Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%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