A032240 Number of identity bracelets of n beads of 3 colors.
3, 3, 1, 3, 12, 37, 117, 333, 975, 2712, 7689, 21414, 60228, 168597, 475024, 1338525, 3788400, 10741575, 30556305, 87109332, 248967446, 713025093, 2046325125, 5883406830, 16944975036, 48880411272, 141212376513
Offset: 1
Keywords
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..200
- C. G. Bower, Transforms (2)
- F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
- F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]
- Index entries for sequences related to bracelets
Crossrefs
Column k=3 of A309528 for n >= 3.
Programs
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Mathematica
m = 3; (* 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,3],{x,0,mx}],x]],{1->3,2->3}] (* Herbert Kociemba, Nov 29 2016 *)
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PARI
a(n)={if(n<3, binomial(3,n), sumdiv(n, d, moebius(n/d)*(3^d/n - if(d%2, 3^((d+1)/2), 2*3^(d/2))))/2)} \\ Andrew Howroyd, Sep 12 2019
Formula
"DHK" (bracelet, identity, unlabeled) transform of 3, 0, 0, 0...
From Herbert Kociemba, Nov 29 2016: (Start)
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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