A243963 a(n) = n*4^n*(-Z(1-n, 1/4)/2 + Z(1-n, 3/4)/2 - Z(1-n, 1)*(1 - 2^(-n))) for n > 0 and a(0) = 0, where Z(n, c) is the Hurwitz zeta function.

0, 0, 2, 3, -8, -25, 96, 427, -2176, -12465, 79360, 555731, -4245504, -35135945, 313155584, 2990414715, -30460116992, -329655706465, 3777576173568, 45692713833379, -581777702256640, -7777794952988025, 108932957168730112, 1595024111042171723, -24370173276164456448

Offset: 0

Author

Paul Curtz, Jun 16 2014

Keywords

• sign

Comments

Previous name was: 0 followed by -(n+1)*A163747(n).

Difference table of a(n):

0, 0, 2, 3, -8, -25,...

0, 2, 1, -11, -17, 121,...

2, -1, -12, -6, 138, 210,...

-3, -11, 6, 144, 72, -3144,...

-8, 17, 138, -72, -3216, -1608,...

25, 121, -210, -3144, 1608,...

a(n) is an autosequence of second kind. Its inverse binomial transform is the signed sequence. Its main diagonal is the first upper diagonal multiplied by 2.

Crossrefs

Cf. A132049, A065619.

Programs

Maple

a := n -> `if`(n=0, 0, n*4^n*(-Zeta(0, 1-n, 1/4)/2 + Zeta(0, 1-n, 3/4)/2 + Zeta(1-n)*(2^(-n)-1))): seq(a(n), n=0..24); # Peter Luschny, Jul 21 2020

Mathematica

a[0] = 0; a[n_] := -n*SeriesCoefficient[(2*E^x*(1 - E^x))/(1 + E^(2*x)), {x, 0, n-1}]*(n-1)!; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Jun 17 2014 *)

Formula

a(n) = 0, 0, followed by (period 4: repeat 1, 1, -1, -1)*A065619(n+2).

a(2n) = (-1)^(n+1)A009752(n). a(2n+1) = (-1)^n*A009843(n+1).

Extensions

New name by Peter Luschny, Jul 21 2020