Consider an alphabet of $n+1$ letters: $\{0,...,n \}$. Let $z$ be a number in base $n+1$ such that it has at most $n$ digits (so the initial/first string of digits can be composed of $0$'s). Let $R_n(z)$ be the sum of these digits. For how many $z$ (for a fixed $n$) is $R_n(z)$ square-free (for every prime, divisible by at most one power thereof)? What is the behaviour of $R$ as $n$ increases (noting that for different values of $n$, $R$ will be defined on different $z$'s; $R_n$ will take strings that are too short and treat them as if they had sufficiently many $0$'s concatenated to their beginning)?
Notation: Let us call $D_n = \{ z \in \mathbb{N}:$ the base-($n+1$) expansion of $z$ has at most $n$ digits $\}$. Let us call $S_n = \{z \in D_n: R_n(z)$ is square-free in $\mathbb{N} \}$.
I am curious about the (asymptotic) ratio of$\frac{|S_n|}{|D_n|}$ (as $n \rightarrow \infty$). My guess is that it is $\frac{6}{\pi^2}$.
What about, say cube-free? Other integer powers-free?