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A056508 Number of periodic palindromic structures of length n using exactly two different symbols.

%I A056508 #25 Jun 29 2018 07:59:47
%S A056508 0,1,1,3,3,6,7,13,15,25,31,50,63,99,127,197,255,391,511,777,1023,1551,
%T A056508 2047,3090,4095,6175,8191,12323,16383,24639,32767,49221,65535,98431,
%U A056508 131071,196743,262143,393471,524287,786697,1048575,1573375,2097151,3146255,4194303
%N A056508 Number of periodic palindromic structures of length n using exactly two different symbols.
%C A056508 For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome. Permuting the symbols will not change the structure.
%C A056508 For odd n, a palindrome cannot be the complement of itself, so a(n) is given by A284855(n,2)/2 - 1. - _Andrew Howroyd_, Apr 08 2017
%D A056508 M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]
%H A056508 Giovanni Resta, <a href="/A056508/b056508.txt">Table of n, a(n) for n = 1..1000</a>
%F A056508 a(n) = A056503(n) - 1.
%F A056508 a(2n + 1) = 2^n - 1. - _Andrew Howroyd_, Apr 07 2017
%e A056508 From _Andrew Howroyd_, Apr 07 2017: (Start)
%e A056508 Example for n=6:
%e A056508 Periodic symmetry means results are either in the form abccba or abcdcb.
%e A056508 There are 3 binary words in the form abccba that start with 0 and contain a 1 which are 001100, 010010, 011110. Of these, 011110 is equivalent to 001100 after rotation.
%e A056508 There are 7 binary words in the form abcdcb that start with 0 and contain a 1 which are 000100, 001010, 001110, 010001, 010101, 011011, 011111. Of these, 011111 is equivalent to 000100, 010001 is equivalent to 001010 and 011011 is equivalent to 010010 from the first set.
%e A056508 There are therefore a total of 7 + 3 - 4 = 6 equivalence classes so a(6) = 6.
%e A056508 (End)
%t A056508 (* b = A164090, c = A045674 *)
%t A056508 b[n_] := (1/4)*(7 - (-1)^n)*2^((1/4)*(2*n + (-1)^n - 1));
%t A056508 c[0] = 1;
%t A056508 c[n_] := c[n] = If[EvenQ[n], 2^(n/2 - 1) + c[n/2], 2^((n - 1)/2)];
%t A056508 a[n_] := If[OddQ[n], b[n]/2, (1/2)*(b[n] + c[n/2])] - 1;
%t A056508 Array[a, 45] (* _Jean-François Alcover_, Jun 29 2018, after _Andrew Howroyd_ *)
%Y A056508 Column 2 of A285012.
%Y A056508 Cf. A052551.
%K A056508 nonn
%O A056508 1,4
%A A056508 _Marks R. Nester_
%E A056508 a(17)-a(45) from _Andrew Howroyd_, Apr 07 2017