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A261501 Number of necklaces with n white beads and 7*n black beads.

%I A261501 #17 Feb 16 2025 08:33:26
%S A261501 1,1,8,85,1128,16451,255704,4141383,69159400,1182125128,20581159608,
%T A261501 363704640476,6506965279992,117626432708864,2145180358634664,
%U A261501 39421026305282660,729242353169440744,13568988503585900648,253785064586356459616,4768543107831461199897
%N A261501 Number of necklaces with n white beads and 7*n black beads.
%H A261501 Alois P. Heinz, <a href="/A261501/b261501.txt">Table of n, a(n) for n = 0..760</a>
%H A261501 F. Ruskey, <a href="http://combos.org/necklace">Necklaces, Lyndon words, De Bruijn sequences, etc.</a>
%H A261501 F. Ruskey, <a href="/A000011/a000011.pdf">Necklaces, Lyndon words, De Bruijn sequences, etc.</a> [Cached copy, with permission, pdf format only]
%H A261501 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Necklace.html">Necklace</a>
%H A261501 Wikipedia, <a href="https://en.wikipedia.org/wiki/Necklace_(combinatorics)">Necklace (combinatorics)</a>
%H A261501 <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>
%F A261501 a(n) = 1/(8*n) * Sum_{d|n} C(8*n/d,n/d) * A000010(d) for n>0, a(0) = 1.
%F A261501 a(n) ~ 2^(24*n-2) / (sqrt(Pi) * 7^(7*n+1/2) * n^(3/2)). - _Vaclav Kotesovec_, Aug 22 2015
%p A261501 with(numtheory):
%p A261501 a:= n-> `if`(n=0, 1, add(binomial(8*n/d, n/d)
%p A261501              *phi(d), d=divisors(n))/(8*n)):
%p A261501 seq(a(n), n=0..25);
%Y A261501 Column k=7 of A261494.
%K A261501 nonn
%O A261501 0,3
%A A261501 _Alois P. Heinz_, Aug 21 2015