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A102899
a(n) = ceiling(n/3)^2 - floor(n/3)^2.
4
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.
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
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Jan 17 2005
STATUS
approved