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A337041 a(n) = exp(-1/5) * Sum_{k>=0} (5*k - 1)^n / (5^k * k!).

%I A337041 #8 Jun 26 2022 12:45:44
%S A337041 1,0,5,25,200,1875,20625,256250,3534375,53515625,881468750,
%T A337041 15667578125,298478828125,6060493750000,130542772265625,
%U A337041 2971013486328125,71193375156250000,1790666151318359375,47145509926611328125,1296156682961425781250,37129279010879638671875
%N A337041 a(n) = exp(-1/5) * Sum_{k>=0} (5*k - 1)^n / (5^k * k!).
%F A337041 G.f. A(x) satisfies: A(x) = (1 - 5*x + x*A(x/(1 - 5*x))) / (1 - 4*x - 5*x^2).
%F A337041 G.f.: (1/(1 + x)) * Sum_{k>=0} (x/(1 + x))^k / Product_{j=1..k} (1 - 5*j*x/(1 + x)).
%F A337041 E.g.f.: exp((exp(5*x) - 1) / 5 - x).
%F A337041 a(0) = 1; a(n) = Sum_{k=1..n-1} binomial(n-1,k) * 5^k * a(n-k-1).
%F A337041 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A005011(k).
%F A337041 a(n) ~ 5^(n - 1/5) * n^(n - 1/5) * exp(n/LambertW(5*n) - n - 1/5) / (sqrt(1 + LambertW(5*n)) * LambertW(5*n)^(n - 1/5)). - _Vaclav Kotesovec_, Jun 26 2022
%t A337041 nmax = 20; CoefficientList[Series[Exp[(Exp[5 x] - 1)/5 - x], {x, 0, nmax}], x] Range[0, nmax]!
%t A337041 a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k] 5^k a[n - k - 1], {k, 1, n - 1}]; Table[a[n], {n, 0, 20}]
%t A337041 Table[Sum[(-1)^(n - k) Binomial[n, k] 5^k BellB[k, 1/5], {k, 0, n}], {n, 0, 20}]
%Y A337041 Cf. A000296, A003577, A005011, A337038, A337039, A337040, A337042, A337043.
%K A337041 nonn
%O A337041 0,3
%A A337041 _Ilya Gutkovskiy_, Aug 12 2020