Proof of Dejean's conjecture for alphabets with 5, 6, 7, 8, 9, 10 and 11 letters

AbstractAxel Thue proved that overlapping factors could be avoided in arbitrarily long words on a two-letter alphabet while, on the same alphabet, square factors always occur in words longer than 3. Françoise Dejean stated an analogous result for three-letter alphabets: every long enough word has a factor, which is a fractional power with an exponent at least 7/4 and there exist arbitrary long words in which no factor is a fractional power with an exponent strictly greater than 7/4. The number 7/4 is called the repetition threshold of the three-letter alphabets.Thereafter, she proposed the following conjecture: the repetition threshold of the k-letter alphabets is equal to k/(k−1) except in the particular cases k=3, where this threshold is 7/4, and k=4, where it is 7/5.For k=4, this conjecture was proved by J.J. Pansiot (1984).In this paper, we give a computer-aided proof of Dejean's conjecture for several other values: 5, 6, 7, 8, 9, 10 and 11