login
a(n+2) = a(n+1) + a(n) + A*t^n, with A = 1 and t = -2.
1

%I #38 Apr 19 2023 02:34:31

%S 0,1,2,1,7,0,23,-9,78,-59,275,-296,1003,-1341,3758,-5775,14367,-24176,

%T 55727,-99521,218350,-405459,861467,-1641144,3414627,-6615125,

%U 13576718,-26592839,54092743,-106717824,215810375,-427778361,861773838,-1713488171,3443252963

%N a(n+2) = a(n+1) + a(n) + A*t^n, with A = 1 and t = -2.

%C With these particular values, a(n) turns out to be Fibonacci(n) + (-2)^n.

%H Vincenzo Librandi, <a href="/A224508/b224508.txt">Table of n, a(n) for n = 0..1000</a>

%H C. N. Phadte and S. P. Pethe, <a href="http://ami.ektf.hu/uploads/papers/finalpdf/AMI_41_from205to210.pdf">On second order non homogeneous recurrence relation</a>, Annales Mathematicae et Infomaticae, 41 (2013), 205-210.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-1,3,2).

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

%t nn = 40; A = 1; t = -2; CoefficientList[Series[(x + x^2 (A - t))/((1 - x*t) (1 - x - x^2)), {x, 0, nn}], x] (* _T. D. Noe_, Sep 21 2013 *) (* or *)

%t LinearRecurrence[{-1,3,2}, {0,1,2}, 35] (* _Georg Fischer_, Jan 26 2022 *)

%Y Cf. A227200.

%K sign,easy

%O 0,3

%A _Chandrakant N Phadte_ and _Yeshwant Shivrai Valaulikar_, Sep 18 2013