20060531, 00:01  #1 
May 2006
100_{2} Posts 
Sylvester's Sequence
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 
20060531, 19:40  #2  
Aug 2005
Brazil
2×181 Posts 
Quote:
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 20060531 at 19:45 

20060531, 21:53  #3 
Aug 2004
New Zealand
11011110_{2} Posts 
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. 
20060601, 12:18  #4 
Aug 2002
Buenos Aires, Argentina
10101000100_{2} Posts 
According to the history stored on the Wikipedia page, that factor was added on May 26th, 2006. So it was already known.

20060601, 14:01  #5 
Jul 2004
Potsdam, Germany
3×277 Posts 
Maybe he found the factor back then (well, 45 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... 
20060601, 18:38  #6  
Aug 2005
Brazil
2·181 Posts 
Quote:


20060601, 20:50  #7  
Aug 2004
New Zealand
DE_{16} Posts 
Quote:


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