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a(0) = 1; a(n) = 5 * Sum_{k=1..n} binomial(n,k) * a(k-1).
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%I #29 Jun 07 2021 14:55:32

%S 1,5,35,265,2195,19625,187755,1909185,20521515,232124745,2752591475,

%T 34108980105,440444019835,5912197332865,82320781521195,

%U 1186703083508025,17680850448587155,271845880552898985,4307188044378111915,70236616096770062945,1177406236243423738475

%N a(0) = 1; a(n) = 5 * Sum_{k=1..n} binomial(n,k) * a(k-1).

%F G.f. A(x) satisfies: A(x) = 1 + 5 * x * A(x/(1 - x)) / (1 - x)^2.

%t a[0] = 1; a[n_] := a[n] = 5 Sum[Binomial[n, k] a[k - 1], {k, 1, n}]; Table[a[n], {n, 0, 20}]

%t nmax = 20; A[_] = 0; Do[A[x_] = 1 + 5 x A[x/(1 - x)]/(1 - x)^2 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%Y Cf. A040027, A094418, A144180, A343523, A343975, A344735, A345077, A345078, A345081.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Jun 07 2021