# The largest possible prime gap?

What is the largest possible prime gap if we observe only 1000-digits numbers? There are few conjectures about this question but is there something that we can say and be absolutely sure of it?

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You're talking about a particular finite interval of numbers. Results and conjectures on prime gaps generally come in the form of either asymptotic growth or densities, so for particulars you either go for a probabilistic idea using heuristics or you crunch everything though a computer given enough time, resources, and power. (Also, you didn't specify binary versus decimal base.) –  anon Sep 17 '11 at 7:51
Well, by Bertrand’s postulate it must be less than $5\times 10^{999}$. –  Brian M. Scott Sep 17 '11 at 7:52

## 1 Answer

If $a=A(n)$ is the $n^{th}$ primorial (that is, the product of the first n primes), then at least $a, a-1,a\pm2, a\pm3,... a\pm p_n,a\pm (p_n+1)$ are composite, giving a prime gap of length at least $2+p_n$. At $n=350, p_n=2357$, the primorial $a=A(n)$ has 1000 digits (its base 10 logarithm is about 999.375). Actually, with Pari/GP it's easy to verify that $a-4152 ... a+3312$ are composite and $a-4153$ and $a+3313$ apparently are prime, giving a prime gap of length 7465.

Wikipedia gives a result of R. Rankin, that (where $g(p)$ denotes the gap after $p$) $g(p) > 2e^\gamma (\ln p) (\ln \ln p) (\ln \ln \ln \ln p)/(\ln \ln \ln p)^3$ infinitely often. For $p\approx 10^{1000}$, that expression evaluates to about 5300. The article also gives a conjectured bound of H. Cramér, that $g(p) = O((\ln p)^2)$, which for 1000-digit numbers would be some multiple of 5 million.

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H. Maier (1985) shows that the expected coefficient of Cramér's asymptotic is $2/e^\gamma\approx1.123,$ so that would suggest 5.95 million. M. Wolf suggests that a closer asymptotic should be W(p)^2, so using the same constant you'd get 5.91 million. Of course both are looking at the largest prime gap up to that point, not restricted to 1000-digit ones, so maybe you'd want to be a smidge smaller. I'd need to play around with probabilistic models a bit to find out roughly how much. So in any case 5 million seems like the right neighborhood. –  Charles Sep 17 '11 at 18:52
@Charles: I believe Andrew Granville proposed that the constant was $2/e^\gamma$. Based on looking at the papers, this is what I made out. Also see (en.wikipedia.org/wiki/…) –  Eric Naslund Sep 17 '11 at 20:41
@Eric Naslund: Granville certainly did suggest that; I thought the number originated with Maier but I could be mistaken. –  Charles Sep 18 '11 at 3:45