cp's OEIS Frontend

A261499 Number of necklaces with n white beads and 5*n black beads.

%I A261499 #17 Feb 16 2025 08:33:26
%S A261499 1,1,6,46,446,4751,54132,642342,7861662,98480332,1256569506,
%T A261499 16273981758,213378976004,2826867619109,37782553160820,
%U A261499 508840821830546,6898459216311582,94070535317459018,1289430373206452136,17755914760643605782,245518560760433583946
%N A261499 Number of necklaces with n white beads and 5*n black beads.
%H A261499 Alois P. Heinz, <a href="/A261499/b261499.txt">Table of n, a(n) for n = 0..850</a>
%H A261499 F. Ruskey, <a href="http://combos.org/necklace">Necklaces, Lyndon words, De Bruijn sequences, etc.</a>
%H A261499 F. Ruskey, <a href="/A000011/a000011.pdf">Necklaces, Lyndon words, De Bruijn sequences, etc.</a> [Cached copy, with permission, pdf format only]
%H A261499 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Necklace.html">Necklace</a>
%H A261499 Wikipedia, <a href="https://en.wikipedia.org/wiki/Necklace_(combinatorics)">Necklace (combinatorics)</a>
%H A261499 <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>
%F A261499 a(n) = 1/(6*n) * Sum_{d|n} C(6*n/d,n/d) * A000010(d) for n>0, a(0) = 1.
%F A261499 a(n) ~ 6^(6*n) / (2 * sqrt(3*Pi) * 5^(5*n+1/2) * n^(3/2)). - _Vaclav Kotesovec_, Aug 22 2015
%p A261499 with(numtheory):
%p A261499 a:= n-> `if`(n=0, 1, add(binomial(6*n/d, n/d)
%p A261499              *phi(d), d=divisors(n))/(6*n)):
%p A261499 seq(a(n), n=0..25);
%Y A261499 Column k=5 of A261494.
%K A261499 nonn
%O A261499 0,3
%A A261499 _Alois P. Heinz_, Aug 21 2015