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A032242 Number of identity bracelets of n beads of 5 colors. 4

%I #34 Sep 13 2019 08:03:45

%S 5,10,10,45,252,1120,5270,23475,106950,483504,2211650,10148630,

%T 46911060,217863040,1017057256,4767774375,22438419120,105960830300,

%U 501928967930,2384170903140,11353241255900

%N Number of identity bracelets of n beads of 5 colors.

%C For n>2 also number of asymmetric bracelets with n beads of five colors. - _Herbert Kociemba_, Nov 29 2016

%H Robert Israel, <a href="/A032242/b032242.txt">Table of n, a(n) for n = 1..1434</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 5, 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)

%p N:= 50: # for a(1)..a(N)

%p 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):

%p S:= series(G,x,N+1):

%p 5,10,seq(coeff(S,x,j),j=3..N); # _Robert Israel_, Jun 24 2019

%t 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 *)

%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,5],{x,0,mx}],x]],{1->5,2->10}] (* _Herbert Kociemba_, Nov 29 2016 *)

%o (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

%Y Column k=5 of A309528 for n >= 3.

%K nonn

%O 1,1

%A _Christian G. Bower_

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