OFFSET
0,3
COMMENTS
a(n) = number of necklaces of n white beads and 2n black beads. - David Callan, Mar 28 2004
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..1000
M. Bousquet and C. Lamathe, Enumeration of solid trees according to edge number and edge degree distribution, Discr. Math., 298 (2005), 115-141.
M. Isachenkov, I. Kirsch, V. Schomerus, Chiral Primaries in Strange Metals, arXiv preprint arXiv:1403.6857 [hep-th], 2014.
FORMULA
From Joerg Arndt, Oct 21 2012: (Start)
a(n) = sum( d divides n, phi(n/d) * C(3*d,d) ) / (3*n) for n>=1, a(0)=1.
a(n) = sum( d divides n, phi(n/d) * C(3*d-1,d) ) / (2*n) for n>=1, a(0)=1.
a(n) = A047996(3*n,n).
(End)
a(n) ~ 3^(3*n) / (2^(2*n+1) * sqrt(3*Pi) * n^(3/2)). - Vaclav Kotesovec, Aug 22 2015
MAPLE
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
MATHEMATICA
a[n_] := DivisorSum[n, EulerPhi[n/#]*Binomial[3#, #]&]/(3n); a[0] = 1; Array[a, 30, 0] (* Jean-François Alcover, Dec 02 2015 *)
PROG
(PARI)
C(n, k)=binomial(n, k);
a(n) = if(n<=0, n==0, sumdiv(n, d, eulerphi(n/d) * C(3*d, d)) / (3*n) );
/* or, second formula: */
/* a(n) = if(n<=0, n==0, sumdiv(n, d, eulerphi(n/d) * C(3*d-1, d)) / (2*n) ); */
/* Joerg Arndt, Oct 21 2012 */
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, May 26 2003
EXTENSIONS
a(0)=1 prepended by Joerg Arndt, Oct 21 2012
STATUS
approved