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#1 |
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"Michael Kwok"
Mar 2006
1,181 Posts |
I was playing around with tpsieve and found five factors at p=10P and two factors at p=100P:
10000006224344089 | 4184379*2^481439+1 10000012821412039 | 6681861*2^482663+1 10000013413645747 | 3946077*2^483496-1 10000016097203239 | 4692531*2^483533+1 10000017044013187 | 1852539*2^482057-1 100000111903220981 | 2995515*2^480627-1 100000274118432799 | 9555297*2^480626-1 Could someone verify these factors (using a software other than tpsieve) to make sure that those k/n pairs really are divisible by those factors? Also, if these factors are indeed valid, what's the max p value of tpsieve? I know it's beyond the optimal TPS sieving depth, but I'd still like to know what tpsieve is capable of. On another note, does anyone want to try beating my records? |
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#2 | |
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A Sunny Moo
Aug 2007
USA
2·47·67 Posts |
Quote:
Mod(number,factor) It returns something like this: %1 = Mod(0, factor) If the 0 had been anything else, the factor would not have divided the number. There's probably a more efficient way to do this, but I'm not particularly savvy with PARI. BTW @gribozavr: you might want to remove these factors from the sieve file next time you do that. It's only a few factors, but hey, no reason to waste them.
Last fiddled with by mdettweiler on 2010-09-18 at 01:39 |
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#3 |
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Mar 2005
Internet; Ukraine, Kiev
11×37 Posts |
Factors saved :)
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#4 | |
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Aug 2006
22·3·499 Posts |
Quote:
Code:
Mod(4184379*2^481439+1, 10000006224344089) Code:
4184379*Mod(2, 10000006224344089)^481439+1 Last fiddled with by CRGreathouse on 2010-09-19 at 18:07 |
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