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A243963
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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.
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1
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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
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OFFSET
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0,3
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COMMENTS
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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.
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LINKS
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FORMULA
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a(n) = 0, 0, followed by (period 4: repeat 1, 1, -1, -1)*A065619(n+2).
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MAPLE
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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
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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