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%I #40 Jan 11 2022 11:26:42
%S 1,1,1,1,0,-3,1,-1,-3,5,1,-2,-2,8,-7,1,-3,0,10,-15,9,1,-4,3,10,-25,24,
%T -11,1,-5,7,7,-35,49,-35,13,1,-6,12,0,-42,84,-84,48,-15,1,-7,18,-12,
%U -42,126,-168,132,-63,17
%N Coefficients of G_i(x) with G_0 = 1, G_1 = 1+x, G_n = (1-2*x)*G_{n-1}+(x-x^2)*G_{n-2}.
%C There are 3 equivalent ways to define this sequence:
%C #1) Recursively defined polynomial sequence:
%C G_0(x)=1, G_1(x)=1+x; G_n=(1-2x)G_{n-1}+(x-x^2)G_{n-2}. One can then form a triangle whose n-th row lists the coefficients of G_n(x) starting from the constant term. The list of these rows forms the sequence. It follows that an explicit formula for G_n is G_n(x)=(1+2x)(1-x)^n-2x(-x)^n.
%C #2) Recursively defined triangle:
%C Define an array g^(i)_n as follows: i) Leftmost column: g^(0)_n=1, n >=0; ii) Define the diagonal sequence D_n = g^(n)_n by D_0=0, D_1=1 and D_n=-2D_{n-1}- D_{n-2}; iii) Boundary conditions: g^(i)_n=0 if i<0, n<0 or n>i; iv) g^(i)_n - g^(i)_{n-1}= -g^(i-1)_{n-1} for n>=2, 1 <=i <=n-1. It follows that g^(i)_n = -(-1)^j(2 C(n,j-1)-C(n,j)) with C(n,j) the binomial coefficient.
%C #3) Polynomial columns:
%C g^(0)_n =1 (n >=0); g^(i)_n= (-1)^i 1/i! (n-(3i-1))(n)_{i-1} for i>=1 and n>=i (g^(i)_n is 0 elsewhere). Here (n)_m is the falling factorial with (n)_0=1 and (n)_{m+1}=(n)_m (n-m).
%C The polynomials G_n = (1+2x)(1-x)^n-2x(-x)^n have interesting limit properties. (All limits are for fixed x as n goes to infinity.)
%C 1) lim G_n(-1) = lim -(2^n-2) = minus infinity
%C 2) lim G_n(x) = minus infinity, for -1 <= x < -1/2
%C 3) lim G_n(-1/2) = lim (1/2)^n =0
%C 4) lim G_n(x) = positive infinity, for -1/2 < x <0
%C 5) lim G_n(0) = 1 (in fact G_n(0)=1 for all n)
%C 6) lim G_n(x) = 0, for 0<x<1
%C 7) lim G_n(1) = lim -(-1)^n 2, diverges by oscillation
%C There are other interesting limits. For example, although for all n, G_n(0)=1, nevertheless lim G_n(1/n) = 1/e.
%D Russell Jay Hendel, "Polynomial Convergence of Recursively Defined Polynomials", West Coast Number Theory Conference, Pacific Grove, 2014.
%H Alois P. Heinz, <a href="/A252840/b252840.txt">Rows n = 0..140, flattened</a>
%H Russell Jay Hendel, <a href="https://www.fq.math.ca/Papers1/53-3/RJHendel03012015.pdf">Coefficient Convergence of Recursively Defined Polynomials</a>, Fibonacci Quart. 53 (2015), no. 3, 247-252.
%H Clark Kimberling, <a href="https://www.fq.math.ca/Papers1/50-4/KimberlingPolynomials2.pdf">Limits of Polynomial Sequences</a>, Fibonacci Quarterly, 50(4), 2012, pp. 294-297.
%F Method #1) G_0(x)=1; G_1(x)=1+x; G_n(x)=(1-2x)G_{n-1}(x) + (x-x^2) G_{n-2}(x). The n-th row of the triangle lists the coefficients of G_n(x)=(1+2x)(1-x)^n-2x(-x)^n.
%F Method #2) g^(0)_n=1; D=g^(n)_n with D_0=1=D_1 and D_n=-2D_{n-1}- D_{n-2}; g^(i)_n =g^(i)_{n-1} - g^(i-1)_{n-1}, n >=2 and 1 <= i <= n-1. Hence g^(i)_n = -(-1)^i (2C(n,i-1)-C(n,i))
%F Method #3) g^(0)_n=1; g^(i)_n= (-1)^i 1/i! (n-(3i-1))(n)_{i-1}, i>=1, n>=i, with (n)_m the falling factorial. For each i, the g^(i)_n are a polynomial of degree i describing the triangle entry in row n and column i with n>=i.
%e G_0(x)=1, G_1(x)=1+x; G_2(x)=(1-2x)G_1(x)+(x-x^2)G_0(x)=1-3x^2. Hence the 0th row is 1; the 1st row is 1,1; the 2nd row is 1,0,-3.
%e Similarly we may give the triangle by columns: For example, the 2nd column is described by the polynomial (-1)^2 1/2! (n-(3*2-1))(n)_1 = -1/2(n-5)n which for n>=2 generates the values -3,-3,-2,0,3,7,12 which are the values of the 2nd column of the triangle starting from the 2nd row.
%e The first few rows of the triangle are
%e 1;
%e 1, 1;
%e 1, 0, -3;
%e 1, -1, -3, 5;
%e 1, -2, -2, 8, -7;
%e 1, -3, 0, 10, -15, 9;
%p g:= proc(n) option remember; `if`(n<2, 1+n*x,
%p expand((1-2*x)*g(n-1)+(x-x^2)*g(n-2)))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(g(n)):
%p seq(T(n), n=0..12); # _Alois P. Heinz_, Dec 22 2014
%t row[n_] := CoefficientList[(1+2x)(1-x)^n-2x(-x)^n, x]; Array[row, 10, 0] // Flatten (* _Jean-François Alcover_, May 24 2016 *)
%o (PARI)
%o /* using FORMULA #2 */
%o /* Function Triangle produces first n rows of coefficients of the recursively defined polynomials */
%o Triangle(n)={
%o m=matrix(n,n,i,j,0);
%o /* Leftmost column is 1*/
%o for(i=1,n,m[i,1]=1);
%o /* Rightmost diagonal are odds with alternating sign */
%o for(i=1,n,m[i,i]=(-1)^i *(2*i-3));
%o /* Body of triangle based on recursive formula */
%o for(i=3,n,for(j=2,i-1,m[i,j]=m[i-1,j]-m[i-1,j-1]));
%o m
%o }
%o /* By calling Triangle(10) one obtains the first 10 rows of the triangle corresponding to the entry in the DATA step */
%K sign,easy,tabl
%O 0,6
%A _Russell Jay Hendel_, Dec 22 2014
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