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a(n) = 6*Zeta(1-n)*n*(2^n-1) - Zeta(-n)*(n+1)*(2^(n+2)-2), for n = 0 the limit is understood.
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%I #19 Apr 25 2016 11:45:33

%S 1,-2,-3,-1,3,3,-9,-17,51,155,-465,-2073,6219,38227,-114681,-929569,

%T 2788707,28820619,-86461857,-1109652905,3328958715,51943281731,

%U -155829845193,-2905151042481,8715453127443

%N a(n) = 6*Zeta(1-n)*n*(2^n-1) - Zeta(-n)*(n+1)*(2^(n+2)-2), for n = 0 the limit is understood.

%C G2(m, n), difference table of a(n):

%C 1, -2, -3, -1, 3, 3, -9, -17, 51,...

%C -3, -1, 2, 4, 0, -12, -8, 68,...

%C 2, 3, 2, -4, -12, 4, 76,...

%C 1, -1, -6, -8, 16, 72,...

%C -2, -5, -2, 24, 56,...

%C -3, 3, 26, 32,...

%C 6, 23, 6,...

%C 17, -17,...

%C -34,...

%C etc.

%C The main diagonal G2(n,n) = 1, -1, 2, -8,... is essentially a signed version of A005439.

%C The first upper diagonal is the main diagonal multiplied by -2. G2(n, n+1) = -2*G2(n, n).

%C G2(m, n) = G2(m, n-1) + G2(m+1, n-1).

%C a(n) = (-1)^n*b(n) of A240485(n).

%C Inverse binomial transform: (-1)^n*A240485(n).

%C a(n) and A240485(n) are reciprocal. Like for instance (-1)^n and 2^n.

%H Vincenzo Librandi, <a href="/A240677/b240677.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) = 3*A226158(n) - A226158(n+1).

%F a(n+3) = -A001469(n+1).

%F a(2n+4) = -3*a(2n+3).

%F a(n) = A240485(n) + 5*A226158(n).

%F E.g.f.: -2*exp(x)*(2*x+exp(x)*(3*x-1)-1)/(exp(x)+1)^2. - _Peter Luschny_, Apr 10 2014

%p A240677 := n -> `if`(n=0, 1, 6*Zeta(1-n)*n*(2^n-1) - Zeta(-n)*(n+1)*(2^(n+2)-2)); seq(A240677(n), n=0..24); # _Peter Luschny_, Apr 11 2014

%t g[0] = 0; g[1] = -1; g[n_] := n*EulerE[n - 1, 0]; a[n_] := 3*g[n] - g[n + 1]; Table[a[n], {n, 0, 25}] (* _Jean-François Alcover_, Apr 10 2014 *)

%o (PARI)

%o x = 'x+O('x^66);

%o A = -2*exp(x)*(2*x+exp(x)*(3*x-1)-1)/(exp(x)+1)^2;

%o Vec( serlaplace(A) ) /* _Peter Luschny_, Apr 10 2014 */

%Y Cf. A240485.

%K sign

%O 0,2

%A _Paul Curtz_, Apr 10 2014