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a(1)=-1; a(2)=-1; a(3)=-2; a(n) = 4*a(n-1) - 3*a(n-2) for n >= 4.
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%I #39 Jan 27 2020 01:36:03

%S -1,-1,-2,-5,-14,-41,-122,-365,-1094,-3281,-9842,-29525,-88574,

%T -265721,-797162,-2391485,-7174454,-21523361,-64570082,-193710245,

%U -581130734,-1743392201,-5230176602,-15690529805,-47071589414,-141214768241,-423644304722,-1270932914165,-3812798742494

%N a(1)=-1; a(2)=-1; a(3)=-2; a(n) = 4*a(n-1) - 3*a(n-2) for n >= 4.

%C Essentially the same as A124302: a(n) = -A124302(n-1).

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-3).

%F From _Philippe Deléham_, Dec 05 2008 [corrected by _Sergei N. Gladkovskii_, Dec 10 2012]: (Start)

%F a(n) = -A124302(n-1).

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

%F G.f.: -1-x*G(0) where G(k) = 1 + 2*x/(1 - 2*x - x*(1-2*x)/(x + (1-2*x)*2/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Dec 10 2012

%p a[1]:=-1: a[2]:=-1: a[3]:=-2: for n from 4 to 29 do a[n]:=4*a[n-1]-3*a[n-2] od: seq(a[n],n=1..29);

%t M = {{1, -1, 0}, {-1, 2, -1}, {0, -1, 1}} v[1] = {1, 0, 0} v[n_] := v[n] = M.v[n - 1] a1 = Table[ -v[n][[1]], {n, 1, 50}]

%Y Cf. A124302.

%K sign,easy,less

%O 1,3

%A _Roger L. Bagula_, Oct 02 2006

%E Edited by _N. J. A. Sloane_, Oct 08 2006