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A297789
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The number of length 2n - 1 strings over the alphabet {0, 1} such that the first half of any initial odd length substring is a permutation of the second half.
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1
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1, 2, 3, 4, 7, 11, 17, 25, 49, 75, 129, 191, 329, 489, 825, 1237, 2473, 3737, 6329, 9435, 16833, 25081, 41449, 61043, 115409, 172441, 290617, 431385, 775641, 1157417, 1938713, 2908069, 5816137, 8786121, 14682489, 21774137, 39391673, 58815073, 97815385
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OFFSET
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1,2
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COMMENTS
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a(n) counts equivalence classes up to swapping the letters of the alphabet.
a(n+1) <= 2*a(n).
Conjecture: lim_{n->infinity} a(n+1)/a(n) exists and is a value in [1, 2]. [The following comment suggests that on the contrary, this limit may not exist. - N. J. A. Sloane, Jan 30 2018, following a comment from Peter Kagey, Jan 29 2017]
a(2^k + 1) = 2 * a(2^k) - 1 for n > 0.
Conjecture: a(n) is odd for all n > 4.
(End)
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LINKS
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EXAMPLE
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For n = 7, one of the a(7) = 17 strings of length 2*7-1 = 13 is "1010110101101" because the first half of every initial odd-length substring is a permutation of the second half.
initial odd substring | first half | second half
----------------------+------------+------------
1 | 1 | 1
101 | 10 | 01
10101 | 101 | 101
1010110 | 1010 | 0110
101011010 | 10101 | 11010
10101101011 | 101011 | 101011
1010110101101 | 1010110 | 0101101
For n = 5, the a(5) = 7 strings are:
101101101,
101101110,
101010110,
101010101,
101011010,
101011001, and
111111111.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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