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#12 |
"Jason Goatcher"
Mar 2005
66638 Posts |
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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. |
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#13 | |
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11×389 Posts |
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#14 | |
"Lucan"
Dec 2006
England
2×3×13×83 Posts |
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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. |
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#15 |
May 2009
Loughborough, UK
22×11 Posts |
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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? |
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#16 |
Nov 2008
91216 Posts |
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Another interesting question: What is the largest k for a factor found by TF?
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#17 | |
May 2009
Loughborough, UK
1011002 Posts |
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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 |
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#19 |
May 2009
Loughborough, UK
22×11 Posts |
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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. |
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#20 |
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6,679 Posts |
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#21 |
Einyen
Dec 2003
Denmark
2×17×101 Posts |
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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 |
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#22 |
Jun 2003
22·32·151 Posts |
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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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