20200826, 18:31  #78 
"Oliver"
Sep 2017
Porta Westfalica, DE
467 Posts 

20200911, 02:15  #79 
"Dana Jacobsen"
Feb 2011
Bangkok, TH
2^{2}×227 Posts 
Arxiv: "Primes in short intervals: Heuristics and calculations" by Granville and Lumley, 10 Sep 2020. Interesting.

20200914, 12:57  #80  
Dec 2008
you know...around...
27E_{16} Posts 
Thanks, Dana. That's quite an interesting read indeed.
Quote:
(The error term may or may not be correctly applied here, but who cares...) Huh?? Now the [$$] tags don't work properly, have to use [TEX] again. I get the feeling this is what especially chapters 6 and 7 in 2009.05000 are pointing toward  I'm still grappling with the connections between u+, c+, delta+, and sigma+ there  but let me paraphrase it in a way that I've worked out by myself. (Great, that prompted my brain to play Depeche Mode on repeat: "Let me show you the world in my eyes..." ) Using Cramér's uniformly distributed probability model, looking for a gap of size (log x)², we want to know the probability P for , which has a series expansion . Considering only odd numbers to be potential prime number candidates, this would turn into and sieving with small primes <=z, where , and since w ~ , P would go down toward zero by "allowing" to sieve primes up to which is just about one Buchstab function away from Granville's conjecture. What I'm not quite sure about is the way that P accumulates over the entirety of x on the number line. If I got this right, the reasoning outlined above assigns the probability to every integer respectively. But aren't we looking at intervals of size (log x)², in each of which Cramér's probability, which is asymptotic to , is in effect? The simple analogy to the series probably comes to mind, which is convergent for m>1. P is even smaller for the modified sieved versions, which in turn would mean we may never see a gap of size (log x)² between primes of the size of x. For now, that's all there is... Last fiddled with by mart_r on 20200914 at 13:04 

20200918, 13:40  #81 
Dec 2008
you know...around...
2·11·29 Posts 
I suppose my rambling theories are, as the saying goes, "not even wrong". Right?

20200918, 17:48  #82  
Aug 2006
3^{2}×5×7×19 Posts 
Quote:
Am I missing something? 

20200918, 18:22  #83 
Dec 2008
you know...around...
638_{10} Posts 
I guess I'm trying to argue that there may be only finitely many gaps of length > (log x)².
So I thought there may be an error in my outlined reasoning (worth elaborating...?) that one of the brilliant minds in this forum could point out to me, or at least tell me whether I'm somewhat on the right path to further enlightenment. Even a simple "wrong" or "right" would be better than nothing at all... Last fiddled with by mart_r on 20200918 at 18:33 
20201003, 18:21  #84 
May 2018
2^{2}·53 Posts 

20201014, 08:19  #85 
Jun 2003
Oxford, UK
2^{2}·3·7·23 Posts 
Here's one for the mathematicians:
Two prime gaps  the first is a first occurrence, the second is the smallest found to date, but probably not first occurrence. 1430 4606937813294064947 1432 84218359021503505748941 What is the formula for determining approximately how many gaps of exactly size 1432 have a lower prime smaller than the smallest found to date? Last fiddled with by robert44444uk on 20201014 at 08:20 
20201014, 13:23  #86  
Dec 2008
you know...around...
2·11·29 Posts 
Quote:
Then again, I'm not a mathematician... ATH has computed the nth occurence of prime gaps of length n here: https://www.mersenneforum.org/showpo...88&postcount=2. With some backoftheenvelope maths I'd guess that there should be a couple of millions gaps of size 1432 below the first currently known occurence. Also, any such formula would only give very rough results for small n. P.S.: Somehow Bobby's post strongly reminds me of a scene in "A Goofy Movie"... 

20201019, 15:57  #87 
Dec 2008
you know...around...
1176_{8} Posts 
Okay, next level.
After some more reading and computations my next question is: can it be shown that Granville's modified intervals are not too sparse in a sense that Cramér BorelCantelli can still be applied?
Now there are asymptotically less coprimes to p# in the interval (p#,p#+p²) than in an average interval of length p² for large enough p. We want to examine all a*p#+(1,p²) to find gaps > p², but this is only reasonable if a isn't too large. How large is too large here? (I suppose) It is reasonable to let where =2e^{y} = 1.1229.... After some clumsy but maybe semilogical scribbles I have Now neither TEX nor [$$] works... see attachment, before I lose my mind Summing p³/p#, with or without any constant factor, over all primes p gives a rather rapidly converging series, which means only a zero proportion of intervals in may have the required scarcity of coprimes to p# (or "zquasiprimes", as they are called in [Pintz 2007: https://projecteuclid.org/download/p...acm/1229619660]). So, can it be that Cramér's/Shanks' lim sup (p_{n+1}p_{n}) ~ log²p after all? (Sorry for mine being so stubborn...;) Some calculations with intervals around p#/d where d is a small primorial such that the number of coprimes to p# in p#/d±p²/2 is as small as possible, revealed no obvious irregularities when sieving with primes > p. But I might gather some more data there, provided I find the time. There might be some impact on the asymptotics above, but again, can it be shown that those impacts are not too huge w.r.t. the density of the critical intervals? I'm trying to dig my way through [BanksFordTao 2019: https://arxiv.org/pdf/1908.08613.pdf], currently on page 5. But this paper is overcrowded with formulae, there's no hope I'll make it through them all before 2025 or so. I was wondering, however, why the sum on top of page 13 is taken over all m and not only a portion of 1/log²x of m in (x,2x], isn't it similar to what I was trying to do in my post #80 above? But maybe I just don't understand it well enough... Is it true that the parity barrier, which isn't addressed in BanksFordTao as far as I can tell, might come in the way of all those "almost surely"? I'd be very grateful for any additional input. Last fiddled with by mart_r on 20201019 at 16:08 Reason: wrong formula in attachment 
20201019, 16:22  #88  
"Oliver"
Sep 2017
Porta Westfalica, DE
467 Posts 
Quote:
\[\sum^a_{b = 1}{\frac{\log^2(b \cdot p\#)}{b \cdot p\#}} \gtrsim \frac{\log^2(p\#)}{p\#} \sum^a_{b = 1}{\frac{1}{b}} \sim \frac{p^2}{p\#} \log a \sim \frac{p^3(\xi  1)}{p\#}\] Code:
\sum^a_{b = 1}{\frac{\log^2(b \cdot p\#)}{b \cdot p\#}} \gtrsim \frac{\log^2(p\#)}{p\#} \sum^a_{b = 1}{\frac{1}{b}} \sim \frac{p^2}{p\#} \log a \sim \frac{p^3(\xi  1)}{p\#} 

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