20090707, 03:52  #12 
"Jason Goatcher"
Mar 2005
6663_{8} Posts 
Well, people probably already thought of this, but if you simply turn the entire k*2^n+c into a raw number, you could probably beat the record rather easily. :)
Depends on whether or not you go by the rule that if k is even then you divide it by 2 and add 1 to n to get the "right" number. 
20090707, 12:43  #13  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11×389 Posts 
Quote:


20090707, 13:55  #14  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
For trial division in ascending order, k is a good measure of the number of factors that need to be tested, but the difficulty of each trial division increases with the size of the factor. 

20090707, 15:09  #15 
May 2009
Loughborough, UK
2^{2}×11 Posts 
The biggest known smallest k is (M426438011)/(2*42643801)
This thread is interesting. What is the average k? How often is k=1? or where is the data so that I can average it myself? 
20090708, 16:39  #16 
Nov 2008
912_{16} Posts 
Another interesting question: What is the largest k for a factor found by TF?

20090709, 10:06  #17  
May 2009
Loughborough, UK
101100_{2} Posts 
Where may I find the data?
Quote:
How many factors should I expect to find? How big a file? The largest factors are missing where Mx is fully factored. How can I reconstitute these factors? Are they tagged somehow as fully factored? The only way I could think of would be to primality test Mx/all_listed_factors_of_Mx 

20090711, 12:55  #19 
May 2009
Loughborough, UK
2^{2}×11 Posts 
Thanks for that.
Will Edgington stores them as k's, so that makes it easier. It looks like the fully factored ones do have the largest factor missing and some of the smaller exponents are missing. 
20090711, 13:35  #20 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6,679 Posts 

20090711, 14:25  #21 
Einyen
Dec 2003
Denmark
2×17×101 Posts 
2*42643801 is not a factor of 2^{42643801}2, same for p=43112609.
Anyway even if they were that would be a 2*k*p factor of 2^{p}2. This thread is about 2*k*p+1 factors of 2^{p}1. Last fiddled with by ATH on 20090711 at 14:26 
20090711, 15:18  #22 
Jun 2003
2^{2}·3^{2}·151 Posts 
Really? How do you figure?
If you agree that M(p) is a factor of M(p), then that is exactly what this thread discusses (hint: 2kp = M(p)1 and 2kp+1 = M(p) are equivalent) 
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