A240485 -id:A240485 - cp's OEIS frontend

A240677 a(n) = 6Zeta(1-n)n*(2^n-1) - Zeta(-n)(n+1)(2^(n+2)-2), for n = 0 the limit is understood.

1, -2, -3, -1, 3, 3, -9, -17, 51, 155, -465, -2073, 6219, 38227, -114681, -929569, 2788707, 28820619, -86461857, -1109652905, 3328958715, 51943281731, -155829845193, -2905151042481, 8715453127443

Offset: 0

Paul Curtz, Apr 10 2014

sign

G2(m, n), difference table of a(n):
1,   -2, -3, -1,   3,   3, -9, -17, 51,...
-3,  -1,  2,  4,   0, -12, -8,  68,...
2,    3,  2, -4, -12,   4, 76,...
1,   -1, -6, -8,  16,  72,...
-2,  -5, -2, 24,  56,...
-3,   3, 26, 32,...
6,   23,  6,...
17, -17,...
-34,...
etc.
The main diagonal G2(n,n) = 1, -1, 2, -8,... is essentially a signed version of A005439.
The first upper diagonal is the main diagonal multiplied by -2. G2(n, n+1) = -2*G2(n, n).
G2(m, n) = G2(m, n-1) + G2(m+1, n-1).
a(n) = (-1)^n*b(n) of A240485(n).
Inverse binomial transform: (-1)^n*A240485(n).
a(n) and A240485(n) are reciprocal. Like for instance (-1)^n and 2^n.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A240485.

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

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 *)

x = 'x+O('x^66);
A = -2*exp(x)*(2*x+exp(x)*(3*x-1)-1)/(exp(x)+1)^2;
Vec( serlaplace(A) )  /* Peter Luschny, Apr 10 2014 */

a(n) = 3*A226158(n) - A226158(n+1).
a(n+3) = -A001469(n+1).
a(2n+4) = -3*a(2n+3).
a(n) = A240485(n) + 5*A226158(n).
E.g.f.: -2*exp(x)*(2*x+exp(x)*(3*x-1)-1)/(exp(x)+1)^2. - Peter Luschny, Apr 10 2014

cp's OEIS Frontend

A240677 a(n) = 6Zeta(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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cp's OEIS Frontend

A240677 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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A240677 a(n) = 6Zeta(1-n)n*(2^n-1) - Zeta(-n)(n+1)(2^(n+2)-2), for n = 0 the limit is understood.