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A261495 Number of necklaces with n white beads and n^2 black beads.

%I A261495 #25 Feb 16 2025 08:33:26
%S A261495 1,1,3,19,245,4751,124936,4141383,166237161,7847250409,426342182761,
%T A261495 26219808548110,1801378010581175,136784412621194274,
%U A261495 11378390032696241010,1029218687419565103111,100592759623604055645649,10565465772302876757883823,1186893721789951847976898669
%N A261495 Number of necklaces with n white beads and n^2 black beads.
%H A261495 Alois P. Heinz, <a href="/A261495/b261495.txt">Table of n, a(n) for n = 0..335</a>
%H A261495 F. Ruskey, <a href="http://combos.org/necklace">Necklaces, Lyndon words, De Bruijn sequences, etc.</a>
%H A261495 F. Ruskey, <a href="/A000011/a000011.pdf">Necklaces, Lyndon words, De Bruijn sequences, etc.</a> [Cached copy, with permission, pdf format only]
%H A261495 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Necklace.html">Necklace</a>
%H A261495 Wikipedia, <a href="https://en.wikipedia.org/wiki/Necklace_(combinatorics)">Necklace (combinatorics)</a>
%H A261495 <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>
%F A261495 a(n) = 1/((n+1)*n) * Sum_{d|n} C((n+1)*n/d,n/d) * A000010(d) for n>0, a(0) = 1.
%F A261495 a(n) ~ exp(n+1/2) * n^(n-5/2) / sqrt(2*Pi). - _Vaclav Kotesovec_, Aug 22 2015
%p A261495 with(numtheory):
%p A261495 a:= n-> `if`(n=0, 1, add(binomial((n^2+n)/d, n/d)
%p A261495              *phi(d), d=divisors(n))/(n^2+n)):
%p A261495 seq(a(n), n=0..20);
%t A261495 a[n_] := If[n==0, 1, DivisorSum[n, Binomial[(n^2+n)/#, n/#]*EulerPhi[#]&]/ (n^2 + n)];
%t A261495 Table[a[n], {n, 0, 20}] (* _Jean-François Alcover_, Mar 25 2017, translated from Maple *)
%o A261495 (PARI) a(n) = if(n<1, 1, sumdiv(n, d, binomial((n^2 + n)/d, n/d) * eulerphi(d)) / (n^2 + n));
%o A261495 for(n=0, 20, print1(a(n),", ")) \\ _Indranil Ghosh_, Mar 25 2017
%Y A261495 Main diagonal of A261494.
%K A261495 nonn
%O A261495 0,3
%A A261495 _Alois P. Heinz_, Aug 21 2015