How to invert the matrix [n choose 2j - i] ? 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.
 A: 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.
A: This is more an idea to explore than a complete answer. 
You may interpret the binomial coefficient $\binom{n}{k}$ as the elementary symmetric function $e_k$ of $1,1,\ldots,1$ ($n$ variables evaluated at $1$). The coefficients of the adjoint matrix of $A_n$ become skew Schur functions of $1,1,\ldots,1$. Then there may be some further simplifications. 
(By the way, this approach gives a nice proof for the value of the determinant of $A_n$: it is the value of the staircase Schur function $s_{(n-1,n-2,\ldots,1,0)}$ evaluated at $1,1,\ldots,1$. Note that the staircase Schur function at $x_1,x_2,\ldots,x_n$ is equal to $\prod_{i \lt j} (x_i+x_j)$).
EDIT: I find that the coefficient $(i,j)$ of the inverse is $(-1)^{i+j} s_{[j]'/(n-i)}(1,1,\ldots,1)/2^{\binom{n}{2}}$, where $[j]$ stands for the partition obtained from $(n-1,n-2,\ldots,1)$ by removing $j$, and $[j]'$ is its conjugate. At this point there is some hope to find a nice formula. First by expressing the skew Schur function as a sum a Schur functions by means of dual Pieri rule:
$$
s_{\lambda/(k)}=\sum s_{\nu}
$$
where the sum is carried over all partitions $\nu$ obtained from $\lambda$ by removing a horizontal strip with $k$ boxes. After that by using the following formula  found in Macdonald, I.3. Ex. 4:
$$
s_{\lambda'}(1,1,\ldots,1)=\prod_{x \in \lambda} \frac{n-c(x)}{h(x)}
$$
where the evaluation is at $(1,1,\ldots,1)$ with $n$ ones, $\lambda'$ is the conjugate of $\lambda$ and $h(x)$ and $c(x)$ are the hook length and content respectively of the box $x$ in the diagram of $\lambda$. 
Hopefully the formulas simplify.
