OFFSET
0,27
COMMENTS
The n-th polynomial, as described in the link, is divisible by x^(n*(n-1)).
So they are here divided by this factor, thus removing the n*(n-1) leading zeros from the n-th row.
The initial degree of the n-th polynomial was n*(2*n-1) (A000384), so, finally the length of the n-th row is n^2 + 1 (A002522).
It appears this sequence is identical to A178666 up to the signs.
LINKS
Maarten De Boeck, The second largest Erdős-Ko-Rado sets of generators of the hyperbolic quadrics Q+(4n+1,q), arXiv:1601.03543 [math.CO], 2016. See page 5.
FORMULA
Pol(n) = (q^(n^2-n)*Prod_{k=1..n} (q^(2*k-1)-1) + 2*Sum_{i=0..n-1} gauss_binomial(2*n,2*i,q)*q^(i^2-i)*Prod_{k=1..i}(q^(2*k-1)-1))/q^(n^2-n), where gauss_binomial(j,k,q) = Prod_{i=1..k} (q^(j+1-i)-1)/(q^i-1)).
EXAMPLE
The rows with leading zeros are:
[1],
[1, 1],
[0, 0, -1, 1, 0, 1, 1],
[0, 0, 0, 0, 0, 0, 1, -1, 0, -1, 1, -1, 1, 0, 1, 1].
The leading zeros are dropped, so the triangle starts:
[1],
[1, 1],
[-1, 1, 0, 1, 1],
[1, -1, 0, -1, 1, -1, 1, 0, 1, 1],
[-1, 1, 0, 1, -1, 1, -1, 1, -2, 1, -1, 1, -1, 1, 0, 1, 1].
MATHEMATICA
gbinom[n_, k_, q_] := Product[(q^(n + 1 - i) - 1)/(q^i - 1), {i, 1, k}];
g[n_] := (q^(n^2 - n) Product[q^(2k-1)-1, {k, 1, n}] + 2 Sum[gbinom[2n, 2i, q] q^(i^2-i) Product[q^(2k-1)-1, {k, 1, i}], {i, 0, n-1}])/q^(n^2-n);
row[n_] := CoefficientList[g[n], q];
Table[row[n], {n, 0, 6}] // Flatten (* Jean-Fran�ois Alcover, Sep 27 2018, from PARI *)
PROG
(PARI) gbinom(n, k, q) = prod(i=1, k, (q^(n+1-i)-1)/(q^i-1));
g(n) = (q^(n^2-n)*prod(k=1, n, q^(2*k-1)-1) + 2*sum(i=0, n-1, gbinom(2*n, 2*i, q)*q^(i^2-i)*prod(k=1, i, q^(2*k-1)-1)))/q^(n^2-n);
row(n) = Vecrev(g(n));
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Michel Marcus, Jan 18 2016
STATUS
approved