The entry OEIS [A139605](https://oeis.org/A139605) (also related OEIS [A145271](https://oeis.org/A145271)) has a matrix computation for the partition polynomials that represent the expansions of iterated derivatives, or vectors in differential geometry, 


$$(g(x)D\_x)^n.$$


The formula section of A139605 contains the matrix formula. Multiply the $n$-th diagonal (with $n=0$ the main diagonal) of the lower triangular Pascal matrix [A007318](https://oeis.org/A007318) by $g\_n = D\_x^n g(x)$ to obtain the matrix $VP$ with $VP\_{n,k} = \binom{n}{k}g\_{n-k} $. Then 
$$(g(x)D)^n = (1, 0, 0,..) [VP \dot \; S]^n (1, D, D^2, ..)^T,$$
where S is the shift matrix [A129185](https://oeis.org/A145271), representing differentiation in the divided powers basis $x^n/n!$. 


Example:


$$(g(x)D\_x)^3$$


$$= (1, 0, 0, 0) [VP \dot \; S]^3 (1, D, D^2, D^3)^T$$


$$= \begin{pmatrix}
1 & 0 & 0 & 0
\end{pmatrix} \begin{pmatrix}
0 & g\_0 & 0 & 0 \\ 
0 & g\_1 & g\_0 & 0\\ 
0 & g\_2 & 2g\_1 & g\_0 \\ 
0 & g\_3 & 3g\_2 & 3g\_1
\end{pmatrix}^3 \begin{pmatrix}
1 \\ 
D \\ 
D^2 \\ 
D^3 \end{pmatrix} $$


$$ = [g\_0g\_1^2 + g\_0^2 g\_2] D + 3 g\_0^2g\_1 D^2 + g\_0^3D^3 $$


And, the pdf [Mathemagical Forests](http://tcjpn.wordpress.com/2008/06/12/mathemagical-forests/) gives a diagrammatic method for creating forests of trees through "natural growth" that represent the partition polynomials.


I have either lost a proof of the validity of this formula or got sidetracked before I developed one.


**Question**:


Can someone prove this conjecture?


**Some background:**


The refined Eulerian numbers (RENs) of [A145271](https://oeis.org/A007318) are related analytically to the compositional inversion of functions and formal generating series and to flow fields generated by tangent vectors. The $n$-th row of RENs are the numerical coefficients of the expansion of $(g(x)\frac{d}{dx})^ng(x)$ in terms of the monomials in the derivatives of $g(x)$, i.e.,


$$g\_k=\frac{d^k}{dx^k}g(x).$$


For example,


$$(g(x)\frac{d}{dx})^3g(x) = 1 g\_0^1 g\_1^3 + 4 g\_0^2 g\_1^1 g\_2^1 + 1 g\_0^3 g\_3^1.$$


With $(\omega,x) = (f(x),f^{(-1)}(\omega))$ and $g(x) = 1/f^{'}(x)$,


$$\exp[t g(x)d/dx]x = \exp[td/d\omega]f^{(-1)}(\omega) = f^{(-1)}(t+\omega)=f^{(-1)}(t+f(x)).$$


Evaluated at the origin of $x$, this gives the compositional inverse 


$$\exp[tg(x)d/dx] x |\_{x=0}=f^{(-1)}(t).$$


See also


1) [MO-Q](https://mathoverflow.net/questions/339061/guises-of-the-refined-eulerian-numbers-generated-by-tangent-vectors-oeis-a1452) Guises of the refined Eulerian numbers generated by tangent vectors


2) [MO-Q](https://mathoverflow.net/questions/214927/important-formulas-in-combinatorics/215203#215203) Important formulas in combinatorics


3) [MO-Q](https://mathoverflow.net/questions/145555/why-is-there-a-connection-between-enumerative-geometry-and-nonlinear-waves/181534#181534) Why is there a connection between enumerative geometry and nonlinear waves?



Expansions of iterated, or nested, derivatives, or vectors--conjectured matrix computation
I have finally written up the proof in detail. It is in my note


* [Darij Grinberg, *Commutators, matrices and an identity of Copeland*](http://www.cip.ifi.lmu.de/~grinberg/algebra/copeland1.pdf), also available as [arXiv:1908.09179v1](https://arxiv.org/abs/1908.09179).


Your result is a particular case of Theorem 4.2. More precisely, you get it from Theorem 4.2 if you set $\mathbb{L}$ to be the ring of differential operators (whatever kind of differential operators you are considering), $\mathbb{K}$ to be the base ring, $a$ to be the differentiation operator (which is your $D$), $x$ to be the "multiplication by $t$" operator (you use $x$ for what I call $t$, but I keep them separate because my ring $\mathbb{L}$ doesn't have to contain the polynomial ring $\mathbb{K}\left[t\right]$), and $h$ to be the identity map.


I derive this from a more general formula (Theorem 2.7), which expresses $\left(ba\right)^n$ as a matrix product when $a$ and $b$ are two arbitrary elements of a noncommutative ring.


I work in a setting that allows the matrices to be finite or infinite as the reader prefers. This is responsible for a lot of the length of the note (as I have to set up an appropriate ring of infinite matrices on which multiplication is defined and associative, but also mess around with partial equalities in the case of finite matrices because the finite version of the shift operator doesn't behave as well as the infinite one). I hope the exposition is not completely wasted; meanwhile I trust you know what to skip and what to skim.


