A302870 Expansion of e.g.f. 1 / Sum_{n >= 0} (n+1)^3*x^n/n!.

1, -8, 101, -1840, 44441, -1340696, 48530653, -2049479216, 98915010545, -5370730092136, 324012625790741, -21502216185516848, 1556657523678767881, -122085765970981019000, 10311495889448094131981, -933128678308256836233136, 90072066063382006331898593

Offset: 0

Author

Seiichi Manyama, Apr 15 2018

Keywords

• sign

Comments

From Vaclav Kotesovec, Apr 15 2018: (Start)

In general, for m>=0, Sum_{k>=0} (k+1)^m * x^k / k! = exp(x) * Sum_{j = 1..m+1} Stirling2(m+1, j) * x^(j-1).

If m tends to infinity, then the real root of the equation Sum_{j = 1..m+1} Stirling2(m + 1, j) * x^(j-1) = 0, with a minimal absolute value, tends to -1/2^m.

(End)

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..345

Crossrefs

Cf. A302189.

Programs

Mathematica

nmax = 20; CoefficientList[Series[1/(E^x*((1 + 7*x + 6*x^2 + x^3))), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Apr 15 2018 *)

Formula

From Vaclav Kotesovec, Apr 15 2018: (Start)

E.g.f: exp(-x)/(1 + 7*x + 6*x^2 + x^3).

a(n) ~ (-1)^n * n! * (371 + 414*r + 74*r^2) * exp(-r) * (7 + 6*r + r^2)^n / 257, where r = -0.16575681568607828288437387419... is the real root of the equation 1 + 7*r + 6*r^2 + r^3 = 0.

(End)