OFFSET
1,2
COMMENTS
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]
From Peter Kagey, Jan 27 2018: (Start)
a(2^k + 1) = 2 * a(2^k) - 1 for n > 0.
Conjecture: a(n) is odd for all n > 4.
(End)
LINKS
Lars Blomberg, Table of n, a(n) for n = 1..61
Li-yao Xia, Mathematics Stack Exchange, Counting particular odd-length strings over a two letter alphabet.
EXAMPLE
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.
CROSSREFS
KEYWORD
nonn
AUTHOR
Peter Kagey, Jan 22 2018
EXTENSIONS
a(28)-a(39) from Lars Blomberg, Feb 02 2018
STATUS
approved