A112030 a(n) = (2 + (-1)^n) * (-1)^floor(n/2).

3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1

Offset

0,1

Comments

The fractions A112031(n)/A112032(n) give the partial sums of a(n)/floor((n+4)/2).

Sum of the two Cartesian coordinates from the image of the point (2,1) after n 90-degree counterclockwise rotations about the origin. - Wesley Ivan Hurt, Jul 06 2013

Links

Table of n, a(n) for n=0..77.

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

Formula

a(n) = A010684(n+1) * (-1)^floor(n/2).

O.g.f.: (3+x)/(1+x^2). - R. J. Mathar, Jan 09 2008

Maple

A112030 := proc(n)
        (2 + (-1)^n) * (-1)^floor(n/2) ;
end proc: # R. J. Mathar, Jul 09 2013

Mathematica

LinearRecurrence[{0, -1}, {3, 1}, 100] (* Jean-François Alcover, Nov 24 2020 *)

Prog

(PARI) a(n)=[3, 1, -3, -1][n%4+1] \\ Charles R Greathouse IV, Aug 21 2011
(Python)
def A112030(n): return (3, 1, -3, -1)[n&3] # Chai Wah Wu, Jan 31 2023

Crossrefs

Cf. A010684, A016116, A112031, A112032, A112033.

Sequence in context: A063062 A066056 A153284 * A010684 A176040 A125768

Adjacent sequences: A112027 A112028 A112029 * A112031 A112032 A112033

Keyword

sign,easy

Author

Reinhard Zumkeller, Aug 27 2005

Status

approved