Let $f\_i=f\_i(x\_1,x\_2,\ldots, x\_n),i=0,1,2, \ldots $ be a family of symmetric polynomials. For the partition $\lambda=(\lambda\_1,\lambda\_2, \ldots, \lambda\_n)$ consider the determinant
$$
D\_\lambda(f)=\left | \begin{array}{lllll}
f\_{\lambda\_1} & f\_{\lambda\_1+1} & f\_{\lambda\_1+2} & \ldots & f\_{\lambda\_1+n-1}\\
f\_{\lambda\_2-1} & f\_{\lambda\_2} & f\_{\lambda\_2+1} & \ldots & f\_{\lambda\_2+n-2}\\
\vdots & \vdots & \vdots & \ldots & \vdots \\
f\_{\lambda\_n-(n-1)} & f\_{\lambda\_n-(n-2)} & f\_{\lambda\_n-(n-3)} & \ldots & f\_{\lambda\_n}
\end{array} \right|.
$$
It is well known (Jacobi−Trudi formulas) that for the elementary symmetric polynomials $e\_i=e\_i(x\_1,x\_2,\ldots, x\_n)$ and for the complete homogeneous symmetric polynomials $h\_i=h\_i(x\_1,x\_2,\ldots, x\_n)$ we have
$$
D\_\lambda(h)=s\_{\lambda}(x\_1,x\_2,\ldots, x\_n) \text{ and } D\_\lambda(e)=s\_{\lambda'}(x\_1,x\_2,\ldots, x\_n), 
$$
where $s\_{\lambda}(x\_1,x\_2,\ldots, x\_n)$ is the Schur polynomial and $\lambda'$ is the conjugate partition.


**Question.** Is there a similar expression for $D\_\lambda(p)$ where $p\_i=p\_i(x\_1,x\_2,\ldots, x\_n)$ is the power sum symmetric polynomials?


By direct calculation for $n=2, \lambda\_2\geq 1$ I got that


$$D\_{(\lambda\_1,\lambda\_2)}(p)=-s\_{(\lambda\_1-1,\lambda\_2-1)} V(x\_1,x\_2)^2$$ and for $n=3,\lambda\_3\geq 2$


$$D\_{(\lambda\_1,\lambda\_2,\lambda\_3)}(p)=-\frac{s\_{(\lambda\_1-1,\lambda\_2-1,\lambda\_3-1)}}{s\_{(1,1,1)}} V(x\_1,x\_2,x\_3)^2$$
and so on.


Here $V$ is the Vandermonde determinant.


I hope that must be a nice formula and for any $n$.



Determinant connection between Schur polynomials and power sum polynomials
I don't know a fully general result, but your pattern for partitions $\lambda$
of length $\leq n$ with $n$-th entry $\lambda\_{n}\geq n-1$ and with $n$
indeterminates persists:



