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A261503 Number of necklaces with n white beads and 9*n black beads.

%I A261503 #17 Feb 16 2025 08:33:26
%S A261503 1,1,10,136,2290,42376,834472,17125354,362345362,7847250409,
%T A261503 173103115760,3875837737520,87857163416200,2012268157890524,
%U A261503 46497242510288804,1082614775186919136,25374686712458407442,598217593341171422090,14176230568954212214234
%N A261503 Number of necklaces with n white beads and 9*n black beads.
%H A261503 Alois P. Heinz, <a href="/A261503/b261503.txt">Table of n, a(n) for n = 0..710</a>
%H A261503 F. Ruskey, <a href="http://combos.org/necklace">Necklaces, Lyndon words, De Bruijn sequences, etc.</a>
%H A261503 F. Ruskey, <a href="/A000011/a000011.pdf">Necklaces, Lyndon words, De Bruijn sequences, etc.</a> [Cached copy, with permission, pdf format only]
%H A261503 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Necklace.html">Necklace</a>
%H A261503 Wikipedia, <a href="https://en.wikipedia.org/wiki/Necklace_(combinatorics)">Necklace (combinatorics)</a>
%H A261503 <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>
%F A261503 a(n) = 1/(10*n) * Sum_{d|n} C(10*n/d,n/d) * A000010(d) for n>0, a(0) = 1.
%F A261503 a(n) ~ 10^(10*n-1/2) / (sqrt(2*Pi) * 9^(9*n+1/2) * n^(3/2)). - _Vaclav Kotesovec_, Aug 22 2015
%p A261503 with(numtheory):
%p A261503 a:= n-> `if`(n=0, 1, add(binomial(10*n/d, n/d)
%p A261503              *phi(d), d=divisors(n))/(10*n)):
%p A261503 seq(a(n), n=0..25);
%Y A261503 Column k=9 of A261494.
%K A261503 nonn
%O A261503 0,3
%A A261503 _Alois P. Heinz_, Aug 21 2015