20121014, 17:09  #133 
Dec 2008
you know...around...
2^{3}×3×5^{2} 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 as described in my document on page 5. Under the assumption that, for any , and (*), if , 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 , and for the denominator this is . So even if the number of primes n drops below at several points (for n= the sequence is in limbo, so to speak), in the long run will by far outnumber , and n<k only finitely often for any arbitrarily large k. I hope this is not too vague. 
20140831, 17:28  #134 
Dec 2008
you know...around...
2^{3}·3·5^{2} 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. Until then, thanks for reading. 
20140831, 19:26  #135  
"Nathan"
Jul 2008
Maryland, USA
3·7·53 Posts 
Quote:


20140831, 20:21  #136 
Dec 2008
you know...around...
2^{3}×3×5^{2} Posts 

20140831, 22:12  #137 
"Nathan"
Jul 2008
Maryland, USA
3×7×53 Posts 

20140902, 15:33  #138 
Dec 2008
you know...around...
2^{3}×3×5^{2} Posts 

20140902, 23:01  #139 
"Nathan"
Jul 2008
Maryland, USA
10001011001_{2} Posts 

20140902, 23:08  #140  
"Nathan"
Jul 2008
Maryland, USA
3·7·53 Posts 
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:


20140903, 03:36  #141 
Jun 2003
1276_{16} Posts 
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.

20140904, 20:15  #142  
Dec 2008
you know...around...
2^{3}×3×5^{2} Posts 
Quote:
Quote:
While I'm here, may I add that Or, equivalently, 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 Do these constants belong to someone here? 

20141210, 02:54  #143 
"W. Byerly"
Aug 2013
1423*2^21790231
137_{8} 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 20141210 at 02:55 