关于k-阿贝尔可避免性的观察与问题
Observations and Problems on k-abelian avoidability
摘要：
我们引入了新的可避免性问题的话，考虑等价关系，k-阿贝尔等价，这是正确的平等和交换平等之间，即阿贝尔平等。对于两个k-阿贝尔等价词，长度为k的不同因子的出现次数一致，前缀（分别为）。后缀）也是相等的。避免单词2-重复的最小字母表的大小，即正方形，是3，对于阿贝尔正方形，是4。因此，对于2-阿贝尔正方形，这个大小必须是3或4。类似地，可以避免2-阿贝尔立方体（即3-重复）的最小字母表的大小是2或3，因为立方体（分别为阿贝尔立方体）在二进制中是可避免的（分别地，三进制）字母表。我们表明，对于2-阿贝尔广场所需的大小是4，在阿贝尔广场的情况下。最长的2-阿贝尔无平方的三进制字的长度是537。2-阿贝尔立方体的问题是开放的。然而，我们有计算证据表明，大小将是2，因为存在长度为100 000的2-阿贝尔无立方二进制字，这意味着2-阿贝尔情况将表现得像单词一样。
英文摘要：
We introduce new avoidability problems for words by considering equivalence relations, k-abelian equivalences, which lie properly in between equality and commutative equality, i.e. abelian equality. For two k-abelian equivalent words the numbers of occurrences of different factors of length k coincide and the prefixies (resp. suffixies) of length k-1 are equal as well. The size of the smallest alphabet avoiding 2-repetitions of words, i.e. squares, is three and for abelian squares it is four. It follows that for 2-abelian squares this size has to be three or four. Similarly, the size of the smallest alphabet where 2-abelian cubes, i.e. 3-repetitions, can be avoided is two or three, because cubes (resp. abelian cubes) are avoidable in binary (resp. ternary) alphabet. We show that for 2-abelian squares the required size is four, as in the case of abelian squares. The longest 2-abelian square-free ternary word is of length 537. The question for 2-abelian cubes is open. Though, we have computational evidence that the size would be two, since there exists 2-abelian cube-free binary word of length 100 000, meaning that the 2-abelian case would behave like that of words.
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