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a(n) = -4*a(n-3) + 11*a(n-2) - a(n-1), a(0) = -5, a(1) = 39, a(2) = -110.
4

%I #18 Oct 05 2024 04:32:25

%S -5,39,-110,559,-1925,8514,-31925,133279,-518510,2112279,-8349005,

%T 33658114,-133946285,537581559,-2145623150,8594805439,-34346986325,

%U 137472338754,-549668410085,2199252081679,-8795493947630,35185940486439,-140733382237085,562960703378434

%N a(n) = -4*a(n-3) + 11*a(n-2) - a(n-1), a(0) = -5, a(1) = 39, a(2) = -110.

%C A153266(n) + a(n) = 4*A001519(n) (apart from initial terms). The generating floretion Z = X*Y with X = 1.5'i + 0.5i' + .25(ii + jj + kk + ee) and Y = 0.5'i + 1.5i' + .25(ii + jj + kk + ee).

%H Vincenzo Librandi, <a href="/A153267/b153267.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-1, 11, -4).

%F a(n) = 2*(-4)^(n+1) + (3/2+1/2*sqrt(5))^(n+1) + (3/2-1/2*sqrt(5))^(n+1).

%F G.f.: -(16*x^2-34*x+5) / ((4*x+1)*(x^2-3*x+1)). - _Colin Barker_, Jun 25 2014

%e a(4) = -1*559 + 11*(-110) - 4*(39) = -1925.

%t CoefficientList[Series[-(16 x^2 - 34 x + 5)/((4 x + 1) (x^2 - 3 x + 1)), {x, 0, 30}], x] (* _Vincenzo Librandi_, Jun 26 2014 *)

%t LinearRecurrence[{-1,11,-4},{-5,39,-110},30] (* _Harvey P. Dale_, Mar 02 2023 *)

%o (PARI) Vec(-(16*x^2-34*x+5)/((4*x+1)*(x^2-3*x+1)) + O(x^100)) \\ _Colin Barker_, Jun 25 2014

%Y Cf. A153266, A153265, A001519.

%K easy,sign

%O 0,1

%A _Creighton Dement_, Jan 02 2009