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#1 | |
Sep 2002
2×131 Posts |
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all b1's are smaller than m so they assume that the largest factor of p-1 is m and they don't reach it. what a waste! Joss |
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#2 | |
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
1177010 Posts |
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Paul |
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#3 |
Sep 2002
10616 Posts |
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So I see!
I tested 2^251-1 b1=2 b2=2 and found the factor 27271151 Tks xilman You're right, b1 is actually b1+m There was no wasted cycle. I pushed the ![]() Joss |
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#4 | |
"Richard B. Woods"
Aug 2002
Wisconsin USA
22·3·641 Posts |
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27271151 - 1 = 2 ร 5^2 ร 41 ร 53 ร 251 |
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#5 |
"Richard B. Woods"
Aug 2002
Wisconsin USA
22·3·641 Posts |
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Pminus1=79299433,660000,3465000,0,0
This will find any factor of 2^79299433-1 where factor = 2 * k * 79299433 + 1 and k has largest factor =< 3465000 and all other factors =< 660000. Actually, there are stage 2 extensions such that some k with largest factor =< (3465000 * a small integer) will be found, in certain cases. Last fiddled with by cheesehead on 2004-11-28 at 13:31 |
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