%I #18 May 07 2024 02:48:20
%S 1,3,6,16,52,184,688,2672,10672,43552,180800,761088,3241088,13937408,
%T 60435968,263962880,1160188672,5127762432,22775636992,101608357888,
%U 455105255424,2045751037952,9225923895296,41731062358016,189275050729472,860630181167104
%N G.f. A(x) satisfies A(x) = (1 + x)^2 + x*A(x)^2 / (1 + x).
%F G.f.: A(x) = 2*(1+x)^2 / (1+sqrt(1-4*x*(1+x))).
%F a(n) = Sum_{k=0..n} binomial(k+2,n-k) * binomial(2*k,k)/(k+1).
%F a(n) ~ 2^(n - 5/4) * (1 + sqrt(2))^(n + 3/2) / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Nov 25 2023
%F D-finite with recurrence (n+1)*a(n) +(-3*n+1)*a(n-1) +2*(-4*n+9)*a(n-2) +4*(-n+4)*a(n-3)=0. - _R. J. Mathar_, Dec 04 2023
%F From _Peter Bala_, May 05 2024: (Start)
%F A(x) = (1 + x)*S(x/(1 + x)), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the g.f. of the large Schröder numbers A006318. Cf. A025227.
%F A333090(n) = [x^n] A(x)^n. (End)
%o (PARI) a(n) = sum(k=0, n, binomial(k+2, n-k)*binomial(2*k, k)/(k+1));
%Y Cf. A006318, A025227, A367640, A333090, A367641.
%K nonn,easy
%O 0,2
%A _Seiichi Manyama_, Nov 25 2023