Generalised Pascal triangles and generalised sequences

[Ralf Stephan]

> 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)).

OK here goes:

First, the row sums. Let A(x,y;j) = Sum{n,k>=0, T(n,k;j)*x^n*y^k }, with
T(n,k;j) = C([n/j],n-k). Then the row sums S(n;j) = Sum{0<=k<=n, T(n,k;j)}
stay the same if we replace C([n/j],n-k) with C([n/j],k).

Further observe that T(jn,k;j)=T(jn+1,k;j)=...=T(jn+j-1,k;j), due to the
properties of the floor function, and this T(jn,k;j) = C(n,k). In other
words, the triangle consists of the sum of j shifted repeated copies
of a sparse version of Pascal's triangle.

Now, see that if we substitute in the power series 1/(1-x)=1+x+x^2+x^3+...
x with x^j, we get a sparse version 1/(1-x^j)=1+x^j+x^(2j)+... with the
nonzero coefficients j spaces apart. The same substitution with 1/(1-2x)=
1+2x+4x^2+8x^3... yields its sparse version 1+2x^j+4x^(2j)+8x^(3j)+...,
as well. And, to make the step from 1D to 2D, we can do that with Pascal's
triangle, too, which has g.f. 1/(1-x(1+y)) to get 1/(1-x^j(1+y)).
Summing all shifted copies results in Sum{i=0..j-1, x^i/(1-x^j(1+y)) },
which is the same as (x^j-1)/(x-1)*1/(1-x^j(1+y)), and substituting 1 for y 
gives the conjectured g.f. for S(n). We could have similarly come to it 
from the well-known row sums of Pascal's triangle.

To the second part, the conjectured g.f. is incorrect, as we will see.
Starting from the above fact that the triangle is composed of sparse
Pascal triangles, we can apply Pascal's recurrence to the diagonals of a
sparse triangle and get a(n)=a(n-j+1)+a(n-j) which yields the g.f. 
denominator 1-x^(j-1)-x^j. Together with start values 1,0,0,0... we
get the full g.f. 1/(1-x^(j-1)-x^j). Like in the first part, the resulting
g.f. is the sum of j shifted ones yielding finally
(x^j-1)/[(x-1)(1-x^(j-1)-x^j)] for the diagonals.

The last part is not as dense as I'd have liked. Any comments welcome.
ralf