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2020-08-26, 18:31   #78
kruoli

"Oliver"
Sep 2017
Porta Westfalica, DE

3·101 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 11100010012 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...

10010000002 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... 26×32 Posts I suppose my rambling theories are, as the saying goes, "not even wrong". Right?
2020-09-18, 17:48   #82
CRGreathouse

Aug 2006

5,923 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... 10010000002 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

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