2020
DOI: 10.1051/ita/2020003
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Avoiding conjugacy classes on the 5-letter alphabet

Abstract: We construct an infinite word w over the 5-letter alphabet such that for every factor f of w of length at least two, there exists a cyclic permutation of f that is not a factor of w. In other words, w does not contain a non-trivial conjugacy class. This proves the conjecture in Gamard et al. [TCS 2018]

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Cited by 2 publications
(2 citation statements)
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“…, x d } be a finite alphabet and we let w be a right-infinite word over Σ. (2) We let A w be the factor algebra of w with base field k = Q. Notice that since V n has a basis given by factors of w of length n, we have…”
Section: Algebraic Interpretation Of Lie Complexitymentioning
confidence: 99%
See 1 more Smart Citation
“…, x d } be a finite alphabet and we let w be a right-infinite word over Σ. (2) We let A w be the factor algebra of w with base field k = Q. Notice that since V n has a basis given by factors of w of length n, we have…”
Section: Algebraic Interpretation Of Lie Complexitymentioning
confidence: 99%
“…The preceding example gives a word in which every factor of length at least two has some cyclic conjugate that is not a factor. Badkobeh and Ochem [2] give an example of such a word over a 5-letter alphabet. In general, the property that L w (n) = 0 for n ≥ i has been studied over various alphabets [14].…”
Section: Examplesmentioning
confidence: 99%