3
$\begingroup$

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 $k \geq 3$?

$\endgroup$
3
$\begingroup$

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 $$ C = \prod_{p} \Big( 1 +\frac{k-1}{p} \Big) \Big( 1- \frac{1}{p}\Big)^{k-1}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.