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%I #14 Mar 17 2023 08:52:39
%S 1,0,0,6,72,720,7560,90720,1270080,20381760,364694400,7125148800,
%T 150186960000,3393726336000,81882210009600,2102315389574400,
%U 57244753133568000,1647544166940672000,49957730917981286400,1591303422125646028800
%N Expansion of e.g.f. exp( (x / (1-x))^3 ).
%F a(n) = n! * Sum_{k=0..floor(n/3)} binomial(n-1,n-3*k)/k!.
%F a(0) = 1; a(n) = (n-1)! * Sum_{k=3..n} (-1)^(k-3) * k * binomial(-3,k-3) * a(n-k)/(n-k)!.
%F From _Vaclav Kotesovec_, Mar 17 2023: (Start)
%F a(n) = 4*(n-1)*a(n-1) - 6*(n-2)*(n-1)*a(n-2) + (n-2)*(n-1)*(4*n - 9)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4).
%F a(n) ~ 3^(1/8) * exp(-1/4 + 5*3^(-1/4)*n^(1/4)/8 - sqrt(3*n)/2 + 4*3^(-3/4)*n^(3/4) - n) * n^(n - 1/8) / 2 * (1 - (409/2560)*3^(1/4)/n^(1/4)). (End)
%t Table[n! * Sum[Binomial[n-1,n-3*k]/k!, {k,0,n/3}], {n,0,20}] (* _Vaclav Kotesovec_, Mar 17 2023 *)
%o (PARI) my(N=20, x='x+O('x^N)); Vec(serlaplace(exp((x/(1-x))^3)))
%o (PARI) a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=3, i, (-1)^(j-3)*j*binomial(-3, j-3)*v[i-j+1]/(i-j)!)); v;
%Y Cf. A000262, A052887, A361576.
%Y Cf. A091695.
%K nonn
%O 0,4
%A _Seiichi Manyama_, Mar 16 2023
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