A099496 a(n) = (-1)^n * Fibonacci(2*n+1).

1, -2, 5, -13, 34, -89, 233, -610, 1597, -4181, 10946, -28657, 75025, -196418, 514229, -1346269, 3524578, -9227465, 24157817, -63245986, 165580141, -433494437, 1134903170, -2971215073, 7778742049, -20365011074, 53316291173, -139583862445, 365435296162, -956722026041

Offset: 0

Author

Paul Barry, Oct 19 2004

Keywords

• easy
• sign

Comments

With interpolated zeros, a Chebyshev transform of A056594, which has g.f. 1/(1+x^2). The image of G(x) under the Chebyshev transform is (1/(1+x^2))G(x/(1+x^2)).

a(n) is the ceiling of the inverse fractional error in approximating phi, the golden section, by the ratio of two successive terms in the Fibonacci series. - Adam Helman (helman(AT)san.rr.com), May 09 2010

Examples

			a(3) = (-1)^3 * Fibonacci(2 * 3 + 1) = -Fibonacci(7) = -13. - _Indranil Ghosh_, Feb 04 2017
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..2386
• Tanya Khovanova, Recursive Sequences
• Index entries for linear recurrences with constant coefficients, signature (-3,-1).

Crossrefs

Cf. A056594, A122367.

Programs

Magma

[(-1)^n*Fibonacci(2*n+1): n in [0..40]]; // Vincenzo Librandi, Jul 04 2015

Maple

seq((-1)^n*combinat:-fibonacci(2*n+1), n=0 .. 100); # Robert Israel, Jul 02 2015

Mathematica

Table[(-1)^n Fibonacci[2 n + 1], {n, 0, 30}] (* Harvey P. Dale, Aug 22 2016 *)

Formula

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

a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)(-1)^k*cos((n-2k)*Pi/2) (with interpolated zeros);

a(n) = Fibonacci(n+1)(-1)^(n/2)(1 + (-1)^n)/2 (with interpolated zeros).

a(n) = -3*a(n-1) - a(n-2), a(0)=1, a(1)=-2. - Philippe Deléham, Nov 03 2008

From Adam Helman (helman(AT)san.rr.com), May 09 2010: (Start)

a(n) = ceiling( phi / (Fibonacci(n+1)/Fibonacci(n) - phi) ).

An exact expression for the inverse fractional error is phi / (Fibonacci(n+1)/Fibonacci(n) - phi) = (phi/sqrt(5)) * ((-1)^n *(phi^2n) - 1). (End)

a(n) = (-1)^n*A122367(n). - R. J. Mathar, Jul 23 2010