I'm trying to do the exercise 3.10 of Computing the Continuous Discretely by Beck-Robbins.
For a polynomial $p(t) = c_d t^d + c_{d-1}t^{d-1} + \cdots + c_0$, let $H_p(z)$ be defined by \begin{equation} \sum_{t \ge 0}p(t) z^t =\frac{H_p(z)}{(1 - z)^{d+1}}. \end{equation}
Consider the linear map $\phi_d : Pol_d \to Pol_d, p \mapsto H_P.$
We want to compute the matrix describing $\phi_d$ for $d = 0, 1, 2,\dots$, then we fix a base $\mathcal{B}=\{1, t, \dots t^d\}$ of $Pol_d.$
For $d=1$, we have \begin{equation} \sum_{t \ge 0} z^t =\frac{1}{1 - z}=\frac{H_1(z)}{(1 - z)^2} \Rightarrow H_1(z)=1-z \end{equation} in general it is easy to check that $H_1(z)=(1-z)^d=\sum_{a=0}^d \binom{d}{a}(-1)^kz^k.$ We derive and so \begin{equation} \sum_{t \ge 0} t z^t =\frac{z}{(1 - z)^2}=\frac{H_t(z)}{(1 - z)^2} \Rightarrow H_t(z)=z \end{equation} in general it is easy to check that $H_t(z)=z(1-z)^{d-1}=\sum_{a=0}^d \binom{d}{a}(-1)^kz^{k+1}.$ Therefore $$[\phi_1]_{\mathcal{B}}=\left[\begin{matrix} 1 & 0 \\ -1 & 1 \end{matrix}\right]$$
For $d=2$, we have seem already that $H_1(z)=(1-z)^2=1-2z+z^2$ and $H_t(z)=z(1-z)=z-z^2$, we derivate again and so
\begin{equation}
\sum_{t \ge 0} t^2 z^t =\frac{z(z+1)}{(1 - z)^3}=\frac{H_{t^2}(z)}{(1 - z)^3} \Rightarrow H_{t^2}(z)=z(z+1)
\end{equation}
in general it is easy to check that $H_{t^2}(z)=z(z+1)(1-z)^{d-2}=\sum_{a=0}^d \binom{d}{a}(-1)^kz^{k+1}(z+1).$ Therefore
$$[\phi_2]_{\mathcal{B}}=\left[\begin{matrix} 1 & 0 &0\\ -2 & 1 & 1\\1 & -1 & 1 \end{matrix}\right].$$
Recursively following the previous process we have observed that in general
$H_{1}=(1-z)^d$
$H_{t}=z(1-z)^{d-1}$
$H_{t^2}=z(z+1)(1-z)^{d-2}$
$H_{t^3}=z(z^2+4z+1)(1-z)^{d-3}$
$H_{t^4}=z(z^3+11z^2+11z+1)(1-z)^{d-3}$
$H_{t^5}=z(z^4+26z^2+66z^2+26z+1)(1-z)^{d-3}$ ect...
The problem is that we haven't found the recurrence. Do you have another idea for this problem?
It looks like you are asking for the formula for $$\sum_{n=0}^\infty n^kz^n.$$ This can be expressed in terms of Eulerian numbers/polynomials: $$\sum_{n=0}^\infty n^kz^n=\frac{\phi_k(z)}{(1-z)^{k+1}}$$ where $$\phi_k(z)=\sum_{j=0}^k E(k,j-1)z^j$$ is an Eulerian polynomial. For $k\ge1$ the Eulerian number $E(k,j)$ counts the number of permutations in $S_k$ with $j$ ascents.