Let $\ \mathbb P\ $ be the set of all natural primes. Pair $\ (p\ q)\ $ are prime neighbors $\ \Leftarrow:\Rightarrow$ $$ \{x\in\mathbb Z: p\le x\le q\}\cap\mathbb P\,\ =\,\ \{p\,\ q\} $$ Prime $\ x\in\mathbb P\ $ is said to be above neighbors $\ (p\ q)\,\ \Leftarrow:\Rightarrow$ $$ 2\cdot p+1\ \le\ x\ \le\ 2\cdot q-1.$$

EXAMPLE: There are no primes above prime neighbors $\ (101\ 103),\ $ i.e. $$ \{x\in\mathbb P:\ 2\cdot101+1\ \le\ x\ \le\ 2\cdot103-1\}\ \cap\ \mathbb P \quad =\quad \emptyset $$

Prime neighbors $\ (p\ q)\ $ are said to be distant $\ \Leftarrow:\Rightarrow\ $ for every prime neighbors $\ (x\ y)\ $ such that $\ x<p\ $ we have $\ y-x<q-p$.

QUESTION (a conjecture):   There is a prime above every pair of distant prime neighbors.

  • 1
    $\begingroup$ Have you numerically confirmed the conjecture? $\endgroup$ Oct 3 at 19:11
  • 2
    $\begingroup$ You are asking if there can be a prime gap larger than 2M with a start below 2p+1 when M is a maximal prime gap starting at p. Seems unlikely! Like a lot of questions about prime gaps, you'll probably be stuck with conjectures about asymptotics. See primerecords.dk/primegaps/maximal.htm for computational records. $\endgroup$ Oct 3 at 19:52
  • $\begingroup$ @ZackWolske, indeed, such q's occur routinely in small bunches: "all" "almost all", "infinitely many", .... $\endgroup$
    – Wlod AA
    Oct 3 at 20:13

1 Answer 1


This is probably true, and will almost certainly be hard to prove. The prime neighbours satisfying your hypothesis grow more quickly than one might expect, and the $p$ are listed in A002386. The corresponding gaps $q-p$ also grow more quickly than one might expect, and are listed in A005250. So an automatic search is easy to do, and shows that the number of primes "above" is also growing. In particular, your question has a positive answer for primes up to $10^{10}$.

  • 2
    $\begingroup$ Dave, thank you for your answer and for your good nature effort. $\endgroup$
    – Wlod AA
    Oct 3 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.