A086344 a(n) = -2*a(n-1) + 4*a(n-2), a(0) = 1, a(1) = 0.

1, 0, 4, -8, 32, -96, 320, -1024, 3328, -10752, 34816, -112640, 364544, -1179648, 3817472, -12353536, 39976960, -129368064, 418643968, -1354760192, 4384096256, -14187233280, 45910851584, -148570636288, 480784678912, -1555851902976, 5034842521600, -16293092655104

Offset: 0

Author

Paul Barry, Jul 17 2003

Keywords

• easy
• sign

Comments

Inverse binomial transform of (1,1,5,5,25,25,.....).

The absolute values are the constant terms of the reduction by x^2->x+1 of the polynomial p(n,x) given for d=sqrt(x+1) by p(n,x)=((x+d)^n-(x-d)^n)/(2d), for n>=1. The coefficient of x under this reduction is given by A103435. See A192232 for a discussion of reduction. - Clark Kimberling, Jun 29 2011

Links

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

Programs

Maple

seq((-2)^n * combinat:-fibonacci(n-1), n = 0 .. 100); # Robert Israel, Oct 02 2014

Mathematica

LinearRecurrence[{-2,4},{1,0},40] (* Harvey P. Dale, Oct 10 2018 *)

Formula

G.f.: (1+2*x)/((1+(1+sqrt(5))*x)(1+(1-sqrt(5))*x)) = ( -1-2*x ) / ( -1-2*x+4*x^2 ).

E.g.f.: exp(-x)*(cosh(sqrt(5)*x)+sinh(sqrt(5)*x)/sqrt(5)).

a(n)=(sqrt(5)-1)^n*(sqrt(5)/10+1/2)+(-sqrt(5)-1)^n*(1/2-sqrt(5)/10).

(-1)^n*a(n) = A063727(n) - 2*A063727(n-1). - R. J. Mathar, Jul 19 2012

(-1)^n*a(n) = sum(k=0..n, binomial(n,k)*(F(n+1)-F(n))), F(n) Fibonacci number A000045. - Peter Luschny, Oct 01 2014

a(n) = (-2)^n *A000045(n-1). - Robert Israel, Oct 02 2014