%I #6 May 25 2014 17:53:41
%S 6,160,2842,44868,681604,10248992,154149762,2327405740,35305388536,
%T 538000530912,8231764528156,126399786937760,1946868985459272,
%U 30066806831424448,465425347391123282,7219408741591089660
%N Sum of all matrix elements of n X n matrix M[i,j]=(2n+i+j)!/(n+i)!/(n+j)!, i,j=1..n.
%C p divides a((p-1)/2) for prime p>2.
%F a(n) = Sum[Sum[(2n+i+j)!/(n+i)!/(n+j)!,{i,1,n}],{j,1,n}].
%t Table[Sum[Sum[(2n+i+j)!/(n+i)!/(n+j)!,{i,1,n}],{j,1,n}],{n,1,20}]
%K nonn
%O 1,1
%A _Alexander Adamchuk_, Jul 04 2006