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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

%I

%S 1,12,804,88680,12386340,1985320512,348219006744,65085592725648,

%T 12753825281316900,2592090993453733200,542345058701093666304,

%U 116192631187950808203648,25387248470938096734043416,5639653178340668177808156480,1270704973262949380127900086640

%N 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).

%H Alois P. Heinz, <a href="/A220119/b220119.txt">Table of n, a(n) for n = 0..200</a>

%H G. Almkvist and W. Zudilin, <a href="http://arxiv.org/abs/math/0402386">Differential equations, mirror maps and zeta values</a>, 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.

%H C. Krattenthaler, T. Rivoal, <a href="http://arxiv.org/abs/0907.2597">Démonstration de l'Observation 2 d'Almkvist et Zudilin</a>, arXiv:0907.2597 [math.NT], 2009.

%F 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.

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

%F From _Gheorghe Coserea_, Aug 21 2016: (Start)

%F 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.

%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)

%e From _Michael B. Porter_, Aug 23 2016: (Start)

%e For n=2, there are 9 terms:

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

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

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

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

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

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

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

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

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

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

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

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

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

%p end:

%p seq(a(n), n=0..20); # _Alois P. Heinz_, Dec 13 2012

%t 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 *)

%o (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);}

%o (PARI)

%o seq(N) = {

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

%o for (n = 3, N,

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

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

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

%o return(concat(1,a));

%o };

%o seq(14) \\ _Gheorghe Coserea_, Aug 21 2016

%Y Cf. A005258, A005259, A013662.

%K nonn,easy

%O 0,2

%A _Michel Marcus_, Dec 11 2012

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Last modified May 11 12:47 EDT 2021. Contains 343791 sequences. (Running on oeis4.)