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Number of necklaces with n white beads and n^2-n black beads.
2

%I #16 Apr 30 2019 08:23:47

%S 1,1,2,10,116,2126,54132,1753074,69159400,3220837534,173103115760,

%T 10551652603526,719578430426044,54297978110913252,4492502634679508204,

%U 404469190271900056316,39370123445405248353744,4120204305690280446004838,461365717080849798202175772

%N Number of necklaces with n white beads and n^2-n black beads.

%H Alois P. Heinz, <a href="/A261496/b261496.txt">Table of n, a(n) for n = 0..335</a>

%H F. Ruskey, <a href="http://combos.org/necklace">Necklaces, Lyndon words, De Bruijn sequences, etc.</a>

%H F. Ruskey, <a href="/A000011/a000011.pdf">Necklaces, Lyndon words, De Bruijn sequences, etc.</a> [Cached copy, with permission, pdf format only]

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Necklace.html">Necklace</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Necklace_(combinatorics)">Necklace (combinatorics)</a>

%H <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>

%F a(n) = 1/(n^2) * Sum_{d|n} C(n^2/d,n/d) * A000010(d) for n>0, a(0) = 1.

%F a(n) ~ exp(n-1/2) * n^(n-5/2) / sqrt(2*Pi). - _Vaclav Kotesovec_, Aug 22 2015

%p with(numtheory):

%p a:= n-> `if`(n=0, 1, add(binomial(n^2/d, n/d)

%p *phi(d), d=divisors(n))/n^2):

%p seq(a(n), n=0..20);

%Y Lower diagonal of A261494.

%Y Cf. A000010.

%K nonn

%O 0,3

%A _Alois P. Heinz_, Aug 21 2015