Abstract

Erdos raised the question whether there exist infinite abelian square-free words over a given alphabet (words in which no two adjacent subwords are permutations of each other). Infinite abelian square-free words have been constructed over alphabets of sizes as small as four. In this paper, we investigate the problem of avoiding abelian squares in partial words (sequences that may contain some holes). In particular, we give lower and upper bounds for the number of letters needed to construct infinite abelian square-free partial words with finitely or infinitely many holes. In the case of one hole, we prove that the minimal alphabet size is four, while in the case of more than one hole, we prove that it is five.

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