> 
> **Theorem 1.** Let $n$ be a positive integer. Let $\lambda=\left( \lambda
> \_{1},\lambda\_{2},\ldots,\lambda\_{n}\right) $ be an integer partition with at
> most $n$ parts. Assume that $\lambda\_{n}\geq n-1$. Consider polynomials in $n$
> indeterminates $x\_{1},x\_{2},\ldots,x\_{n}$. For each nonnegative integer $k$,
> we set
> \begin{align\*}
> p\_{k}:=x\_{1}^{k}+x\_{2}^{k}+\cdots+x\_{n}^{k}.
> \end{align\*}
> (This is the $k$-th power-sum symmetric polynomial in $x\_{1},x\_{2}
> ,\ldots,x\_{n}$ when $k>0$. We have $p\_{0}=n$.) Define the $n\times n$-matrix
> \begin{align\*}
> P:=\left( p\_{\lambda\_{i}-i+j}\right) \_{1\leq i\leq n,\ 1\leq j\leq
> n}=\left(
> \begin{array}
> [c]{cccc}
> p\_{\lambda\_{1}} & p\_{\lambda\_{1}+1} & \cdots & p\_{\lambda\_{1}+n-1}\\
> p\_{\lambda\_{2}-1} & p\_{\lambda\_{2}} & \cdots & p\_{\lambda\_{2}+n-2}\\
> \vdots & \vdots & \ddots & \vdots\\
> p\_{\lambda\_{n}-n+1} & p\_{\lambda\_{n}-n+2} & \cdots & p\_{\lambda\_{n}}
> \end{array}
> \right) .
> \end{align\*}
> Let $\mu=\left( \mu\_{1},\mu\_{2},\ldots,\mu\_{n}\right) $ be the partition
> defined by
> \begin{align\*}
> \mu\_{i}=\lambda\_{i}-\left( n-1\right) \ \ \ \ \ \ \ \ \ \ \text{for each
> }i\in\left\{ 1,2,\ldots,n\right\} .
> \end{align\*}
> (This is indeed a partition, since $\mu\_{n}=\underbrace{\lambda\_{n}}\_{\geq
> n-1}-\left( n-1\right) \geq0$.) Let $s\_{\mu}$ be the corresponding Schur
> polynomial in the $n$ indeterminates $x\_{1},x\_{2},\ldots,x\_{n}$. Furthermore,
> let
> \begin{align\*}
> V\_{n}:=\prod\_{1\leq i<j\leq n}\left( x\_{i}-x\_{j}\right)
> \end{align\*}
> be the Vandermonde determinant. Then,
> \begin{align\*}
> \det P=\left( -1\right) ^{n\left( n-1\right) /2}s\_{\mu}\cdot V\_{n}^{2}.
> \end{align\*}
> 
> 
> 


*Proof.* Let $A\_{\mu}$ be the $n\times n$-matrix
\begin{align\*}
\left( x\_{j}^{\mu\_{i}+n-i}\right) \_{1\leq i\leq n,\ 1\leq j\leq n}=\left(
\begin{array}
[c]{cccc}
x\_{1}^{\mu\_{1}+n-1} & x\_{2}^{\mu\_{1}+n-1} & \cdots & x\_{n}^{\mu\_{1}+n-1}\\
x\_{1}^{\mu\_{2}+n-2} & x\_{2}^{\mu\_{2}+n-2} & \cdots & x\_{n}^{\mu\_{2}+n-2}\\
\vdots & \vdots & \ddots & \vdots\\
x\_{1}^{\mu\_{n}+n-n} & x\_{2}^{\mu\_{n}+n-n} & \cdots & x\_{n}^{\mu\_{n}+n-n}
\end{array}
\right) .
\end{align\*}
It is then well-known that
\begin{equation}
s\_{\mu}=\dfrac{\det\left( A\_{\mu}\right) }{V\_{n}}
.
\label{darij1.eq.slam=frac}
\tag{1}
\end{equation}
Indeed, this is the alternant formula for Schur polynomials. For a proof, see,
e.g., Corollary 2.6.7 in the lecture notes [Darij Grinberg and Victor Reiner,
*Hopf Algebras in Combinatorics*,
arXiv:1409.8356v7](https://arxiv.org/abs/1409.8356v7). (The notations in those
notes are not quite ours. Namely, our matrix $A\_{\mu}$ is the transpose of the
matrix whose determinant is $a\_{\mu+\rho}$ in the notes, whereas our $V\_{n}$
is $a\_{\rho}$ in these notes. Corollary 2.6.7 has to be applied to $\mu$
instead of $\lambda$.)


