Generalised Pascal triangles and generalised sequences

[Paul Barry]

Would anybody care to prove/disprove the following result (or find a
reference to it):

1. The triangle binomial(floor(n/j), n-k) has row sums with g.f.
sum{i=0..j-1, x^i}/(1-2x^j), and diagonal sums with g.f. 
sum{i=0..j-1, x^(2i)}/(1-x^(2j-1)-x^(2j)).

2. The triangle binomial(floor((j*n-(j-1)*k-(j-1))/j,n-k) has row 
sums with g.f. 1/(1-x-x^j) and diagonal sums with g.f.
(1+x^(2j-1))/((1+x)(1-x-x^(2j)).

3. The second triangle is the inverse of
(-1)^(n-k)*binomial(floor(n/j), n-k).

Note that Pascal's triangle is the case j=1.

Best wishes,
Paul Barry