%I #8 Jan 17 2013 09:05:19
%S 0,1,2,11,32,125,418,1511,5248,18601,65250,230099,809248,2849989,
%T 10030018,35311375,124293632,437545489,1540200002,5421774299,
%U 19085364000,67183428301,236495292002,832498651511,2930516834432,10315851565625
%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+2); 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.
%F Conjecture: a(n) = 2*a(n-1)+6*a(n-2)-2*a(n-3)-a(n-4). G.f.: x^2*(x^2+1) / (x^4+2*x^3-6*x^2-2*x+1). [_Colin Barker_, Jan 17 2013]
%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)=2+x+x^2 -> 3+2x
%e p(3,x)=6x+3x^2+x^3 -> 4+11x.
%e From these, we read
%e A192346=(1,0,3,4,...) and A192347=(1,1,2,11...)
%t (See A192346.)
%Y Cf. A192232, A192346.
%K nonn
%O 1,3
%A _Clark Kimberling_, Jun 28 2011