inverse of a matrix  What is the inverse of the following matrix ? 
$$
\begin{bmatrix}
\binom{N}{0} &\binom{N+1}{0}  &...   &\binom{2N-1}{0} \\ 
\binom{N}{1} &\binom{N+1}{1}  &...   &\binom{2N-1}{1}\\ 
 ...& ... &...  &... \\ 
\binom{N}{N-1} &\binom{N+1}{N-1}  &...   &\binom{2N-1}{N-1} 
\end{bmatrix}
$$

Your matrix, call it $A_N$, can be factorized as $B_N C_N$, where
$$B_N = \left[ \binom{N}{i-j} \right]_{i,j=0,\dots,N-1}$$
and
$$C_N = \left[ \binom{j}{j-i} \right]_{i,j=0,\dots,N-1}.$$
For example, for $N=5$ this will be
$$
\left(
\begin{array}{lllll}
1 & 1 & 1 & 1 & 1 \\
5 & 6 & 7 & 8 & 9 \\
10 & 15 & 21 & 28 & 36 \\
10 & 20 & 35 & 56 & 84 \\
5 & 15 & 35 & 70 & 126
\end{array}
\right)
=
\left(
\begin{array}{lllll}
1 & 0 & 0 & 0 & 0 \\
5 & 1 & 0 & 0 & 0 \\
10 & 5 & 1 & 0 & 0 \\
10 & 10 & 5 & 1 & 0 \\
5 & 10 & 10 & 5 & 1
\end{array}
\right)
\left(
\begin{array}{lllll}
1 & 1 & 1 & 1 & 1 \\
0 & 1 & 2 & 3 & 4 \\
0 & 0 & 1 & 3 & 6 \\
0 & 0 & 0 & 1 & 4 \\
0 & 0 & 0 & 0 & 1
\end{array}
\right).
$$
The inverses of these factors are
$$B_N^{-1} = \left[ (-1)^{i+j} \binom{N-1+i-j}{N-1} \right]_{i,j=0,\dots,N-1}$$
and
$$C_N^{-1} = \left[ (-1)^{i+j} \binom{j}{j-i} \right]_{i,j=0,\dots,N-1}.$$
Then $A_N^{-1}=C_N^{-1} B_N^{-1}$ of course, but I don't know if that simplifies to something nice. (I didn't find any hits in OEIS.)
In any case, the fact that $B_N$ and $C_N$ have determinant one (since they are triangular with ones on the diagonal) explains why $A_N^{-1}$ has integer entries.
Edit:
Here's the picture promised in my comment. It illustrates the situation for $N=5$.

Entry $(i,j)$ (counting from zero) in the matrix $A_5 = B_5 C_5$ equals the number of paths through the directed graph from source number $i$ on the left to sink number $j$ on the right.
Incidentally, the network shows that $B_N$ can be factorized further, just for fun:
$$
B_5 =
\left(
\begin{array}{lllll}
1 & 0 & 0 & 0 & 0 \\
4 & 1 & 0 & 0 & 0 \\
6 & 4 & 1 & 0 & 0 \\
4 & 6 & 4 & 1 & 0 \\
1 & 4 & 6 & 4 & 1
\end{array}
\right)
\left(
\begin{array}{lllll}
1 & 0 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 \\
1 & 2 & 1 & 0 & 0 \\
1 & 3 & 3 & 1 & 0 \\
1 & 4 & 6 & 4 & 1
\end{array}
\right).
$$
(The first factor corresponds to the first four blue arrows, and the second factor to the next four blue arrows.)
