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Old 2009-11-25, 10:20   #562
Andi47
 
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Quote:
Originally Posted by Andi47 View Post
So - do we use 13e or 14e?
p-1: B1=5e8, B2=5e13, no factor
p+1: 3 runs at B1=1e8, B2=6e12, no factor.

Running some more ECM until it is decided, whether we use 13e or 14e.
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Old 2009-11-25, 11:17   #563
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4000 @ 4e7 completed, no factor.

Use 13e. I'll take 2M-6M.
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Old 2009-11-25, 11:53   #564
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Quote:
Originally Posted by fivemack View Post
4000 @ 4e7 completed, no factor.

Use 13e. I'll take 2M-6M.
reserving 6M - 7M
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Old 2009-11-25, 13:35   #565
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taking 7M-9M
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Old 2009-11-29, 19:51   #566
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Line 2472 is done. Line 2473, after only some very early prelim ECM (two Quick ECMs and partial aliqueit at B1=5e4) looks like 2^5*c169.
If this c169 is semiprime, what might happen to the guide?
Edit: Never mind, found a p23, c147 remaining. Survived 2 Quick ECMs and 100@5e4 so far.
Edit 2: Finished all 214@5e4, (25 digits) will also finish the 430@25e4 (30 digits) and do some P+-1.
Edit 3: Finished 430@25e4. (30 digits) Also did P-1, max B1 100M, and P+1, max B1 50M. All in DB. Not working on it any more ATM.

Last fiddled with by Mini-Geek on 2009-11-29 at 20:43
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Old 2009-11-29, 20:50   #567
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Code:
Using B1=43000000, B2=388112953420, polynomial Dickson(30), sigma=1315477237
Step 1 took 138222ms
Step 2 took 77110ms
********** Factor found in step 2: 4123290685142646964541600067761
Found probable prime factor of 31 digits: 4123290685142646964541600067761
Composite cofactor 83203877652763456216344080057649507874810835149655593949890433813292280209307613162194128906010127548847166175620653 has 116 digits
I will do the c116
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Old 2009-11-29, 21:04   #568
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Quote:
Originally Posted by jrk View Post
Code:
Using B1=43000000, B2=388112953420, polynomial Dickson(30), sigma=1315477237
Step 1 took 138222ms
Step 2 took 77110ms
********** Factor found in step 2: 4123290685142646964541600067761
Found probable prime factor of 31 digits: 4123290685142646964541600067761
Composite cofactor 83203877652763456216344080057649507874810835149655593949890433813292280209307613162194128906010127548847166175620653 has 116 digits
I will do the c116
43e6 - opening eggs with nuclear warheads?

I have found the same c31 with B1=3e6 after 68 curves ~5 minutes ago. (I planned to run it overnight, but I stopped ECMing now, hence you started GNFS)

Last fiddled with by Andi47 on 2009-11-29 at 21:05
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Old 2009-11-29, 21:54   #569
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c116 = p37.p37.p43

p37 = 3134160372909851723612253077515467493
p37 = 3636840079737787436938569640509488477
p43 = 7299584610785326987792600494916936890870973
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Old 2009-11-29, 23:25   #570
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I'm doing the c109 now.
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Old 2009-11-30, 03:41   #571
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Quote:
Originally Posted by jrk View Post
I'm doing the c109 now.
Done.

Code:
prp54 factor: 481989758002837826944000482973711890329883965855795003
prp56 factor: 16617620933533935114464391289780146092392771482229686557
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Old 2009-11-30, 06:01   #572
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line 2476 c121 finished:

Code:
Using B1=3000000, B2=5706890290, polynomial Dickson(6), sigma=3159156024
Step 1 took 8056ms
Step 2 took 4090ms
********** Factor found in step 2: 13432838311461612250464777552146733002407
Found probable prime factor of 41 digits: 13432838311461612250464777552146733002407
Probable prime cofactor 130982517230425503437968877198264838192630336209630023456287838747078505224065181 has 81 digits
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