mersenneforum.org  

Go Back   mersenneforum.org > Math Stuff > Other Mathematical Topics

Reply
 
Thread Tools
Old 2012-10-14, 17:09   #133
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

23×3×52 Posts
Default

I was running out of ideas as of late, but here's a small suggestion on how to treat the uniqueness problem, i.e. to prove that, if such a constant exists, then there are infinitely many of them.

Essentially, the number of "immortal" primes after each stage with the corresponding prime p and the current number of primes n should be asymptotic to \frac{2n}{\sqrt p} as described in my document on page 5.

Under the assumption that, for any \eps>0, \lim_{i\to\infty}\frac{\sum_{x=2}^i x^{\frac 1 2 -\eps}}{\sum_{x=2}^i x-\theta(x)}<1 and \lim_{i\to\infty}\frac{\sum_{x=2}^i x^{\frac 1 2 +\eps}}{\sum_{x=2}^i x-\theta(x)}>1 (*), if \lim_{p\to\infty}\frac{2n}{\sqrt p}=\infty, then y is not unique.

(*) One of my pet ideas, sorry for bringing it up all the time despite it may still be wrongly phrased.


Under average conditions (nextprime(p+1)=p+log(p) and such), the absolute growth of the numerator of the last mentioned fraction after each step is \frac{2n}{\sqrt p}, and for the denominator this is \frac{\frac 1 2 \log p}{\sqrt p}. So even if the number of primes n drops below \frac 1 2 \sqrt p at several points (for n=\frac 1 2 \sqrt p the sequence is in limbo, so to speak), in the long run \frac{2n}{\sqrt p} will by far outnumber \frac{\frac 1 2 \log p}{\sqrt p}, and n<k only finitely often for any arbitrarily large k.


I hope this is not too vague.
mart_r is offline   Reply With Quote
Old 2014-08-31, 17:28   #134
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

23·3·52 Posts
Default Gravedigging

Some consolidation work and major improvements on minor details.

This version should be closer to something publicly acceptable.

I guess I should ask about the most important flaws that this "paper" indubitably still contains, and maybe some hints how to try to fix them.

Until then, thanks for reading.
Attached Files
File Type: zip Project p#Y.zip (244.7 KB, 54 views)
mart_r is offline   Reply With Quote
Old 2014-08-31, 19:26   #135
NBtarheel_33
 
NBtarheel_33's Avatar
 
"Nathan"
Jul 2008
Maryland, USA

3·7·53 Posts
Default

Quote:
For y >= 1.5, floor (2#*y) must be an odd prime 2m-1...
Not necessarily! What if y = 2, or indeed what if y is any other integer greater than 2?
NBtarheel_33 is offline   Reply With Quote
Old 2014-08-31, 20:21   #136
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

23×3×52 Posts
Default

Quote:
Originally Posted by NBtarheel_33 View Post
Not necessarily! What if y = 2, or indeed what if y is any other integer greater than 2?
Floor(p#*y) must be prime by definition. And apart from 2, all other primes are odd. If y>=2 then 2#*y is already >= 4, thus an odd prime.
mart_r is offline   Reply With Quote
Old 2014-08-31, 22:12   #137
NBtarheel_33
 
NBtarheel_33's Avatar
 
"Nathan"
Jul 2008
Maryland, USA

3×7×53 Posts
Default

Quote:
Originally Posted by mart_r View Post
Floor(p#*y) must be prime by definition. And apart from 2, all other primes are odd. If y>=2 then 2#*y is already >= 4, thus an odd prime.
Floor(p#*y) is the largest integer less than or equal to the product of p# and y, correct? Or is the * not multiplication?
NBtarheel_33 is offline   Reply With Quote
Old 2014-09-02, 15:33   #138
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

23×3×52 Posts
Default

Quote:
Originally Posted by NBtarheel_33 View Post
Floor(p#*y) is the largest integer less than or equal to the product of p# and y, correct?
That is most definitely correct.
mart_r is offline   Reply With Quote
Old 2014-09-02, 23:01   #139
NBtarheel_33
 
