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%I #42 Aug 12 2022 09:18:21
%S 1,2,3,4,7,11,17,25,49,75,129,191,329,489,825,1237,2473,3737,6329,
%T 9435,16833,25081,41449,61043,115409,172441,290617,431385,775641,
%U 1157417,1938713,2908069,5816137,8786121,14682489,21774137,39391673,58815073,97815385
%N 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.
%C a(n) counts equivalence classes up to swapping the letters of the alphabet.
%C a(n+1) <= 2*a(n).
%C 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]
%C From _Peter Kagey_, Jan 27 2018: (Start)
%C a(2^k + 1) = 2 * a(2^k) - 1 for n > 0.
%C Conjecture: a(n) is odd for all n > 4.
%C (End)
%H Lars Blomberg, <a href="/A297789/b297789.txt">Table of n, a(n) for n = 1..61</a>
%H Li-yao Xia, Mathematics Stack Exchange, <a href="https://math.stackexchange.com/a/2624346/121988">Counting particular odd-length strings over a two letter alphabet</a>.
%e 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.
%e initial odd substring | first half | second half
%e ----------------------+------------+------------
%e 1 | 1 | 1
%e 101 | 10 | 01
%e 10101 | 101 | 101
%e 1010110 | 1010 | 0110
%e 101011010 | 10101 | 11010
%e 10101101011 | 101011 | 101011
%e 1010110101101 | 1010110 | 0101101
%e For n = 5, the a(5) = 7 strings are:
%e 101101101,
%e 101101110,
%e 101010110,
%e 101010101,
%e 101011010,
%e 101011001, and
%e 111111111.
%K nonn
%O 1,2
%A _Peter Kagey_, Jan 22 2018
%E a(28)-a(39) from _Lars Blomberg_, Feb 02 2018