A115360 Period 6: repeat [1,-1,1,0,0,0].

1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0, 1, -1, 1

Offset

0,1

Comments

Diagonal sums of number triangle A115359.

The partial sum operator applied twice gives A001399. - Gregory L. Simay, Sep 30 2017

Links

Table of n, a(n) for n=0..80.

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

Formula

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

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

a(n) = cos(2*Pi*n/3)/3 - sin(2*Pi*n/3)/sqrt(3) + cos(Pi*n)/2 + 1/6.

a(n) = Sum_{k=0..floor(n/2)} if(n-k=k, 1, 0) OR if(n-k=2k+1, -1, 0).

a(n) = (1/2)*((-1)^n + i^(4-2*(n mod 3))), where i=sqrt(-1). - Bruno Berselli, Oct 31 2012

a(n) = (floor(n/6) - floor((n-3)/6))*(-1)^n. - Wesley Ivan Hurt, Sep 08 2015

|a(n)-a(n-1)| = A164965(n). a(n)+a(n-1) = A079978(n). - R. J. Mathar, Aug 11 2021

a(n) = sign((n-4) mod 3) - sign((n-4) mod 2). - Wesley Ivan Hurt, Feb 04 2022

Mathematica

LinearRecurrence[{-1, 0, 1, 1}, {1, -1, 1, 0}, 100] (* Vincenzo Librandi, Sep 09 2015 *)

Prog

(Magma) &cat[[1, -1, 1, 0, 0, 0]: n in [0..15]]; // Vincenzo Librandi, Sep 09 2015

Crossrefs

Cf. A001399, A115359.

Sequence in context: A195053 A267136 A117908 * A088911 A179763 A257341

Adjacent sequences: A115357 A115358 A115359 * A115361 A115362 A115363

Keyword

sign,easy

Author

Paul Barry, Jan 21 2006

Status

approved