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A032242
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Number of identity bracelets of n beads of 5 colors.
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4
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5, 10, 10, 45, 252, 1120, 5270, 23475, 106950, 483504, 2211650, 10148630, 46911060, 217863040, 1017057256, 4767774375, 22438419120, 105960830300, 501928967930, 2384170903140, 11353241255900
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OFFSET
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1,1
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COMMENTS
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For n>2 also number of asymmetric bracelets with n beads of five colors. - Herbert Kociemba, Nov 29 2016
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LINKS
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FORMULA
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"DHK" (bracelet, identity, unlabeled) transform of 5, 0, 0, 0...
More generally, gf(k) is the g.f. for the number of asymmetric bracelets with n beads of k colors.
gf(k): Sum_{n>=1} mu(n)*( -log(1-k*x^n)/n - Sum_{i=0..2} binomial(k,i)x^(n*i)/(1-k*x^(2*n)) )/2. (End)
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MAPLE
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N:= 50: # for a(1)..a(N)
G:= add(1/2*numtheory:-mobius(n)*(-log(1-5*x^n)/n - add(binomial(5, i)*x^(n*i)/(1-5*x^(2*n)), i=0..2)), n=1..N):
S:= series(G, x, N+1):
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MATHEMATICA
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m=5; (* asymmetric bracelets of n beads of m colors *) Table[Sum[MoebiusMu[d](m^(n/d)/n - If[OddQ[n/d], m^((n/d+1)/2), ((m+1)m^(n/(2d))/2)]), {d, Divisors[n]}]/2, {n, 3, 20}] (* Robert A. Russell, Mar 18 2013 *)
mx=40; gf[x_, k_]:=Sum[MoebiusMu[n]*(-Log[1-k*x^n]/n-Sum[Binomial[k, i]x^(n i), {i, 0, 2}]/(1-k x^(2n)))/2, {n, mx}]; ReplacePart[Rest[CoefficientList[Series[gf[x, 5], {x, 0, mx}], x]], {1->5, 2->10}] (* Herbert Kociemba, Nov 29 2016 *)
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PROG
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(PARI) a(n)={if(n<3, binomial(5, n), sumdiv(n, d, moebius(n/d)*(5^d/n - if(d%2, 5^((d+1)/2), 3*5^(d/2))))/2)} \\ Andrew Howroyd, Sep 12 2019
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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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