The [Cauchy Identity](https://en.wikipedia.org/wiki/Schur_polynomial#The_Cauchy_identity) 
$$ \sum\_{\nu}s\_{\nu}(x)s\_{\nu}(y) = \prod\_{j,k=1}^{\infty}\frac{1}{1-x\_{j}y\_{k}} $$
expresses the sum over all integer partitions of the product of pairs of [Schur polynomials](https://en.wikipedia.org/wiki/Schur_polynomial) as the double product in the right hand side above. Since Schur polynomials verify
$$ s\_{\nu}(x\_{1},...,x\_{n})=0 $$
if $l(\nu)>n$, specializing all but $n$ variables from each set in the Cauchy identity to zero we obtain
$$ \sum\_{l(\nu)\leq n}s\_{\nu}(x\_{1},...,x\_{n})s\_{\nu}(y\_{1},...,y\_{n}) = \prod\_{j,k=1}^{n}\frac{1}{1-x\_{j}y\_{k}}. $$
Note that the sum is now restricted to partitions of length not greater than $n$.


My question is the following: is there a simple way to express the sum
$$ \sum\_{\nu\_{1}\leq n}s\_{\nu}(x)s\_{\nu}(y), $$
where the sum is now restricted to partitions with first part less than or equal to $n$? (possibly specializing variables, as above).



Cauchy identity, with sum restricted over partitions with first part $\leq n$
The sum (in any number of variables) is equal to the determinant
$$\det(B\_{j-i})\_{1\le i,j\le n},$$
where
$$B\_i=\sum\_{l=0}^\infty e\_{l+i}(x)e\_l(y),$$
and $e\_l$ is the elementary symmetric function, with $e\_l=0$ for $l<0$.
This follows by applying the involution $\omega$ in $x$ and $y$ to Theorem 16 of my paper [*Symmetric functions and P-recursiveness*](http://www.sciencedirect.com/science/article/pii/009731659090060A), Journal of Combinatorial Theory, Series A 53 (1990), 257–285. The theorem follows fairly easily from the Cauchy-Binet formula and the Jacobi-Trudi formula. 


