Average prime value in n factorial. I was wondering about the (weighted) average prime value in the factorisation of $n!$. $\\$
 If we call $f(n)$ the average prime value in $n!$, then $f$ seems to increase rather linear.
Is there a clear explanation for that? Anything known about asymptotic behavior? I know that the number of primes smaller then $x$ is approximately $x/\log(x)$ 
, but here im somewhat intrigued. A friend made a plot for me of $f(n)$. 
Check the following link:
EDIT: Since my question wasnt clear to everyone: I meant the weighted average of primes. I mean explicitly the fraction $$\frac{\sum_{p\leq n}p \sum_{k\geq 1}{[n/p^k]}}{\sum_{p\leq n}\sum_{k\geq 1}[n/p^k]} $$
https://sagecell.sagemath.org/?q=smlmgo
Thanks for any suggestions in advance. 
 A: More a comment than an answer and lots of handwaving here, but I think it is OK.  As $n$ becomes large, a prime $p$ will divide $n!$ about $\frac n{p-1}$ times. It will never reach that fraction, but will converge on it from below.  There is a switch in the order of limits here that I have not justified, but then you are asking about $$\frac {\sum p_i\frac n{p_i-1}}{\sum \frac n{p_i-1}}$$ where the sums are taken over the primes $p_i$up to some limit.  Now we can divide out the $n$ and concentrate on the primes.  This presumes that the ratio is not perturbed too much by the fact that the larger primes have not approached their limiting density as well as the smaller ones.  I think this is the biggest threat to the argument.  We know that $p_i \approx i(\log i + \log (\log (i))-1)$ so this becomes the study of the asymptotics of $$\frac {\sum i(\log i + \log (\log (i))-1)\frac 1{i(\log i + \log (\log (i))-1)-1}}{\sum \frac 1{i(\log i + \log (\log (i))-1)-1}}$$  I couldn't get Alpha to help, but someone with Mathematica will find this easy.  I hope this prompts somebody to work on it further.  
Added:  Daniel Fischer shows in a comment that this leads to the average being approximately $$\dfrac{n}{(\log n)(\log\log n)}$$
