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A032281
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Number of bracelets (turnover necklaces) of n beads of 2 colors, 9 of them black.
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6
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1, 1, 5, 12, 35, 79, 185, 375, 750, 1387, 2494, 4262, 7105, 11410, 17930, 27407, 41107, 60335, 87154, 123695, 173173, 238957, 325845, 438945, 585265, 772252, 1009868, 1308742, 1682660, 2146420, 2718806, 3419924, 4274905
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OFFSET
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9,3
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COMMENTS
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Also number of non-equivalent necklaces of 9 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in the case k=9 (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 (2,3,-6,-6,6,13,-2,-18,-1,11,3,0,0,-3,-11,1,18,2,-13,-6,6,6,-3,-2,1).
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FORMULA
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"DIK[ 9 ]" (necklace, indistinct, unlabeled, 9 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) =(1/3)*s(n,0,9) + (n-3)*(n-6)*s(n,0,3)/162 + (n-2)(n-4)*(n-6)*(n-8)*(945 + (n-1)*(n-3)*(n-5)*(n-7))/725760, if n is even; a(n) = (1/3)*s(n,0,9) + (n-3)*(n-6)*s(n,0,3)/162 +(n-1)*(n-3)*(n-5)*(n-7)*(945 + (n-2)*(n-4)*(n-6)*(n-8))/725760, if n is odd. - Vladimir Shevelev, Apr 23 2011
G.f.: (1/2)*x^9*((1+x)/(1-x^2)^5 + 1/9*(1/(1-x)^9 - 2/(-1+x^3)^3 - 6/(-1+x^9))).
G.f.: k=9, 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] (End)
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MATHEMATICA
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k = 9; 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=9; 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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