Reserved for MF - Sequence 4788

[yoyo]

I queued some.

[yoyo]

Nothing found so far: http://www.rechenkraft.net/yoyo/y_status_ecm.php

[RichD]

Wow! Thanks for all the curves.

A good polynomial was found by Alfred. Not sure how Batalov found his post. Maybe it was accidentally deleted after the critique. It did sieve better than VBCurtis which was posted later.

This might be a nice job for NFS@Home...

[Batalov]

No, no.
There was no post.

Instead of a post (or a post with quote), "Alfred" pressed the "Report the post" button, and the system sent an email to all moderators. Way to go. We have nothing else to do other than read "complaints" about the posts which are not even complaints. Or reposting posts which were not even posts. But hey, we are nice -- we can do even that.

[RichD]

So Alfred was reporting his results, but not by posting a post, but by reporting a post.

So he used report a post, to report his results, instead of posting a post.

I get it. (I think.)

[Batalov]

A 50-50 chance to crack the 2^3*3^2 driver here, hmm?
(No one seems to be excited...)

[LaurV]

Originally Posted by Batalov

A 50-50 chance to crack the 2^3*3^2 driver here, hmm?
(No one seems to be excited...)

You know, first I was thinking exactly to the same thing, hehe. Then I realized that if you consider the 3 (mod 4) factor in front and the fact that the cofactor is not prime (and it has 164 digits, and also can split in more than two ways), then the chances are much much smaller... somewhere close to one in billions... many billions...

[Batalov]

Originally Posted by LaurV

... consider the 3 (mod 4) factor in front and ... many billions...

Maybe we are looking at different "in front"s?
...97 is 1 (mod 4) and ...21 is 1 (mod 4).
As Ostap Bender used to say, "I can determine that even without a stethoscope"

[RichD]

Originally Posted by LaurV

... (and it has 164 digits, and also can split in more than two ways) ...

Not very likely since yoyo has taken it to a full t60. Anyway Team sieve #43 is on the board. Come join in on the fun. Hopefully I didn't have another brain fart this time.

[LaurV]

Originally Posted by Batalov

Maybe we are looking at different "in front"s?
...97 is 1 (mod 4) and ...21 is 1 (mod 4).
As Ostap Bender used to say, "I can determine that even without a stethoscope"

No, we were looking to the same. It was a typo on my side about that 97 being 3 mod 4 (in fact was not a typo, I was idiot in the morning, as usual, but in the tests I put 97, grrr)... Also, the cofactor is 2 mod 3 and 1 mod 4, which is 5 mod 12. Your error (if it was not a joke about 50-50, I took it as a joke) might come from the fact that you forgot to take out the primes, or composites which contains 3, or are 1 mod 3, for all those you get higher chances (especially adding the primes, two primes which are 1 mod 4... but we know the cofactor is not prime). As the digits go higher the chances to get rid of the 2^3 are lower. Also, please remark that in my test there are a lot of multiples of 97 in those k's, therefore artificially increasing the chance of losing the driver (you get 97 squared, so you have two times a perfect square in front, this always adds a bit more than 1% chances (1/97) from which you can't get rid off, but in reality the chances are much lower, you have to decrease 0.01 from the final result [edit: in bold red]). The test takes only composites over a number of digits, which are 5 mod 12 and counts them. When going higher, factoring goes slower, therfore the breaks.

gp > cnt=0; digits=5; kcn=0; n=12*10^digits; for(k=1,10^6, p=n+12*k+5; if(!isprime(p), kcn++; a=2^3*3^2*97*p; s=sigma(a)-a; f=factorint(s); if(f[1,2]!=3,cnt++; printf("...%d : %2.7f...%c",k,1.0*cnt/kcn,13))))
...999999 : 0.2972856...
gp > cnt=0; digits=10; kcn=0; n=12*10^digits; for(k=1,10^6, p=n+12*k+5; if(!isprime(p), kcn++; a=2^3*3^2*97*p; s=sigma(a)-a; f=factorint(s); if(f[1,2]!=3,cnt++; printf("...%d : %2.7f...%c",k,1.0*cnt/kcn,13))))
 ...134719 : 0.2059614...
  *** user interrupt after 34,942 ms.

gp > cnt=0; digits=40; kcn=0; n=12*10^digits; for(k=1,10^5, p=n+12*k+5; if(!isprime(p), kcn++; a=2^3*3^2*97*p; s=sigma(a)-a; f=factorint(s); if(f[1,2]!=3,cnt++; printf("...%d : %2.7f...%c",k,1.0*cnt/kcn,13))))
 ...337 : 0.0701220...
  ***  user interrupt after 1min, 42,441 ms.
break>

[RichD]

2^3 * 3^4 * ... * C178

t50 by batalov posted here.

Any more curves should be 11e7 or higher.