Reserved for MF - Sequence 4788 - Page 86

Greebley · 2010-07-15, 14:28

Quote:

Originally Posted by Raman

This is the first sequence to hit up with iteration (index) number 9000
Let us continue with that until it acquires a driver

What are the chances of it to acquire a downdriver at this point
at 171 digits
with a factorization such as
8 times some prime of form 1 (mod 4)?

Well the chance of getting a random number of 170 digits prime is 1 in 170*ln(10) (ln(10) is approx 2.3)

However in this case we know the number isn't divisible by 2, 3, or 5 (so its 1,7,11,13,17,19,23,29 mod 30) which should be 15/4 times more likely to be prime. However only 1/2 the primes are 1 mod 4. This give 15/8 * 1/(170*ln(10)) or 1 in 8*34/3*ln(10).

The final answer becomes approximately 1 in 209.

2 is obvious and 3,5 follow because the 2^3 factor means we can't add a 3 or 5 factor (the same way we can't drop it with 2^3*3*5).

I am not completely sure I have considered all possible factors, but I think for all other primes, the chances are even that they will divide making the number effectively random for all factors above 5.

Batalov · 2010-07-17, 07:06

Someone jestfully posted a p56 factor :-)

firejuggler · 2010-07-17, 07:07

the C169 have been broken,
c169=p56*c113, and it is not me
I guess that a ggnfs is in order?
yhis will end with a p56*p56*p56 or something of that order
running poly select... will take about 2 hours

frmky · 2010-07-17, 07:49

Quote:

Originally Posted by firejuggler

running poly select... will take about 2 hours

Don't bother. Nothing slips past you guys!

firejuggler · 2010-07-17, 08:02

hrmmm... well, ended in a p55*p56*p59... wich is quite a nice 3 way split

Batalov · 2010-07-17, 08:13

I have 2M rels for the c108... too late?

Greg is probably cranking on a 400-cpu cluster: one shot of sieving ...and pop goes the matrix.

firejuggler · 2010-07-17, 08:14

not yet. we need 5M relation for this size, right?
hmm the guide is 2^3*7.... I hope for the drop of the 7 and 2^3...
but i feel that it will acquire the 2^2*7 driver... which is bad

10metreh · 2010-07-17, 08:48

Quote:

Originally Posted by firejuggler

not yet. we need 5M relation for this size, right?
hmm the guide is 2^3*7.... I hope for the drop of the 7 and 2^3...
but i feel that it will acquire the 2^2*7 driver... which is bad

2^3 can't be lost when there is a factor of 7 unless the 7 is raised to an even power (or the number is 2^3 * 7^n where n == 1 mod 4).

frmky · 2010-07-17, 08:49

Quote:

Originally Posted by Batalov

Greg is probably cranking on a 400-cpu cluster: one shot of sieving ...and pop goes the matrix.

Actually only 24 cores. And I put in a bunch of ECM first, so you may be ahead of me...

Batalov · 2010-07-17, 09:57

This factor of 7 is bad news. It drives the seq up, and will stay, at least for this iteration. Now, c148 will need some night ecming, eh?

firejuggler · 2010-07-17, 15:39

since my prediction was somewhat accurate for the 169... lets play again..
a p58*p91 for 4788:i2548

I have a question. For those number with 130+ digits, is it safe to assume that ignoring the lower bound B1 (ie : 1e4 to 1e6)
and go with a B1 of 11e6 and above will allow us to factor the cofactor faster ? Or does the 'speed' increase is too low to be usefull (1e4 is a matter of milliseconds.. but atm, for the c148, a B1 set to 25e4 take me 3.5 sec each iteration, skipping the 400 iterations would 'save' me about 23 minutes)?

2010-07-17, 07:07	#938
firejuggler Apr 2010 Over the rainbow 2×1,303 Posts	the C169 have been broken, c169=p56c113, and it is not me I guess that a ggnfs is in order? yhis will end with a p56p56p56 or something of that order running poly select... will take about 2 hours Last fiddled with by firejuggler on 2010-07-17 at 07:23*

2010-07-17, 08:13	#941
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9,497 Posts	I have 2M rels for the c108... too late? Greg is probably cranking on a 400-cpu cluster: one shot of sieving ...and pop goes the matrix. Last fiddled with by Batalov on 2010-07-17 at 08:16

2010-07-17, 08:14	#942
firejuggler Apr 2010 Over the rainbow 2×1,303 Posts	not yet. we need 5M relation for this size, right? hmm the guide is 2^37.... I hope for the drop of the 7 and 2^3... but i feel that it will acquire the 2^27 driver... which is bad Last fiddled with by firejuggler on 2010-07-17 at 08:33

2010-07-17, 15:39	#946
firejuggler Apr 2010 Over the rainbow 5056₈ Posts	since my prediction was somewhat accurate for the 169... lets play again.. a p58p91 for 4788:i2548 I have a question. For those number with 130+ digits, is it safe to assume that ignoring the lower bound B1 (ie : 1e4 to 1e6) and go with a B1 of 11e6 and above will allow us to factor the cofactor faster ? Or does the 'speed' increase is too low to be usefull (1e4 is a matter of milliseconds.. but atm, for the c148, a B1 set to 25e4 take me 3.5 sec each iteration, skipping the 400 iterations would 'save' me about 23 minutes)? Last fiddled with by firejuggler on 2010-07-17 at 16:13*

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2010-07-17, 07:06	#937
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9,497 Posts	Someone jestfully posted a p56 factor :-)

2010-07-17, 08:02	#940
firejuggler Apr 2010 Over the rainbow 2×1,303 Posts	hrmmm... well, ended in a p55p56p59... wich is quite a nice 3 way split

2010-07-17, 09:57	#945
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9,497 Posts	This factor of 7 is bad news. It drives the seq up, and will stay, at least for this iteration. Now, c148 will need some night ecming, eh?