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A342510 a(n) = k where Z_k is the largest Zimin word that n (read as a binary word) does not avoid. 3

%I #24 Mar 18 2021 19:12:02

%S 1,1,1,1,1,2,1,2,2,2,2,2,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,

%T 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,

%U 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,2

%N a(n) = k where Z_k is the largest Zimin word that n (read as a binary word) does not avoid.

%C Zimin words are defined recursively: Z_1 = A, Z_2 = ABA, Z_3 = ABACABA, and Z_{i+1} = Z_i a_{i+1} Z_i.

%C Every Zimin word Z_i is an "unavoidable" word, meaning that every sufficiently long string over a finite alphabet contains a substring that is an instance of Z_i. A word w is instance of a Zimin word Z_i if there's a nonerasing monoid homomorphism from Z_i to w.

%C a(n) >= 2 for all n >= 2^4.

%C a(n) >= 3 for all n >= 2^28.

%C For any fixed k, a(n) >= k for sufficiently large n, however there exists a value of a(n) = 3 with n >= 2^10482.

%C The first occurrence of k is when n = A001045(2^k), that is, the binary expansion of n is "1010101...01" with 2^k - 1 bits.

%H Peter Kagey, <a href="/A342510/b342510.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>.

%e For n = 10101939, the binary representation is "100110100010010010110011", and the substring "0010010010" is an instance of the Zimin word Z_3 = ABACABA with A = "0", B = "01", and C = "01".

%e No substring is an instance of the Zimin word Z_4 = ABACABADABACABA, so a(10101939)= 3.

%o (Python)

%o def sd(w): # sesquipower degree

%o return 1 + max([0]+[sd(w[:i]) for i in range(1, (len(w)+1)//2) if w[:i] == w[-i:]])

%o def a(n):

%o w = bin(n)[2:]

%o return max(sd(w[i:j]) for i in range(len(w)) for j in range(i+1, len(w)+1))

%o print([a(n) for n in range(87)]) # _Michael S. Branicky_, Mar 15 2021

%Y Cf. A001045.

%Y Cf. A342511, A342512.

%K nonn,base

%O 0,6

%A _Peter Kagey_, Mar 14 2021

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