A319196 a(n) = 2^(n-1)*Fibonacci(n-3), n >= 0.

1, -1, 2, 0, 8, 16, 64, 192, 640, 2048, 6656, 21504, 69632, 225280, 729088, 2359296, 7634944, 24707072, 79953920, 258736128, 837287936, 2709520384, 8768192512, 28374466560, 91821703168, 297141272576, 961569357824, 3111703805952, 10069685043200, 32586185310208, 105451110793216

Offset

0,3

Comments

This sequence gives the elements M^n(2, 2) of the matrix M = [[3, 1], [1, -1]].

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.

For a similar signed sequence see A087205.

Links

Table of n, a(n) for n=0..30.

Index entries for linear recurrences with constant coefficients, signature (2,4).

Formula

a(n) = 2^(n-1)*F(n-3), n >= 0, with F = A000045 with F(-1) = 1, F(-2) = -1 and F(-3) = 1.

G.f: ((1-3*x)/(1- 2*x - (2*x)^2).

a(n) = 2*(a(n-1) + 2*a(n-2)), n >= 2, with a(0) = 1 and a(1) = -1.

a(n) = 2^(n-1)*(phi^(n-3) - (1 - phi)^(n-3))/(2*phi - 1) with the golden section phi = A001622.

Mathematica

Table[2^(n-1) Fibonacci[n-3], {n, 0, 40}] (* Vincenzo Librandi, Oct 09 2018 *)
LinearRecurrence[{2, 4}, {1, -1}, 40] (* Harvey P. Dale, Mar 29 2020 *)

Prog

(Magma) [2^(n-1)*Fibonacci(n-3): n in [0..30]]; // Vincenzo Librandi, Oct 09 2018

Crossrefs

Cf. A000045, A001622, A049310, A085449, A063782, A087205, A319053.

Sequence in context: A128063 A171552 A087205 * A308536 A241682 A076687

Adjacent sequences: A319193 A319194 A319195 * A319197 A319198 A319199

Keyword

sign,easy

Author

Wolfdieter Lang, Oct 09 2018

Status

approved