It's probably me. 8000@1e7 running, and a polsel.

I'll do the GNFS for this one; it'll be over by Christmas. Probably even by Christmas Eve.

sieved 2M-16.5M with
[code]
# norm 4.431184e-13 alpha -7.190056 e 3.778e-11
skew: 1381889.42
c0: 13935895326125594374428769951334655
c1: 6832990194435681055086947697
c2: -76268592758638424003951
c3: -29988262496226737
c4: 49802168836
c5: 660
Y0: -353206674018659220179872166
Y1: 561766051832977
n: 3628612227069559920455070746827882655714694792773312751322418651421798373899250166140599596858424814418370603853288254380696130063491713
lpbr: 27
lpba: 27
mfbr: 54
mfba: 54
alambda: 2.5
rlambda: 2.5
alim: 10000000
rlim: 10000000
[/code]

and should probably have used 28-bit lp to get better yield (IE this run was a bit slower than a slightly larger number that I did a year ago with 28/13)

[code]
Wed Dec 23 17:58:15 2009 found 3500504 hash collisions in 15421358 relations
Wed Dec 23 17:59:09 2009 found 3639407 duplicates and 11845280 unique relations
Wed Dec 23 18:09:52 2009 matrix is 1626466 x 1626692 (470.4 MB) with weight 121859512 (74.91/col)
Wed Dec 23 18:09:52 2009 sparse part has weight 107055126 (65.81/col)
Wed Dec 23 18:09:52 2009 matrix includes 64 packed rows
Wed Dec 23 21:55:17 2009 BLanczosTime: 13774
Wed Dec 23 22:12:06 2009 sqrtTime: 1009
Wed Dec 23 22:12:06 2009 prp57 factor: 615246762379545008872051386151610963316901978007615471813
Wed Dec 23 22:12:06 2009 prp79 factor: 5897816045444012068057492946558964042767474423553931349916095671773447719452301
[/code]

running ecm on C125 of step 2493 now

2493 down, set 8000@1e7 going on C136 of 2494 (factor 519158260471434103953487553 divides C163, factorization.ath.cx not presently responding)

C121 of 2495 has had 200@1e7, I'll be giving it some more but probably it's GNFS-worthy. Someone else can do this one.

Some bad news: there is a 1 in 2 chance that 4788 will pick up a 3 on the next line (assuming the c121 splits into 2 primes).

[QUOTE=fivemack;199768]C121 of 2495 has had 200@1e7, I'll be giving it some more but probably it's GNFS-worthy. Someone else can do this one.[/QUOTE]

Can you please post the c121? The DB seems to be down...

[QUOTE=Andi47;199774]Can you please post the c121? The DB seems to be down...[/QUOTE]

Here it is:
[code]
3942887076714143202481073805784961368619711215403426727623016228578779996971144714931046107405659926915911799176550312767
[/code]

[QUOTE=Andi47;199774]Can you please post the c121? The DB seems to be down...[/QUOTE]

It's up now.

[QUOTE=R. Gerbicz;199776]Here it is:
[code]
3942887076714143202481073805784961368619711215403426727623016228578779996971144714931046107405659926915911799176550312767
[/code][/QUOTE]

Thanks. I will do a p-1 followed by GNFS

edit: started p-1 and polsel in parallel

[quote=10metreh;199770]
Some bad news: there is a 1 in 2 chance that 4788 will pick up a 3 on the next line (assuming the c121 splits into 2 primes).[/quote]

Just I analyzed about that, simply

c171 = 2[sup]4[/sup] . 3559. 220681 . 592772569 . 7547146554874360724158181221231 . c121

s(c171) = (1+2+4+8+16) (1+3559) (1+220681) (1+592772569) (1+7547146554874360724158181221231) (1+p) (1+q) - c171

c171 = 1 mod 3, all the factors 3559, 220681, 592772569, 7547146554874360724158181221231 are 1 mod 3, 31 is 1 mod 3.

c121 is 1 mod 3. If it splits up into 2 factors, it can split up as
(1 mod 3 × 1 mod 3) or (2 mod 3 × 2 mod 3). Each has equally 50% chance.

When it splits up as (2 mod 3 × 2 mod 3) then (1+p) will be divisible by 3, then sigma(c171) - c171 will not be divisible by 3. When it splits up into (1 mod 3 × 1 mod 3), then
s(c171) = (1 . 2[sup]6[/sup] - 1) mod 3 = 1 - 1 = 0 mod 3, then it will pick up the factor 3.

Suppose that it splits up into 3 factors, then a factor of 2 mod 3, will not allow to pick up a 3 as stated above. Even if all the 3 factors are 1 mod 3, then s(c171) = (1 . 2[sup]7[/sup] - 1) mod 3 = 2 - 1 = 1 mod 3, then it will not pick up a 3 anywhere at all.

To come up to a conclusion, the next iteration [B]will[/B] acquire up the factor 3 if and only if
the [B]c121 splits up into two (1 mod 3) factors[/B].
Does [B]not[/B] pick up a 3, when the [B]c121 splits up into two (2 mod 3) factors[/B].

Right, man?