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A056493
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Number of primitive (period n) periodic palindromes using a maximum of two different symbols.
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7
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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
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1,1
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COMMENTS
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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.
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REFERENCES
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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]
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LINKS
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FORMULA
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Sum_{d|n} 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
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EXAMPLE
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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
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MATHEMATICA
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mx=40; gf[x_, k_]:=Sum[ MoebiusMu[n]*Sum[Binomial[k, i]x^(n i), {i, 0, 2}]/( 1-k x^(2n)), {n, mx}]; CoefficientList[Series[gf[x, 2], {x, 0, mx}], x] (* Herbert Kociemba, Nov 29 2016 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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