%I #33 Mar 20 2019 19:12:45
%S 1,1,3,10,43,201,1038,5538,30667,173593,1001603,5864750,34769374,
%T 208267320,1258579654,7663720710,46976034379,289628805623,
%U 1794932468571,11175157356522,69864075597643,438403736549145,2760351032959050,17433869214973754,110420300879752990
%N a(n) = (1/(3*n))*Sum_{d|n, d even} phi(2*n/d)*binomial(3d/2,d).
%C a(n) = number of necklaces of n white beads and 2n black beads. - _David Callan_, Mar 28 2004
%H Alois P. Heinz, <a href="/A082936/b082936.txt">Table of n, a(n) for n = 0..1000</a>
%H M. Bousquet and C. Lamathe, <a href="http://dx.doi.org/10.1016/j.disc.2004.11.015">Enumeration of solid trees according to edge number and edge degree distribution</a>, Discr. Math., 298 (2005), 115-141.
%H M. Isachenkov, I. Kirsch, V. Schomerus, <a href="http://arxiv.org/abs/1403.6857">Chiral Primaries in Strange Metals</a>, arXiv preprint arXiv:1403.6857 [hep-th], 2014.
%F From _Joerg Arndt_, Oct 21 2012: (Start)
%F a(n) = sum( d divides n, phi(n/d) * C(3*d,d) ) / (3*n) for n>=1, a(0)=1.
%F a(n) = sum( d divides n, phi(n/d) * C(3*d-1,d) ) / (2*n) for n>=1, a(0)=1.
%F a(n) = A047996(3*n,n).
%F (End)
%F a(n) ~ 3^(3*n) / (2^(2*n+1) * sqrt(3*Pi) * n^(3/2)). - _Vaclav Kotesovec_, Aug 22 2015
%p with(numtheory): f := proc(n) local t1,d; t1 := 0; for d from 1 to n do if n mod d = 0 then if d mod 2 = 0 then t1 := t1+phi(n/d)*binomial(3*d/2,d) fi; fi; od; 2*t1/(3*n); end; # use with n even
%t a[n_] := DivisorSum[n, EulerPhi[n/#]*Binomial[3#, #]&]/(3n); a[0] = 1; Array[a, 30, 0] (* _Jean-François Alcover_, Dec 02 2015 *)
%o (PARI)
%o C(n, k)=binomial(n,k);
%o a(n) = if(n<=0, n==0, sumdiv(n, d, eulerphi(n/d) * C(3*d,d)) / (3*n) );
%o /* or, second formula: */
%o /* a(n) = if(n<=0, n==0, sumdiv(n, d, eulerphi(n/d) * C(3*d-1,d)) / (2*n) ); */
%o /* _Joerg Arndt_, Oct 21 2012 */
%Y Cf. A003239.
%Y Column k=2 of A261494.
%K nonn
%O 0,3
%A _N. J. A. Sloane_, May 26 2003
%E a(0)=1 prepended by _Joerg Arndt_, Oct 21 2012