a(n) = Sum_{i=0..n} Sum_{j=0..n} Sum_{k=0..n} binomial(n, i)*binomial(n, j) *binomial(n, k)*binomial(3n, i+j+k). -
Benoit Cloitre, Mar 08 2005
O.g.f. (with a(0):=1): (cb(x^(1/3)) + sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))+1+2*x^(1/3)))/3, with cb(x):=1/sqrt(1-4*x) (o.g.f. of
A000984) and P(x):=P(-1/2,4*x) = 1/sqrt(1+4*x+16*x^2) (o.g.f. of
A116091, with P(x,z) the o.g.f. of the Legendre polynomials). -
Wolfdieter Lang, Mar 24 2011
D-finite with recurrence n*(3n-1)*(3n-2)*a(n) = 8*(6n-5)*(6n-1)*(2n-1)*a(n-1). -
R. J. Mathar, Sep 17 2012
a(n) = hypergeom([-3*n, -3*n], [1], 1). -
Peter Luschny, Mar 19 2018
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k.
a(n) = [(x*y)^(3*n)] (1 + x + y)^(6*n). Cf.
A001448. (End)
Conjecture: a(n) = [x^n] G(x)^(2*n), where G(x) = (1 + x)*(1 - 6*x + x^2)/(2*x) + (x^2 - 1)*sqrt(1 - 14*x + x^2)/(2*x) = 1 + 10*x + 81*x^2 + 720*x^3 + .... The algebraic function G(x) satisfies the quadratic equation x*G(x)^2 - (1 - 5*x - 5*x^2 + x^3)*G(x) + (1 + x)^4 = 0. Cf.
A001450. -
Peter Bala, Oct 27 2022
a(n) = Sum_{k = 0..3*n} binomial(3*n+k-1, k). -
Peter Bala, Jun 04 2024
O.g.f: 3F2(1/6,1/2,5/6; 1/3,2/3 ; 64*x). -
R. J. Mathar, Jan 11 2025
