References to A091844

a(1) = 4. To get a(n+1), write the string a(1)a(2)...a(n) as xy^k for words x and y (where y has positive length) and k is maximized, i.e. k = the maximal number of repeating blocks at the end of the sequence so far. Then a(n+1) = max(k,4).
4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 5, 4, 4, 4

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