This is a noncommutative version of these three previous questions:


[differential operator power coefficients](https://mathoverflow.net/questions/80828/differential-operator-power-coefficients)


[Сlosed formula for $(g\partial)^n$](https://mathoverflow.net/questions/337330/%D0%A1losed-formula-for-g-partialn)


[A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$?](https://mathoverflow.net/questions/415617/a-leibniz-like-formula-for-fx-fracddxn-fx)


Let $(R,\partial)$ be a noncommutative differential ring (unitality does not play a role), i.e.
we have $\forall a,b \in R$:
$$
\partial (ab) = \partial(a) b + a \partial (b).
$$
Fix a generic element ("regular function") $f \in R$. I am interested in an explicit description of the iterations
$$
(R\_f \partial)^n f,\ n \in \mathbb{N},
$$
where $R\_f$ denotes the multiplication operator by $f$ *from the right*. For example:
$$
(R\_f \partial) f = \partial(f) f
$$
and
$$
(R\_f \partial)^2 f = (R\_f \partial) (\partial(f) f) = \partial^2(f) f^2 + \partial(f)^2 f
$$
My question is thus:



> 
> Is there a known explicit description for the words made out of the letters $\partial^n(f),\dotsc,\partial(f),f$ involved in the expansion of $(R\_f \partial)^n f$ and the coefficients in front of these words? Has this been investigated anywhere?
> 
> 
> 


Ideally, I am looking for a description similar to Comtet's theorem in the commutative case cited by [Gjergji Zaimi](https://mathoverflow.net/a/80873) in [the first link mentioned above](https://mathoverflow.net/questions/80828/differential-operator-power-coefficients).


**Aside:** My setting is actually slightly more complicated than this. I only have a derivation "with a twist":
$$
\partial (ab) = \partial(a) \varphi(b) + a \partial(b),
$$
where $\varphi$ is an (injective, non-unital) ring endomorphism with $[\varphi,\partial]$ not being very illuminating, which only seems to complicate the combinatorics even further.



The combinatorics of $(f \partial)^n$ in the noncommutative setting?
In fact, Comtet's formula works almost directly in noncommutative case under an appropriate ordering of the products. Here is just a bit deeper look under the hood.


In umbral form $(R\_f \partial)^n f$ can be written as polynomial
$$f\_n(x\_1,\dots,x\_n) := x\_1(x\_1+x\_2)(x\_1+x\_2+x\_3)\cdots(x\_1+\dots+x\_n),$$
where in the expansion of $f\_n(x\_1,\dots,x\_n)$ each monomial $x\_1^{a\_1}x\_2^{a\_2}\cdots x\_n^{a\_n}$ corresponds to
$(\partial^{a\_1}f)(\partial^{a\_2}f)\cdots(\partial^{a\_n}f)f$.


The coefficient of $x\_1^{a\_1}x\_2^{a\_2}\cdots x\_n^{a\_n}$ in $f\_n(x\_1,\dots,x\_n)$ is nonzero only if the exponents satisfy the inequalities
$$a\_n + a\_{n-1} + \dots + a\_{n-k+1} \leq k,\qquad k\in\{1,2,\dots,n-1\},$$
with the total degree being $n$:
$$a\_n + a\_{n-1} + \dots + a\_1 = n,$$
in which case this coefficient is given by the formula:
\begin{split}
&\binom{1}{a\_n}\binom{2-a\_n}{a\_{n-1}}\binom{3-a\_n-a\_{n-1}}{a\_{n-2}}\cdots \binom{n-a\_n-a\_{n-1}-\dots-a\_2}{a\_1} \\
&=\frac{(2-a\_n)(3-a\_n-a\_{n-1})\cdots (n-a\_n-a\_{n-1}-\dots-a\_2)}{a\_1!a\_2!\cdots a\_n!}.
\end{split}




---


PS. It's worth to notice connection of $f\_n(x\_1,\dots,x\_n)$ to other combinatorial objects, such as $q$-factorial (for $x\_i=q^{i-1}$) and the generating function for Stirling numbers of first kind (for $x\_2=\dots=x\_n=1$).


