This question is a follow-up on another MO query here.

Question. For $r\geq$ an integer, is it true that there exists homogeneous symmetric polynomial $P_r(x_1,\dots,x_n)$ with positive coefficients such that $$(-1)^r\sum_{k=1}^nx_k^{2r-1}\prod_{\ell\neq k}^{1,n}\frac1{x_{\ell}^2-x_k^2} =\frac{P_r(x_1,\dots,x_n)}{\prod_{j=1}^nx_j}\prod_{i<j}^{1.n}\frac1{x_i+x_j} \,\, ?$$ I would be content with $0\leq r\leq n$. Is there an explicit formulation of $P_r$?

Remark. The factor $\prod_jx_j$, in the denominator of the RHS, is needed only when $r=0$.

For example, when $n=3, r=0$, the RHS is $\frac{x_1+x_2+x_3}{x_1x_2x_3(x_1+x_2)(x_1+x_3)(x_2+x_3)}$.

If $n=4, r=2$ then the RHS takes the form $$\frac{x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4}{(x_1+x_2)(x_1+x_3)(x_1+x_4)(x_2+x_3)(x_2+x_4)(x_3+x_4)}.$$


Denote by $\delta_n$ the partition $(n-1,n-2,\dots,1)$ of $\binom{n}2$, and the associated schur function is $$s_{\delta_n}(x_1,\dots,x_n)=\prod_{i<j}(x_i+x_j).$$ Consider now the ratio of determinants $$\begin{vmatrix} x_1^{2r-1} & x_2^{2r-1} &\cdots & x_n^{2r-1} \\ x_1^{2n-4} & x_2^{2n-4} &\cdots & x_n^{2n-4} \\ \vdots &\vdots & \ddots & \vdots\\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ 1 & 1& \cdots &1 \end{vmatrix} \cdot \begin{vmatrix} x_1^{2n-2} & x_2^{2n-2} &\cdots & x_n^{2n-2} \\ x_1^{2n-4} & x_2^{2n-4} &\cdots & x_n^{2n-4} \\ \vdots &\vdots & \ddots & \vdots\\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ 1 & 1& \cdots &1 \end{vmatrix}^{-1}$$ This evaluates to $\sum_{k=1}^nx_k^{2r-1}\prod_{\ell\neq k}^{1,n}\frac1{x_{\ell}^2-x_k^2}$. Proof: Expand the first determinant by the first row, and use the fact that the second is just a Vandermonde that evaluates to $\prod_{i<j} (x_j^2-x_i^2)$.

Therefore your polynomial $P_r(x_1,x_2,\dots,x_n)$ is equal to $$(-1)^r \begin{vmatrix} x_1^{2r} & x_2^{2r} &\cdots & x_n^{2r} \\ x_1^{2n-3} & x_2^{2n-3} &\cdots & x_n^{2n-3} \\ \vdots &\vdots & \ddots & \vdots\\ x_1^3 & x_2^3 & \cdots & x_n^3 \\ x_1 & x_2& \cdots &x_n \end{vmatrix} \cdot \begin{vmatrix} x_1^{2n-2} & x_2^{2n-2} &\cdots & x_n^{2n-2} \\ x_1^{2n-4} & x_2^{2n-4} &\cdots & x_n^{2n-4} \\ \vdots &\vdots & \ddots & \vdots\\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ 1 & 1& \cdots &1 \end{vmatrix}^{-1}\cdot s_{\delta_n}(x_1,x_2,\dots,x_n)$$ When $0\le r\le n$ we can use the definition of Schur polynomials as ratios of determinants to simplify this expression as $$s_{\lambda_{r,n}}(x_1,x_2,\dots,x_n)$$ Where the partition $\lambda_{r,n}$ is given by $(n-2,n-3,\dots,r+1,r,r,r,r-1,\dots,2,1)$. Therefore, not only is $P_{r}(x_1,x_2,\dots,x_n)$ symmetric with positive integer coefficients, but we also get a combinatorial interpretation of these coefficients as counting semistandard tableaux of shape $\lambda_{r,n}$.

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  • $\begingroup$ I liked your approach and even more the interpretation. Wonderful! One point: when $r=0$, there is a factor $\prod_{i=1}^nx_i$ in the denominator. Can you account for that? $\endgroup$ – T. Amdeberhan Feb 28 '17 at 22:22
  • $\begingroup$ Yes, I already did, that's why I increased the exponents by 1, in the first determinant. Combinatorially this just means that when $r>0$ the partition admits a tableaux where every label $1,2,...,n$ appears, which is easy since when $r>0$ then $\lambda_{r,n}$ has a column of length $n$, but not when $r=0$ which is why the $\prod x_i$ remains in the denominator for that case. $\endgroup$ – Gjergji Zaimi Feb 28 '17 at 22:27
  • $\begingroup$ Okay, thanks. You might like to look into my formulation of an answer at the link; perhaps it is possible to write one determinant for all before splitting into a sum. The link is mathoverflow.net/questions/263265/… $\endgroup$ – T. Amdeberhan Feb 28 '17 at 22:52

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