A189924 a(n) = abs(Stirling1(n+2,2)) - abs(Stirling1(n+2,3)).

1, 2, 5, 15, 49, 140, -64, -8540, -146124, -2124936, -30374136, -445116672, -6793958016, -108691150464, -1826654613120, -32257962443520, -598196854045440, -11635261535301120, -237044583523514880, -5050811716879104000

Offset: 0

Author

Wolfdieter Lang, Jun 21 2011

Keywords

• sign
• easy

Comments

This is the fourth (k=3) column sequence in triangle A094645 without leading zeros.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..445

Crossrefs

Cf. A081047 (column k=2).

Programs

Magma

[Abs(StirlingFirst(n+2,2)) - Abs(StirlingFirst(n+2,3)): n in [0..30]]; // G. C. Greubel, Jan 13 2018

Mathematica

Table[Abs[StirlingS1[n+2,2]]-Abs[StirlingS1[n+2,3]],{n,0,20}] (* Harvey P. Dale, May 21 2015 *)

PARI

a(n)=abs(stirling(n+2,2))-abs(stirling(n+2,3)) \\ Charles R Greathouse IV, Jun 27 2011

Formula

a(n) = abs(Stirling1(n+2,2)) - abs(Stirling1(n+2,3)), with the unsigned Stirling1 numbers abs(Stirling1(n,k)) = A132393(n,k).

E.g.f.: (1/2)*(2-log(1-x)^2)/(1-x)^2 (from differentiating three times (1-x)*((-log(1-x))^3)/3!).