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Expansion of e.g.f. 1 / (1 - 5 * log(1 + x))^(1/5).
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%I #10 Sep 11 2023 09:02:42

%S 1,1,5,50,720,13650,320370,8967720,291538080,10795026840,448484788680,

%T 20658543923280,1044915105622800,57572197848878400,

%U 3432143603792520000,220109018869587398400,15110184224165199667200,1105545474191480800492800,85881534014930659599571200

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

%F a(n) = Sum_{k=0..n} Stirling1(n,k) * A008548(k).

%F a(n) ~ n! * exp(1/25) / (Gamma(1/5) * 5^(1/5) * n^(4/5) * (exp(1/5) - 1)^(n + 1/5)). - _Vaclav Kotesovec_, Aug 14 2021

%F a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k-1) * (5 - 4*k/n) * (k-1)! * binomial(n,k) * a(n-k). - _Seiichi Manyama_, Sep 11 2023

%t nmax = 18; CoefficientList[Series[1/(1 - 5 Log[1 + x])^(1/5), {x, 0, nmax}], x] Range[0, nmax]!

%t Table[Sum[StirlingS1[n, k] 5^k Pochhammer[1/5, k], {k, 0, n}], {n, 0, 18}]

%Y Cf. A006252, A008548, A320343, A346984, A346987, A347020, A347021, A347023.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Aug 11 2021