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.



How to invert the matrix [n choose 2j - i] ?
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.


