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A220119

a(n) = Sum_{0<=j<=n, 0<=k<=n} binomial(n,j)^2 * binomial(n,k)^2 * binomial(n+j,n) * binomial(n+k,n) * binomial(j+k,n).

1, 12, 804, 88680, 12386340, 1985320512, 348219006744, 65085592725648, 12753825281316900, 2592090993453733200, 542345058701093666304, 116192631187950808203648, 25387248470938096734043416, 5639653178340668177808156480, 1270704973262949380127900086640

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OFFSET

0,2

LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..200

G. Almkvist and W. Zudilin, Differential equations, mirror maps and zeta values, arXiv:math/0402386 [math.NT], 2004; also in: Mirror Symmetry V, N. Yui, S.-T. Yau, and J.D. Lewis (eds.), AMS/IP Studies in Adv. Math. 38 (2007), Intern. Press & Amer. Math. Soc., 481--515.

C. Krattenthaler, T. Rivoal, Démonstration de l'Observation 2 d'Almkvist et Zudilin, arXiv:0907.2597 [math.NT], 2009.

FORMULA

Recurrence: n^5*a(n) = 3*(2*n-1)*(3*n^2-3*n+1)*(15*n^2-15*n+4)*a(n-1) + 3*(n-1)^3*(3*n-4)*(3*n-2)*a(n-2) for n > 1.

a(n) ~ sqrt(6)*(5+3*sqrt(3)) * (135+78*sqrt(3))^n/(16*(Pi*n)^(5/2)). - Vaclav Kotesovec, Aug 13 2013

From Gheorghe Coserea, Aug 21 2016: (Start)

0 = x^5*(27*x^2 + 270*x - 1)*y''''' + x^4*(405*x^2 + 3375*x - 10)*y'''' + x^3*(1752*x^2 + 11502*x - 25)*y''' + x^2*(2412*x^2 + 11259*x - 15)*y'' + x*(816*x^2 + 2130*x - 1)*y' + 12*x*(2*x + 1)*y, where y is the g.f.

lim b(n)/a(n) = zeta(4) (= A013662), where b(n) satisfies the same recurrence relation as a(n) with the initial conditions b(0)=0, b(1)=13, b(2)=13923/16, b(3)=62195315/648. (End)

EXAMPLE

From Michael B. Porter, Aug 23 2016: (Start)

For n=2, there are 9 terms:

j=0, k=0: 1^2 * 1^2 * 1 * 1 * 0 = 0

j=0, k=1: 1^2 * 2^2 * 1 * 3 * 0 = 0

j=0, k=2: 1^2 * 1^2 * 1 * 6 * 1 = 6

j=1, k=0: 2^2 * 1^2 * 3 * 1 * 0 = 0

j=1, k=1: 2^2 * 2^2 * 3 * 3 * 1 = 144

j=1, k=2: 2^2 * 1^2 * 3 * 6 * 3 = 216

j=2, k=0: 1^2 * 1^2 * 6 * 1 * 1 = 6

j=2, k=1: 1^2 * 2^2 * 6 * 3 * 3 = 216

j=2, k=2: 1^2 * 1^2 * 6 * 6 * 6 = 216

so a(2) = 0 + 0 + 6 + 0 + 144 + 216 + 6 + 216 + 216 = 804. (End)

MAPLE

a:= proc(n) option remember; `if`(n<2, 1+11*n,

(3*(2*n-1)*(15*n^2-15*n+4)*(3*n^2-3*n+1)* a(n-1)

+3*(3*n-2)*(3*n-4)*(n-1)^3 *a(n-2)) / n^5)

end:

seq(a(n), n=0..20); # Alois P. Heinz, Dec 13 2012

MATHEMATICA

Table[Sum[Sum[Binomial[n, j]^2*Binomial[n, k]^2*Binomial[n+j, n]*Binomial[n+k, n]*Binomial[j+k, n], {j, 0, n}], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Aug 13 2013 *)

PROG

(PARI) a(n) = {v = 0; for (j=0, n, for (k=0, n, v += binomial(n, j)^2* binomial(n, k)^2*binomial(n+j, n)*binomial(n+k, n)*binomial(j+k, n); ); ); return (v); }

(PARI)

seq(N) = {

my(a = vector(N)); a[1] = 12; a[2] = 804;

for (n = 3, N,

my(t1 = 3*(2*n-1)*(3*n^2-3*n+1)*(15*n^2-15*n+4)*a[n-1],

t2 = 3*(n-1)^3*(3*n-4)*(3*n-2)*a[n-2]);

a[n] = (t1 + t2)/n^5);

return(concat(1, a));

};

seq(14) \\ Gheorghe Coserea, Aug 21 2016

CROSSREFS

Cf. A005258, A005259, A013662.