A118405 Row sums of triangle A118404.

1, 0, 0, -2, 4, -6, 12, -26, 52, -102, 204, -410, 820, -1638, 3276, -6554, 13108, -26214, 52428, -104858, 209716, -419430, 838860, -1677722, 3355444, -6710886, 13421772, -26843546, 53687092, -107374182, 214748364, -429496730, 858993460, -1717986918, 3435973836, -6871947674

Offset

0,4

Links

Table of n, a(n) for n=0..35.

Index entries for linear recurrences with constant coefficients, signature (-2,-1,-2).

Formula

G.f.: A(x) = (1+x)^2/(1+x^2)/(1+2*x).

From Paul Curtz, Oct 31 2018: (Start)

a(n) + a(n+2) = 1, -2, 4, -8, ... = A122803(n).

a(2n+2) = -2*a(2n+1) = 4*A015521(n). (End)

a(n) = -2a(n-1) - a(n-2) - 2a(n-3). - Charles R Greathouse IV, Nov 06 2018

5*a(n) = (-2)^n + 2*A056594(n-1) + 4*A056594(n). - R. J. Mathar, Jan 27 2021

Maple

seq(coeff(series((1+x)^2/(1+x^2)/(1+2*x), x, n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Oct 31 2018

Mathematica

Total /@ Table[SeriesCoefficient[(-1)^k/((1 + x^2) (1 + x)^(k - 1)), {x, 0, n - k}], {n, 0, 35}, {k, 0, n}] (* Michael De Vlieger, Oct 31 2018 *)
LinearRecurrence[{-2, -1, -2}, {1, 0, 0}, 40] (* Harvey P. Dale, Aug 31 2020 *)

Prog

(PARI) a(n)=polcoeff((1+x)^2/(1+x^2)/(1+2*x+x*O(x^n)), n, x)
(PARI) a(n)=([0, 1, 0; 0, 0, 1; -2, -1, -2]^n*[1; 0; 0])[1, 1] \\ Charles R Greathouse IV, Nov 06 2018

Crossrefs

Cf. A118404, A118406.

Cf. A015521, A122803.

Sequence in context: A355699 A340638 A177905 * A332345 A095912 A143474

Adjacent sequences: A118402 A118403 A118404 * A118406 A118407 A118408

Keyword

sign,easy

Author

Paul D. Hanna, Apr 27 2006

Status

approved