Given an integer partition $\lambda\vdash n$, define a polynomial in $N$ whose roots are the negatives of the contents of the partition, $$ [N]_\lambda=\prod_{\square \in \lambda}(N+c(\square)).$$ This polynomial is closely related to the value of a Schur function evaluated at the $N\times N$ identity matrix. On the other hand, given $\nu\vdash m$ and $\rho\vdash k$ contained in $\nu$, Jacobi-Trudi applied to a skew-Schur function leads to a determinant of binomial coefficients $$s_{\nu/\rho}(1_N)=\det_{1\le,i,j\le m}\left({N+\nu_i-i-\rho_j+j-1 \choose \nu_i-i-\rho_j+j}\right).$$ The final ingredient I need for my question is another determinant of binomials, $$A_{\lambda\rho}=\det_{1\le,i,j\le k}\left({\rho_i-i \choose \lambda_j-j}\right).$$

Now, in the course of some physics calculation, I arrived at the quantity $$ E_{\lambda\nu}(N)=\sum_{\lambda\subset\rho\subset\nu} A_{\lambda\rho}s_{\nu/\rho}(1_N).$$ I thought this was as far as I could push it, but experimentation convinced me that, as a function of $N$, this guy satisfies $$ E_{\lambda\nu}(N)\propto [N]_{\nu/\lambda}.$$ It is very surprising to me that this sum should factor like this.

For example, if $\nu=(2,2,1)$ and $\lambda=(1)$, the six terms in the sum are $$\{\frac{1}{24}N(N^2-1)(5N-6),-\frac{1}{2}N^2(N-1) ,\frac{1}{3}N(N^2-1) ,\frac{1}{2}N(N-1),-N^2,N\}.$$ When all these are added, the result is proportional to $N(N-2)(N^2-1)=[N]_{(2,2,1)/(1)}$.

Moreover, I think I know the proportionality constant when $\nu$ and $\lambda$ are both hooks: $$E_{\lambda\nu}(N)= \frac{1}{(m-n)!}{m-n \choose m-n-\ell(\nu)+\ell(\lambda)}[N]_{\nu/\lambda}.$$

I would appreciate some help proving a conjecture related to combinatorics and representation theory.

Given an integer partition $\lambda\vdash n$, define a polynomial in $N$ whose roots are the negatives of the contents of the partition, $$ [N]_\lambda=\prod_{\square \in \lambda}(N+c(\square)).$$ This polynomial is closely related to the value of a Schur function evaluated at the $N\times N$ identity matrix. On the other hand, given $\nu\vdash m$ and $\rho\vdash k$ contained in $\nu$, Jacobi-Trudi applied to a skew-Schur function leads to a determinant of binomial coefficients $$s_{\nu/\rho}(1_N)=\det_{1\le,i,j\le m}\left({N+\nu_i-i-\rho_j+j-1 \choose \nu_i-i-\rho_j+j}\right).$$ The final ingredient I need for my question is another determinant of binomials, $$A_{\lambda\rho}=\det_{1\le,i,j\le k}\left({\rho_i-i \choose \lambda_j-j}\right).$$

Now, in the course of some physics calculation, I arrived at the quantity $$ E_{\lambda\nu}(N)=\sum_{\lambda\subset\rho\subset\nu} A_{\lambda\rho}s_{\nu/\rho}(1_N).$$ I thought this was as far as I could push it, but experimentation convinced me that, as a function of $N$, this guy satisfies $$ E_{\lambda\nu}(N)\propto [N]_{\nu/\lambda}.$$ It is very surprising to me that this sum should factor like this.

The question is how to prove the above conjecture.

For example, if $\nu=(2,2,1)$ and $\lambda=(1)$, the six terms in the sum are $$\{\frac{1}{24}N(N^2-1)(5N-6),-\frac{1}{2}N^2(N-1) ,\frac{1}{3}N(N^2-1) ,\frac{1}{2}N(N-1),-N^2,N\}.$$ When all these are added, the result is proportional to $N(N-2)(N^2-1)=[N]_{(2,2,1)/(1)}$.

