Let $P\subseteq \mathbb{N}$ be the set of primes, and for any integer $n>1$ let $L(n) = \max\{p \in P: p \mid n\}$ be the largest prime divisor of $n$. Moreover, for $n \in \mathbb{N}$ with $n>1$ we let $M(n)$ to be the median of the set $$\{L(m)/m : m\in \mathbb{N} \land 1 < m \leq n\}.$$
Does $\lim_{n\to\infty}M(n)$ exist? If yes, is its value known?
The number of prime divisors of $n$ grows typically as $\log \log n$. Suppose $n$ has $k$ prime factors. Now $n/L(n)$ has only $k-1$ prime factors, so $$ k-1 \approx \log \log \frac{n}{L(n)} = \log \log n + \log\left( 1-\frac{\log L(n)}{\log n}\right) = k + \log\left( 1-\frac{\log L(n)}{\log n}\right) $$ and hence $$ \log\left( 1-\frac{\log L(n)}{\log n}\right) \approx -1 $$ i.e., typically we will have $$ \log L(n) \approx \left( 1-\frac{1}{e}\right)\log n $$ or indeed $$ L(n) \approx n^{1-\frac{1}{e}} = n^{0.632...}. $$ Note that this means that typically the largest prime divisor of $n$ will be bigger than $\sqrt{n}$, which is a nice isolated statement as well.
As mathworker21 noted, it suffices to show that the average of $L(m)$ for $1 < m \leq n$ tends to $0$ as $n \to \infty$. Let $A(n)$ be this average. Thus in the typical case $$ A(n) = \frac{1}{n} \sum_{m=1}^n \frac{L(m)}{m} \approx \frac{1}{n} \sum_{m=1}^n \frac{1}{m^{\frac{1}{e}}} = \frac{1}{n} H_{n,\frac{1}{e}} $$ where $H_{n,\frac{1}{e}}$ is the $n$th generalized harmonic number of order $\frac{1}{e}$. As $n\to \infty$, an asymptotic expansion of this right-hand side yields $$ A(n) \approx \frac{1}{n} H_{n,\frac{1}{e}} = \frac{e}{n^{1/e}(e-1)} + \frac{\zeta(\frac{1}{e})}{n} + O\left( \frac{1}{n^2}\right) \to 0 $$ as $n \to \infty$. Hence also $\lim_{n \to \infty} M(n) = 0$.
