Smallest constant so that there are at least $n/\log_2{n}$ primes between $n$ and a constant multiple of $n$ What is the smallest known $c$ so that for any $n\geq 2$ there are at least $n/\log_2{n}$ primes between $n$ and $cn$ (inclusive)?
The prime number theorem seems to give an asymptotic result so I am not sure how to get a particular constant from it.
I am interested in answers that depend on well known conjectures as well as answers that are known to be correct.
Following a comment, the notation $\log_2{n}$ refers to taking the logarithm base $2$.
[Cross-posted from math.se where $c=13/6$ was suggested as a possibility.]
 A: Indeed, the suggestion given in the other thread is quite appropriate.  Use a lower bound from Dusart for $\pi(cn)$, and an upper bound for $\pi(n)$, and you want the difference between these bounds to be at least $n/\log_2 n$.  Using results from Dusart which apply for $n \gt $ three billion (primarily that $\pi(n) \leq (n/ \ln n)[1 + 1/ \ln n + 2.334 / (\ln n)^2]$ ), rewriting $1/ \ln n$ as $\epsilon$ and $1/ \ln cn$ as $\epsilon/(1 + \epsilon \ln c)$, after the dust settles one wants $c$ that satisfies
$$c \frac{1}{1 + \epsilon \ln c} [1 + \frac{\epsilon}{1 + \epsilon \ln c} 
+ \frac{2\epsilon^2}{(1+ \epsilon \ln c)^2}] > \ln 2 + 1 + \epsilon + 2.334 \epsilon^2,$$
where we have $\epsilon \lt 1/20$.  Thus, look for a $c$ that works for the first three billion $n$ (I haven't checked, but $c=13/6$ seems like a good candidate), and make sure that the above equation is satisfied for this $c$ and $\epsilon$ not too large (which it is for $c=13/6$, since the left hand side is at least 2 and the right hand side is less than 2.
Gerhard "Just A Matter Of Computation" Paseman, 2015.05.15
A: Assuming I made no mistake, $c=11/5$ is the smallest such value.
Let $c=11/5$. Using Theorem 1 of Rosser-Schoenfeld, we see easily that
$$\pi(cn)-\pi(n-1)\geq(\log 2)\frac{n}{\log n},\qquad n\geq 74,$$
where $\log$ is the natural logarithm as usual in analytic number theory. We can verify the same inequality for $2\leq n\leq 73$ by computer. Finally, for $n=5$ the above inequality means that there are at least $3$ primes in $[5,5c]$, whence $c\geq 11/5$ is necessary.
