Reserved for MF - Sequence 4788 - Page 67

Andi47 · 2010-01-05, 10:41

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

unconnected · 2010-01-05, 10:45

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

Andi47 · 2010-01-05, 11:12

Quote:

Originally Posted by unconnected

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

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

Please correct me if I'm wrong.

10metreh · 2010-01-05, 18:55

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

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

Greebley · 2010-01-05, 20:40

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.

10metreh · 2010-01-05, 20:43

Quote:

Originally Posted by Greebley

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.

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

Greebley · 2010-01-05, 20:56

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.

Andi47 · 2010-01-06, 10:48

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

fivemack · 2010-01-06, 11:42

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.

bsquared · 2010-01-06, 14:03

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

Andi47 · 2010-01-06, 14:17

Quote:

Originally Posted by bsquared

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

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

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

2010-01-05, 20:56	#733
Greebley May 2009 Dedham Massachusetts USA 34B₁₆ Posts	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. Last fiddled with by Greebley on 2010-01-05 at 20:57

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2010-01-05, 10:41	#727
Andi47 Oct 2004 Austria 2×17×73 Posts	I have started 1000 curves at 11e6. (44 done so far, no factor)

2010-01-05, 10:45	#728
unconnected May 2009 Russia, Moscow A21₁₆ Posts	2300@11e6 are done with no result. This is about 0.5*t45, how much ECM'ing we need?

2010-01-05, 20:40	#731
Greebley May 2009 Dedham Massachusetts USA 3×281 Posts	I feel escape form 2^335 is different from escape from 2^33^25 because 2^33^25 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. 2^53^27 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 2^335 driver.

2010-01-06, 10:48	#734
Andi47 Oct 2004 Austria 2×17×73 Posts	p-1: B1=2e9, B2=1e15, no factor

2010-01-06, 11:42	#735
fivemack (loop (#_fork)) Feb 2006 Cambridge, England 3·2,141 Posts	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.

2010-01-06, 14:03	#736
bsquared "Ben" Feb 2007 2²·3·293 Posts	1600 @ 43e6 completed. 800 @ 11e7 scheduled for tonight