NBtarheel_33's Avatar
 
"Nathan"
Jul 2008
Maryland, USA

100010110012 Posts
Default

Quote:
Originally Posted by mart_r View Post
That is most definitely correct.
Then how are you claiming that quantity is prime?
NBtarheel_33 is offline   Reply With Quote
Old 2014-09-02, 23:08   #140
NBtarheel_33
 
NBtarheel_33's Avatar
 
"Nathan"
Jul 2008
Maryland, USA

3·7·53 Posts
Default

Quote:
Originally Posted by mart_r View Post
Floor(p#*y) must be prime by definition.
Definition? Or do you mean *by assumption*? In other words, are you restricting your choices of p and y such that floor(p#y) must be prime, and ignoring all other candidates for p and y?

Quote:
And apart from 2, all other primes are odd. If y>=2 then 2#*y is already >= 4, thus an odd prime.
This claim is false. Counterexample: Let y = 6. Then floor(2#y) = floor(2*6) = floor(12) = 12, neither odd nor prime.
NBtarheel_33 is offline   Reply With Quote
Old 2014-09-03, 03:36   #141
axn
 
axn's Avatar
 
Jun 2003

127616 Posts
Default

Quote:
Originally Posted by NBtarheel_33 View Post
Definition? Or do you mean *by assumption*? In other words, are you restricting your choices of p and y such that floor(p#y) must be prime, and ignoring all other candidates for p and y?
He is trying to prove that there are no y >= 1.5 that can satisfy the conjecture using Reductio. Assume there exists y >= 1.5 that satisfies the conjecture. By definition (from the conjecture), floor(p#*y) must be prime for all primes. That leads to the conclusion that there exists an m such that m*p#-1 is prime for all primes, which is then considered as absurd, and hence y (if it exists) must be < 1.5. QED.
axn is offline   Reply With Quote
Old 2014-09-04, 20:15   #142
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

23×3×52 Posts
Default

Quote:
Originally Posted by NBtarheel_33 View Post
Definition? Or do you mean *by assumption*? In other words, are you restricting your choices of p and y such that floor(p#y) must be prime, and ignoring all other candidates for p and y?
Restricting the choices for y, that is.

Quote:
Originally Posted by NBtarheel_33 View Post
This claim is false. Counterexample: Let y = 6. Then floor(2#y) = floor(2*6) = floor(12) = 12, neither odd nor prime.
That simply means y=6 is not a solution.


While I'm here, may I add that
\lim_{p\to\infty} \int_0^p \frac{dx}{\log x}-\sum_{x=2}^p \frac{1}{\log x}=0.24323834289098075541505913546...

Or, equivalently,
\lim_{p\to\infty} \int_a^p \frac{dx}{\log x}=\sum_{x=2}^p \frac{1}{\log x}
for a=1.54983177364507780342792218629...

There is yet another constant arising in the computation which I can make neither head nor tail of: The value above, 0.243..., can be approximated quite well for integer p with
\int_0^p \frac{dx}{\log x}-\sum_{x=2}^p \frac{1}{\log x}+\frac{1}{2\log x}-\frac{0.1734474150838006...}{x(\log_2 x)^2}

Do these constants belong to someone here?
mart_r is offline   Reply With Quote
Old 2014-12-10, 02:54   #143
Trilo
 
Trilo's Avatar
 
"W. Byerly"
Aug 2013
1423*2^2179023-1

1378 Posts
Default

Have you made any progress on calculating a y value that floor(p#*y) is prime? I am interested in your work.

Last fiddled with by Trilo on 2014-12-10 at 02:55
Trilo is offline   Reply With Quote
Reply

Thread Tools


All times are UTC. The time now is 19:59.

Sat Oct 31 19:59:25 UTC 2020 up 51 days, 17:10, 2 users, load averages: 1.95, 1.86, 1.83

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.