A207816 Number of distinct necklaces with n red, n green, n blue and n white beads.

1, 6, 318, 30804, 3941598, 586637256, 96197661156, 16875655269948, 3111284141045598, 595909785174057204, 117634021777132574568, 23797087019979071174580, 4912693780461352534397604, 1031629572413246016139181544, 219809927417367534490107035244, 47426945434432859336092700072304

Offset: 0

Author

Thotsaporn Thanatipanonda, Feb 20 2012

Keywords

• nonn

Examples

			For n=1, a(1)=6 since for four beads necklaces with each bead from each of the four colors say (R,G,B,W), we can arrange as following, [R,G,B,W], [R,G,W,B], [R,B,G,W], [R,B,W,G], [R,W,G,B] and [R,W,B,G].
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..400

Crossrefs

Cf. A003239, A118644, A137729.

Column k=4 of A208183. - Alois P. Heinz, Feb 24 2012

Programs

Maple

with(combinat): with(numtheory):
# This formula comes from Polya Counting Theorem:
# Z(C_n) = add(phi(d)*(a_d)^(n/d), d in divisors(n))/n;
PolyaBrace:= proc(S) option remember; local n, s, d;
               n:= add(s, s=S);
               add(phi(d) *PolyaCoeff(d, S), d=divisors(n))/n
             end:
# Find coeff of prod(a[i]^s[i], i=1..n) of a_d^(n/d) (symmetric function)
PolyaCoeff:= proc(d, S) option remember; local n, pow, s;
               n:= add(s, s=S);
               pow:= n/d;
               if {seq(s mod d, s = S)} = {0}
                  then multinomial(pow, seq(s/d, s = S))
                  else 0
               fi:
             end:
a:= n-> `if`(n=0, 1, PolyaBrace([n$4])):
seq(a(n), n=0..20);

Mathematica

a[n_] := DivisorSum[n, EulerPhi[n/#] (4#)!/(#!^4 * 4n)&]; a[0]=1;
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Mar 24 2017, after Alois P. Heinz *)

Formula

a(n) = Sum_{d|n} phi(n/d)*(4*d)!/(d!^4*4*n) if n>0 and a(0) = 1. - Alois P. Heinz, Feb 24 2012

a(n) ~ 2^(8*n-5/2) / (Pi^(3/2) * n^(5/2)). - Vaclav Kotesovec, Aug 23 2015