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A342510
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a(n) = k where Z_k is the largest Zimin word that n (read as a binary word) does not avoid.
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3
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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, 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, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2
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OFFSET
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0,6
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COMMENTS
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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.
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.
a(n) >= 2 for all n >= 2^4.
a(n) >= 3 for all n >= 2^28.
For any fixed k, a(n) >= k for sufficiently large n, however there exists a value of a(n) = 3 with n >= 2^10482.
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.
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LINKS
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EXAMPLE
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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".
No substring is an instance of the Zimin word Z_4 = ABACABADABACABA, so a(10101939)= 3.
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PROG
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(Python)
def sd(w): # sesquipower degree
return 1 + max([0]+[sd(w[:i]) for i in range(1, (len(w)+1)//2) if w[:i] == w[-i:]])
def a(n):
w = bin(n)[2:]
return max(sd(w[i:j]) for i in range(len(w)) for j in range(i+1, len(w)+1))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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