Number of $k$-free integers of bounded radical

Question

Let $n \in \mathbb{N}$. Define the radical $R(n)$ of $n$ by

$$\displaystyle R(n) = \prod_{p | n} p.$$

In other words, $R(n)$ is the largest square-free number which divides $n$.

For an integer $k \geq 2$ we say that an integer $n$ is $k$-free (generalization of square-free) if $p | n$ implies that $p^k \nmid n$.

Define

$$\displaystyle S_k(X) = \# \{n \in \mathbb{N} : n \text{ is } k\text{-free}, R(n) \leq X\}.$$

When $k = 2$ this is just the count for square-free numbers up to $X$, and it is well-known that $S_2(X) = \frac{6}{\pi^2} X + O(\sqrt{X})$.

How does one obtain the asymptotic expression for $S_k(X)$ (if one exists) for $k \geq 3$?

Answer 1

If $m=p_1\cdots p_{\ell}$ is square-free, then the $k$-free integers $n$ that have $m$ as a radical are given by $$ \prod_{j=1}^{\ell} p_j^{a_j} $$ with $1\le a_j \le (k-1)$. Clearly there are $(k-1)^{\ell}$ such integers $n$. Therefore the problem amounts to evaluating $$ \sum_{\substack{m \le x \\ m \text{ square-free}}} (k-1)^{\omega(m)}, $$ where $\omega(m)$ denotes the number of prime factors of $m$. In other words, we are asked to compute the average of the multiplicative function $f$ given by $f(p)= (k-1)$ and $f(p^r) =0$ for larger prime powers. The standard argument comparing the Dirichlet series for this with $\zeta(s)^{k-1}$ produces the asymptotic formula $$ \sim C x \frac{(\log x)^{k-2}}{(k-2)!}, $$ with the constant $C$ given by $$ C = \prod_{p} \Big( 1 +\frac{k-1}{p} \Big) \Big( 1- \frac{1}{p}\Big)^{k-1}. $$