OFFSET
0,20
COMMENTS
From Tilman Piesk, Feb 21 2016: (Start)
The sum of each row is 0. The even rows are symmetric; in the odd rows numbers with the same absolute value and opposed signum are symmetric to each other.
The odd rows where n mod 4 = 3 have the central value 0.
The even rows where n mod 4 = 0 have positive central values. They form the sequence A269298 and are also the rows maximal values.
A086376 contains the maximal values of each row, A160089 the maximal absolute values, and A086394 the absolute parts of the minimal values.
Rows of this triangle can be used to efficiently calculate values of A026807.
(End)
LINKS
Alois P. Heinz, Rows n = 0..40, flattened
Dorin Andrica and Ovidiu Bagdasar, On some results concerning the polygonal polynomials, Carpathian Journal of Mathematics (2019) Vol. 35, No. 1, 1-11.
Tilman Piesk, Rows n = 0..40 as left-aligned and centered table
FORMULA
T(n,k) = [x^k] Product_{i=1..n} (1-x^i).
T(n,k) = T(n-1, k) + (-1)^n*T(n-1, n*(n+1)/2-k), n > 1. - Gevorg Hmayakyan, Feb 09 2017 [corrected by Giuliano Cabrele, Mar 02 2018]
EXAMPLE
For n=2 the corresponding polynomial is (1-x)*(1-x^2) = 1 -x - x^2 + x^3.
Irregular triangle starts:
k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
n
0 1
1 1 -1
2 1 -1 -1 1
3 1 -1 -1 0 1 1 -1
4 1 -1 -1 0 0 2 0 0 -1 -1 1
5 1 -1 -1 0 0 1 1 1 -1 -1 -1 0 0 1 1 -1
MAPLE
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))
(expand(mul(1-x^i, i=1..n))):
seq(T(n), n=0..10); # Alois P. Heinz, Dec 22 2013
MATHEMATICA
Table[If[k == 0, 1, Coefficient[Product[(1 - x^i), {i, n}], x^k]], {n, 0, 6}, {k, 0, (n^2 + n)/2}] // Flatten (* Michael De Vlieger, Mar 04 2018 *)
PROG
(PARI) row(n) = pol = prod(i=1, n, 1 - x^i); for (i=0, poldegree(pol), print1(polcoeff(pol, i), ", ")); \\ Michel Marcus, Dec 21 2013
(Python)
from sympy import poly, symbols
def a231599_row(n):
if n == 0:
return [1]
x = symbols('x')
p = 1
for i in range(1, n+1):
p *= poly(1-x**i)
p = p.all_coeffs()
return p[::-1]
# Tilman Piesk, Feb 21 2016
CROSSREFS
KEYWORD
AUTHOR
Marc Bogaerts, Nov 11 2013
STATUS
approved