[quote=Raman;221303]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)?[/quote]

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.

Someone jestfully posted a p56 factor :-)

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

[QUOTE=firejuggler;221749]
running poly select... will take about 2 hours[/QUOTE]
Don't bother. Nothing slips past you guys! :smile:

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

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

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

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

[QUOTE=firejuggler;221758]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[/QUOTE]

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).

[QUOTE=Batalov;221757]
Greg is probably cranking on a 400-cpu cluster: one shot of sieving ...and pop goes the matrix.[/QUOTE]
Actually only 24 cores. And I put in a bunch of ECM first, so you may be ahead of me...

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?

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)?