Does $$\frac{1}{n-1}\sum_{i=2}^{n}{\frac{\log{\text{gpf}(i)}}{\log{i}}}$$ have a lower bound as $n\rightarrow\infty$? Here, $n\in\mathbb N$ with $n>1$, and gpf returns the greatest prime factor of an integer.
Empirically, it seems to get down to around $2/3$ quickly and then really slows down, although seems to decrease on average. Here's a plot for $n\leq 500$.
Basically, I'm interested in the long-term behavior of gpfs as a proportion of their integer as the integer grows. I tried looking at just $\frac{\log{\text{gpf}(n)}}{\log{n}}$ first, but it seems way too chaotic to identify any kind of trend.
I'm wondering if there's a theorem (or even conjecture or heuristic) which predicts whether this very slowly decreases without bound (aside from a trivial $0$ lower bound, I guess), or whether it's bounded below. In particular, I'm curious whether there's any chance the mean could dip below $0.5$ eventually, which my gut says is unlikely, but my gut is not a number theorist.
The average converges to the Golomb-Dickman Constant. A helpful heuristic argument is given in Prime Numbers and Computer Methods for Factorization by Hans Riesel on pages 157-158.
Let $N$ be any number with exactly $s$ (not necessarily distinct) prime factors. Then, since the number of prime factors $\omega(n) \approx \ln(\ln(n))$ it follows that $s-1 \approx \ln(\ln(\frac{n}{P(n)}))$ where $P(n) = \text{gpf}(n)$
$$\implies s-1 \approx \ln(\ln(\frac{n}{P(n)})) = \ln(\ln(n) -\ln(P(n)))$$ $$=\ln(\ln(n)) + \ln(1-\frac{\ln(P(n))}{\ln(n)})=s+ \ln(1-\frac{\ln(P(n))}{\ln(n)})$$
Therefore, heuristically, $$-1 \approx \ln(1-\frac{\ln(P(n))}{\ln(n)})$$ $$\frac{\ln(P(n))}{\ln(n)} \approx 1- \frac1e$$
It follows that $\displaystyle \sum_{i=2}^{n}{\frac{\log({P(i)})}{\log({i})}}\approx n \cdot (1- \frac1e)$
Of course this isn't rigorous, but I think it gives an intuitive feel as to why your average exists and is bigger than $0$. There exist way more rigorous approaches to the problem out there, but I think that this argument is quite neat
Edit: I also found this answer to be very useful