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I found two consecutive prime numbers $401!-3463$ and $401!+4021$. These have a difference of $7664$. Is there some kind of technique that is known in order to find consecutive prime numbers with sufficiently large gaps?

I just used the fact that for all positive integers $n$, it follows that all the integers on the interval $[n!+2, n!+n]$ are composite, which can make great prime gaps. Are there other more efficient techniques?

Apologies if this question is too broad. I was just curious :)

Carl Christian
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Mr Pie
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  • I am not aware of better techniques, $n$# is no better. Of course, if we go into larger ranges, we will find larger gaps, but finding large primes and the corresponding gap is unfortunately time-consuming. Do you want both primes to be proven ? This makes the task even harder because proving the primality of huge primes takes usually very long. – Peter Dec 25 '19 at 18:22
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    I upvoted the question, nevertheless it would be nice to be a bit more specific. Which magnitude should the primes have ? Are probable primes enough ? And perhaps some other aspects. – Peter Dec 25 '19 at 18:24
  • An expert of prime gaps, but unfortunately often very busy, is Dana Jacobsen. – Peter Dec 25 '19 at 18:25
  • Similar question: Finding large prime gaps without using brute force albeit with no answer. – TheSimpliFire Dec 25 '19 at 19:56
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    I helped Dana speed up a semiprime and twin prime search code before if memory serves me. –  Dec 25 '19 at 20:16
  • You'll note only prime addons for the most part are the ones that fail at smaller than nextprime(n)^2. same trick works for subtraction. now if only factorials had pstterns in their quotient with 6... –  Dec 25 '19 at 20:25
  • @Peter probable primes suffice. They may be prime! However it would be ideal if both primes were proven, but you are right - it would take a long time to do as such. – Mr Pie Dec 25 '19 at 22:48
  • @Peter I have never heard of Dana until now, so thanks for telling me! – Mr Pie Dec 25 '19 at 22:51
  • @reuns concerning Cramer's conjecture, what does that mean? What is a "zero-mean exponential random variable"? Apologies for my lack in knowledge. – Mr Pie Dec 27 '19 at 05:48
  • @RoddyMacPhee I care about those conjectures :) – Mr Pie Dec 27 '19 at 05:49
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    Cramer is saying that given $p_k$ prime, for $a>0$ not too large, the event "$p_k+a$ is prime" is independent from one $a$ to the other and the probability that it is prime is $1/\log p_k$, ie. $g(k)=p_{k+1}-p_k$ behaves like a binomial random variable of parameter $1/\log p_k$, in this model when $p_k$ gets large then $\frac{g(k)}{\log k}$ converges to an exponential random variable $\log \ P(\frac{g(k)}{\log k}> A) \approx \log \prod_{m\le A\log k} (1-\frac1{\log k}) \approx -A$. In particular $g(k)/\log k$ is unbounded but to get $g(k)/\log k > n$ you need to check about $e^n$ values of $k$ – reuns Dec 27 '19 at 16:39
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    Perhaps this will help: https://math.stackexchange.com/questions/2311652/im-trying-to-find-the-longest-consecutive-set-of-composite-numbers/2311764#2311764 – Mastrem Dec 29 '19 at 21:19
  • @Mastrem thank you! :) – Mr Pie Dec 30 '19 at 01:56

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