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

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 07:51

3Well, by [Bertrand’s postulate](http://en.wikipedia.org/wiki/Bertrand%27s_postulate) it must be less than $5\times 10^{999}$. – Brian M. Scott Sep 17 '11 at 07:52
1 Answers
If $a=A(n)$ is the $n^{th}$ primorial (that is, the product of the first n primes), then at least $a,a\pm2, a\pm3,... a\pm p_n,a\pm (p_n+1)$ are composite, giving a prime gap of length at least $p_n1$. 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 $a4152 ... a+3312$ are composite and $a4153$ 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 1000digit numbers would be some multiple of 5 million.
For a broader discussion with numerous references, see the Prime Gaps page at wolfram.com. For morerecent results on lower bounds for maximal prime gaps, see the arXiv.org paper or preprint Long gaps between primes by Ford, Green, Konyagin, Maynard, and Tao, and see Tao's commentary at terrytao.wordpress, and see additional results / background at the same site.
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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 1000digit 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

1@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 (http://en.wikipedia.org/wiki/Cram%C3%A9r%27s_conjecture#Cram.C3.A9rGranville_conjecture) – 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 03:45


@BradGraham, it isn't (eg 2*3*51 = 29) so I have edited to correct that error. Thanks. – James Waldby  jwpat7 Aug 05 '15 at 21:54