Let $p_i$ be $i^{th}$ prime number (for example $p_4=7$). Is there $m=m(n, k)$ such that $n \leq \min \{p_{m+i}-p_{m+i-1} : 1 \leq i \leq k\}$ for every $n$ and $k$? Even, consider the problem for $k=2$.

  • $\begingroup$ For $k=1$, $a!+2, a!+3, ..., a!+a$ make an arbitrary large gap, when $a$ is an arbitrary large number. $\endgroup$ – Arash Ahadi Dec 4 '16 at 3:48
  • $\begingroup$ This is a consequence of the prime k-tuples conjecture. Based on recent work involving k=2 (Tao, Pintz, Maynard and others) it is likely that m will be exponential in n and k. Gerhard "A Prime Time For Mathematics" Paseman, 2016.12.03. $\endgroup$ – Gerhard Paseman Dec 4 '16 at 3:59
  • $\begingroup$ AFAIK, the OP is asking whether there are arbitrarily long run of arbitrarily large prime gaps, which is not exactly the question the above comments are answering. $\endgroup$ – Fan Zheng Dec 4 '16 at 4:07
  • $\begingroup$ I am sure that there was a good intention but to me, this question does not make any sense whatsoever. $\endgroup$ – Włodzimierz Holsztyński Dec 5 '16 at 7:27
  • 1
    $\begingroup$ @WłodzimierzHolsztyński: The question makes full sense. As Fan Zheng remarked, it can be reformulated in plain English as: are there arbitrarily long runs of arbitrarily large prime gaps? Then answer is yes, see my response. $\endgroup$ – GH from MO Dec 5 '16 at 11:33

Yes, this was proved in a strong quantitative form by Maier in his paper Chains of large gaps between consecutive primes, Adv. in Math. 39 (1981), 257-269. Recently, Pintz strengthened this result with Maynard's method, see Theorem 4' in this arXiv preprint.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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