Given $n\in\Bbb N$, we know that $\ln(n\#)\sim n$ where $n\#=\prod_{i=1}^{\pi(n)}p_i$ is primorial function with $p_i$ being $i$th prime while $\pi(n)$ being number of primes below $n$. How many distinct factors in range $[\frac{\sqrt{n\#}}2,\frac{3\sqrt{n\#}}2]$ does $n\#$ contain?
I think it is more simple than that. You have $\ln(n\#)\sim n$. We have $\pi(n)$ numbers atmost. Note that $\ln(n\#)-\ln(\frac{n}2\#)\sim n$. This implies if we seek products from primes from $n$ to $\frac{n}2$ we should have reasonable estimate. Number of primes from $n$ to $\frac{n}2$ is $\pi(n)-\pi(\frac{n}2)\sim \frac{n}{\ln n}$. We seek to multiply half of these at a time to get an estimate of $\binom{\frac{n}{\ln n}}{\frac{n}{2\ln n}}\sim 2^{\frac{n}{\ln n}}\sqrt{\frac{2\ln n}{n\pi}}$.
Hence, we have atleast $ 2^{\frac{n}{\ln n}}\sqrt{\frac{2\ln n}{n\pi}}$ distinct factors in range $[\frac{\sqrt{n\#}}2,\frac{3\sqrt{n\#}}2]$. Is my estimate anything reasonable?
A really rough estimate is to look at the total number of factors ($2^{\pi(n)}$), and assume that each factor has a logarithm with a random value in the range $(0,n)$. The size of your interval mapped logarithmically is $\ln(3)$, so say each factor has a chance of $ln(3)/n$ of being valid. Then $\frac{\ln(3)}n 2^{\pi(n)}$ is a really, really rough estimate. Note, however, that there are only $2^{\pi(n)}$ possible factors, so this is a trivial upper bound.
As far as more precise estimates are concerned, we do not even know enough about prime distribution to get very satisfactory results on estimating $\ln(n\#)$ or prime gaps, so there does not seem to be much hope for an estimate with an exact range or an estimate that is very accurate at all.