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

1, 0, 5, 25, 200, 1875, 20625, 256250, 3534375, 53515625, 881468750, 15667578125, 298478828125, 6060493750000, 130542772265625, 2971013486328125, 71193375156250000, 1790666151318359375, 47145509926611328125, 1296156682961425781250, 37129279010879638671875

Offset

0,3

Links

Table of n, a(n) for n=0..20.

Formula

G.f. A(x) satisfies: A(x) = (1 - 5*x + x*A(x/(1 - 5*x))) / (1 - 4*x - 5*x^2).

G.f.: (1/(1 + x)) * Sum_{k>=0} (x/(1 + x))^k / Product_{j=1..k} (1 - 5*j*x/(1 + x)).

E.g.f.: exp((exp(5*x) - 1) / 5 - x).

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

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A005011(k).

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

Mathematica

nmax = 20; CoefficientList[Series[Exp[(Exp[5 x] - 1)/5 - x], {x, 0, nmax}], x] Range[0, nmax]!
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}]
Table[Sum[(-1)^(n - k) Binomial[n, k] 5^k BellB[k, 1/5], {k, 0, n}], {n, 0, 20}]

Crossrefs

Cf. A000296, A003577, A005011, A337038, A337039, A337040, A337042, A337043.

Sequence in context: A294213 A330198 A346269 * A254335 A193939 A352075

Adjacent sequences: A337038 A337039 A337040 * A337042 A337043 A337044

Keyword

nonn

Author

Ilya Gutkovskiy, Aug 12 2020

Status

approved