%I #12 Sep 12 2022 03:05:52
%S 1,0,0,0,24,60,240,1260,108864,1149120,12160800,138045600,5605649280,
%T 122049607680,2378318604480,45712559692800,1529842399303680,
%U 47673689320857600,1382823169839820800,38831806109898547200,1378613101427645184000
%N E.g.f. satisfies A(x) = 1/(1 - x * A(x))^(x^3 * A(x)).
%F a(n) = n! * Sum_{k=0..floor(n/4)} (n-2*k+1)^(k-1) * |Stirling1(n-3*k,k)|/(n-3*k)!.
%t m = 21; (* number of terms *)
%t A[_] = 0;
%t Do[A[x_] = 1/(1 - x*A[x])^(x^3*A[x]) + O[x]^m // Normal, {m}];
%t CoefficientList[A[x], x]*Range[0, m - 1]! (* _Jean-François Alcover_, Sep 12 2022 *)
%o (PARI) a(n) = n!*sum(k=0, n\4, (n-2*k+1)^(k-1)*abs(stirling(n-3*k, k, 1))/(n-3*k)!);
%Y Cf. A184949, A349556, A356970.
%Y Cf. A356968.
%K nonn
%O 0,5
%A _Seiichi Manyama_, Sep 07 2022