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Expansion of e.g.f. 1/(1 - log(1 + x)^5).
2

%I #10 Sep 20 2024 17:14:06

%S 1,0,0,0,0,120,-1800,21000,-235200,2693880,-28690200,210447600,

%T 1465952400,-123513355680,4155643171680,-114924516470400,

%U 2886135295680000,-66750668391381120,1375830884058456960,-22036006671394705920,70186623981895296000,16180846322732941893120

%N Expansion of e.g.f. 1/(1 - log(1 + x)^5).

%F a(0) = 1; a(n) = 120 * Sum_{k=1..n} binomial(n,k) * Stirling1(k,5) * a(n-k).

%F a(n) = Sum_{k=0..floor(n/5)} (5*k)! * Stirling1(n,5*k).

%t With[{nn=30},CoefficientList[Series[1/(1-Log[1+x]^5),{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, Sep 20 2024 *)

%o (PARI) my(N=30, x='x+O('x^N)); Vec(serlaplace(1/(1-log(1+x)^5)))

%o (PARI) a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=120*sum(j=1, i, binomial(i, j)*stirling(j, 5, 1)*v[i-j+1])); v;

%o (PARI) a(n) = sum(k=0, n\5, (5*k)!*stirling(n, 5*k, 1));

%Y Cf. A006252, A354229.

%Y Cf. A353200, A354135, A354232.

%K sign

%O 0,6

%A _Seiichi Manyama_, May 20 2022