<s>[INST] Fraction of prime number between consecutive powers of two A $n$-bit integer is an integer $x$ such that $2^{n-1} \le x < 2^n$. In [1] it is claimed without proof that a corollary of the prime number theorem is that:

For any $n > 1$, the fraction of $n$-bit integers that are prime is at least $\frac{1}{3n}$.

Can somebody point me to a proof of this?
It is (perhaps) worth noticing that a direct application of the prime number theorem allows to prove an asymptotic version of the above (with a slightly better constant).
$$
\begin{align*}
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg( \pi(2^n-1) - \pi(2^{n-1})\bigg) &= 
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg(  \pi(2^n) - \pi(2^{n-1}) \bigg) \\&=
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}}  \bigg( \frac{2^n}{n \ln 2}  - \frac{2^{n-1}}{(n-1)\log2} \bigg)\\ & =
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}}  \bigg( \frac{2^n}{2n \ln 2 } \cdot\frac{n-2}{n-1}   \bigg) = 1.
\end{align*}
$$
Therefore $\frac{\pi(2^n-1) - \pi(2^{n-1})}{2^{n-1}} \sim \frac{1}{n \ln 2}$, and for every $\varepsilon >0$ there is some $n_0$ such that, for all $n \ge n_0$, the fraction of $n$-bit integers that are prime is at least $\frac{1-\varepsilon}{n \ln 2}$.
A not-so-elegant way to prove the the claim would be that of choosing $\varepsilon =1-\frac{\ln 2}{3} \approx 0.7689$ and finding an upper bound on $n_0$ that lies in the range for which $\pi(\cdot)$ has been computed. Are such upper bounds known?
[1] Jonathan Katz, Yehuda Lindell. Introduction to Modern Cryptography (3rd edition).  CRC Press. ISBN 9781351133012.
 [/INST] This is just a more detailed write-up of the strategy proposed by Marty Cohen's answer.
We can prove the stronger claim: for $n \ge 1$ there are at least $\frac{2^{n-2}}{2n}$ $n$-bit prime numbers (i.e., the fraction of $n$-bit prime numbers is at least $\frac{1}{2n}$).
According to [1], for $x \ge 55$,
$\dfrac{x}{\log x+2}
\lt \pi(x)
\lt \dfrac{ n}{\log x-4}
$.
Defining $L(n) = \dfrac{2^n}{n\log 2+2}-\dfrac{ 2^{n-1}}{(n-1)\log 2-4}$ we have:
$
\pi(2^n -1 )-\pi(2^{n-1}) = \pi(2^n)-\pi(2^{n-1}) \ge L(n). 
$
As a hint that we are going towards the right direction, we notice that $L(n)$ is roughly:
$$
\dfrac{ 2^n}{n\log 2}-\dfrac{ 2^{n-1}}{(n-1)\log 2} \approx
\dfrac{2^n}{n\log 2}-\dfrac{ 2^{n-1}}{n\log 2}
= \dfrac{2^{n-1}}{n\log 2} > \dfrac{2^{n-1}}{n}.
$$
More formally, let $\ell(n) = \dfrac{2n \log 2}{n \log 2 + 2} - \dfrac{n \log 2}{(n-1) \log 2 - 4}$:
$$
L(n) =  \dfrac{2^n}{n\log 2+2}-\dfrac{ 2^{n-1}}{(n-1)\log 2-4}
= \dfrac{2^{n-1}}{n \log2} \left( \dfrac{2n \log 2}{n \log 2 + 2} - \dfrac{n \log 2}{(n-1) \log 2 - 4} \right) = \dfrac{2^{n-1}}{n \log2} \cdot \ell(n).
$$
Taking the derivative of $\ell(n)$ we see that:
$$
\dfrac{d \ell(n)}{dn} = \dfrac{4 \log 2}{(2 + n \log 2)^2} + \frac{(4 + \log 2) \log 2}{(n \log 2  - 4 - \log 2)^2} \ge 0,
$$
therefore $\ell(n)$ is defined for all $n \ge 10$ and monotonically increasing. Evaluating $\ell(10)$ we see that $\ell(10) \ge \frac{1}{2}$. Therefore:
$$
L(n) \ge \dfrac{2^{n-1}}{n \log2} \cdot \dfrac{1}{2} \ge \frac{2^{n-1}}{2n}.
$$
This proves the claim for $n \ge 10$. For $ 2 \le n < 10$ we can exhaustively inspect $\pi(2^n -1 )-\pi(2^{n-1})$.
[1] "Explicit Bounds for Some Functions of Prime Numbers",
Barkley Rosser
American Journal of Mathematics
Vol. 63, No. 1 (Jan., 1941), pp. 211-232
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