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%I #19 Dec 14 2024 07:40:57
%S 1,-1,2,0,8,16,64,192,640,2048,6656,21504,69632,225280,729088,2359296,
%T 7634944,24707072,79953920,258736128,837287936,2709520384,8768192512,
%U 28374466560,91821703168,297141272576,961569357824,3111703805952,10069685043200,32586185310208,105451110793216
%N a(n) = 2^(n-1)*Fibonacci(n-3), n >= 0.
%C This sequence gives the elements M^n(2, 2) of the matrix M = [[3, 1], [1, -1]].
%C Motivation to look into these matrix powers came from A319053. M^n[1, 1] = A063782 and M^n(1, 2) = M^n(2, 1) = A085449(n). Proof by Cayley-Hamilton, using S(n, -I) = (-I)^n*F(n+1), and S(n, x) from A049310 and F = A000045.
%C For a similar signed sequence see A087205.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,4).
%F a(n) = 2^(n-1)*F(n-3), n >= 0, with F = A000045 with F(-1) = 1, F(-2) = -1 and F(-3) = 1.
%F G.f.: (1-3*x)/(1-2*x-(2*x)^2).
%F a(n) = 2*(a(n-1) + 2*a(n-2)), n >= 2, with a(0) = 1 and a(1) = -1.
%F a(n) = 2^(n-1)*(phi^(n-3) - (1 - phi)^(n-3))/(2*phi - 1) with the golden section phi = A001622.
%t Table[2^(n-1) Fibonacci[n-3], {n, 0, 40}] (* _Vincenzo Librandi_, Oct 09 2018 *)
%t LinearRecurrence[{2,4},{1,-1},40] (* _Harvey P. Dale_, Mar 29 2020 *)
%o (Magma) [2^(n-1)*Fibonacci(n-3): n in [0..30]]; // _Vincenzo Librandi_, Oct 09 2018
%Y Cf. A000045, A001622, A049310, A085449, A063782, A087205, A319053.
%K sign,easy
%O 0,3
%A _Wolfdieter Lang_, Oct 09 2018