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%I #8 Jun 29 2019 13:33:27
%S 1,1,5,38,386,4904,74776,1330272,27046848,618653280,15723024864,
%T 439559609664,13405656582336,442915145716224,15759326934391296,
%U 600783539885546496,24430204949876794368,1055516761826050203648,48286612866726631489536,2331682676308057000255488
%N Expansion of e.g.f. (1 + log(1 - x))/(1 + 2*log(1 - x)).
%F a(0) = 1; a(n) = Sum_{k=1..n} |Stirling1(n,k)| * 2^(k-1) * k!.
%F a(n) ~ n! * exp(n/2) / (4 * (exp(1/2) - 1)^(n+1)). - _Vaclav Kotesovec_, Jun 29 2019
%t nmax = 19; CoefficientList[Series[(1 + Log[1 - x])/(1 + 2 Log[1 - x]), {x, 0, nmax}], x] Range[0, nmax]!
%t Join[{1}, Table[Sum[Abs[StirlingS1[n, k]] 2^(k - 1) k!, {k, 1, n}], {n, 1, 19}]]
%Y Cf. A001710, A002866, A008275, A011782, A050351, A088500, A320349, A308878.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Jun 29 2019
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