Let $B$ be the $n\times n$-matrix
\begin{align\*}
\left( x\_{i}^{j-1}\right) \_{1\leq i\leq n,\ 1\leq j\leq n}=\left(
\begin{array}
[c]{cccc}
1 & x\_{1} & \cdots & x\_{1}^{n-1}\\
1 & x\_{2} & \cdots & x\_{2}^{n-1}\\
\vdots & \vdots & \ddots & \vdots\\
1 & x\_{n} & \cdots & x\_{n}^{n-1}
\end{array}
\right) .
\end{align\*}
The Vandermonde determinant formula yields
\begin{align\*}
\det B & =\prod\_{1\leq i<j\leq n}\underbrace{\left( x\_{j}-x\_{i}\right)
}\_{=-\left( x\_{i}-x\_{j}\right) }=\prod\_{1\leq i<j\leq n}\left( -\left(
x\_{i}-x\_{j}\right) \right) \\
& =\left( -1\right) ^{n\left( n-1\right) /2}\underbrace{\prod\_{1\leq
i<j\leq n}\left( x\_{i}-x\_{j}\right) }\_{=V\_{n}}=\left( -1\right) ^{n\left(
n-1\right) /2}V\_{n}.
\end{align\*}


However, we have
\begin{equation}
A\_{\mu}B=P.
\label{darij1.eq.AB=P}
\tag{2}
\end{equation}
(Indeed, for any $i,j\in\left\{ 1,2,\ldots,n\right\} $, the $\left(
i,j\right) $-th entry of the matrix $A\_{\mu}B$ is
\begin{align\*}
\sum\_{k=1}^{n}\underbrace{x\_{k}^{\mu\_{i}+n-i}x\_{k}^{j-1}}\_{\substack{=x\_{k}
^{\mu\_{i}+n-i+j-1}=x\_{k}^{\lambda\_{i}-i+j}\\\text{(since }\mu\_{i}=\lambda
\_{i}-\left( n-1\right) \text{ and}\\\text{thus }\mu\_{i}+n-i+j-1=\lambda
\_{i}-\left( n-1\right) +n-i+j-1=\lambda\_{i}-i+j\text{)}}} & =\sum\_{k=1}
^{n}x\_{k}^{\lambda\_{i}-i+j}\\
& =x\_{1}^{\lambda\_{i}-i+j}+x\_{2}^{\lambda\_{i}-i+j}+\cdots+x\_{n}^{\lambda
\_{i}-i+j}=p\_{\lambda\_{i}-i+j},
\end{align\*}
which happens to be precisely the $\left( i,j\right) $-th entry of the
matrix $P$. Thus, \eqref{darij1.eq.AB=P} follows.)


Now, the two matrices $A\_{\mu}$ and $B$ are square matrices. Hence,
\begin{align\*}
\det\left( A\_{\mu}B\right) & =\underbrace{\det\left( A\_{\mu}\right)
}\_{\substack{=s\_{\mu}V\_{n}\\\text{(by \eqref{darij1.eq.slam=frac})}}
}\cdot\underbrace{\det B}\_{=\left( -1\right) ^{n\left( n-1\right) /2}
V\_{n}}\\
& =s\_{\mu}V\_{n}\cdot\left( -1\right) ^{n\left( n-1\right) /2}V\_{n}=\left(
-1\right) ^{n\left( n-1\right) /2}s\_{\mu}\cdot V\_{n}^{2}.
\end{align\*}
In view of \eqref{darij1.eq.AB=P}, we can rewrite this as
\begin{align\*}
\det P=\left( -1\right) ^{n\left( n-1\right) /2}s\_{\mu}\cdot V\_{n}^{2}.
\end{align\*}
This proves Theorem 1. $\blacksquare$


The claim of Theorem 1 can further be rewritten by observing that (in $n$
indeterminates $x\_{1},x\_{2},\ldots,x\_{n}$) we have
\begin{align\*}
s\_{\lambda}=s\_{\mu}\cdot\left( x\_{1}x\_{2}\cdots x\_{n}\right) ^{n-1}
\end{align\*}
(because the entries of $\lambda$ are the respective entries of $\mu$ plus
$n-1$). The product $x\_{1}x\_{2}\cdots x\_{n}$ can also be rewritten as
$s\_{\left( 1^{n}\right) }$, where $\left( 1^{n}\right) $ is the partition
$\left( 1,1,\ldots,1\right) $ with $n$ entries.


