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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.

%I A243963 #18 Dec 25 2023 17:20:14
%S A243963 0,0,2,3,-8,-25,96,427,-2176,-12465,79360,555731,-4245504,-35135945,
%T A243963 313155584,2990414715,-30460116992,-329655706465,3777576173568,
%U A243963 45692713833379,-581777702256640,-7777794952988025,108932957168730112,1595024111042171723,-24370173276164456448
%N 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.
%C A243963 Previous name was: 0 followed by -(n+1)*A163747(n).
%C A243963 Difference table of a(n):
%C A243963 0,    0,    2,     3,    -8,   -25,...
%C A243963 0,    2,    1,   -11,   -17,   121,...
%C A243963 2,   -1,  -12,    -6,   138,   210,...
%C A243963 -3, -11,    6,   144,    72, -3144,...
%C A243963 -8,  17,  138,   -72, -3216, -1608,...
%C A243963 25, 121, -210, -3144,  1608,...
%C A243963 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.
%F A243963 a(n) = 0, 0, followed by (period 4: repeat 1, 1, -1, -1)*A065619(n+2).
%F A243963 a(2n) = (-1)^(n+1)A009752(n). a(2n+1) = (-1)^n*A009843(n+1).
%p A243963 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
%t A243963 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 *)
%Y A243963 Cf. A132049, A065619.
%K A243963 sign
%O A243963 0,3
%A A243963 _Paul Curtz_, Jun 16 2014
%E A243963 New name by _Peter Luschny_, Jul 21 2020