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#1 |
May 2006
22 Posts |
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I have completely factored the 10th Sylvester number, with "ecm -one -I 1 -c 0 1"
Using B1=29896574, B2=106305143196, polynomial Dickson(12), sigma=1602805018 Step 1 took 834376ms Step 2 took 239259ms ********** Factor found in step 2: Sylvester(10) = 2287 * 2271427 * 21430986826194127130578627950810640891005487 * P156 see http://www.research.att.com/~njas/sequences/A091335 |
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#2 | |
Aug 2005
Brazil
2×181 Posts |
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EDIT: Yes, I was. I did my calculations again and now they're suddenly correct! What a strange thing... EDIT2: Congratulations (I forgot to say), but shouldn't you start at a higher B1 than 1? It should've saved you some curves... Last fiddled with by fetofs on 2006-05-31 at 19:45 |
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#3 |
Aug 2004
New Zealand
2×5×23 Posts |
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There's a wiki for the sequence that includes kenta's factor already :-)
http://en.wikipedia.org/wiki/Sylvester%27s_sequence These numbers arise naturally from Euclid's proof of infinite prime. In this regard they are similiar to the Euclid Mullin sequence. |
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#4 |
Aug 2002
Buenos Aires, Argentina
2×761 Posts |
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According to the history stored on the Wikipedia page, that factor was added on May 26th, 2006. So it was already known.
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#5 |
Jul 2004
Potsdam, Germany
3·277 Posts |
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Maybe he found the factor back then (well, 4-5 days earlier), and only wrote the Mersenneforum posting some days later?
After all, he is a new user, so it sounds reasonable to me that someone pointed him to this forum after some time... |
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#6 | |
Aug 2005
Brazil
2·181 Posts |
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#7 | |
Aug 2004
New Zealand
2·5·23 Posts |
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