OFFSET
0,6
COMMENTS
There are 3 equivalent ways to define this sequence:
#1) Recursively defined polynomial sequence:
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.
#2) Recursively defined triangle:
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.
#3) Polynomial columns:
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).
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.)
1) lim G_n(-1) = lim -(2^n-2) = minus infinity
2) lim G_n(x) = minus infinity, for -1 <= x < -1/2
3) lim G_n(-1/2) = lim (1/2)^n =0
4) lim G_n(x) = positive infinity, for -1/2 < x <0
5) lim G_n(0) = 1 (in fact G_n(0)=1 for all n)
6) lim G_n(x) = 0, for 0<x<1
7) lim G_n(1) = lim -(-1)^n 2, diverges by oscillation
There are other interesting limits. For example, although for all n, G_n(0)=1, nevertheless lim G_n(1/n) = 1/e.
REFERENCES
Russell Jay Hendel, "Polynomial Convergence of Recursively Defined Polynomials", West Coast Number Theory Conference, Pacific Grove, 2014.
LINKS
Alois P. Heinz, Rows n = 0..140, flattened
Russell Jay Hendel, Coefficient Convergence of Recursively Defined Polynomials, Fibonacci Quart. 53 (2015), no. 3, 247-252.
Clark Kimberling, Limits of Polynomial Sequences, Fibonacci Quarterly, 50(4), 2012, pp. 294-297.
FORMULA
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.
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))
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.
EXAMPLE
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.
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.
The first few rows of the triangle are
1;
1, 1;
1, 0, -3;
1, -1, -3, 5;
1, -2, -2, 8, -7;
1, -3, 0, 10, -15, 9;
MAPLE
g:= proc(n) option remember; `if`(n<2, 1+n*x,
expand((1-2*x)*g(n-1)+(x-x^2)*g(n-2)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(g(n)):
seq(T(n), n=0..12); # Alois P. Heinz, Dec 22 2014
MATHEMATICA
row[n_] := CoefficientList[(1+2x)(1-x)^n-2x(-x)^n, x]; Array[row, 10, 0] // Flatten (* Jean-François Alcover, May 24 2016 *)
PROG
(PARI)
/* using FORMULA #2 */
/* Function Triangle produces first n rows of coefficients of the recursively defined polynomials */
Triangle(n)={
m=matrix(n, n, i, j, 0);
/* Leftmost column is 1*/
for(i=1, n, m[i, 1]=1);
/* Rightmost diagonal are odds with alternating sign */
for(i=1, n, m[i, i]=(-1)^i *(2*i-3));
/* Body of triangle based on recursive formula */
for(i=3, n, for(j=2, i-1, m[i, j]=m[i-1, j]-m[i-1, j-1]));
m
}
/* By calling Triangle(10) one obtains the first 10 rows of the triangle corresponding to the entry in the DATA step */
CROSSREFS
KEYWORD
AUTHOR
Russell Jay Hendel, Dec 22 2014
STATUS
approved