20171016, 18:18  #12 
"Robert Gerbicz"
Oct 2005
Hungary
1,429 Posts 
Using (one) Mertens theorem I see the slightly better: for every eps>0 and for k>k0(eps) it is true that p1<(1+eps)*sqrt(k*log(k)). Some quick computation confirms this (say for k=2e6).

20171016, 21:01  #13  
May 2003
13·19 Posts 
Quote:


20171016, 21:25  #14  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20171016, 21:41  #15 
May 2003
13·19 Posts 
If a factor has been forbidden, is that only tentative until all the factorizations are done?

20171016, 22:54  #16 
Apr 2006
97 Posts 
Forget about the word "forbidden". Just consider the proof as a case analysis. We look for an odd perfect N < 10^2000.
First, we consider the case "127 divides N". So, we explore the tree for 127, which is made of the subtrees for "127^2  N", "127^4  N", and so on. After exploring all the tree for "127 divides N" without finding any N < 10^2000, we consider the other case, that is, "127 does not divide N". The assumption "127 does not divide N" is not enough to conclude that N > 10^2000. So we continue and we consider now the case "19 divides N". So we suppose both "127 does not divide N" and "19 divides N" and we explore the subtrees for "19^2  N", "19^4  N", ... (Notice that the subtree for "19^2  N" is quite short). After exploring all the tree for "19 divides N" without finding any N < 10^2000, we consider the other case, that is, "19 does not divide N". The assumptions "127 does not divise N" and "19 does not divide N" are not enough to conclude that N > 10^2000. So we continue and we consider now the case "7 divides N". We suppose gcd(N,127*19)=1 and "7 divides N" and we explore the subtrees for "7^2  N", "7^4  N", ... (Notice that the subtree for "7^2  N" is quite short). After that , we know that gcd(N,127*19*7)=1, which is still not enough to conclude that N > 10^2000. We continue in this manner until there only remains the case gcd(N,127*19*7*11*331*31*97*61*13*398581*1093*3*5*307*17)=1. Now this is sufficient to conclude that N > 10^2000. 
20171021, 21:07  #17  
"William"
May 2003
New Haven
2^{3}·5·59 Posts 
Quote:
The first factor tree is built assuming 127 divides the OPN. It turns out that no OPN less than the threshold is possible if 127 divides the OPN. The next factor tree is built assuming 19 divides the OPN. At this time we know that 127 does not divide our OPN, so any time we reach a branch that would have 127 divide the OPN, we can end that branch. In this sense, 127 is forbidden in the factor tree for 19. The third factor tree assumes 7 divides the OPN  but we can trim any branch that has 127 or 19 in it because we have already proven that our OPN cannot be divisible by either  hence 127 and 19 are now forbidden in building the third factor tree. There are factor trees testing for an OPN divisible by 3 and 7. Neither is forbidden until after that factor tree has shown they are impossible. So 3^n and 7^n composites show up in the t1200 file because their factors would simplify the corresponding factor trees (and possibly some of the other factor trees that are created before 3 and 7). 

20171126, 23:23  #18 
May 2003
60B_{16} Posts 
Just saw this thread. Pascal, thank you for your kind words about my paper!

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