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 2012-10-14, 17:09 #133 mart_r     Dec 2008 you know...around... 23×3×52 Posts 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
2014-08-31, 17:28   #134
mart_r

Dec 2008
you know...around...

23·3·52 Posts
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.

Attached Files
 Project p#Y.zip (244.7 KB, 54 views)

2014-08-31, 19:26   #135
NBtarheel_33

"Nathan"
Jul 2008
Maryland, USA

3·7·53 Posts

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?

2014-08-31, 20:21   #136
mart_r

Dec 2008
you know...around...

23×3×52 Posts

Quote:
 Originally Posted by NBtarheel_33 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.

2014-08-31, 22:12   #137
NBtarheel_33

"Nathan"
Jul 2008
Maryland, USA

3×7×53 Posts

Quote:
 Originally Posted by mart_r 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?

2014-09-02, 15:33   #138
mart_r

Dec 2008
you know...around...

23×3×52 Posts

Quote:
 Originally Posted by NBtarheel_33 Floor(p#*y) is the largest integer less than or equal to the product of p# and y, correct?
That is most definitely correct.

2014-09-02, 23:01   #139
NBtarheel_33

"Nathan"
Jul 2008
Maryland, USA

100010110012 Posts

Quote:
 Originally Posted by mart_r That is most definitely correct.
Then how are you claiming that quantity is prime?

2014-09-02, 23:08   #140
NBtarheel_33

"Nathan"
Jul 2008
Maryland, USA

3·7·53 Posts

Quote:
 Originally Posted by mart_r 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.

2014-09-03, 03:36   #141
axn

Jun 2003

127616 Posts

Quote:
 Originally Posted by NBtarheel_33 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.

2014-09-04, 20:15   #142
mart_r

Dec 2008
you know...around...

23×3×52 Posts

Quote:
 Originally Posted by NBtarheel_33 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 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?

 2014-12-10, 02:54 #143 Trilo     "W. Byerly" Aug 2013 1423*2^2179023-1 1378 Posts 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

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