Is There a Sequence on Four Symbols in Which No Two Adjacent Segments Are Permutations of One Another?

It has long been known (see [1–3, 5, 6, 10–12, 15, 16]) that there exist sequences on 3 symbols which contain no 2 identically equal consecutive segments, and sequences on 2 symbols which contain no 3 identically equal consecutive segments. Indeed, Axel Thue obtained these results around 1906. See [6] for a brief account of the contexts of the various independent rediscoveries of these results, and see [7,14] for an account of other properties of these sequences. Let X be a set and let s = x1x2x3 be a sequence on X . Then for i+1 k, s[i+1;k] = xi+1xi+2 xk is a segment of s, and the segments s[i+ 1; j];s[ j + 1;k] are consecutive. The segments s[i+ 1; j] and s[p+ 1;q] are identically equal if k i = q p and xi+1 = xp+1;xi+2 = xp+2; : : : ;xk = xq or, in other words, if s[i+1;k] = s[p+1;q] in X , the free semigroup generated by the set X . An interesting situation arises when we allow the symbols within a segment to commute with each other. It will be convenient to use the following terminology. Given a set X and a sequence s on X , we regard segments of s as elements of X . (Thus the results mentioned above say that there exist sequences on 3 symbols without 2nd powers as segments, and sequences on 2 symbols without 3rd powers.) Now let X denote the free commutative semigroup generated by X , and let α : X 7! X be the natural homomorphism (α(x) = x for x 2 X). If s has k consecutive segments f1; : : : ; fk such that α( f1) = = α( fk), then we say that s has a kth power mod α . In this language, the question of the title is: Does there exist a sequence on four symbols without 2nd powers mod α? It is an easy matter to verify that every sequence on 3 symbols contains 2nd powers mod α , and that every sequence on 2 symbols has 3rd powers mod α . For example, if X = fx;yg, one can show by examining all cases that the longest elements of X which do not contain a 3rd power mod α are xxyyxyyxx;xxyyxyyxy; and a few others. Evdomikov [4] constructed a sequence on 25 symbols without 2nd powers mod α , and conjectured that perhaps 5 symbols would suffice. Justin [8], with a remarkable half-page proof, constructed a sequence on 2 symbols without 5th powers mod α . This sequence is obtained by successive iterations of the transformation x 7! xxxxy and y 7! xyyyy, starting with x. Thus the first few iterations give x;xxxxy;(xxxxy)4xyyyy; [(xxxxy)4xyyyy]4xxxxy(xyyyy)4. Then in 1970 a paper appeared [13] in which P. A. B. Pleasants gave a construction of a sequence on 5 symbols without 2nd powers mod α . Pleasants’ sequence, which extends to infinity in both directions, is constructed by