Abstract
A word is called a palindrome if it is equal to its reversal. In the paper we consider a k-abelian modification of this notion. Two words are called k-abelian equivalent if they contain the same number of occurrences of each factor of length at most k. We say that a word is a k-abelian palindrome if it is k-abelian equivalent to its reversal. A question we deal with is the following: how many distinct palindromes can a word contain? It is well known that a word of length n can contain at most n + 1 distinct palindromes as its factors; such words are called rich. On the other hand, there exist infinite words containing only finitely many distinct palindromes as their factors; such words are called poor. It is easy to see that there are no abelian poor words, and there exist words containing Θ(n 2) distinct abelian palindromes. We analyze these notions with respect to k-abelian equivalence. We show that in the k-abelian case there exist poor words containing finitely many distinct k-abelian palindromic factors, and there exist rich words containing Θ(n 2) distinct k-abelian palindromes as their factors. Therefore, for poor words the situation resembles normal words, while for rich words it is similar to the abelian case.
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