Let $k$ be a field of characteristic $p>0$, and let $\mathcal{D}=\mathcal{D}_k(k[x])$ be the ring of differential operators on $k[x]$. That is, we define $$ \mathcal{D}_{\leq -1}:=\{0\}\subseteq\operatorname{End}_k(k[x])\\ \text{for }n\in\mathbb{Z}_{\geq 0}:\quad\mathcal{D}_{\leq n}:=\{D\in\operatorname{End}_k(k[x])\ |\ \forall f\in k[x]:\ [D,\mu_f]\in\mathcal{D}_{\leq n-1}\}\\ \mathcal{D}:=\bigcup_{n\geq -1}\mathcal{D}_{n} $$ where $\mu_\bullet$ denotes multiplication by the element in the subscript, and $[\bullet,\bullet]$ is the Lie-bracket (i.e. $[f,g]=f\circ g-g\circ f$).
Now for $n\in\mathbb{Z}_{\geq 0}$, let $\partial_n:k[x]\to k[x]$ be defined by $\partial_n(x^m):=\binom{m}{n}x^{m-n}$, extended by $k$-linearity; it is not hard to verify that $\partial_n\in\mathcal{D}_{\leq n}$. Then is it true that $\mathcal{D}$ is generated as a $k$-algebra by $\{\partial_1,\partial_p,\partial_{p^2},\ldots\}\cup\mathcal{D}_{\leq 0}$?
Playing around a bit I think that every $\partial_n$ can be expressed in terms of $\{\partial_{p^i}\}_{i\geq 0}$, so the question reduces to whether $\{\partial_n\}_{n\geq 0}\cup\mathcal{D}_{\leq 0}$ generates $\mathcal{D}$. Is that the case?
This question is motivatet by trying to understand the ring of differential operators in positive characteristic. Note that in characteristic zero, $\mathcal{D}$ is generated by $\mathcal{D}_{\leq 1}$ and can then be described fairly concretely.
Note that $f \partial_n \in D_{\leq n}$ kills $x^0, \dots, x^{n-1}$ and sends $x^n$ to $f$. Given $D \in D_{\leq n}$, we can therefore adjust $D$ by an element of $k[x] + k[x] \partial_1 + \cdots + k[x] \partial_n$ so that $D x^i = 0$ for $i \leq n$. It therefore suffices to prove that the following claim.
Claim: If $D \in \mathcal{D}_{\leq n}$ and $D x^i = 0$ for $i \leq n$ then $D = 0$.
Proof: We will prove $D x^k = 0$ by induction on $n$ and $k$. We may assume $n \geq 0$ and $k > n$. Now $$ D x^k = x D x^{k-1} + [D, x] x^{k-1}.$$ By induction on $k$, $D x^{k-1} = 0$. Since $[D, x] x^i = 0$ for $i \leq n-1$, $[D, x] x^{k-1} = 0$ by induction on $n$. Hence $D x^k = 0$.