|
%I #24 Oct 10 2018 17:58:40
%S 1,0,4,-8,32,-96,320,-1024,3328,-10752,34816,-112640,364544,-1179648,
%T 3817472,-12353536,39976960,-129368064,418643968,-1354760192,
%U 4384096256,-14187233280,45910851584,-148570636288,480784678912,-1555851902976,5034842521600,-16293092655104
%N a(n) = -2*a(n-1) + 4*a(n-2), a(0) = 1, a(1) = 0.
%C Inverse binomial transform of (1,1,5,5,25,25,.....).
%C 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
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (-2,4).
%F G.f.: (1+2*x)/((1+(1+sqrt(5))*x)(1+(1-sqrt(5))*x)) = ( -1-2*x ) / ( -1-2*x+4*x^2 ).
%F E.g.f.: exp(-x)*(cosh(sqrt(5)*x)+sinh(sqrt(5)*x)/sqrt(5)).
%F a(n)=(sqrt(5)-1)^n*(sqrt(5)/10+1/2)+(-sqrt(5)-1)^n*(1/2-sqrt(5)/10).
%F (-1)^n*a(n) = A063727(n) - 2*A063727(n-1). - _R. J. Mathar_, Jul 19 2012
%F (-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
%F a(n) = (-2)^n *A000045(n-1). - _Robert Israel_, Oct 02 2014
%p seq((-2)^n * combinat:-fibonacci(n-1), n = 0 .. 100); # _Robert Israel_, Oct 02 2014
%t LinearRecurrence[{-2,4},{1,0},40] (* _Harvey P. Dale_, Oct 10 2018 *)
%K easy,sign
%O 0,3
%A _Paul Barry_, Jul 17 2003
|
