Can pairs of consecutive primes with large merit be found efficiently?

Question

The merit of a prime gap between the consecutive primes $p_{n}$ and $p_{n+1}$ is defines as $$m:=\frac{p_{n+1}-p_n}{\ln p_n}$$

How can I find efficiently, lets say, a pair of consecutive random primes with $100$ digits with a merit of at least $20$ ?

Brute force seems to be not very efficient. Even to find a pair with merit $10$ is difficult to reach this way. Should I use multiples of relative large primorials, or is there a better way?

Answer 1

This is an unsolved problem so the best we can hope for at the moment are conjectures and estimations. Heuristic arguments by Marek Wolf show, that the pair of consecutive primes separated by a distance $d$ appears for the first time at

$$ p_d \approx \sqrt d \exp \Big(\frac{1}{2}\sqrt {\ln^2 d +4d}\Big) $$

which is at least numerically the best conjecture we have on this topic. Using g this we can work out the approximate location of the first merit of a size $m$. Since

$$ m = \frac{p_d}{\ln p_d} \approx \sqrt d $$

or $d \approx m^2$, the first occurrence of a merit of $m$ will occur approximately at

$$ m \exp \Big(\frac{1}{2}\sqrt {2\ln^2 m +4m^2}\Big) \approx me^m + \frac{1}{2}e^m \ln^2 m $$

But for finding the exact value the above formula is not of much use.

Reference:

• First occurrence of a given gap between consecutive primes, 1997