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A056343
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Number of bracelets of length n using exactly three different colored beads.
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6
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0, 0, 1, 6, 18, 56, 147, 411, 1084, 2979, 8043, 22244, 61278, 171030, 477929, 1345236, 3795750, 10758902, 30572427, 87149124, 248991822, 713096352, 2046303339, 5883433409, 16944543810, 48879769575
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OFFSET
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1,4
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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=3 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=3 is the number of colors. (End)
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EXAMPLE
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For a(4)=6, the arrangements are ABAC, ABCB, ACBC, AABC, ABBC, and ABCC. Only the last three are chiral, their reverses being AACB, ACBB, and ACCB respectively. - Robert A. Russell, Sep 26 2018
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MATHEMATICA
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t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]);
T[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
a[n_] := T[n, 3];
k=3; 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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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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