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A032282
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Number of bracelets (turnover necklaces) of n beads of 2 colors, 11 of them black.
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6
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1, 1, 6, 16, 56, 147, 392, 912, 2052, 4262, 8524, 16159, 29624, 52234, 89544, 148976, 242086, 384111, 597506, 911456, 1367184, 2017509, 2934559, 4209504, 5963464, 8347612, 11558232, 15837472, 21493712, 28903332
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OFFSET
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11,3
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COMMENTS
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Also number of non-equivalent necklaces of 11 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in case k=11 (see our comment to A032279).
(End)
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REFERENCES
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N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (5,-5,-15,35,1,-65,45,45,-65,1,36,-20,0,20,-36,-1,65,-45,-45,65,-1,-35,15,5,-5,1).
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FORMULA
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"DIK[ 11 ]" (necklace, indistinct, unlabeled, 11 parts) transform of 1, 1, 1, 1...
Put s(n,k,d)=1, if n==k(mod d), and s(n,k,d)=0, otherwise. Then
a(n) = 5*s(n,0,11)/11+(3840*C(n-1,10)+11*(n-2)*(n-4)*(n-6)(n-8)*(n-10))/84480, if n is even;
a(n) = 5*s(n,0,11)/11+(3840*C(n-1,10)+11*(n-1)*(n-3)*(n-5)*(n-7)*(n-9))/84480, if n is odd.
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G.f.: 1/22*x^11*(1/(1-x)^11 + 11/((-1+x)^6*(1+x)^5) - 10/(-1+x^11)).
G.f.: k=11, x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^floor[(k+2)/2])/2. [edited by Petros Hadjicostas, Jul 18 2018]
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MATHEMATICA
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k = 11; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
k=11; CoefficientList[Series[x^k*(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2, {x, 0, 50}], x] (* Herbert Kociemba, Nov 04 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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STATUS
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approved
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