A102899 a(n) = ceiling(n/3)^2 - floor(n/3)^2.

0, 1, 1, 0, 3, 3, 0, 5, 5, 0, 7, 7, 0, 9, 9, 0, 11, 11, 0, 13, 13, 0, 15, 15, 0, 17, 17, 0, 19, 19, 0, 21, 21, 0, 23, 23, 0, 25, 25, 0, 27, 27, 0, 29, 29, 0, 31, 31, 0, 33, 33, 0, 35, 35, 0, 37, 37, 0, 39, 39, 0, 41, 41, 0, 43, 43, 0, 45, 45, 0, 47, 47, 0, 49, 49, 0, 51, 51, 0, 53, 53, 0

Offset

0,5

Comments

If n is a multiple of 3, then a(n) = 0, and if n is of the form 3k+r, with r = 1 or 2, then a(n) = 2*k + 1. - Antti Karttunen, Apr 14 2022

References

Maria Paola Bonacina and Nachum Dershowitz, Canonical Inference for Implicational Systems, in Automated Reasoning, Lecture Notes in Computer Science, Volume 5195/2008, Springer-Verlag.

Links

G. C. Greubel, Table of n, a(n) for n = 0..5000

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

Formula

G.f.: x*(1+x+x^3+x^4)/(1-2*x^3+x^6).

a(n) = A011655(n)*A004396(n).

a(n) = (2/3)*floor((2*n+1)/3)*(1-cos(2*Pi*n/3)).

From M. F. Hasler, Dec 13 2007: (Start)

a(n) = |A120691(n+1)| for n>0.

a(n) = ([n/3]*2 + 1)*dist(n,3Z). (End)

a(n) = 2*sin(n*Pi/3)*(4*n*sin(n*Pi/3)-sqrt(3)*cos(n*Pi))/9. - Wesley Ivan Hurt, Sep 24 2017

a(n) = 2*a(n-3) - a(n-6), for n > 5. - Chai Wah Wu, Jul 27 2022

Mathematica

LinearRecurrence[{0, 0, 2, 0, 0, -1}, {0, 1, 1, 0, 3, 3}, 90] (* G. C. Greubel, Dec 09 2022 *)

Prog

(PARI) A102899(n)=(n\3*2+1)*(0<n%3) \\ M. F. Hasler, Dec 13 2007
(Magma) I:=[0, 1, 1, 0, 3, 3]; [n le 6 select I[n] else 2*Self(n-3) - Self(n-6): n in [1..91]]; // G. C. Greubel, Dec 09 2022
(SageMath)
def A102899(n): return (1+2*(n//3))*((n%3)>0)
[A102899(n) for n in range(91)] # G. C. Greubel, Dec 09 2022

Crossrefs

Cf. A003417, A004396, A011655, A120691, A353314, A353327.

Sequence in context: A256119 A217552 A128046 * A353327 A072689 A021972

Adjacent sequences: A102896 A102897 A102898 * A102900 A102901 A102902

Keyword

easy,nonn

Author

Paul Barry, Jan 17 2005

Status

approved