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%I #28 Dec 10 2022 01:29:32
%S 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,
%T 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,
%U 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
%N a(n) = ceiling(n/3)^2 - floor(n/3)^2.
%C 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
%D Maria Paola Bonacina and Nachum Dershowitz, Canonical Inference for Implicational Systems, in Automated Reasoning, Lecture Notes in Computer Science, Volume 5195/2008, Springer-Verlag.
%H G. C. Greubel, <a href="/A102899/b102899.txt">Table of n, a(n) for n = 0..5000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).
%F G.f.: x*(1+x+x^3+x^4)/(1-2*x^3+x^6).
%F a(n) = A011655(n)*A004396(n).
%F a(n) = (2/3)*floor((2*n+1)/3)*(1-cos(2*Pi*n/3)).
%F From _M. F. Hasler_, Dec 13 2007: (Start)
%F a(n) = |A120691(n+1)| for n>0.
%F a(n) = ([n/3]*2 + 1)*dist(n,3Z). (End)
%F 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
%F a(n) = 2*a(n-3) - a(n-6), for n > 5. - _Chai Wah Wu_, Jul 27 2022
%t LinearRecurrence[{0,0,2,0,0,-1}, {0,1,1,0,3,3}, 90] (* _G. C. Greubel_, Dec 09 2022 *)
%o (PARI) A102899(n)=(n\3*2+1)*(0<n%3) \\ _M. F. Hasler_, Dec 13 2007
%o (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
%o (SageMath)
%o def A102899(n): return (1+2*(n//3))*((n%3)>0)
%o [A102899(n) for n in range(91)] # _G. C. Greubel_, Dec 09 2022
%Y Cf. A003417, A004396, A011655, A120691, A353314, A353327.
%K easy,nonn
%O 0,5
%A _Paul Barry_, Jan 17 2005
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