Consider this functions below.
$$f(n)=\sum_{k=2}^{n}gpf(k)$$ $$g(n)=\prod_{2}^{n}gpf(k)$$
where $gpf$ is the greatest prime factor function.(For example, $gpf(30)=5$)
Is it possible to find an asymptotic formula for $f(n)$ and $g(n)$ for larger $n$?
Or lets simplify the question. Assume $p\leq n$ is a prime number. For how many $m\leq n$, $gpf(m)=p$ is true?
On
I can give you some bounds for $g$. We have $g(x) \ge \prod_{p^k \le x} p = \exp(\psi(x))$ where $\psi(x)$ is the second Chebyshev function. This has known asymptotics $ \psi(x) = x + O(x/\log x)$, so $g(x) \ge \exp(x + O(x/\log x))$.
On the other side, there's the obvious $gpf(n) \le n$, which leads to $g(x) \le x! < \exp(x \log x)$.
Numerically, it looks to me like $\log(g(x))/(x \log x)$ approaches a constant near $0.6$ as $x \to \infty$. Here's a plot:
The asymptotics of $f(n)$ is given by $$ f(n)\sim \frac{\zeta(2)}{2}\frac{n^2}{\log(n)}. $$ More generally, let $b_m(n)$ be the $m$-th power of the greatest prime factor in the prime factorization of $n$. Then Jakimczuk proved in this paper that $$ \sum_{k=2}^nb_m(k)\sim \frac{\zeta(m+1)}{m+1}\frac{n^{m+1}}{\log(n)}. $$