How many numbers between a prime $p$ and $p^2$ have some prime factor $>p$? For $p=2$ the answer is all of them (given that between here can only reasonably mean strictly between).  For $p=3$ or 5 or 7 the answer is just under half of them.  But for $p=11$, unless I slipped up it is just 29 out of 110.
I am afraid the answer as $p$ rises may be embarrassingly obvious, but don't see the pattern, so I want to ask.  
I am not looking for an exact count.  That is why I did not report the exact count for $p=3,5,7$ though of course I found it.
 A: As RideTheWavelet noted in comments below the OP, the exact answer is given by
$$\sum_{p\lt q\le p^2}\left\lfloor p^2\over q\right\rfloor$$
where the sum is over primes $q$ between $p$ and $p^2$. An upper bound is therefore
$$p^2\sum_{p\lt q\lt p^2}{1\over q}$$
Since 
$$\sum_{q\lt x}{1\over q}\approx\ln(\ln x)$$
for large $x$, we get $p^2(\ln(\ln(p^2)-\ln(\ln p))=p^2\ln2$ as an approximate upper bound for large $p$. For a lower bound we have
$$p^2\sum_{p\lt q\lt p^2}{1\over q}-\sum_{p\lt q\lt p^2}1\approx p^2\ln2-(\pi(p^2)-\pi(p))\approx p^2\ln2-{p^2\over2\ln p}=p^2\left(\ln2-{1\over2\ln p} \right)$$
Again, the approximations here are trustworthy only for large $p$.  However, we can at least look at them for a small prime like $p=11$. They suggest $121\ln2\approx83.87$ as an upper bound and $121(\ln2-1/(2\ln11))\approx58.64$ as a lower bound.  This doesn't come close to the OP's count of $29$. But that count is inaccurate:  There are $25$ primes between $11$ and $121$, and the correct count turns out to be
$$\sum_{11\lt q\lt121}\left\lfloor121\over q\right\rfloor=\left\lfloor121\over 13\right\rfloor+\left\lfloor121\over 17\right\rfloor+\cdots+\left\lfloor121\over 113\right\rfloor=9+7+\cdots+1=60$$
So that untrustworthy lower bound, in this case, is surprisingly accurate!
A: We know the number of primes in $[p,p^2]$ is gonna be $\pi(p^2)-\pi(p)$. However we also need to include numbers of the form $kq$, where $q$ is prime and $p<q<p^2$. For each such $q$, there are roughly $(p^2-p)/q$ such $k$'s. 
The $n$'th prime $p_n\approx n\ln n+n\ln\ln n-n$. If $q$ is prime, then it's index is $\pi(q)$. So we're summing:
$$\sum_{p<q<p^2, q:\mbox{ prime }}\frac{p^2-p}{q}\approx (p^2-p)\sum_{r=\pi(p)}^{\pi(p^2)}\frac{1}{r\ln r}\approx (p^2-p)[\log(\log(\pi(p^2))-\log(\log(\pi(p))].$$
