|
%I #16 Oct 21 2023 11:07:49
%S 1,2,-18,270,-4902,98538,-2110794,47227846,-1090742094,25806364434,
%T -622267199554,15236456140542,-377814588773622,9468373002766074,
%U -239434464005544570,6101951612867546166,-156561081975745809566,4040863076496835880226
%N G.f. satisfies A(x) = 1 + x/A(x)^3*(1 + 1/A(x)^3).
%F G.f.: A(x) = 1/B(-x) where B(x) is the g.f. of A363304.
%F a(n) = (-1)^(n-1) * (1/n) * Sum_{k=0..n} binomial(n,k) * binomial(4*n+3*k-2,n-1) for n > 0.
%F a(n) ~ c*(-1)^(n+1)*256^n*27^(-n)*4F3([-n, 4*n/3, (4n-1)/3, (4*n+1)/3], [n, n+1/3, n+2/3], -1)*n^(-3/2), with c = (1/8)*sqrt (3/(2*Pi)). - _Stefano Spezia_, Oct 21 2023
%o (PARI) a(n) = if(n==0, 1, (-1)^(n-1)*sum(k=0, n, binomial(n, k)*binomial(4*n+3*k-2, n-1))/n);
%Y Cf. A364398, A364399.
%Y Cf. A363304.
%K sign
%O 0,2
%A _Seiichi Manyama_, Jul 22 2023
|
