A240485 a(n) = -Zeta(1-n)*n*(2^(n+1) - 4) - Zeta(-n)*(n+1)*(2^(n+2) - 2), for n = 0 the limit is understood.

1, 3, 2, -1, -2, 3, 6, -17, -34, 155, 310, -2073, -4146, 38227, 76454, -929569, -1859138, 28820619, 57641238, -1109652905, -2219305810, 51943281731, 103886563462, -2905151042481, -5810302084962, 191329672483963, 382659344967926, -14655626154768697

Offset: 0

Author

Paul Curtz, Apr 06 2014

Keywords

• sign

Comments

Let G(m, n) denote the difference table of a(n):

1, 3, 2, -1, -2, 3, 6, -17, -34,...

2, -1, -3, -1, 5, 3, -23, -17,...

-3, -2, 2, 6, -2, -26, 6,...

1, 4, 4, -8, -24, 32,...

3, 0, -12, -16, 56,...

-3, -12, -4, 72,...

-9, 8, 76,...

17, 68,...

51,...

a(n) = G(0, n).

The main diagonal G(n, n) = 1, -1, 2, -8, 56, -608,... is essentially a signed version of A005439.

The first upper diagonal is the main diagonal multiplied by 3. G(n, n+1) = 3*G(n, n).

G(m, n) = G(m, n-1) + G(m+1,n-1).

Inverse binomial transform: after 1, 2, -3, A110501(n+1) is interleaved with 3*A110501(n+1), signed two by two. I. e. b(n) = 1, 2, -3, 1, 3, -3, -9, 17, 51,... . a(n+2) + b(n+2) = -1, 0, 1, 0, -3, 0, 17,... = A226158(n+2).

This is particular to the Genocchi numbers. If the first upper diagonal is proportional to the main diagonal (1, -1, 2, -8,...), the sequence and the inverse binomial transform are simply connected to the Genocchi numbers.

Crossrefs

Cf. A001469, A005439, A110501, A226158, A230324.

Programs

Maple

A240485 := proc(n) if n = 0 then 1 elif n = 1 then 3 else
m := 2*iquo(n-1, 2) + 2; -2^irem(n-1, 2)*m*euler(m-1, 0) fi end:
seq(A240485(n), n=0..27); # Peter Luschny, Apr 09 2014

Mathematica

a[n_] := Which[n == 0, 1, n == 1, 3, True, m = 2*Quotient[n-1, 2]+2; -2^Mod[n-1, 2]*m*EulerE[m-1, 0]]; Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Apr 09 2014, after Peter Luschny *)

Sage

def A240485(n):
    if n < 3: return [1,3,2][n]
    m = 2*((n+1)//2)
    b = 2*(1-2^m)*bernoulli(m)
    if is_even(n): b = 2*b
    return (-1)^ceil((n^2+1)/2)*b
[A240485(n) for n in (0..24)]  # Peter Luschny, Apr 08 2014

Formula

a(2*n+1) = a(2*n+2)/2 for n > 0.

-a(2*n+2)/2 = A226158(2*n+2) = A001469(n+1) = (2*n+2)*E(2*n+1, 0) where E(n, x) are the Euler polynomials.

a(n) = -2*A226158(n) - A226158(n+1).

E.g.f.: (2*exp(x)*(3*x+exp(x)*(2*x+1)+1))/(exp(x)+1)^2. - Peter Luschny, Apr 10 2014