%I #19 Sep 08 2022 08:45:15
%S 0,1,-1,-2,3,1,-4,1,3,-2,-1,1,0,1,-1,-2,3,1,-4,1,3,-2,-1,1,0,1,-1,-2,
%T 3,1,-4,1,3,-2,-1,1,0,1,-1,-2,3,1,-4,1,3,-2,-1,1,0,1,-1,-2,3,1,-4,1,3,
%U -2,-1,1,0,1,-1,-2,3,1,-4,1,3,-2,-1,1,0,1,-1,-2,3,1,-4,1,3,-2,-1,1,0,1,-1,-2,3,1,-4,1,3,-2,-1,1,0,1,-1,-2,3,1,-4,1,3,-2
%N G.f.: x*(1-x^2)/((1+x^2)*(1+x+x^2)).
%H G. C. Greubel, <a href="/A098554/b098554.txt">Table of n, a(n) for n = 0..1000</a>
%H G. I. Lehrer and G. B. Segal, <a href="https://doi.org/10.1007/PL00004831">Homology stability for classical regular semisimple varieties</a>, Math. Zeit., 236 (2001), 251-290; see Th. 7.12.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-1,-2,-1,-1).
%F Let b(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(0^(n-2k)-(-1)^(n-2k)). Then a(n) = b(n) - b(n-2), or a(n) = Sum_{j=0..n} b(n-j)*(binomial(1, j/2)*(-1)^(j/2)*(1+(-1)^j)/2). The g.f. is obtained from the g.f. x/(1+x) of 0^n-(-1)^n by applying the transformation G(x)->((1-x^2)/(1+x^2))G(x/(1+x^2)). - _Paul Barry_, Oct 26 2004
%F a(n) = (-1)^n*(A112553(n-1) - A112553(n-3)). - _R. J. Mathar_, Sep 27 2014
%F a(0)=0, a(1)=1, a(2)=-1, a(3)=-2, a(n) = a(n-1) - 2*a(n-2) - a(n-3) - a(n-4). - _Harvey P. Dale_, Jan 16 2016
%t CoefficientList[Series[x*(1-x^2)/((1+x^2)*(1+x+x^2)),{x,0,110}],x] (* or *) LinearRecurrence[{-1,-2,-1,-1},{0,1,-1,-2},110] (* _Harvey P. Dale_, Jan 16 2016 *)
%o (PARI) x='x+O('x^30); concat([0], Vec(x*(1-x^2)/((1+x^2)*(1+x+x^2)))) \\ _G. C. Greubel_, Jan 17 2018
%o (Magma) I:=[0,1,-1,-2]; [n le 4 select I[n] else -Self(n-1) - 2*Self(n-2) -Self(n-3) - Self(n-4): n in [1..30]]; // _G. C. Greubel_, Jan 17 2018
%K sign,easy
%O 0,4
%A _N. J. A. Sloane_, Oct 26 2004