%I #10 Dec 15 2024 06:45:55
%S 1,2,10,62,394,2562,16966,113794,770458,5254658,36046470,248449104,
%T 1719175846,11935608518,83100064834,579994824042,4056746450106,
%U 28428354905268,199550820571858,1402832286126650,9875127071717694,69599814539512900,491081313666879968,3468458841769675496
%N a(n) = Sum_{k=0..floor(n/2)} binomial(2*n+k-1,k) * binomial(2*n+k,n-2*k).
%F a(n) = [x^n] 1/( 1/(1 + x) - x^2 )^(2*n).
%F a(n) == 0 (mod 2) for n>0.
%o (PARI) a(n) = sum(k=0, n\2, binomial(2*n+k-1, k)*binomial(2*n+k, n-2*k));
%Y Cf. A379022, A379085.
%K nonn
%O 0,2
%A _Seiichi Manyama_, Dec 15 2024