

A056493


Number of primitive (period n) periodic palindromes using a maximum of two different symbols.


7



2, 1, 2, 3, 6, 7, 14, 18, 28, 39, 62, 81, 126, 175, 246, 360, 510, 728, 1022, 1485, 2030, 3007, 4094, 6030, 8184, 12159, 16352, 24381, 32766, 48849, 65534, 97920, 131006, 196095, 262122, 392364, 524286, 785407, 1048446, 1571310, 2097150, 3143497
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OFFSET

1,1


COMMENTS

For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome.
Also number of aperiodic necklaces (Lyndon words) with two colors that are the same when turned over.


REFERENCES

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for a pdf file of Chap. 2]


LINKS



FORMULA

Sum_{dn} mu(d)*b(n/d), where b(n) = A029744(n+1). [Corrected by Petros Hadjicostas, Oct 15 2017. The original formula referred to a previous version of sequence A029744 that had a different offset.]
More generally, let gf(k) be the g.f. for the number of necklaces with reflectional symmetry but no rotational symmetry and beads of k colors. Then gf(k): Sum_{n >= 1} mu(n)*Sum_{i=0..2} binomial(k,i)*x^(n*i)/(1  k*x^(2*n)).  Herbert Kociemba, Nov 29 2016
G.f.: Sum_{n >= 1} mu(n)*x^n*(2 + 3*x^n)/(1  2*x^(2*n)). The g.f. by _Herbet Kociemba_ above, with k = 2, becomes Sum_{n>=1} mu(n)*(x^n + 1)^2/(1  2*x^(2*n)). The two formulae differ by the "undetermined" constant Sum_{n >= 1} mu(n).  Petros Hadjicostas, Oct 15 2017


EXAMPLE

a(1) = 2 with aaa... and bbb..., a(2) = 1 with ababab..., a(3) = 2 with aabaab... and abbabb..., a(4) = 3 with aaabaaab... and aabbaabb... and abbbabbb....  Michael Somos, Nov 29 2016


MATHEMATICA

mx=40; gf[x_, k_]:=Sum[ MoebiusMu[n]*Sum[Binomial[k, i]x^(n i), {i, 0, 2}]/( 1k x^(2n)), {n, mx}]; CoefficientList[Series[gf[x, 2], {x, 0, mx}], x] (* Herbert Kociemba, Nov 29 2016 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



