Let $s\_{\lambda}$ denote the schur function and $\lambda$ is the partion of an integer. The schur function written in power sum symmetric basis apper as following. $\chi$ denote the character. 
\begin{equation}
s\_\lambda(p\_1,p\_2,p\_3,\ldots) = \sum\_{\nu} \frac{\chi^\lambda\_\nu}{z\_\nu} p\_\nu = \sum\_{\rho=(1^{r\_1},2^{r\_2},3^{r\_3},\dots)}\chi^\lambda\_\rho \prod\_k \frac{p^{r\_k}\_k}{r\_k! k^{r\_k} }
\end{equation}


If lambda is a partition of type $(l,1^{d-l})$ we call it a hook partition. We set $p\_2,p\_3,\ldots$ to zero then 
$$ s\_{\lambda}(p\_1,0,0,\ldots):=\frac{1}{dd!} \binom{d-1}{\ell-1}p\_1^d$$
We get the above equation by replacing $\chi^{(l,1^{d-l})}\_{\nu}$ interms of binomial coffecient and $\chi$ is nonzero only when $\nu$ is hook so such over $d$ hooks. 


I want to study the following generating function 
$\sum\_{d=1}^{\infty}\sum\_{l=1}^{d}(-1)^{d-l}s\_{l,1^{d-l}}(p\_1,p\_2\ldots)x^d$.


Now if we set $p\_2,p\_3,\ldots=0$ then the above generating function becomes
\begin{align}
\sum\_{d=1}^{\infty}\sum\_{l=1}^{d}s\_{l,1^{d-l}}(p\_1,0,0,\dots)x^d&=\sum\_{d=1}^{\infty}\frac{1}{dd!}\sum\_{\ell=1}^{d} (-1)^{(\ell+1)}\binom{d-1}{\ell-1}x^{d} \\
&=\sum\_{d=1}^{\infty}\frac{1}{d^2}\sum\_{l=1}^{d}(-1)^{l+1}\frac{1}{(\ell-1)!(d-\ell)!}
\end{align}
Let define 
$$P(x):=1+\sum\_{d=1}^{\infty}\frac{1}{d!}x^dp\_1^d$$,
$$Q(x):=x+\sum\_{d=1}^{\infty}(-1)^{d}\frac{1}{d!}x^{(d+1)}p\_1^{(d+1)}$$.
$$R(x):=\sum\_{d=1}^{\infty}\frac{1}{d^2}x^d$$.


The generating function $\sum\_{d=1}^{\infty}\sum\_{l=1}^{d}s\_{l,1^{d-l}}(p\_1,0,0,\dots)x^d$ can be written as Cauchyproduct of $P(x)$ and $Q(x)$ that is $P(x)Q(x)$and Hadamard product of $R(x)$ and $P(x)R(x)$.


My questions are following


Let we have the sequence $p\_1,p\_2,0,0\ldots$ we have generating function
$$ \sum\_{d=1}^{\infty}\sum\_{l=1}^{d}s\_{l,1^{d-l}}(p\_1,p\_2,0,\ldots)x^d .$$ Does their exist $P(x)$,$Q(x)$ and $R(x)$ such that we could write it as above? If not why not ? 


If we take the closer look 
$$P(x):=1+\sum\_{d=1}^{\infty}s\_{d}(p\_1,0\ldots)x^d$$ and


$$ Q(x):=x+\sum\_{d=1}^{\infty}s\_{1^d}(p\_1,0,\ldots)$$


If we change $P(x):=1+\sum\_{d=1}^{\infty}s\_{d}(p\_1,p\_2,0\ldots)x^d$ and 
$Q(x):=x+\sum\_{d=1}^{\infty}s\_{1^d}(p\_1,p\_2,0,\ldots)$ what is $R(x)$ ? such that it will give me the decomposition.



Decompostion of hook schur function in terms of cauchy product of holonomic functions
It follows from the Cauchy identity and Exercise 7.43 of *Enumerative
Combinatorics*, vol. 2, that
 $$ 1+ (u+t)\sum\_{1\leq l\leq d} s\_{l,1^{d-l}}(x)u^{l-1} t^{d-l} =
 \prod\_i\frac{1+tx\_i}{1-ux\_i} $$
 $$ \qquad = \exp \sum\_{n\geq 1}\frac 1n p\_n(x)(u^n-(-t)^n). $$
If the decomposition you want actually exists then it should follow
from the above formula, though I did not try to do this.


