Let $\lambda=(\lambda_1,\lambda_2,\cdots,\lambda_d)$ be a partition, with $|\lambda|=n$. Let $\nu=\nu(\lambda):=(\lambda_1-1,\lambda_2,\cdots,\lambda_d).$ In other words, $\nu$ is obtained from $\lambda$ by removing the right-most box on the first row. Is there a nice formula for:
$\sum_{\lambda \vdash n}s_{\lambda}(\mathbf{x})s_{\nu(\lambda)}(\mathbf{y}),$
where $s_\lambda(\mathbf{x})$ is the corresponding Schur function, and we can assume that the sum is over all $\lambda$ which give a valid partition $\nu(\lambda)$? The usual Cauchy identity that I know doesn't apply here unfortunately since $\nu$ cannot be written as $\lambda\backslash \mu$, for some partition $\mu$. On the other hand, $\nu\subset \lambda$.