I have started 1000 curves at 11e6. (44 done so far, no factor)

2300@11e6 are done with no result. This is about 0.5*t45, how much ECM'ing we need?

[QUOTE=unconnected;200932]2300@11e6 are done with no result. This is about 0.5*t45, how much ECM'ing we need?[/QUOTE]

My rough guess is somewhere between 0.5*t55 = 9k @ 11e7 and 1*t55 ~ 17.5k @ 11e7

Please correct me if I'm wrong.

I've found this from earlier in our work on 4788:
[code]2333. 267978784508632868343114847060812948279629292222915880423759712858412228867957672971392000334670854838983335804438635643938788616600711408 = 2^4 * 3^2 * 92989028981 * 20012724099852599089965372134485155116134772186454249553781779673226527290856944625213723452411520667787389702557824640299547
2334. 481989619367342300734926329299579333889808536162741292564431101669679385519428605078705127063681626407548814866268364104159634219199843400 = 2^3 * 3^2 * 5^2 * 113 * 1613 * 1469103417254906821891838525901519591281166652227511390565077811335259014361034774239982074917602330156553759267494027765310350877
2335. 1152029206656316235471974481634606578356690329167224215707869782740824424411095903792101012459097585754639615021880805093048397651097766560 = 2^5 * 3^2 * 5 * 32003 * 1699413013515872926645731409301488559658444959 * 14709955753239154289970165983314290789323403834840450008018750264105583016407344451037[/code]

This escape from 2^3 * 3 * 5 is the highest ever, at 138 digits.

I feel escape form 2^3*3*5 is different from escape from 2^3*3^2*5 because 2^3*3^2*5 can never become a driver without changing the power of 2. This is because there are 2 factors of 3 - one from 2^3 and one from 5. 2^5*3^2*7 is also this way with sigma(2^5)=63 - which is 2 factors of 3 as well.

So if you have a square term on the 3 it shouldn't count as the 2^3*3*5 driver.

[QUOTE=Greebley;200985]I feel escape form 2^3*3*5 is different from escape from 2^3*3^2*5 because 2^3*3^2*5 can never become a driver without changing the power of 2. This is because there are 2 factors of 3 - one from 2^3 and one from 5. 2^5*3^2*7 is also this way with sigma(2^5)=63 - which is 2 factors of 3 as well.

So if you have a square term on the 3 it shouldn't count as the 2^3*3*5 driver.[/QUOTE]

I thought that if the 5 is squared, the 3^2 could drop back to 3^1.

Actually that is true, but rare enough that I still don't consider them the same.

The driver concept doesn't really fully cover higher powers adequately to really say.

p-1: B1=2e9, B2=1e15, no factor

2000@4e7 complete, no factor.

Will run 2000@1e8 over the weekend (I have real-work to do on the dual-E5520 machine until then); I think it might now be sensible for jrk to start a polynomial search.

1600 @ 43e6 completed. 800 @ 11e7 scheduled for tonight

[QUOTE=bsquared;201046]1600 @ 43e6 completed. 800 @ 11e7 scheduled for tonight[/QUOTE]

I have done 1 curve each at 4e7 and 1e8 to read the number of required curves for each digit level (used for the calculation below)

So we currently have a total of:

[CODE]
900 @ 1e6
2300 @ 11e6
2001 @ 40e6
1600 @ 43e6
1 @ 1e8[/CODE]

this sums up to ~322.8% of t45 = 52.3% of t50 = 7.65% of t55