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a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) with initial terms 1,0,-2,-1,0.
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%I #24 Dec 14 2023 05:16:09

%S 1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,

%T 0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,

%U 0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2,-1,0,0,1,0,-2

%N a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) with initial terms 1,0,-2,-1,0.

%C Partial sums of A101675. Partial sums are A101677.

%C Sequence has period 6. - _Ralf Stephan_, May 16 2007

%H G. C. Greubel, <a href="/A101676/b101676.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,-1,1,-1,1).

%F G.f.: (1 - x - x^2)/((1 - x)*(1 + x^2 + x^4)).

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

%t LinearRecurrence[{1, -1, 1, -1, 1},{1, 0, -2, -1, 0},105] (* _Ray Chandler_, Sep 03 2015 *)

%o (PARI) x='x+O('x^100); Vec((1-x-x^2)/((1-x)*(1+x^2+x^4))) \\ _G. C. Greubel_, Sep 07 2018

%o (Magma) I:=[1, 0, -2, -1, 0]; [n le 5 select I[n] else Self(n-1) - Self(n-2) +Self(n-3) -Self(n-4) +Self(n-5): n in [1..100]]; // _G. C. Greubel_, Sep 07 2018

%K sign,easy

%O 0,3

%A _Paul Barry_, Dec 11 2004