Abstract:
We consider a recently defined notion of $k$-abelian equivalence of words by giving some basic results and concentrating on avoidability problems. This equivalence relation counts the numbers of factors of length $k$ for a fixed natural number $k$. We ask for the size of the smallest alphabet for which $k$-abelian squares and cubes can be avoided, respectively. For $2$-abelian squares this is four – as in the case of abelian words, while for $2$-abelian cubes we have only strong evidence that the size is two – as it is in the case of words. In addition, we point out a few properties of morphisms supporting the view that it might be difficult to find solutions to our questions by simply iterating a morphism.
Key words and phrases:combinatorics on words, $k$-abelian equivalence, avoidability.