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A192351 Coefficient of x in the reduction (by x^2->x+1) of polynomial p(n,x) identified in Comments.

%I A192351 #13 Jan 01 2018 13:17:00
%S A192351 0,1,2,20,56,320,1120,5312,20608,90880,368640,1577984,6522880,
%T A192351 27578368,114909184,483328000,2020573184,8480555008,35502817280,
%U A192351 148874461184,623609118720,2614000353280,10952269365248,45901678641152,192340840939520
%N A192351 Coefficient of x in the reduction (by x^2->x+1) of polynomial p(n,x) identified in Comments.
%C A192351 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 A192351 For an introduction to reductions of polynomials by substitutions such as x^2->x+1, see A192232.
%H A192351 Robert Israel, <a href="/A192351/b192351.txt">Table of n, a(n) for n = 0..1606</a>
%F A192351 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 A192351 Confirmation of conjecture by _Robert Israel_, Jan 01 2018: (Start)
%F A192351 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 A192351 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 A192351 note that the numerator of G(z)-g(z) is divisible by x^2-x-1. (End)
%e A192351 The first four polynomials p(n,x) and their reductions are as follows:
%e A192351 p(0,x)=1 -> 1
%e A192351 p(1,x)=x -> x
%e A192351 p(2,x)=5+x+x^2 -> 6+2x
%e A192351 p(3,x)=15x+3x^2+x^3 -> 4+20x.
%e A192351 From these, we read
%e A192351 A192350=(1,0,6,4,...) and A192351=(0,1,2,20...)
%p A192351 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 A192351 map(f, [$0..50]); # _Robert Israel_, Jan 01 2018
%t A192351 (See A192350.)
%Y A192351 Cf. A192232, A192350.
%K A192351 nonn
%O A192351 0,3
%A A192351 _Clark Kimberling_, Jun 28 2011
%E A192351 Offset corrected by _Robert Israel_, Jan 01 2018