20131114, 05:21  #1 
Sep 2008
Kansas
6364_{8} Posts 
Team sieve #41: C165 from 3366:i2098
Tracking Aliquot Sequence 3366 index 2098.
Code:
# sieve with ggnfs lasieve4I15e on a side from Q=14M to 27M # aq3366:2098 n: 325051015990535995874007588250630424661613002093251050951251894916362694976876466038809353011636859335168409712756235986441762310332466382944321915205259453868831101 # norm 5.420993e016 alpha 6.929648 e 7.006e013 rroots 5 skew: 9356889.28 c0: 1926024418625074017922955379536534110520 c1: 1894119714203966541640748866185150 c2: 1214709679644137937069462919 c3: 2325548176534019814 c4: 13139381174804 c5: 232560 Y0: 67465745109864492336315060899031 Y1: 266382034364015119 rlim: 42000000 alim: 42000000 lpbr: 30 lpba: 30 mfbr: 60 mfba: 87 rlambda: 2.6 alambda: 3.6 Postprocessing by: ??? Run the siever using the following command: Code:
gnfslasieve4I15e a 3366_2098.poly o <output_file_name> f <start_of_range> c <length_of_range> (I like to watch the progress.) In case a restart is needed in the range, add the R switch. (All range numbers must be full decimal representations. i.e., 21M = 21000000) Since ggnfs is single thread, multiple copies need to be run at the same time to utilize multiple cores. Be sure to use a different output_file_name for each range. Something like 3408_22M as signifying the starting point. The output files can be placed at any of the file sharing services: DropBox, RapidShare, SendSpace, etc. This is a nice poly. Expect yields greater than 4:1 in some regions. To start things off I'll take 1418M. 
20131114, 08:25  #2 
"Frank <^>"
Dec 2004
CDP Janesville
2×1,061 Posts 

20131114, 15:08  #3 
Dec 2012
2×139 Posts 
I'm not sure how much to take. I can do 1M in under a week. Is 2M okay?
If so, I'll take 18M20M. Last fiddled with by Jayder on 20131114 at 15:18 
20131114, 15:38  #4 
"Ed Hall"
Dec 2009
Adirondack Mtns
2×1,847 Posts 
I'll do 20M22M.

20131115, 19:38  #5 
"Ed Hall"
Dec 2009
Adirondack Mtns
111001101110_{2} Posts 
20M22M are on SendSpace:
3366i2098_20M21M.rels.zip 3366i2098_21M22M.rels.zip I had a lot of trouble with this task. Half of my machines segfaulted out of the sieving. I think everything is cleaned up, but you might want to test them separately. They have been run through remdups4 and show as 4521249 and 4522756 unique, respectively. remdups4 showed a total of 90445 duplicates. I'll run 22M24M. 
20131116, 03:00  #6 
Sep 2008
Kansas
3316_{10} Posts 

20131117, 01:50  #7 
"Ed Hall"
Dec 2009
Adirondack Mtns
7156_{8} Posts 
22M24M are on SendSpace:
3366i2098_22M23M.rels.zip 3366i2098_23M24M.rels.zip They have been run through remdups4 and show as 4440508 and 4432062 unique, respectively. remdups4 showed a total of 81892 duplicates. Since I seem to be going along well, I'll run 24M26M. How many relations are we shooting for? I seem to work it out to about 93M relations needed for a c165. But, I only see about 58M coming from the range we're working. Am I that far off? 
20131117, 03:20  #8  
Sep 2008
Kansas
2^{2}·829 Posts 
Quote:
I'm at 72% with 4.8:1 yield. 

20131117, 03:30  #9  
Sep 2008
Kansas
2^{2}×829 Posts 
Quote:
Edit: So we have to go to Q=40 or 42M (or more?). Another week and a half. :) Last fiddled with by RichD on 20131117 at 03:34 

20131117, 04:22  #10 
"Frank <^>"
Dec 2004
CDP Janesville
2·1,061 Posts 
There's a mistake with the first link....I got the second one, though....

20131117, 14:59  #11 
"Ed Hall"
Dec 2009
Adirondack Mtns
7156_{8} Posts 
Sorry about that!
The correct link should be: http://www.sendspace.com/file/uep62s I will repost all the (hopefully correct) links in my next set of rels... 
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