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A056342
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Number of bracelets of length n using exactly two different colored beads.
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6
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0, 1, 2, 4, 6, 11, 16, 28, 44, 76, 124, 222, 378, 685, 1222, 2248, 4110, 7683, 14308, 27010, 50962, 96907, 184408, 352696, 675186, 1296856, 2493724, 4806076, 9272778, 17920858, 34669600, 67159048, 130216122, 252745366, 490984486, 954637556, 1857545298, 3617214679, 7048675958, 13744694926, 26818405350
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OFFSET
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1,3
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COMMENTS
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Turning over will not create a new bracelet.
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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 pdf file of Chap. 2]
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LINKS
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FORMULA
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a(n) = (k!/4) * (S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/2n) * Sum_{d|n} phi(d) * S2(n/d,k), where k=2 is the number of colors and S2 is the Stirling subset number A008277.
G.f.: (k!/4) * x^(2k-2) * (1+x)^2 / Product_{i=1..k} (1-i x^2) - Sum_{d>0} (phi(d)/2d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=2 is the number of colors. (End)
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EXAMPLE
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For a(6)=11, the arrangements are AAAAAB, AAAABB, AAABAB, AAABBB, AABAAB, AABBBB, ABABAB, ABABBB, ABBABB, ABBBBB, and AABABB, the last being chiral. Its reverse is AABBAB. - Robert A. Russell, Sep 26 2018
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MATHEMATICA
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a[n_] := (1/4)*(Mod[n, 2] + 3)*2^Quotient[n, 2] + DivisorSum[n, EulerPhi[#]*2^(n/#)&]/(2*n) - 2; Array[a, 41] (* Jean-François Alcover, Nov 05 2017 *)
k=2; Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#, k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n, 1, 30}] (* Robert A. Russell, Sep 26 2018 *)
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PROG
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(PARI) a(n) = my(k=2); (k!/4)*(stirling(floor((n+1)/2), k, 2) + stirling(ceil((n+1)/2), k, 2)) + (k!/(2*n))*sumdiv(n, d, eulerphi(d)*stirling(n/d, k, 2)); \\ Michel Marcus, Sep 28 2018
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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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