Determinant connection between Schur polynomials and power sum polynomials 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 $n$.
 A: 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. (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.
