Kostant showed that the subring of $\mathcal U(\mathfrak{sl}_2)$ generated by the divided powers $e^c/c!$ and $f^a/a!$ has a $\mathbb Z$-basis given by the elements $\frac{f^a}{a!}\binom hb \frac{e^c}{c!}$ for $a,b,c\in\mathbb N$.

Is there an analogous basis for the $\mathbb Z[q,q^{-1}]$-subalgebra of $\mathcal U_q(\mathfrak{sl}_2)$ generated by $K^{\pm 1}$ and the divided powers $e^c/[c]_q!$ and $f^a/[a]_q!$?

I believe (see e.g. Equation (8) in pg 238 of Jantzen, [Lectures on Quantum Groups](https://doi.org/10.1090/gsm/006)) the question boils down to the following:

Write

\begin{gather*}
[K;n] := \frac{Kq^n-K^{-1}q^{-n}}{q-q^{-1}}, \\
{K;a \brack n} := \frac{[K;a][K;a-1]\dotsm[K;a-n+1]}{[n]_q!}.
\end{gather*}

What is a $\mathbb Z[q,q^{-1}]$-basis for the $\mathbb Z[q,q^{-1}]$-algebra $A$ generated by $K^{\pm 1}$ and ${K;a\brack n}$?

I guess this would be a $q$-analogue of the fact that the subring $R$ of $\mathbb Q[h]$ generated by $\binom hn$ has a $\mathbb Z$-basis generated by the $\binom hn$ themselves.

The ring $R$ is the set of polynomials $f\in\mathbb Q[h]$ such that $f(\mathbb Z)\subseteq \mathbb Z$. Is there a similar interpretation for the $\mathbb Z[q,q^{-1}]$-algebra generated by $K^{\pm 1}$ and ${K;0\brack n}$?