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%I #6 Aug 10 2021 09:04:17
%S 1,15,140,1050,6951,42399,239800,1164570,2553551,-54771717,
%T -1384474728,-23286667950,-339924740609,-4554547609233,
%U -56481301888144,-630768487283886,-5665064764515849,-18095553874845909,924820173031946752,35413415495503624986
%N E.g.f.: log( 1 + (exp(x) - 1)^5 / 5! ).
%F a(n) = Stirling2(n,5) - (1/n) * Sum_{k=1..n-1} binomial(n,k) * Stirling2(n-k,5) * k * a(k).
%F a(n) ~ -(n-1)! * 2^(1+n) * 5^n * cos(n*arctan((2*arctan(sqrt(10 - 2*sqrt(5))/(1 + sqrt(5) + 2^(7/5)/15^(1/5)))) / log(1 + 3^(1/5)*5^(7/10)/2^(2/5) + 15^(1/5)/2^(2/5) + 2^(6/5)*15^(2/5)))) / (100*arctan(sqrt(10 - 2*sqrt(5))/(1 + sqrt(5) + 2^(7/5)/15^(1/5)))^2 + (5*log(1 + 3^(1/5)*5^(7/10)/2^(2/5) + 15^(1/5)/2^(2/5) + 2^(6/5)*15^(2/5)))^2)^(n/2). - _Vaclav Kotesovec_, Aug 10 2021
%t nmax = 24; CoefficientList[Series[Log[1 + (Exp[x] - 1)^5/5!], {x, 0, nmax}], x] Range[0, nmax]! // Drop[#, 5] &
%t a[n_] := a[n] = StirlingS2[n, 5] - (1/n) Sum[Binomial[n, k] StirlingS2[n - k, 5] k a[k], {k, 1, n - 1}]; Table[a[n], {n, 5, 24}]
%Y Cf. A000481, A327506, A346955, A346974, A346975, A346976.
%K sign
%O 5,2
%A _Ilya Gutkovskiy_, Aug 09 2021
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