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2020-08-26, 18:31   #78
kruoli

"Oliver"
Sep 2017
Porta Westfalica, DE

2×13×19 Posts

Quote:
 Originally Posted by mart_r The TEX expression doesn't work and I can't figure out why. Anyway it's in the graph in the attachment.
It works with the "fancy" TeX engine after messing a bit with the command:
$(\log\frac{\#\text{gaps with }M>k}{2^{21}}+k) \cdot 2n$

 2020-09-11, 02:15 #79 danaj   "Dana Jacobsen" Feb 2011 Bangkok, TH 22×227 Posts Arxiv: "Primes in short intervals: Heuristics and calculations" by Granville and Lumley, 10 Sep 2020. Interesting.
2020-09-14, 12:57   #80
mart_r

Dec 2008
you know...around...

12128 Posts

Thanks, Dana. That's quite an interesting read indeed.

Quote:
 Originally Posted by mart_r The density of gaps with merit >=M between consecutive primes appears to be on average $e^{-M+\frac{a_M}{\log p}}$ for integer M with the following values aM
The behaviour of aM is only a prelude to a known line of probabilistic reasoning, isn't it?
$(1-\frac{1}{\log x})^{M(\log x)}\hspace{3}=\hspace{3}(1-\frac{M}{2\hspace{1}\log x}+O(\frac{1}{(\log x)^2}))\hspace{1}e^{-M}
$

(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 $(1-\frac{1}{\log x})^{(\log x)^2}$, which has a series expansion $e^{-(\log x+\frac{1}{2}+\frac{1}{3\hspace{1}\log x}+...)}\hspace{3}=\hspace{3}e^{-\sum_{n=1}^\infty \frac{(\log x)^{2-n}}{n}}$. Considering only odd numbers to be potential prime number candidates, this would turn into $(1-\frac{2}{\log x})^{\frac{1}{2}(\log x)^2}\hspace{3}=\hspace{3}e^{-\sum_{n=1}^\infty \frac{2^{n-1}\hspace{1}(\log x)^{2-n}}{n}} $ and sieving with small primes <=z, where $w=\prod_{primes\hspace{1}q}^z \frac{q}{q-1}$, $(1-\frac{w}{\log x})^{\frac{1}{w}(\log x)^2}\hspace{3}=\hspace{3}e^{-\sum_{n=1}^\infty \frac{w^{n-1}\hspace{1}(\log x)^{2-n}}{n}} $ and since w ~ $(log x)\hspace{1}e^{\gamma}$, P would go down toward zero by "allowing" to sieve primes up to $z=x^{e^{-\gamma}}$ 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 $\frac{1}{x\hspace{1}\sqrt{e}}$, is in effect? The simple analogy to the series $\sum_{n=2}^\infty \frac{1}{n\hspace{1}(\log n)^m}$ 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 2020-09-14 at 13:04  2020-09-18, 13:40 #81 mart_r Dec 2008 you know...around... 2·52·13 Posts I suppose my rambling theories are, as the saying goes, "not even wrong". Right? 2020-09-18, 17:48 #82 CRGreathouse Aug 2006 32·5·7·19 Posts Quote:  Originally Posted by mart_r I suppose my rambling theories are, as the saying goes, "not even wrong". Right? I wouldn't say that, but I don't really understand what you're trying for here. The point of the paper was to give a heuristic which significantly improves upon Cramer; why would you analyze it with Cramer's model? It's like analyzing a black hole with Newtonian physics. Am I missing something?  2020-09-18, 18:22 #83 mart_r Dec 2008 you know...around... 2×52×13 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 2020-09-18 at 18:33 2020-10-03, 18:21 #84 Bobby Jacobs May 2018 21310 Posts Quote:  Originally Posted by mart_r I suppose my rambling theories are, as the saying goes, "not even wrong". Right? Left.  2020-10-14, 08:19 #85 robert44444uk Jun 2003 Oxford, UK 78D16 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 2020-10-14 at 08:20 2020-10-14, 13:23 #86 mart_r Dec 2008 you know...around... 2×52×13 Posts Quote:  Originally Posted by robert44444uk 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? Hmm. Page 15 in https://arxiv.org/pdf/2002.02115.pdf talks about the n-th ﬁrst-occurrence gap between primes in progression, but I don't think that's quite useful w.r.t. this question. I'm sure some arXiv paper has investigated this, but I can't find it at the moment I'm afraid. Then again, I'm not a mathematician... ATH has computed the n-th occurence of prime gaps of length n here: https://www.mersenneforum.org/showpo...88&postcount=2. With some back-of-the-envelope 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"...  2020-10-19, 15:57 #87 mart_r Dec 2008 you know...around... 2·52·13 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 $\cap$ Borel-Cantelli 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 $a where $\xi$=2e-y = 1.1229.... After some clumsy but maybe semi-logical scribbles I have $\sum_{a=1}^\infty \frac {\log^2(a\hspace{1}p#)}{a\hspace{1}p#}\hspace{2}\gtrsim\hspace{2}\frac{\log^2(p)}{p#}\sum_{a=1}^\infty \frac{1}{a}\hspace{2}\sim\hspace{2}\frac{p^2}{p#}\log{a}\hspace{2}\sim\hspace{2}\frac{p^3(\xi-1)}{p#}$ 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 $\mathbb{N}$ may have the required scarcity of coprimes to p# (or "z-quasiprimes", as they are called in [Pintz 2007: https://projecteuclid.org/download/p...acm/1229619660]). So, can it be that Cramér's/Shanks' lim sup (pn+1-pn) ~ 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 [Banks-Ford-Tao 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 Banks-Ford-Tao 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. Attached Thumbnails   Last fiddled with by mart_r on 2020-10-19 at 16:08 Reason: wrong formula in attachment
2020-10-19, 16:22   #88
kruoli

"Oliver"
Sep 2017
Porta Westfalica, DE

49410 Posts

Quote:
 Originally Posted by mart_r Now neither TEX nor [] works... see attachment, before I lose my mind
Here you go:

$\sum^a_{b = 1}{\frac{\log^2(b \cdot p\#)}{b \cdot p\#}} \gtrsim \frac{\log^2(p\#)}{p\#} \sum^a_{b = 1}{\frac{1}{b}} \sim \frac{p^2}{p\#} \log a \sim \frac{p^3(\xi - 1)}{p\#}$

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\#}
Of course, you might omit the \cdot if you do not want them.

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