particular, for every $n$, there are at least $1.35^n$ cube-free words in $\mathcal{A}_1\times\mathcal{A}_2\times\dots\times \mathcal{A}_n$. We also prove that if the list of alphabets is computable then one of these words is computable and its $n$th letter can be computed in time polynomial in $n$.