Closed form for sequence A145271 I would like to know if there is a simple formula or method
of expanding the expression given by
$\left[g(x) \frac{d}{dx}\right]^n g(x)$
where $n$ is a positive integer, without having to resort to
actually carrying out the differentiation. The aim is like
that of expanding $(x + y)^n$. For this one, there is a simple 
formula (goes like combination $n$ take $r$) to obtain each 
numerical coefficient and each term in the expansion can simply 
be written without the need to actually carry out the 
multiplication. One can likewise simply use Pascal's triangle.
Attempt at solution
Ignoring the numerical coefficients, one can simply expand the
expression above using Ferrer's diagram. Take for instance, 
$n = 4$. The corresponding set of Ferrer's diagrams is
$
\begin{align*}
\begin{array}[t]{l}
\bigcirc\\
\bigcirc\\
\bigcirc\\
\bigcirc
\end{array}
\qquad
\begin{array}[t]{ll}
\bigcirc & \bigcirc\\
\bigcirc\\
\bigcirc\\
\end{array}
\qquad
\begin{array}[t]{ll}
\bigcirc & \bigcirc\\
\bigcirc & \bigcirc
\end{array}
\qquad
\begin{array}[t]{lll}
\bigcirc & \bigcirc &\bigcirc\\
\bigcirc
\end{array}
\qquad
\begin{array}[t]{llll}
\bigcirc & \bigcirc &\bigcirc&\bigcirc
\end{array}
\end{align*}
$
The number of columns of each Ferrer's diagram corresponds 
to the number of $g$ (or its derivatives) factors plus one
for each term in the expansion. Thus, the expansion for $n = 4$
takes the form
$\left[g(x) \frac{d}{dx} \right]^{n=4}g(x)
:
( )( ) + ( )( )( ) + ( )( )( ) + ( )( )( )( ) + ( )( )( )( )( )
$
The length of each column for a given Ferrer's diagram 
corresponds to the order of derivatives. We can then improve the
schematic relation above as
$
\begin{align*}
\left[g(x) \frac{d}{dx}\right]^{n=4}g(x)
&:
(g'''')( ) + (g''')(g' )( ) + (g'')(g'')( ) + (g'')(g')(g')( ) + (g')(g')(g')(g')( )
\\
&=
g''''() + g'''g'() + g''^2() + g''g'^2() + g'^4()
\end{align*}
$
The remaining factor to fill in () is $g$ to a certain power. For each term, the exponent is simply the sum of orders of derivative of each factor minus the number of factors with derivatives plus 1. For instance, for $g'''g'( )$, the sum of the orders of derivatives is 4 (just count the number of primes) and the number of factors with derivatives is 2. As such, $g'''g'() = g'''g'g^{4 - 2 + 1} = g'''g'g^3$. Similarly, $g''g'^2( ) = g''g'^2g^{4 - 3 + 1} = g''g'^2g^2$ One can then simply write down (ignoring the numerical coefficients) the expansion as
$
\begin{align*}
\left[g(x) \frac{d}{dx} \right]^{n=4}g(x)
&:
g''''g^4 + g'''g'g^3 + g''^2g^3 + g''g'^2g^2 + g'^4g
\end{align*}
$
The problem now is the coefficient of each term in the
expansion. The sequence of coefficients is actually 
sequence number A145271 (OEIS). For $n = 4$, this sequence
is simply $\{1, 7, 4, 11, 1\}$ so that
$
\begin{align*}
\left[g(x) \frac{d}{dx} \right]^{n=4}g(x)
&=
g''''g^4 + 7g'''g'g^3 + 4g''^2g^3 + 11g''g'^2g^2 + g'^4g
\end{align*}
$
but I cannot find a simple procedure to reproduce this sequence 
without actually expanding the original expression above through
differentiation; hence, this post. Any diagrammatic method,
closed form expression, or a null statement (eg., such a
closed form expression does not exist) would be greatly
appreciated.
[Note: I am not a mathematician and is not very familiar
with the area of Combinatorics. I encountered this problem
while trying to solve a differential equation in physics.]
