A133131 a(4n) = 3n+1, a(4n+1) = -3n, a(4n+2) = -3n-3, a(4n+3) = 3n+2.

1, 0, -3, 2, 4, -3, -6, 5, 7, -6, -9, 8, 10, -9, -12, 11, 13, -12, -15, 14, 16, -15, -18, 17, 19, -18, -21, 20, 22, -21, -24, 23, 25, -24, -27, 26, 28, -27, -30, 29, 31, -30, -33, 32, 34, -33, -36, 35, 37, -36, -39, 38, 40, -39, -42, 41, 43, -42, -45, 44, 46, -45, -48, 47, 49, -48, -51, 50, 52, -51, -54, 53, 55, -54, -57, 56

Offset

0,3

Comments

Second differences of 0, 0, 1, 2, 0, 0, 3, 4 .. .

Links

Table of n, a(n) for n=0..75.

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

Formula

G.f.: (1+x^4+x^3-x^2+x)/((x+1)*(x^2+1)^2). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009

a(n) = (-2*(-1)^n+(7+6*n+3*(-1)^n)*cos(n*Pi/2)+(13-6*n+9*(-1)^n)*sin(n*Pi/2) )/8. - Wesley Ivan Hurt, May 07 2021

a(n) = - a(n-1) - 2*a(n-2) - 2*a(n-3) - a(n-4) - a(n-5). - Wesley Ivan Hurt, Feb 14 2026

Mathematica

CoefficientList[Series[(1 + x^4 + x^3 - x^2 + x)/((x + 1)*(x^2 + 1)^2), {x, 0, 80}], x]
LinearRecurrence[{-1, -2, -2, -1, -1}, {1, 0, -3, 2, 4}, 80] (* Wesley Ivan Hurt, Feb 14 2026 *)

Crossrefs

Sequence in context: A335507 A372826 A195459 * A261985 A026923 A058708

Adjacent sequences: A133128 A133129 A133130 * A133132 A133133 A133134

Keyword

sign

Author

Paul Curtz, Sep 20 2007

Status

approved