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%I #17 Mar 18 2021 19:12:27
%S 1,1,3,3,6,1,6,1,1,1,2,2,10,2,1,3,3,2,4,2,3,4,4,4,1,2,3,4,1,4,3,6,6,4,
%T 6,3,6,6,5,4,5,6,7,6,5,8,6,7,3,3,5,4,4,6,6,7,2,4,5,7,3,7,6,10,10,7,9,
%U 5,10,8,7,5,9,9,10,8,8,9,7,7,8,8,11,8,9
%N a(n) is the number of substrings of the binary representation of n that are instances of the Zimin word Z_k, where k = A342510(n).
%C This value of k is chosen so that Z_k is the largest Zimin word that the binary expansion of n does not avoid.
%H Peter Kagey, <a href="/A342512/b342512.txt">Table of n, a(n) for n = 0..8191</a>
%H Peter Kagey, <a href="https://codegolf.stackexchange.com/q/220679/53884">Matching ABACABA-type patterns</a>, Code Golf Stack Exchange.
%H Danny Rorabaugh, <a href="http://arxiv.org/abs/1509.04372">Toward the Combinatorial Limit Theory of Free Words</a>, arXiv preprint arXiv:1509.04372 [math.CO], 2015.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Sesquipower">Sesquipower</a>.
%F a(n) = A342511(n, A342510(n)).
%e For n = 121, the binary expansion is "1111001", which avoids the Zimin word Z_3 = ABACABA, but does not avoid the Zimin word Z_2 = ABA. In particular, there are a(121) = 7 substrings that are instances of Z_2:
%e (111)1001 with A = 1 and B = 1,
%e 1(111)001 with A = 1 and B = 1,
%e (1111)001 with A = 1 and B = 11,
%e 111(1001) with A = 1 and B = 00,
%e 11(11001) with A = 1 and B = 100,
%e 1(111001) with A = 1 and B = 1100, and
%e (1111001) with A = 1 and B = 11100.
%Y Cf. A342510, A342511.
%K nonn,base,look
%O 0,3
%A _Peter Kagey_, Mar 14 2021
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