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Old 2006-08-06, 15:08   #67
xilman
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Quote:
Originally Posted by xilman
Looking very much like approx 20M lines, each of length 54M (so the skewness is 27/20, close to the correct value of 1.337) will do it. The factorbases run to 50M on each side with LPB=500M on each side. Circa 43M relations will be needed.
20M lines will generate 46M relations, or thereabouts, which is ample.


Paul
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Old 2006-08-07, 12:16   #68
R.D. Silverman
 
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Quote:
Originally Posted by hlaiho
I'm sorry for my stupid expectations about 2LM numbers.

However, I will ask, how big composite is too big with GNFS for NFSNET currently.
Are the numbers 2,1574L and 2,1586L, both c156, too big?
11,251+, c151 is smaller and 2,1658L, c147 is even smaller.

Is there any good candidate that NFSNET could do?

Heikki
Nobody said they were stupid. It is quite possible that you had
an idea for good polynomials that noone else had thought of.
Which is why I said "teach me"; I could not find reasonable polynomials
for the numbers you suggested.

The numbers you suggest for doing via GNFS are within range, but
there is no reason to do them *at this time*. Generally, participants
prefer to do numbers that are among the first few holes. Look at
the Cunningham 'wanted' lists; they are on Sam Wagstaff's website.
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Old 2006-08-25, 12:51   #69
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Originally Posted by Wacky View Post
A month later and we are at 69%. That's 20% in the last month. And at the present rate, we should finish this one at the end of the month.
It is nearing the end of the month......

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Old 2006-08-25, 15:32   #70
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Quote:
Originally Posted by R.D. Silverman View Post
It is nearing the end of the month......

From PM dated Monday, the 21st:

Quote:
Originally Posted by Wacky
We should expect to finish sieving on 3,479+ about the 27th or 28th
(another week).
Bruce
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Old 2006-09-10, 14:37   #71
xilman
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Sieving finished a while back and Richard perforned some preliminary filtering before turning the surviving relations over to me.

I was able to turn them into a 5.03M matrix of weight 349M.

That matrix is currently 40 hours into a computation predicted to take 323 hurs, or just under two weeks.


Paul
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Old 2006-09-17, 08:52   #72
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Originally Posted by xilman View Post
Sieving finished a while back and Richard perforned some preliminary filtering before turning the surviving relations over to me.

I was able to turn them into a 5.03M matrix of weight 349M.

That matrix is currently 40 hours into a computation predicted to take 323 hurs, or just under two weeks.
Now at something over 60% complete. It looks like it will finish about the time I'm visiting Harvard. Completing the factorization will most likely have to wait until I get back.


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Old 2006-09-22, 08:08   #73
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Now at something over 60% complete. It looks like it will finish about the time I'm visiting Harvard. Completing the factorization will most likely have to wait until I get back.


Paul
Squeaked in in the nick of time. I'm actually already on my way to the airport while typing this:

Code:
Factorization completed after 2048.45 seconds, at Fri Sep 22 08:23:07 2006
Original number had 197 digits:
98843386509177521595904605338374556546613001655483917613534171106190384350917469364953354041535092335695935657974170411872241239891377134460477922256385857005513727420340442274468415195491759667831
Probable prime factor 1 has 100 digits:
8472423314580350450217632140650154894206616336671035361679713874206239480866924866454101040705885621
Probable prime factor 2 has 98 digits:
11666483465135187963585838119336434764875738256006740572675427278475860680359350759067856017206011
Not an ECM miss this time.

More details either from Wacky or from me when I return, which ever comes first.

Paul
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Old 2006-09-22, 10:30   #74
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Wow, nice factors!

Alex
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Old 2006-09-23, 15:47   #75
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I know I'm not a math person, but how can the two factors of a number collectively have more digits than their original product like in the example where the original number had 197 digits and the two factors had a combined total of 198?
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Old 2006-09-23, 16:21   #76
rogue
 
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Quote:
Originally Posted by Jwb52z View Post
I know I'm not a math person, but how can the two factors of a number collectively have more digits than their original product like in the example where the original number had 197 digits and the two factors had a combined total of 198?
Think 2*3=6.
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Old 2006-10-02, 11:28   #77
R.D. Silverman
 
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Quote:
Originally Posted by xilman View Post
Squeaked in in the nick of time. I'm actually already on my way to the airport while typing this:

Code:
Factorization completed after 2048.45 seconds, at Fri Sep 22 08:23:07 2006
Original number had 197 digits:
98843386509177521595904605338374556546613001655483917613534171106190384350917469364953354041535092335695935657974170411872241239891377134460477922256385857005513727420340442274468415195491759667831
Probable prime factor 1 has 100 digits:
8472423314580350450217632140650154894206616336671035361679713874206239480866924866454101040705885621
Probable prime factor 2 has 98 digits:
11666483465135187963585838119336434764875738256006740572675427278475860680359350759067856017206011

Paul
I just finished 2,1478L.

2,1478L C170 = p79.p91

p79 = 2690689930879440037372552876789965290601761240871891720842427997721986219162721

The linear algebra for 2,1490L is in progress. 2,1526L has been sieved.
I am now sieving 5,314+.

Any info on 5,311-?????
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