%I #15 Feb 14 2022 10:44:58
%S 0,1,4,22,155,1333,13541,158688,2107682,31291894,513590170,9234669420,
%T 180534475832,3812852144788,86517295628188,2099170738243328,
%U 54233876338638192,1486517654443664016,43084555863325589232,1316588795487600071904,42306543064537291007424,1426115146736949130634400
%N Expansion of e.g.f. -log(1 + log(1 - x))/(1 + log(1 - x)).
%H Seiichi Manyama, <a href="/A302548/b302548.txt">Table of n, a(n) for n = 0..418</a>
%F a(n) = Sum_{k=1..n} |Stirling1(n,k)|*H(k)*k!, where H(k) is the k-th harmonic number.
%F a(n) ~ sqrt(2*Pi) * log(n) * n^(n + 1/2) / (exp(1)-1)^(n+1). - _Vaclav Kotesovec_, Jun 23 2018
%e E.g.f.: A(x) = x + 4*x^2/2! + 22*x^3/3! + 155*x^4/4! + 1333*x^5/5! + 13541*x^6/6! + ...
%p H:= proc(n) H(n):= 1/n +`if`(n=1, 0, H(n-1)) end:
%p a:= n-> add(abs(Stirling1(n, k))*H(k)*k!, k=1..n):
%p seq(a(n), n=0..23); # _Alois P. Heinz_, Jun 21 2018
%t nmax = 21; CoefficientList[Series[-Log[1 + Log[1 - x]]/(1 + Log[1 - x]), {x, 0, nmax}], x] Range[0, nmax]!
%t Table[Sum[Abs[StirlingS1[n, k]] HarmonicNumber[k] k!, {k, 0, n}], {n, 0, 21}]
%Y Cf. A000254, A001008, A002805, A003713, A007840, A073596, A222058, A300490, A302547.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Jun 20 2018
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