Newbie advice - Page 3

Belteshazzar · 2011-03-02, 14:45

Got it- sieving took ~9 days. p59*p78, factors reported to factordb.

Well, having read up some more about the algorithms in the meantime, my curiosity is satisfied for now, and I'll be glad to have my laptop freed up. I may visit again in the future if I end up doing more number theory or get hardware more fit for the task (say, a Bulldozer desktop and a CUDA-capable card).

Thanks!

Batalov · 2011-03-05, 05:58

Now, 109!+2 is factored and 109 is not a member of the sequence, then 110!+2 is factored and 110 is a member. Next candidates are 112 and 114.

Input number is (2+110!)/446 (176 digits)
Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=664425596
Step 1 took 33854ms
Step 2 took 13759ms
********** Factor found in step 2: 144476918413758184246036836336545025029
Found probable prime factor of 39 digits: 144476918413758184246036836336545025029
Probable prime cofactor ((2+110!)/446)/144476918413758184246036836336545025029 has 138 digits

sean · 2011-03-06, 19:03

I have run 17900 curves with b1=110e6 and 4000 curves with b1=300e6 on 114!+1 without finding a factor.
114!+2 is easily factored and has an even number of factors.

Batalov · 2011-03-06, 20:58

Yes, both 114!+1 and 115!+1 are tough nuts on this road.
(This was only 'easy' until the +2 side caught up with the +1 bleeding edge.) After them, there are 118, 119, 120, etc etc...

sean · 2011-03-07, 02:24

It has been brought to my attention that 101 has not been definitively ruled out of this sequence. I do not appear to have a complete factorization for 101!+2 (and likely never had one). The remaining C145 almost certainly splits into two primes which would eliminate 101.

Anyone want to have a crack at it?

chris2be8 · 2011-03-09, 17:56

If you post or PM the C145 I'll have a go. It'll take me about 3 weeks though.

Chris K

sean · 2011-03-09, 18:55

Quote:

Originally Posted by chris2be8

If you post or PM the C145 I'll have a go. It'll take me about 3 weeks though.

Code:

101!+2 = 2.76651.42727279519.C145

C145=1439037017974459737209176591173154490623757641646671713778278703958022205299519888479141257083536507005774582500092299135064595968911493780619229

I've run over 5000 curves with b1=11e6, and a few with b1=110e6 without breaking it, so it should yield a good size factor.

chris2be8 · 2011-03-10, 17:22

I'll run ecm at 43e6 for a few days. T50 is about the right amount of ecm to to run before starting gnfs. If anyone wants to help just post how many you've run and I'll post when we have enough.

Chris K

richs · 2011-03-11, 06:01

167!+2 = 2.p13.p14.3693165342354759209.p256

chris2be8 · 2011-03-11, 16:56

Quote:

Originally Posted by sean

I've run over 5000 curves with b1=11e6, and a few with b1=110e6 without breaking it, so it should yield a good size factor.

How many curves did you run with b1=110e6?

Chris K

sean · 2011-03-13, 19:09

Quote:

Originally Posted by chris2be8

How many curves did you run with b1=110e6?

Only 60.

2011-03-06, 19:03	#25
sean Aug 2004 New Zealand DF₁₆ Posts	I have run 17900 curves with b1=110e6 and 4000 curves with b1=300e6 on 114!+1 without finding a factor. 114!+2 is easily factored and has an even number of factors. Last fiddled with by sean on 2011-03-06 at 19:05

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2011-03-02, 14:45	#23
Belteshazzar Feb 2011 2⁵ Posts	Got it- sieving took ~9 days. p59*p78, factors reported to factordb. Well, having read up some more about the algorithms in the meantime, my curiosity is satisfied for now, and I'll be glad to have my laptop freed up. I may visit again in the future if I end up doing more number theory or get hardware more fit for the task (say, a Bulldozer desktop and a CUDA-capable card). Thanks!

2011-03-05, 05:58	#24
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·47·101 Posts	Now, 109!+2 is factored and 109 is not a member of the sequence, then 110!+2 is factored and 110 is a member. Next candidates are 112 and 114. Input number is (2+110!)/446 (176 digits) Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=664425596 Step 1 took 33854ms Step 2 took 13759ms ********** Factor found in step 2: 144476918413758184246036836336545025029 Found probable prime factor of 39 digits: 144476918413758184246036836336545025029 Probable prime cofactor ((2+110!)/446)/144476918413758184246036836336545025029 has 138 digits

2011-03-06, 20:58	#26
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·47·101 Posts	Yes, both 114!+1 and 115!+1 are tough nuts on this road. (This was only 'easy' until the +2 side caught up with the +1 bleeding edge.) After them, there are 118, 119, 120, etc etc...

2011-03-07, 02:24	#27
sean Aug 2004 New Zealand 223 Posts	It has been brought to my attention that 101 has not been definitively ruled out of this sequence. I do not appear to have a complete factorization for 101!+2 (and likely never had one). The remaining C145 almost certainly splits into two primes which would eliminate 101. Anyone want to have a crack at it?

2011-03-09, 17:56	#28
chris2be8 Sep 2009 2²×523 Posts	If you post or PM the C145 I'll have a go. It'll take me about 3 weeks though. Chris K

2011-03-10, 17:22	#30
chris2be8 Sep 2009 82C₁₆ Posts	I'll run ecm at 43e6 for a few days. T50 is about the right amount of ecm to to run before starting gnfs. If anyone wants to help just post how many you've run and I'll post when we have enough. Chris K

2011-03-11, 06:01	#31
richs "Rich" Aug 2002 Benicia, California 2·659 Posts	167!+2 = 2.p13.p14.3693165342354759209.p256