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a(n) = [x^n] -3/(2*x - 1)^5.
1

%I #3 May 29 2021 05:19:53

%S 3,30,180,840,3360,12096,40320,126720,380160,1098240,3075072,8386560,

%T 22364160,58490880,150405120,381026304,952565760,2353397760,

%U 5752750080,13927710720,33426505728,79586918400,188114534400,441660211200,1030540492800,2390853943296

%N a(n) = [x^n] -3/(2*x - 1)^5.

%F a(n) = 2*(n + 4)*a(n - 1) / n for n >= 2.

%F a(n) = 2^(n - 3)*(n + 1)*(n + 2)*(n + 3)*(n + 4).

%F a(n) = n! * [x^n] (2*x^4 + 16*x^3 + 36*x^2 + 24*x + 3)*exp(2*x).

%F a(n) ~ 2^(n - 3) * n^4.

%p a := n -> 2^(n - 3)*(n + 1)*(n + 2)*(n + 3)*(n + 4): seq(a(n), n = 0..25);

%Y Row sums of A344565.

%K nonn

%O 0,1

%A _Peter Luschny_, May 29 2021