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 2009-07-07, 03:52 #12 jasong     "Jason Goatcher" Mar 2005 66638 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.
2009-07-07, 12:43   #13
TimSorbet
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11×389 Posts

Quote:
 Originally Posted by jasong 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.
We're talking about the k in 2kp+1 (it's proven that all factors of Mersenne numbers are of this form, for some integer k), not about the k in k*b^n+c.

2009-07-07, 13:55   #14
davieddy

"Lucan"
Dec 2006
England

2×3×13×83 Posts

Quote:
 Originally Posted by Uncwilly We all know that it is not the size of the factor that counts, rather the size of the k.
Yes and no.
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.

 2009-07-07, 15:09 #15 plandon   May 2009 Loughborough, UK 22×11 Posts The biggest known smallest k is (M42643801-1)/(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?
 2009-07-08, 16:39 #16 10metreh     Nov 2008 91216 Posts Another interesting question: What is the largest k for a factor found by TF?
2009-07-09, 10:06   #17
plandon

May 2009
Loughborough, UK

1011002 Posts

Where may I find the data?

Quote:
 At the bottom of the GIMPS search status page you will find links to several files:
There are no links at the bottom of that page and the link to factors.zip gives a 505 error.

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

 2009-07-09, 10:23 #18 kar_bon     Mar 2006 Germany 2,999 Posts look here at the 1st third of the page, there you'll find files named results.pfk.split..bz2 with all factors of Mersenne numbers.
 2009-07-11, 12:55 #19 plandon   May 2009 Loughborough, UK 22×11 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.
2009-07-11, 13:35   #20
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

6,679 Posts

Quote:
 Originally Posted by plandon The biggest known smallest k is (M42643801-1)/(2*42643801)
Why not p=43112609? Wouldn't that be larger?

2009-07-11, 14:25   #21
ATH
Einyen

Dec 2003
Denmark

2×17×101 Posts

Quote:
 Originally Posted by plandon The biggest known smallest k is (M42643801-1)/(2*42643801)
Quote:
 Originally Posted by retina Why not p=43112609? Wouldn't that be larger?
2*42643801 is not a factor of 242643801-2, same for p=43112609.

Anyway even if they were that would be a 2*k*p factor of 2p-2. This thread is about 2*k*p+1 factors of 2p-1.

Last fiddled with by ATH on 2009-07-11 at 14:26

2009-07-11, 15:18   #22
axn

Jun 2003

22·32·151 Posts

Quote:
 Originally Posted by ATH 2*42643801 is not a factor of 242643801-2, same for p=43112609.
Really? How do you figure?

Quote:
 Originally Posted by ATH Anyway even if they were that would be a 2*k*p factor of 2p-2. This thread is about 2*k*p+1 factors of 2p-1.
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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