 A: There is no simple procedure. Even in the simpler case $\left[x\frac{d}{dx}\right]^n g(x)$, the formula involves the Stirling numbers of the second kind $S(n,k)$ : 
$\left[x\frac{d}{dx}\right]^n g(x) = \sum_{k=0}^{n}S(n,k)x^k\frac{d^kg}{dx^k}$
Eq.10 in : http://mathworld.wolfram.com/DifferentialOperator.html
A: The OEIS entry A145271 now contains a graded matrix computation for the sequence of partition polynomials, involving Pascal's triangle: Multiply the $n$-th diagonal (with $n=0$ the main diagonal) of the lower triangular Pascal matrix by $g_n = D_x^n g(x)$ to obtain the matrix $VP$ with $VP_{n,k} = \binom{n}{k}g_{n-k} $. Then $(g(x)D_x)^n g(x) = (1, 0, 0,..,0) [VP \dot \; S]^n (g_0, g_1, g_2, .., g_n)^T$, where S is the shift matrix A129185, representing differentiation in the divided powers basis $x^n/n!$. 
Example:
$$(g(x)D_x)^3 g(x)$$
$$= (1, 0, 0, 0) [VP \dot \; S]^3 (g_0, g_1, g_2, g_3)^T$$
$$= \begin{pmatrix}
1 &  0 &  0 & 0
\end{pmatrix} \begin{pmatrix}
0 & g_0 & 0 & 0 \\ 
0 & g_1 & g_0 & 0\\ 
0 & g_2 & 2g_1 & g_0 \\ 
0 & g_3 & 3g_2 & 3g_1
\end{pmatrix}^3 \begin{pmatrix}
g_0 \\ 
g_1 \\ 
g_2 \\ 
g_3 \end{pmatrix} $$
$$ = g_0g_1^3 + 4 g_0^2g_1g_2 + g_0^3g_3$$
$$= \begin{pmatrix}
1 &  0 &  0 & 0
\end{pmatrix} \begin{pmatrix}
0 & g_0 & 0 & 0 \\ 
0 & g_1 & g_0 & 0\\ 
0 & g_2 & 2g_1 & g_0 \\ 
0 & g_3 & 3g_2 & 3g_1
\end{pmatrix}^4 \begin{pmatrix}
0 \\ 
1 \\ 
0 \\ 
0 \end{pmatrix} $$
This is related to the partition polynomials characterized by Comtet in the reference in A139605.
And, the pdf Mathemagical Forests gives a diagrammatic method for creating forests of trees through "natural growth" that represent the partition polynomials.
Edit (2019): See an update in MQ-QExpansion of iterated derivatives ... .
A: A145271, which I authored, contains in the Crossref Section a list of several Lagrange inversion formulas (LIFs) with different reps of $g(x)$ for obtaining the series for the formal compositional inverse of a convergent series, or analytic function, about the origin. These LIFs have nonrecursive formulas for the coefficients of each partition at each order of the series. To convert from any of these expressions to the partitions of A145271, you can use either A133314 (signed A049019) or A263633, which have explicit multinomial formulas that can be gleaned from one in A049019.
For example, to find the inverse series for $f(x)$ with $f(0) = 0$ and $f'(0)=1,$ the classic LIF A134685 can be used with
$$g(x) = 1/f'(x)= 1 / [1 + f_2 x + f_3 \frac{x^2}{2!} + ...],$$
so
$$ 1 + f_2 x + f_3 \frac{x^2}{2!} + ... = 1 / [1 + g_1 x + g_2 \frac{x^2}{2!} + ...]$$
with $f_n$ given by A133314 with $b_n = f^{(n+1)}(0)  = f_{n+1}$ and $a_n = g_n$.
Then the first few coefficients of the reciprocal are
$$ f_1 = g_0 = 1$$
$$ f_2 = -g_1 $$
$$ f_3 = -g_2 + 2g_1^2$$
$$ f_4 = -g_3 + 6g_2g_1 - 6 g_1^3.$$
Plugging these expressions into A134685 for, say, the third order term gives
$$ 3f_2^2 - f_3 = 3g_1^2 + g_2 - 2g_1^2= g_1^2 + g_2$$ 
in agreement with A145271.
