Dividing the $k$th column of the lower triangular matrix $T$ (OEIS A111492) in Andrej Bauer's answer by $(k-1)!$ for each column generates A135278 (the $f$-vectors, or face-vectors for the $n$-simplexes). Then ignoring the first column gives A104712, so $T$ acting on the column vector $(-0,d,-d^2/2!,d^3/3!,...)$ gives the Euler classes for hypersurfaces of degree $d$ in $CP^n$. (See Daniel Dugger, A Geometric Introduction to K-Theory, pg. 168.)

$T$ also has relations to the number of permutations of the symmetric group $S_n$ that are pure $k$-cycles, colored forests of "naturally-grown" trees, disposition of flags on flagpoles, the colorings of the vertices of the complete graphs $K_n$, encoded in their chromatic polynomials (see A130534), and the commutator $[log(D), x^nD^n]=d(x^nD^n)/d(xD)$ for $D=d/dx$ (cf. A238363).

**Update** (Apr 26 and May 20 2014):

The Vandermonde matrix $V_n$ is intimately connected to the $(n-1)$-simplex and its edge projection onto a plane, the complete graphs $K_n$. There are several definitions in use, so to be definite let

$$V_n=V_n(x_1,x_2,...,x_n) = \left[ \begin{array}{} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots & x_n \\ \vdots & \vdots & \ddots & \vdots\\ x^{n-1}_1 & x^{n-1}_2 & \cdots & x^{n-1}_{n}\end{array} \right]$$

and determinant

$$|V_n|=|V_n(x_1, \ldots, x_n)| = \prod_{1 \leq i < j \leq n} (x_j - x_i).$$

To obtain a generating function for the rows of $T$ from Chervov's operator, first note the action of the generalized shift/dilation operator $exp(t:xd/dx:)f(x)=f((1+t)x)$ (a generalization of $e^{td/dx}f(x)=f(x+t))$, where $(:x_i\frac{d}{dx_i}:)^n=x_i^n(\frac{d}{dx_i})^n$, i.e., the power distributes over the expressions between colons. Also let $p_k(x_1,...,x_n)$ be the power sum symmetric polynomial. Then act on $|V_n|$ with a sum of $exp(t:x_id/dx_i:)$ obtaining

$$W_n(x_1 , \ldots , x_n;t)=\sum_{k\geq0} \frac{t^k}{k!}\sum_{i=1}^{n} x_i^k \frac{d^k}{dx_i^k} |V_n|$$

$$=exp[t \cdot p.(:x_1\frac{d}{dx_1}:,...,:x_n\frac{d}{dx_n}:)]|V_n|$$

$$=\sum_{i=1}^{n}exp(t :x_i\frac{d}{dx_i}:) |V_n(x_1 , \ldots , x_n)|$$

$$= |V_n((1+t)x_1 ,x_2, \ldots , x_n)|+|V_n(x_1 ,(1+t)x_2, \ldots , x_n)|+ \ldots+|V_n(x_1 ,x_2,\ldots,(1+t) x_n)|$$

$$= [1+(1+t)+(1+t)^2+\;...+(1+t)^{n-1}]\; |V_n(x_1 , \ldots , x_n)|$$

$$= \frac{(1+t)^n-1}{t}\; |V_n(x_1 , \ldots , x_n)|.$$

Therefore, $$G_n(t)=\frac{W_n(x_1 , \ldots , x_n;t)}{|V_n(x_1 , \ldots , x_n)|}=\frac{(1+t)^n-1}{t},$$

which gives an exponential generating function for the rows of the matrix $T$ (OEIS A111492, A238363) in Bauer's guess that is in agreement with Serre's answer, and an ordinary generating function for the f-polynomials (f-vectors) of the number of k-faces of the $(n-1)$-simplex (OEIS A135278).

For example,

$$G_4(t)=\frac{(1+t)^4-1}{t}=4+6t+8 \frac{t^2}{2!}+6 \frac{t^3}{3!}=4+6t+4t^2+t^3.$$

$V_n,K_n,G_n$ are associated to the $(n−1)$-simplex, and the $3$-simplex is the tetrahedron with $4$ vertices, $6$ edges, $4$ triangles, $1$ polyhedron. (The number of factors in the product formula for $|V_n|$ is given by the number of edges of $K_n$ (OEIS A000217). See also the MO-Qs Cyclotomic Polynomials in Combinatorics and my notes on The Vandermonde Matrix and Goin' with the Flow at my website.)

There is another expression for $G_n(t)$:

$G_n(t)=\left[ \begin{array}{} 1 & t & \frac{t^2}{2!} & \cdots & \frac{t^{n-1}}{(n-1)!}\end{array} \right]|V_n|^{-1}V_n(:x_1 \frac{d}{dx_1}:,...,:x_n \frac{d}{dx_n}:) |V_n| \left[ \begin{array}{} 1 \\ 1 \\ \; \vdots\\ 1\end{array} \right]$

since

$\left[ \begin{array}{} 1 & t & \frac{t^2}{2!} & \cdots & \frac{t^{n-1}}{(n-1)!}\end{array} \right]V_n\left[ \begin{array}{} 1 \\ 1 \\ \; \vdots\\ 1\end{array} \right]=\sum_{k<n} \frac{t^k}{k!}\sum_{i=1}^{n} x_i^k=\sum_{k<n} \frac{t^k}{k!}p_k(x_1,...,x_n),$

and when acting on a polynomial of degree $\leq (n-1)$, the finite operator sum $\sum_{k<n} \frac{t^k}{k!}p_k(:x_1d/dx_1:,...,:x_nd/dx_n:)$ is equivalent to $exp[t \cdot p.(:x_1\frac{d}{dx_1}:,...,:x_n\frac{d}{dx_n}:)]$ above.

The row polynomials of $T$ are given by replacing $t^j/j!$ by $t^j$ in the operator expression.

(Edit 6/2014)

Derivatives of the generating function $W_n(x_1,...,x_n;t)$ generate the inverse of $V_n$:

Note that $D^k_{t=-1}f[(1+t)x]=x^k f^{(k)}(0)=x^k D^k_{x=0}f(x)$, so acting on the two different expressions for $W_n$ gives, for $k=0,...,n-1$,

$$D^k_{t=-1} W_n(x_1,...,x_n;t)= \sum_{i=1}^n x_i^k D^k_{x_i=0} |V_n(x_1,...,x_n) |= k! |V_n(x_1,...,x_n)|.$$

Writing out the determinants in matrix form, you can identify the coefficients with the Cramer's rule soln. to the elements of the inverse of $V_n$. Each equation is then proportional to the inner product of a column covector of the adjugate matrix with a row vector of $V_n$.

matrix, I was confused. You probably received my confused answer by email while I was fixing it. – Andrej Bauer Oct 19 '11 at 7:35