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A056345
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Number of bracelets of length n using exactly five different colored beads.
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4
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0, 0, 0, 0, 12, 150, 1200, 7905, 46400, 255636, 1346700, 6901725, 34663020, 171786450, 843130688, 4110958530, 19951305240, 96528492700, 466073976900, 2247627076731, 10832193571460, 52194109216950
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OFFSET
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1,5
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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=5 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=5 is the number of colors. (End)
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EXAMPLE
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For a(5)=12, pair up the 24 permutations of BCDE, each with its reverse, such as BCDE-EDCB. Precede the first of each pair with an A, such as ABCDE. These are the 12 arrangements, all chiral. If we precede the second of each pair with an A, such as AEDCB, we get the chiral partner of each. - Robert A. Russell, Sep 27 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, 5];
k=5; 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 27 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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