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Old 2020-09-14, 12:57   #80
mart_r's Avatar
Dec 2008
you know...around...

599 Posts

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

Originally Posted by mart_r View Post
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}<br />
(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}}<br />

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}}<br />

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
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