The combinatorics of $(f \partial)^n$ in the noncommutative setting?

Question

This is a noncommutative version of these three previous questions:

differential operator power coefficients

Сlosed formula for $(g\partial)^n$

A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$?

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 in the first link mentioned above.

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.

Answer 1

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}

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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 $x_2=\dots=x_n=1$).