TITLE: How to invert the matrix [n choose 2j - i] ?
QUESTION [13 upvotes]: In a certain model of a stat-physics type, one encounters a matrix 
$$
A_n:=\left[\binom{n}{2j-i}\right]_{i,j=1}^{n-1}.
$$
The determinant of this matrix (equal to $2^{\binom n2}$) counts the number of all possible configurations, and our understanding of the model would greatly increase if we would know the inverse of this matrix. So the question is,

is there a closed-form expression for the inverse of the matrix $A_n$?

A little more information about the matrix:
its eigenvalues are $2^i\colon i=1,\dots,n-1$. The eigenvectors are not orthogonal to each other (and $A_n$ is not a symmetric matrix, for sure), but the vectors corresponding to even $i$'s are orthogonal to the ones which correspond to the odd $i$'s (i.e., the whole space orthogonally decomposes into the "even" and the "odd" parts).
PS I would appreciate any help or reference. I've looked through Krattenthaler's seminal "Advanced determinant calculus", but didn't find such a matrix there.

REPLY [9 votes]: You probably know the following already.  Oh, well.
Anyway, Eric Nordenstam and I answered exactly this problem.  
It is a special case of Theorem 1 in our preprint, http://arxiv.org/abs/1201.4138 and it also appears in the proceedings of FPSAC 2012.  Our method was inspired (and suggested) by Krattenthaler: We guessed the answer from empirical data, and then proved our guess was right by multiplying the matrices out and confirming that the answer was the identity matrix.