I guess that the number of integers $x$ which satisfy the condition $x\*p(x) \leq n$ is $O(n^{2/3})$ or $O(n^{3/4} / \ln n$), but I cannot prove it. I just write a program to count the number. The results are listed.  

Note: $p(x)$ means the largest prime factor of $x$.  

n the numbers  

$10^5$ 1894  

$10^6$ 9108  

$10^7$ 44948  

$10^8$ 228102 



How many integers $x$ satisfy that $x*p(x) \leq n$, where $p(x)$ means the largest prime factor of $x$?
Quick observation. Choose $\alpha$ close to 1 and consider the number for which $x\leqslant n^{\alpha}=:N$, $p(x)\leqslant n^{1-\alpha}=N^{(1-\alpha)/\alpha}$. The number of such $x$ grows as $\rho(\frac\alpha{1-\alpha})N$, where $\rho$ is [Dickman's function](https://en.wikipedia.org/wiki/Dickman_function). It already implies that the number of $x$ satisfying $xp(x)\leqslant n$ grows faster than $n^{\alpha}$ for any $\alpha<1$.


