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Coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x)=(x+2)(x+4)...(x+2n).
2

%I #8 May 08 2014 15:19:04

%S 0,1,7,58,583,6959,96510,1527125,27170285,537109100,11682187715,

%T 277285358845,7132907069640,197684330603485,5872470327374035,

%U 186153757195641730,6272161769194950475,223842624694659656675,8435226009748039509150

%N Coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x)=(x+2)(x+4)...(x+2n).

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

%F Conjecture: a(n) +(-4*n+1)*a(n-1) +(4*n^2-6*n+1)*a(n-2)=0. - _R. J. Mathar_, May 08 2014

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

%e p(0,x)=1

%e p(1,x)=x+2 -> x+2

%e p(2,x)=(x+2)(x+4) -> 9+7x

%e p(3,x)=(x+2)(x+4)(x+6) -> 61+58x

%e From these, read

%e A192939=(1,2,9,61,...) and A192940=(0,1,7,58,...)

%t (See A192939.)

%Y Cf. A192232, A192744, A192939.

%K nonn

%O 0,3

%A _Clark Kimberling_, Jul 13 2011