Eigenvectors of a matrix with entries involving combinatorics In the question Eigenvalues of a matrix with entries involving combinatorics No_way asked about eigenvectors of $n\times n$ matrix $M$ with entries \begin{eqnarray*}
M_{ij}=(-1)^{i+j}F(n, l, i, j), 
\end{eqnarray*}
where $F(n,l,i,j)$ is the cardinality of the set
\begin{eqnarray*}
\{(k_1, \cdots, k_n)\in\mathbb{Z}^{n}|0\leq k_r\leq l-1\text{ for }1\leq r\leq n\text{, }k_1+\cdots+k_n=lj-i\}.
\end{eqnarray*}
These eigenvalues are known to be $1, l, l^2, \cdots, l^{n-1}$.
Let's remove signs and consider the matrix $M$ with $M_{ij}=F(n, l, i, j)$. According to my numerical experiments eigenvectors do not depend on $l$ for $l\ge 2$ and they are polynomials. 

Q1: Why do eigenvectors not depend on $l$?

For $l=2$ we have $M_{ij}=\binom n{2j-i},$ and first examples are (eigenvectors of $M$ are rows of $V$)
$$n=2,\qquad M=\left(
\begin{array}{cc}
 2 & 0 \\
 1 & 1 \\
\end{array}
\right),\qquad V=\left(
\begin{array}{cc}
 1 & 1 \\
 0 & 1 \\
\end{array}
\right);$$
$$n=3,\qquad M=\left(
\begin{array}{ccc}
 3 & 1 & 0 \\
 1 & 3 & 0 \\
 0 & 3 & 1 \\
\end{array}
\right),\qquad V=\left(
\begin{array}{ccc}
 1 & 1 & 1 \\
 -1 & 1 & 3 \\
 0 & 0 & 1 \\
\end{array}
\right);$$
$$n=4,\qquad M=\left(
\begin{array}{cccc}
 4 & 4 & 0 & 0 \\
 1 & 6 & 1 & 0 \\
 0 & 4 & 4 & 0 \\
 0 & 1 & 6 & 1 \\
\end{array}
\right),\qquad V=\left(
\begin{array}{cccc}
 1 & 1 & 1 & 1 \\
 -1 & 0 & 1 & 2 \\
 2 & -1 & 2 & 11 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right);$$
$$n=5,\qquad M=\left(
\begin{array}{ccccc}
 5 & 10 & 1 & 0 & 0 \\
 1 & 10 & 5 & 0 & 0 \\
 0 & 5 & 10 & 1 & 0 \\
 0 & 1 & 10 & 5 & 0 \\
 0 & 0 & 5 & 10 & 1 \\
\end{array}
\right),\qquad V=\left(
\begin{array}{ccccc}
 1 & 1 & 1 & 1 & 1 \\
 -3 & -1 & 1 & 3 & 5 \\
 11 & -1 & -1 & 11 & 35 \\
 -3 & 1 & -1 & 3 & 25 \\
 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right).$$
Denote by $v_m=(v_m(1),\ldots,v_m(n))$ rows of $V$ ($0\le m\le n-1$). They defined up to multiplicative constant and $v_m(k)=\mu_m P_m(k)$ where $P_m(x)$ are some special polynomials of degree $m$. In particular for $m=0,1,2,3,4$ we have $$P_0(x)=1,\quad P_1(x)=2x-n,\quad P_2(x)=3x^2-3nx+\frac{n(3n-1)}{4},$$
$$P_3(x)=4x^3-6nx^2+n(3n-1)x-\frac{n^2(n-1)}{2},$$
$$P_4(x)=5x^4-10nx^3+\frac{5n(3n-1)}{2}x^2-\frac{5n^2(n-1)}{2}x+\frac{n(15n^3-30n^2+5n+2)}{48}.$$

Q2: What is the generating function for these polynomials?


In fact for a fixed $n$, the matrices $M(l, n)$ for $l>0$ commute with each other and thus are simultaneously diagonalisable. For your second question, if $\{p_j(y)\}$ is a sequence of polynomials satisfying 
\begin{eqnarray}
\left(\frac{t}{\sinh t}\right)^y=\sum_{j=0}^\infty p_j(y)t^{2j}.
\end{eqnarray}
then the $i$-th entry of an eigenvector corresponding to the eigenvalue $l^{n-k-1}$ is 
\begin{eqnarray}
(-1)^{i-1}\sum_{j=0}^{\lfloor\frac{k}{2}\rfloor}\frac{p_j(n)}{(k-2j)!}(n-2i)^{k-2j}.
\end{eqnarray}
I believe from here we can work out what the generating function of your $P_m(x)$ is.
