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A192459 Coefficient of x in the reduction by x^2->x+2 of the polynomial p(n,x) defined below in Comments.

%I A192459 #17 Jul 28 2021 04:27:50
%S A192459 1,3,17,133,1315,15675,218505,3485685,62607195,1250116875,27468111825,
%T A192459 658579954725,17109329512275,478744992200475,14354443912433625,
%U A192459 459128747151199125,15604187119787140875,561558837528374560875,21332903166207470462625
%N A192459 Coefficient of x in the reduction by x^2->x+2 of the polynomial p(n,x) defined below in Comments.
%C A192459 The polynomial p(n,x) is defined by recursively by p(n,x)=(x+2n)*p(n-1,x) with p[0,x]=x.  For an introduction to reductions of polynomials by substitutions such as x^2->x+2, see A192232.
%C A192459 Let transform T take the sequence {b(n), n>=1} to the sequence {c(n)} defined by: c(n) = det(M_n), where M_n denotes the n X n matrix with elements M_n(i,j) = b(2*j) for i>j and M_n(i,j) = b(i+j-1) for i<=j. Conjecture: a(n) = abs(c(n+1)), where c(n) denotes transform T of triangular numbers (A000217). - _Lechoslaw Ratajczak_, Jul 26 2021
%F A192459 a(n) = (1/3)*(2^(n+1)*(n+1)! + (2n-1)!!). - _Vaclav Potocek_, Feb 04 2016
%e A192459 The first four polynomials p(n,x) and their reductions are as follows:
%e A192459 p(0,x)=x -> x
%e A192459 p(1,x)=x(2+x) -> 2+3x
%e A192459 p(2,x)=x(2+x)(4+x) -> 14+17x
%e A192459 p(3,x)=x(2+x)(4+x)(6+x) -> 118+133x.
%e A192459 From these, read
%e A192459 A192457=(1,2,14,118,...) and A192459=(1,3,17,133,...)
%t A192459 (See A192457.)
%Y A192459 Cf. A192232, A192457.
%K A192459 nonn
%O A192459 0,2
%A A192459 _Clark Kimberling_, Jul 01 2011