%I #30 Sep 12 2019 21:12:33
%S 3,3,1,3,12,37,117,333,975,2712,7689,21414,60228,168597,475024,
%T 1338525,3788400,10741575,30556305,87109332,248967446,713025093,
%U 2046325125,5883406830,16944975036,48880411272,141212376513
%N Number of identity bracelets of n beads of 3 colors.
%C For n>2 also number of asymmetric bracelets with n beads of three colors. - _Herbert Kociemba_, Nov 29 2016
%H Andrew Howroyd, <a href="/A032240/b032240.txt">Table of n, a(n) for n = 1..200</a>
%H C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%H F. Ruskey, <a href="http://combos.org/necklace">Necklaces, Lyndon words, De Bruijn sequences, etc.</a>
%H F. Ruskey, <a href="/A000011/a000011.pdf">Necklaces, Lyndon words, De Bruijn sequences, etc.</a> [Cached copy, with permission, pdf format only]
%H <a href="/index/Br#bracelets">Index entries for sequences related to bracelets</a>
%F "DHK" (bracelet, identity, unlabeled) transform of 3, 0, 0, 0...
%F From _Herbert Kociemba_, Nov 29 2016: (Start)
%F More generally, gf(k) is the g.f. for the number of asymmetric bracelets with n beads of k colors.
%F 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)
%t 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 *)
%t 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 *)
%o (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
%Y Column k=3 of A309528 for n >= 3.
%K nonn
%O 1,1
%A _Christian G. Bower_