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A032241
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Number of identity bracelets of n beads of 4 colors.
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5
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4, 6, 4, 15, 72, 266, 1044, 3780, 14056, 51132, 188604, 693845, 2572920, 9566046, 35758628, 134134080, 505159200, 1908539864, 7233104844, 27486455049, 104713295712, 399817073946, 1529746919604
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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 four 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 4, 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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MATHEMATICA
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m = 4; (* 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/(2 d))/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, 4], {x, 0, mx}], x]], {1->4, 2->6}] (* Herbert Kociemba, Nov 29 2016 *)
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PROG
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(PARI) a(n)={if(n<3, binomial(4, n), sumdiv(n, d, moebius(n/d)*(4^d/n - if(d%2, 4^((d+1)/2), 5*4^(d/2)/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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