Actually, I think I know the proportionality constant when $\nu$ and $\lambda$ are both hooks: $$E_{\lambda\nu}(N)= \frac{1}{(m-n)!}{m-n \choose m-n-\ell(\nu)+\ell(\lambda)}[N]_{\nu/\lambda}.$$

Given an integer partition $\lambda\vdash n$, define a polynomial in $N$ whose roots are the negatives of the contents of the partition, $$ [N]_\lambda=\prod_{\square \in \lambda}(N+c(\square)).$$ This polynomial is closely related to the value of a Schur function evaluated at the $N\times N$ identity matrix. On the other hand, given $\nu\vdash m$ and $\rho\vdash k$ contained in $\nu$, Jacobi-Trudi applied to a skew-Schur function leads to a determinant of binomial coefficients $$s_{\nu/\rho}(1_N)=\det_{1\le,i,j\le m}\left({N+\nu_i-i-\rho_j+j-1 \choose \nu_i-i-\rho_j+j}\right).$$ The final ingredient I need for my question is another determinant of binomials, $$A_{\lambda\rho}=\det_{1\le,i,j\le k}\left({\rho_i-i \choose \lambda_j-j}\right).$$

Now, in the course of some physics calculation, I arrived at the quantity $$ E_{\lambda\nu}(N)=\sum_{\lambda\subset\rho\subset\nu} A_{\lambda\rho}s_{\nu/\rho}(1_N).$$ I thought this was as far as I could push it, but experimentation convinced me that, as a function of $N$, this guy satisfies $$ E_{\lambda\nu}(N)\propto [N]_{\nu/\lambda}.$$ It is very surprising to me that this sum should factor like this.

For example, if $\nu=(2,2,1)$ and $\lambda=(1)$, the six terms in the sum are $$\{\frac{1}{24}N(N-1)(5N-6)(N+1) ,-\frac{1}{2}N^2(N-1) ,\frac{1}{3}N(N+1)(N-1) ,\frac{1}{2}N(N-1),-N^2,N\}.$$ When all these are added, the result is proportional to $N(N-2)(N^2-1)=[N]_{(2,2,1)/(1)}$.

Moreover, I think I know the proportionality constant when $\nu$ and $\lambda$ are both hooks: $$E_{\lambda\nu}(N)= \frac{1}{(m-n)!}{m-n \choose m-n-\ell(\nu)+\ell(\lambda)}[N]_{\nu/\lambda}.$$

Given an integer partition $\lambda\vdash n$, define a polynomial in $N$ whose roots are the negatives of the contents of the partition, $$ [N]_\lambda=\prod_{\square \in \lambda}(N+c(\square)).$$ This polynomial is closely related to the value of a Schur function evaluated at the $N\times N$ identity matrix. On the other hand, given $\nu\vdash m$ and $\rho\vdash k$ contained in $\nu$, Jacobi-Trudi applied to a skew-Schur function leads to a determinant of binomial coefficients $$s_{\nu/\rho}(1_N)=\det_{1\le,i,j\le m}\left({N+\nu_i-i-\rho_j+j-1 \choose \nu_i-i-\rho_j+j}\right).$$ The final ingredient I need for my question is another determinant of binomials, $$A_{\lambda\rho}=\det_{1\le,i,j\le k}\left({\rho_i-i \choose \lambda_j-j}\right).$$

Now, in the course of some physics calculation, I arrived at the quantity $$ E_{\lambda\nu}(N)=\sum_{\lambda\subset\rho\subset\nu} A_{\lambda\rho}s_{\nu/\rho}(1_N).$$ I thought this was as far as I could push it, but experimentation convinced me that, as a function of $N$, this guy satisfies $$ E_{\lambda\nu}(N)\propto [N]_{\nu/\lambda}.$$ It is very surprising to me that this sum should factor like this.

For example, if $\nu=(2,2,1)$ and $\lambda=(1)$, the six terms in the sum are $$\{\frac{1}{24}N(N^2-1)(5N-6),-\frac{1}{2}N^2(N-1) ,\frac{1}{3}N(N^2-1) ,\frac{1}{2}N(N-1),-N^2,N\}.$$ When all these are added, the result is proportional to $N(N-2)(N^2-1)=[N]_{(2,2,1)/(1)}$.

Moreover, I think I know the proportionality constant when $\nu$ and $\lambda$ are both hooks: $$E_{\lambda\nu}(N)= \frac{1}{(m-n)!}{m-n \choose m-n-\ell(\nu)+\ell(\lambda)}[N]_{\nu/\lambda}.$$