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%I #14 Mar 02 2015 23:55:08
%S 1,11,136,1787,24376,341048,4859968,70223483,1025790616,15116164136,
%T 224365547968,3350371999928,50287277411008,758124098549696,
%U 11473331826459136,174221578556572283,2653437885092286808,40520013896165905928
%N A036917/8 (omitting leading term of A036917).
%C It appears that a(n) == 16^n/Pi^3 * Integrate[x=0..1, x^n*F(x)*F(1-x)], where F(x) = Pi/2 * hypergeometric([1/2, 1/2], [1], x) (== elliptic K(sqrt(x))). - _Vladimir Reshetnikov_, Jan 20 2011
%F G.f.: 7/8 + (1/2)*(K(16x)/pi)^2, where K(x) is the elliptic integral of the first kind (as defined in Mathematica). - _Emanuele Munarini_, Mar 12 2011
%F a(n) = (1/8)*sum(binomial(2k,k)^2*binomial(2n-2k,n-k)^2, k=0..n) for n >= 1. - _Emanuele Munarini_, Mar 12 2011
%p seq(add(binomial(2*k, k)^2*binomial(2*(n-k), n-k)^2, k=0..n)/8, n=1..12); # _Emanuele Munarini_, Mar 12 2011
%t Table[Sum[Binomial[2k, k]^2 Binomial[2n-2k,n-k]^2, {k, 0, n}]/8, {n, 1, 12}] (* _Emanuele Munarini_, Mar 12 2011 *)
%o (Maxima) makelist(sum(binomial(2*k,k)^2*binomial(2*(n-k),n-k)^2,k,0,n)/8,n,1,12); /* _Emanuele Munarini_, Mar 12 2011 */
%K nonn,easy
%O 1,2
%A _N. J. A. Sloane_, Oct 23 2000
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