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