cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A032282 Number of bracelets (turnover necklaces) of n beads of 2 colors, 11 of them black.

Original entry on oeis.org

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
Offset: 11

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Comments

From Vladimir Shevelev, Apr 23 2011: (Start)
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)

References

  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Column k=11 of A052307.

Programs

  • Mathematica
    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 *)

Formula

"DIK[ 11 ]" (necklace, indistinct, unlabeled, 11 parts) transform of 1, 1, 1, 1...
From Vladimir Shevelev, Apr 23 2011: (Start)
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.
(End)
From Herbert Kociemba, Nov 05 2016: (Start)
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]
(End)