How many $2048$ bit primes with the form $2q+1$ are there? Suppose $p,q$ are prime and $p = 2q+1$. Assuming that the likelihood of $q$ and $2q+1$ being prime are independent, how many $2048$ bit primes with the form $2q+1$ do you expect exist?
 A: A few broad hints, to leave you something to do for yourself:


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*We say that 'the probability' of a number $x$ being prime is $\frac1{\ln x}$, in the sense that the Prime Number Theorem states that there are approximately $\frac{x}{\ln x}$ primes less than $x$ — and more to the point, the number of primes between $x$ and $x+n$ is approximately $\frac{n}{\ln x}$, when $n\in O(x)$ (you can convince yourself that the PNT implies that this holds to first order whenever $n\lt cx$ for arbitrary constant $c$).  Given this, you should be able to figure out (approximately) how many primes $q$ there are between $2^{2046}$ and $2^{2047}-1$ (i.e., those for which $2q+1$ will be a number in your original $2048$-bit range).

*Once you have an approximate count of the $2047$-bit primes, you can multiply this number by the probability that a random number between $2^{2047}$ and $2^{2048}-1$ is prime — which is, of course, very similar to the (probability) value you just computed above.  But there's one subtle catch here: the numbers you're testing aren't 'generic' numbers in the relevant range, they're specifially odd numbers; it turns out that this has an effect on the probability.  Essentially, the probability that an odd number is prime is twice the probability that an arbitrary number is prime; intuitively, this is because there are $n$ numbers in e.g. the interval $(x, x+n)$ (of which $\approx\frac{n}{\ln x}$ are prime, as mentioned above), but there are only $\frac n2$ odd numbers in that interval and all the primes will be found among them.
