Basic sequence of $S_{-y}(\delta)$

Question

Let $\delta$ be a delta-operator with associated basic sequence $p_0=1,p_1,p_2,p_3,...$ and consider the shift map $S_{-y}:K[x]\to K[x]$ given by $f(x)\mapsto f(x-y)$, where $f$ is a polynomial whose coefficients are in a field of characteristic $0$.

Is there a closed form or recursive form for the new delta operator $S_{-y}\circ \delta?$

After some experimentation with $\delta=d/dx$, ie the usual derivative which has basic sequence $\{x^n\}$, I found that $q_n(x)=x(x+ny)^{n-1}$ (ie a form of the Abel polynomials) is the new basic sequence for $n>0$, with $q_0\equiv 1$. I'm wondering if a similar relation holds for any $\delta$ operator. Thanks!

See here for a description of the delta operator.

Edit:

Here is the definition of basic sequence: $\{q_n\}$ is a basic sequence for $\delta$ if:

1. $q_0\equiv 1$
2. $q_n(0)=0$ for all $n\geq 0$
3. $q_n=n\delta(q_{n-1})$ for all $n\geq 1$.

Answer 1

We use two results without proof [See On the Foundations of Combinatorial Theory. VIII. Finite Operator Calculus by G-C Rota, D. Kahaner, A. Odlyzko, Academic Press 1973]

{Theorem 1} $Q$ is a delta-operator iff $Q=DP$ for some invertible, shift invariant operator $P$, where $D=\frac{d}{dx}$ is the usual derivative operator.

{Theorem 2} If $p_n(x)$ is a basic sequence for delta operator $Q=DP$, then for $n>0$, $$p_n(x)=xP^{-n}x^{n-1}$$

Onto the problem

let $Q=\delta$, $\tilde{Q}=\tilde{\delta}$. Then, by Theorem 1 and the definition of $\tilde{Q}$, (noting that $D,\tau_{-y}$ and $P$ all commute as shift-invariant operators): $$Q=DP\implies \tilde{Q}=\tau_{-y}\delta =\tau_{-y}DP=D\tau_{-y}P$$ Henceforth, let $\tilde{P}=\tau_{-y}P$. By commutativity, $$\tilde{P}^{-n}=(\tau_{-y}P)^{-n}=P^{-n}\tau_{ny}$$

By theorem 2, the basic polynomials, $\{\tilde{p}_n\}$ of $\tilde{Q}$ are: $$\tilde{p}_n(x)=x\tilde{P}^{-n}x^{n-1}$$ Also by Theorem 2, we have that: $$p_n(x)=xP^{-n}x^{n-1}\iff \frac{p_n(x)}{x}=P^{-n}(x^{n-1}) \iff \frac{p_n(x+ny)}{x+ny}=P^{-n}(x+ny)^{n-1}=P^{-n}\tau_{ny} x^{n-1}=\tilde{P}^{-n}x^{n-1}$$ and hence, combining the above two equations, we find that: $$\tilde{p}_n(x)=x\tilde{P}^{-n}x^{n-1}=x\frac{p_n(x+ny)}{x+ny}$$ as desired.