20201019, 16:36  #89 
Dec 2008
you know...around...
2^{5}×19 Posts 
(Testing with [$])
\(\sum^a_{b = 1}{\frac{\log^2(b \cdot p\#)}{b \cdot p\#}} \gtrsim \frac{\log^2(p\#)}{p\#} \sum^a_{b = 1}{\frac{1}{b}} \sim \frac{p^2}{p\#} \log a \sim \frac{p^3(\xi  1)}{p\#}\) Strange, it didn't work before... But thanks! 
20201020, 15:04  #90 
Dec 2008
you know...around...
2^{5}×19 Posts 
You're lucky that I don't post here every time I have a new idea
But the formula above really needs some tweaking. The impact on the output is a constant factor as I expected, but it's better to have some maths to back it up: If we're bothering to check all intervals (1,p²) after a*p# for all a<=p#, i.e. our last checked interval is p#²+(1,p²), the density of critical intervals per p is \(\sum^a_{b=1} {\frac { \log ^2 (b \cdot p#)} {b \cdot p#}} \hspace{2} \lt \hspace{2} \frac { \log^2 (a \cdot p#)}{p#} \sum^a_{b=1} \frac {1}{b} \hspace{2} \sim \hspace{2} \frac {(2p)^2}{p#} \log a \hspace{2} \sim \hspace{2} \frac {4p^3}{p#}\) And about p#/q#, for any c>1 and q=p/c, \(lim_{q \to \infty} \frac {q#}{p#}=0\) and we still have zero density of critical intervals. Right? Please tell me I'm right this time. Recap Re: what am I trying to do? I'm trying to show that the critical intervals in which large gaps, i.e. gaps of size 2e^{y}*(log x)² could appear, are too scarce for applying a global argument of Cramér, Granville et al. I have most terrible problems with (1.10) on page 5 in 1908.08613, it looks like the sieve of Eratosthenes of size N including trying to sieve with primes in the range N^(1/2,e^{y}). No wonder one might eventually arrive at Granville's gap bound with \(\mathcal R\). Local statistics applied on a global scale. The authors are surely aware of that, but I don't see any way to reconcile theorem 1.1 with conjecture 1.2 using (1.10). It must be somewhere inbetween the chapters 3 and 7  starting to read chapter 3 I get lost very quickly. *Sigh* May I ask whether any mod could kindly render the expressions correctly, the tags don't seem to like me I'm afraid... Thanks very much in advance. Last fiddled with by mart_r on 20201020 at 15:12 
20201020, 15:26  #91 
"Oliver"
Sep 2017
Porta Westfalica, DE
16F_{16} Posts 
\[\sum^a_{b=1} {\frac { \log ^2 (b \cdot p\#)} {b \cdot p\#}} \lt \frac { \log^2 (a \cdot p\#)}{p\#} \sum^a_{b=1} \frac {1}{b} \sim \frac {(2p)^2}{p\#} \log a \sim \frac {4p^3}{p\#}\]
Hm... \hspace is not supported by our $$, and the TEX tag ignores it. The other problem is: # needs to be escaped. 
20201020, 16:58  #92  
Dec 2008
you know...around...
2^{5}·19 Posts 
Quote:
Thanks! 

20201023, 14:23  #93 
Dec 2008
you know...around...
2^{5}·19 Posts 
Sticking with BorelCantelli, the same lemma that proves (?) the existence of gaps > log^{2}p in \(\mathcal{R}\), could be used against conjecture 1.2 in 1908.08613 in the sense that the probability to find an interval that would make it possible to find such a large gap at random is equal to zero. Cf. page 13: ''we (...) restrict to special values of m, namely m \(\equiv\) b mod Q''. Or am I missing something here?
Oh Mr Silverman, where art thou? 
20201023, 16:56  #94  
Aug 2006
2×2,969 Posts 
Quote:


20201023, 18:24  #95 
Dec 2008
you know...around...
2^{5}×19 Posts 
I'm aware of Maier's paper, and I understand it for the most part (though I have to examine his 1981 paper again...). The main result is, with respect to the problem of gaps >= log^{2}p, that there are infinitely many instances where there are, by a constant factor, less than log(p) primes in such an interval. I'm not sure yet how farreaching this argument is in light of prime gaps of this size, but I do keep that in mind of course.

20201024, 01:24  #96 
Feb 2017
Nowhere
2^{4}×3×79 Posts 
Hmm. You can use \; to insert white space. The pound or hashtag sign, I could find no good remedy.
I tried \# wrapped in tex tags. "Preview post" got me a message: In case that didn't display for you, it was text, in red, mimeTeX failed to render your expression inside a red box. I tried "#", but that didn't work, either. No error message, though. I did find one thing that (sort of) works  \sharp gives a sharp sign. Wrapping p\sharp in tex tags gives Last fiddled with by Dr Sardonicus on 20201024 at 01:26 Reason: fignix stopy 
20201028, 03:59  #98  
"Seth"
Apr 2019
2^{3}·5^{2} Posts 
Quote:
I've noticed that expected gap is largest at d=1, but chance of large merit is generally maximized with d=7#,11#,13#, I wrote a tool that calculates expected gap and chance of large (>merit 15 gap). I count the number of X, (X, m*P#) = 1 for 0 <= X <= SL (sieve length) for 0 <= m <=d (m, d) = 1. I do this in two phases to make the computation more efficient but it doesn't matter for the result. I calculate expected gap and average count of X for all the m mults and then choose the number with the smallest count remaining (equivalent to largest insufficient percent) I also measure throwing in a large prime to avoid overlapping effort with other contributors. I played around with choosing a subset of factors (e.g. 2*5*7, 2*3*7, 3*5*7 but it didn't seem to help much) Code:
d optimizer for P = 907#  large prime=877  sl=10884 (12.0 merit) Optimizing D  d = 1 * 2#  881 remaining, 4856 avg gap  sl insufficient 0.000% of time Optimizing D  d = 1 * 3#  611 remaining, 4537 avg gap  sl insufficient 0.017% of time Optimizing D  d = 1 * 5#  544 remaining, 3545 avg gap  sl insufficient 0.044% of time Optimizing D  d = 1 * 7#  541 remaining, 2774 avg gap  sl insufficient 0.045% of time Optimizing D  d = 1 * 11#  564 remaining, 2331 avg gap  sl insufficient 0.032% of time Optimizing D  d = 1 * 13#  593 remaining, 2039 avg gap  sl insufficient 0.021% of time Optimizing D  d = 1 * 17#  622 remaining, 1857 avg gap  sl insufficient 0.013% of time Optimizing D  d = 1 * 19#  650 remaining, 1728 avg gap  sl insufficient 0.009% of time Optimizing D  d = 1 * 23#  675 remaining, 1639 avg gap  sl insufficient 0.006% of time Optimizing D  d = 877 * 2#  882 remaining, 4813 avg gap  sl insufficient 0.000% of time Optimizing D  d = 877 * 3#  614 remaining, 4492 avg gap  sl insufficient 0.016% of time Optimizing D  d = 877 * 5#  547 remaining, 3507 avg gap  sl insufficient 0.040% of time Optimizing D  d = 877 * 7#  543 remaining, 2743 avg gap  sl insufficient 0.041% of time Optimizing D  d = 877 * 11#  566 remaining, 2306 avg gap  sl insufficient 0.029% of time Optimizing D  d = 877 * 13#  594 remaining, 2018 avg gap  sl insufficient 0.019% of time Optimizing D  d = 877 * 17#  623 remaining, 1838 avg gap  sl insufficient 0.012% of time 

20201028, 21:30  #99 
May 2018
2×3^{2}×11 Posts 

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