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Old 2011-07-03, 18:19   #1266
Mini-Geek
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"Tim Sorbera"
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Quote:
Originally Posted by Mini-Geek View Post
I realized I made a rather large mistake in that math: it ignores that 3 can not be a cofactor of a downdriver, since otherwise it would be the 2*3 perfect driver. ... The new and improved script is attached.
More trivia: there is a 5.01% chance of it ending by 163.6 digits, and a 95.05% chance of it ending by 66.2 digits.
(the first line without the downdriver was ~10^165.0; from the time of that posting we only had a ~2.55% chance of breaking by then) Either we were very unlucky, (i.e. about 1 in 39 such runs would end so soon) or my numbers are very off. But anyway, with the breaking being 2*7^2*p, I noticed one more thing that made it a little more inaccurate: I didn't take into account that 2*n^2*p can break it, too! I accounted for this by multiplying the probability of each line breaking by 1 + (sum 1/prime(k)^2, k=3 to infinity) (about 1.091). I was looking around a little for more accurate calculations than mine, and only found the Losing Downdriver thread, which seems to have last settled on numbers very different from mine: 1/420 for 100 digits. I calculate 1.09113630893*3/2/log(10)/100~=1/141. To simply state each, they (mainly Greebley post #6 with modifications for squares by CRGreathouse post #12) say about 1/(4.2*digits) I say 1/(1.4*digits). A factor of 3 different. I think it's because I'm accounting for the fact that the downdriver's cofactor can't have a factor of 2 or 3 by definition.
Does anyone have an automated way to look through many downdrivers and see if my numbers hold up to real test data? Just looking at this one, it'd seem I'm way off, and my memory seems to also say it's lost far easier, but I'd like something far more concrete.

The new script, taking into account the squares, is attached.
Attached Files
File Type: zip downdriverprobability.zip (689 Bytes, 81 views)

Last fiddled with by Mini-Geek on 2011-07-03 at 18:26
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Old 2011-07-05, 01:40   #1267
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Quote:
Originally Posted by jrk View Post
Preliminary poly for line 2696:
It's obsolete now, but in case anyone happens to collect these, here's the best one I found:
Code:
# sieve with ggnfs siever 14e on alg side from Q=12M to ??? (TBD)
# aq4788:2696
n: 8755586957647299753989940018066954131853563140628725006016461530329394193683073655733471708006846371139255974755721298957099259657401475696647916915828783
# norm 6.055972e-15 alpha -7.175530 e 3.223e-12 rroots 3
skew: 2748791.27
c0:  22518556851938655674731450005532544565
c1:  49007811576301845607236669717695
c2:  12280205727986637086832355
c3: -20291335212629008203
c4: -848640062100
c5:  765072
Y0: -408993941175622024607029493362
Y1:  1200827227936261913
rlim: 24000000
alim: 24000000
lpbr: 29
lpba: 29
mfbr: 58
mfba: 58
rlambda: 2.5
alambda: 2.5
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Old 2011-07-05, 01:42   #1268
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+400 curves @ 8e7 for line 2703
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Old 2011-07-07, 16:46   #1269
bchaffin
 
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I did the c142 from 2703, and four more terms. Now at 2708 with a c157. Except for one 2^2*5 it continues to decrease...

I've run 1000 @ 1M and 3000 at 10M so far. I'll keep running curves but this one is too big for me to sieve on my own.
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Old 2011-07-07, 17:44   #1270
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I'll start poly selection on line 2708 as usual.
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Old 2011-08-11, 16:04   #1271
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Done:

Code:
prp64 factor: 4038843625314075842900554513884322322635236034496696017887771851
prp93 factor: 288180620682225170824303650548350247409615780738733199886385422296800168275435704019909533659
Sequence still decreasing... I've done 1000 curves at 1e6 on the C153 at iteration 2709.
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Old 2011-08-14, 02:00   #1272
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I've done 3k curves at 10M and 1800 curves at 100M on the C153 of iteration 2709. I'm run with it a while longer and do the nfs.
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Old 2011-08-14, 02:48   #1273
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+960 curves @ B1=8e7
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Old 2011-08-14, 03:04   #1274
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Quote:
Originally Posted by bsquared View Post
Done:

Code:
prp64 factor: 4038843625314075842900554513884322322635236034496696017887771851
prp93 factor: 288180620682225170824303650548350247409615780738733199886385422296800168275435704019909533659
Sequence still decreasing... I've done 1000 curves at 1e6 on the C153 at iteration 2709.
At least it's still alive in terms of not getting a driver or an upwardly increasing guide!
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Old 2011-08-14, 23:26   #1275
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Quote:
Originally Posted by Batalov View Post
E = mc2 = , where m=2 and c=7...
No need to worry about a 2 · 7^2 ...
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Old 2011-08-15, 06:03   #1276
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Quote:
Originally Posted by biwema View Post
No need to worry about a 2 · 7^2 ...
There is a need to worry - it killed the downdriver.
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