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Old 2020-10-19, 16:36   #89
mart_r
 
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(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!
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Old 2020-10-20, 15:04   #90
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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 in-between 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 2020-10-20 at 15:12
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Old 2020-10-20, 15:26   #91
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\[\sum^a_{b=1} {\frac { \log ^2 (b \cdot p\#)} {b \cdot p\#}} \lt \frac { \log^2 (a \cdot p\#)}{p\#} \sum^a_{b=1} \frac {1}{b} \sim \frac {(2p)^2}{p\#} \log a \sim \frac {4p^3}{p\#}\]

Hm... \hspace is not supported by our $$, and the TEX tag ignores it.

The other problem is: # needs to be escaped.
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Old 2020-10-20, 16:58   #92
mart_r
 
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Quote:
Originally Posted by kruoli View Post
\[\sum^a_{b=1} {\frac { \log ^2 (b \cdot p\#)} {b \cdot p\#}} \lt \frac { \log^2 (a \cdot p\#)}{p\#} \sum^a_{b=1} \frac {1}{b} \sim \frac {(2p)^2}{p\#} \log a \sim \frac {4p^3}{p\#}\]

Hm... \hspace is not supported by our $$, and the TEX tag ignores it.

The other problem is: # needs to be escaped.
So hspace is causing the trouble. I figured it was something like that, but it's hard to discern when the whole formula isn't displayed. And # or \# doesn't go with TEX, although the mimeTeX editor doesn't have problems with it, that was a bit easier to figure out since it's where the "[?]" appeared. It's a bit unnerving when most of the time to write the post is used for the TEX formatting...

Thanks!
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Old 2020-10-23, 14:23   #93
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Sticking with Borel-Cantelli, the same lemma that proves (?) the existence of gaps > log2p 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?
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Old 2020-10-23, 16:56   #94
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Quote:
Originally Posted by mart_r View Post
Sticking with Borel-Cantelli, the same lemma that proves (?) the existence of gaps > log2p 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?
Borel-Cantelli gives good intuition for where to start, but the whole point of the paper is that the primes have more subtle behavior than the Cramer distribution would have you believe (which we've known since at least 1985).
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Old 2020-10-23, 18:24   #95
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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 >= log2p, 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 far-reaching this argument is in light of prime gaps of this size, but I do keep that in mind of course.
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Old 2020-10-24, 01:24   #96
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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

p\sharp

Last fiddled with by Dr Sardonicus on 2020-10-24 at 01:26 Reason: fignix stopy
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Old 2020-10-24, 03:52   #97
LaurV
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Quote:
Originally Posted by Dr Sardonicus View Post
\#
p\sharp
Works as it should.
The forum uses mathjax, which is not exactly \(\TeX\)

\(a\ \#\ b\)

Last fiddled with by LaurV on 2020-10-24 at 03:54
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Old 2020-10-28, 03:59   #98
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Quote:
Originally Posted by robert44444uk View Post
I was thinking about mart_r's approach this morning. Would this be an improvement?
  • choose the range of integers L
  • reduce L to L' using small primes up to p[n] where p[n]# is the largest primorial smaller than L. The set L' comprises those integers in L which are coprime to p[n]#
  • choose a target prime p[q], which would give prime gaps of approximately 20 merit over L, if p[q]# is used
  • run the primes p[n+1]...p[q] individually over L' for each of the mods and rank them in order of the difference between the best performing and worst performing mod - your approach described above
  • collect, say, the top fifteen p and collect for each, the mods which are within say 3 of the best for that p
  • carry out a 15-nested loop to determine which group of mods, one for each of the 15 primes, provides the most efficient reduction to L'
  • Go with this mod solution for those 15 primes, and reduce L' to take out those which are not coprime to each of these primes. The new set is L"
  • Repeat the prescription with another 15 primes, excluding those up to p[n] and the 15 already tested, which show the greatest variation of performing mods applied to L"
  • Repeat until all primes up to p[q] are treated

If 15 was too many, given the permutations possible, a smaller collection would suffice.

I won't have time to do more than think now for the next three weeks as I am in deepest rural Uganda from tomorrow. No computer, no internet. Bliss.
After an offline discussion wanted to comment on what I do here.

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
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Old 2020-10-28, 21:30   #99
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Quote:
Originally Posted by mart_r View Post
P.S.: Somehow Bobby's post strongly reminds me of a scene in "A Goofy Movie"...
What scene are you talking about? I have watched "A Goofy Movie" before, and I cannot recall a scene where the word "left" was used in the way I used it.
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