A267313 Expansion of x*(-1 + 2*x + 3*x^2 - 2*x^3 + x^4)/((1 - x)^3*(1 + x + x^2)^2).

0, -1, 1, 4, 0, 5, 11, 4, 12, 21, 11, 22, 34, 21, 35, 50, 34, 51, 69, 50, 70, 91, 69, 92, 116, 91, 117, 144, 116, 145, 175, 144, 176, 209, 175, 210, 246, 209, 247, 286, 246, 287, 329, 286, 330, 375, 329, 376, 424, 375, 425, 476, 424, 477, 531, 476, 532, 589, 531, 590, 650, 589, 651, 714, 650, 715

Offset

0,4

Comments

First differences are -1, 2, 3, -4, 5, 6, -7, 8, 9, -10, 11, 12, ... - N. J. A. Sloane, May 20 2019

Links

N. J. A. Sloane, Table of n, a(n) for n = 0..20000

Ilya Gutkovskiy, Graphs of first N terms for various values of N

Index entries for linear recurrences with constant coefficients, signature (1,0,2,-2,0,-1,1).

Formula

G.f.: x*(-1 + 2*x + 3*x^2 - 2*x^3 + x^4)/((1 - x)^3*(1 + x + x^2)^2).

a(n) = Sum_{k = 0..n} (-1)^(k mod 3)*k.

a(n) = Sum_{k = 0..n} -(-1)^A010882(k)*k.

From G. C. Greubel, Feb 03 2016: (Start)

a(n+7) = a(n+6) + 2*a(n+4) - 2*a(n+3) - a(n+1) + a(n).

E.g.f.: (1/18)*exp(-x/2)*((3*x^2+6*x-4)*exp(3*x/2) + 4*(1-3*x)*cos(sqrt(3)*x/2) - 4*sqrt(3)*(1+x)*sin(sqrt(3)*x/2)). (End)

Example

a(0) = 0;
a(1) = 0 - 1 = -1;
a(2) = 0 - 1 + 2 = 1;
a(3) = 0 - 1 + 2 + 3 = 4;
a(4) = 0 - 1 + 2 + 3 - 4 = 0;
a(5) = 0 - 1 + 2 + 3 - 4 + 5 = 5;
a(6) = 0 - 1 + 2 + 3 - 4 + 5 + 6 = 11;
a(7) = 0 - 1 + 2 + 3 - 4 + 5 + 6 - 7 = 4;
a(8) = 0 - 1 + 2 + 3 - 4 + 5 + 6 - 7 + 8 = 12;
a(9) = 0 - 1 + 2 + 3 - 4 + 5 + 6 - 7 + 8 + 9 = 21, etc.

Mathematica

Table[Sum[(-1)^Mod[k, 3] k, {k, 0, n}], {n, 0, 65}]
LinearRecurrence[{1, 0, 2, -2, 0, -1, 1}, {0, -1, 1, 4, 0, 5, 11}, 65]
CoefficientList[Series[x (1 - 2 x - 3 x^2 + 2 x^3 - x^4) / (x^7 - x^6 - 2 x^4 + 2 x^3 + x - 1), {x, 0, 70}], x] (* Vincenzo Librandi, Jan 13 2016 *)

Prog

(PARI) Vec(x*(1-2*x-3*x^2+2*x^3-x^4)/(x^7-x^6-2*x^4+2*x^3+x-1) + O(x^100)) \\ Altug Alkan, Jan 25 2016

Crossrefs

Cf. A010882, A130472.

Sequence in context: A355921 A164108 A064520 * A108174 A134530 A351571

Adjacent sequences: A267310 A267311 A267312 * A267314 A267315 A267316

Keyword

sign,easy

Author

Ilya Gutkovskiy, Jan 13 2016

Status

approved