A192351 - OEIS

%I #13 Jan 01 2018 13:17:00

%S 0,1,2,20,56,320,1120,5312,20608,90880,368640,1577984,6522880,

%T 27578368,114909184,483328000,2020573184,8480555008,35502817280,

%U 148874461184,623609118720,2614000353280,10952269365248,45901678641152,192340840939520

%N Coefficient of x in the reduction (by x^2->x+1) of polynomial p(n,x) identified in Comments.

%C To define the polynomials p(n,x), let d=sqrt(x+5); then p(n,x)=(1/2)((x+d)^n+(x-d)^n).  These are similar to polynomials at A161516.

%C For an introduction to reductions of polynomials by substitutions such as x^2->x+1, see A192232.

%H Robert Israel, <a href="/A192351/b192351.txt">Table of n, a(n) for n = 0..1606</a>

%F Conjecture: a(n) = 2*a(n-1)+12*a(n-2)-8*a(n-3)-16*a(n-4). G.f.: x*(4*x^2+1) / (16*x^4+8*x^3-12*x^2-2*x+1). [_Colin Barker_, Jan 17 2013]

%F Confirmation of conjecture by _Robert Israel_, Jan 01 2018: (Start)

%F The polynomials p(n,x) have g.f. G(z) = (1-x*z)/(1-2*x*z-5*z^2-x*z^2+x^2*z^2).

%F The reductions mod x^2-x-1 have g.f. g(z) = (1+x*z-2*z-6*z^2+4*x*z^3)/(1-2*z-12*z^2+8*z^3+16*z^4):

%F note that the numerator of G(z)-g(z) is divisible by x^2-x-1. (End)

%e The first four polynomials p(n,x) and their reductions are as follows:

%e p(0,x)=1 -> 1

%e p(1,x)=x -> x

%e p(2,x)=5+x+x^2 -> 6+2x

%e p(3,x)=15x+3x^2+x^3 -> 4+20x.

%e From these, we read

%e A192350=(1,0,6,4,...) and A192351=(0,1,2,20...)

%p f:= gfun:-rectoproc({a(n) = 2*a(n-1)+12*a(n-2)-8*a(n-3)-16*a(n-4),a(0)=0,a(1)=1,a(2)=2,a(3)=20},a(n),remember):

%p map(f, [$0..50]); # _Robert Israel_, Jan 01 2018

%t (See A192350.)

%Y Cf. A192232, A192350.

%K nonn

%O 0,3

%A _Clark Kimberling_, Jun 28 2011

%E Offset corrected by _Robert Israel_, Jan 01 2018