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A261500 Number of necklaces with n white beads and 6*n black beads.

%I A261500 #17 Feb 16 2025 08:33:26
%S A261500 1,1,7,64,735,9276,124936,1753074,25366335,375677659,5667212132,
%T A261500 86775157140,1345153548264,21069043965984,332927800269694,
%U A261500 5301031234085664,84967018635587775,1369846562874360887,22199151536133457885,361411377745122110422,5908312923795257331460
%N A261500 Number of necklaces with n white beads and 6*n black beads.
%H A261500 Alois P. Heinz, <a href="/A261500/b261500.txt">Table of n, a(n) for n = 0..800</a>
%H A261500 F. Ruskey, <a href="http://combos.org/necklace">Necklaces, Lyndon words, De Bruijn sequences, etc.</a>
%H A261500 F. Ruskey, <a href="/A000011/a000011.pdf">Necklaces, Lyndon words, De Bruijn sequences, etc.</a> [Cached copy, with permission, pdf format only]
%H A261500 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Necklace.html">Necklace</a>
%H A261500 Wikipedia, <a href="https://en.wikipedia.org/wiki/Necklace_(combinatorics)">Necklace (combinatorics)</a>
%H A261500 <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>
%F A261500 a(n) = 1/(7*n) * Sum_{d|n} C(7*n/d,n/d) * A000010(d) for n>0, a(0) = 1.
%F A261500 a(n) ~ 7^(7*n-1/2) / (2 * sqrt(3*Pi) * 6^(6*n) * n^(3/2)). - _Vaclav Kotesovec_, Aug 22 2015
%p A261500 with(numtheory):
%p A261500 a:= n-> `if`(n=0, 1, add(binomial(7*n/d, n/d)
%p A261500              *phi(d), d=divisors(n))/(7*n)):
%p A261500 seq(a(n), n=0..25);
%Y A261500 Column k=6 of A261494.
%K A261500 nonn
%O A261500 0,3
%A A261500 _Alois P. Heinz_, Aug 21 2015