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A032282
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Number of bracelets (turn over necklaces) of n beads of 2 colors, 11 of them black.
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3
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 11,3
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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)
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REFERENCES
| H. Gupta, Enumeration of incongruent cyclic k-gons, Indian J. Pure and Appl. Math., 10 (1979), no.8, 964-999.
V. Shevelev, Necklaces and convex k-gons, Indian J. Pure and Appl. Math., 35 (2004), no. 5, 629-638.
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LINKS
| C. G. Bower, Transforms (2)
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
V. Shevelev,Spectrum of permanent's values and its extremal magnitudes in Lambda_n^3 and Lambda_n(alpha,beta,gamma)(Cf. Section 5)
Index entries for sequences related to bracelets
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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),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)
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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 (russell(AT)post.harvard.edu), Sep 27 2004
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CROSSREFS
| Sequence in context: A026086 A175659 A192000 * A163302 A084057 A091649
Adjacent sequences: A032279 A032280 A032281 * A032283 A032284 A032285
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KEYWORD
| nonn
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AUTHOR
| Christian G. Bower (bowerc(AT)usa.net)
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