User blog comment:Primussupremus/Harvey Friedmans word puzzle./@comment-5529393-20170308083219

You can correspond each string of letters to an ordinal. For example aaaaa = 5. With two letters a and b, each b can represent a power of w, so for example aabbaaabababba = w^6 * 2 + w^4 * 3 + w^3 + w^2 + 1. Then if you add the letter c, each c will separate powers of w^w, then d will separate powers of w^w^2, and so on. In general, strings of n letters will represent ordinals below w^w^(n-1). So the 26 letter alphabet will represent strings up to w^w^25. So, if we start from zz, follow it with the three letter sequence corresponding to the largest ordinal less than zz (which is yyz), follow that with the four letter sequence corresponding to the largest ordinal less than the one corresponding to yyz (which is xxyz), and so on, the sequence will continue for about H_{w^w^25}(2) steps, or F_{w^25}(2) written in the fast-growing hierarchy. It is possible to prove (using complicated mathematics) an upper bound for the longest possible such sequence such that no string is contained in a later one is of about this size. This will also be an upper bound to your question (since if we have a string that satisfies your constraints, we can pluck a sequence of strings that satisfies mine; for example, from abbbaaaaaaa we can pluck bb, bba, baaa, aaaaa).

Going the other direction, Friedman proved that a lower bound for the three letter problem is F_w(7198), and I can see how to extend this proof to prove that a lower bound for the n letter problem F_{w^(n-2)}(7198) (or possibly 7197 or 7196, but pretty close). So we have a lower bound of about F_{w^24}(7198), and an upper bound of about F_{w^25}(2).