191214

Appears in sequences

• Start with any initial string of n numbers s(1), ..., s(n), with s(1) = 2, other s(i)'s = 2 or 3 (so there are 2^(n-1) starting strings). The rule for extending the string is this as follows: To get s(n+1), write the string s(1)s(2)...s(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. Then a(n) = number of starting strings for which k > 1.at n=18A093370
• Integer part of the area of integer triangle [A001611(n), A001611(n+1), A001611(n+2)].at n=18A097281
• a(n) = 2^n - A122536(n).at n=17A121880