20220811, 12:22  #672 
"Ed Hall"
Dec 2009
Adirondack Mtns
2·5·7·73 Posts 
In looking over the tables I'm seeing what I'll call stragglers that are within reach, although some are bordering on discomfort. 293^65, as mentioned, and 14536^38 fall into the comfortable straggler category, while 263^69 falls into the uncomfortable straggler category. I have started work on all three. They should complete their bases when terminated.

20220811, 12:24  #673 
May 2007
Kansas; USA
2^{3}×5×17^{2} Posts 
Post #1 duplicated itself. The top part is correct. The bottom part is what it was like last night.
I find stragglers often. Most of the "random" terminations you see of mine is me running additional ECM on some decent looking sequences and they happen to crack. Often that's all that is needed to make it much easier to terminate. Last fiddled with by gd_barnes on 20220811 at 12:27 
20220811, 12:50  #674 
May 2007
Kansas; USA
10110100101000_{2} Posts 
35^107, 44^94, and 59^93 terminate

20220811, 13:19  #675 
"Ed Hall"
Dec 2009
Adirondack Mtns
2·5·7·73 Posts 
Thanks for helping me keep post #1 straight. I figured the stragglers I mentioned were already ECM'd by you and deemed outside your reach, so I'm bringing them down to <140. I'll be tied up shortly, so I've queued the two mentioned and we can discuss further work later.

20220811, 14:02  #676 
May 2007
Kansas; USA
10110100101000_{2} Posts 
306^64 terminates.
That completes base 306! Nice work, Ed, to bring that one down. I just realized this. The 59^93 that I just terminated brought base 59 down to only 59^95 remaining. It's 166/138/5. ECM to only t35. That one you might consider. It's not often we get close like that by chance on bases between 50 and 100. With 100 exponents each, they get large in size. We've already done the largest two exponents on this one. There's a good chance that 59^95 could be easier than your 263^69, which is 164/160/7. ECM is done to t40 on it. If the C160 doesn't crack in ECM, it would be tough. Last fiddled with by gd_barnes on 20220811 at 14:10 
20220811, 21:11  #677 
May 2007
Kansas; USA
2^{3}×5×17^{2} Posts 
392^61 terminates

20220811, 22:05  #678 
"Ed Hall"
Dec 2009
Adirondack Mtns
1001111110110_{2} Posts 
263^69 is now at 155/153, so I'll let it finish (<140) and add 59^95. After that, I'll probably pick out some more stragglers. I also want to change my scripts a bit, but I'm going to be busy with other stuff tomorrow, so I don't know if I'll queue more sequences or work on the scripts if I have time at all..

20220812, 05:49  #679  
May 2007
Kansas; USA
2^{3}·5·17^{2} Posts 
28^112 and 277^63 terminate
293^65 terminated by Ed. That completes base 293! Pretty good for a base that was just recently initialized. I've been taking on some tougher tasks and working on some that are current size < 150 digits in case we decide to tackle that group. Both of the ones that I terminated here were in that group. Most of the rest are pretty tough for me. I'll take one more last look over the list of them and see if there are any more that I can tackle. Quote:
I'm curious if you'll encounter some tough C150+'s on 59^95. Last fiddled with by gd_barnes on 20220812 at 06:24 

20220812, 11:54  #680 
May 2007
Kansas; USA
11560_{10} Posts 
59^95, 263^69, 1152^52, and 14536^38 terminate
Thank you, Ed, for the great reductions! This completes bases 59, 263, and 14536!! Including earlier bases 293 and 306, that completes 5 bases in a single 24hour period!! Last fiddled with by gd_barnes on 20220812 at 12:27 Reason: add one more 
20220812, 12:22  #681 
May 2007
Kansas; USA
2^{3}×5×17^{2} Posts 
Edited the last post for a 5th base completion in 24 hours. Signing off now. :)
Last fiddled with by gd_barnes on 20220812 at 12:28 Reason: 4th > 5th 
20220812, 13:06  #682 
"Ed Hall"
Dec 2009
Adirondack Mtns
5110_{10} Posts 
I added a progress file to my scripts and it shows ending status for each step of a sequence. Here's 293^65:
Code:
293^65 (151/138) 293^65 (150/137) 293^65 (149/124) 293^65 (149/112) 293^65 (148/139) 293^65 (147/135) 293^65 (147/142) 293^65 (146/130) 293^65 (145/127) 293^65 (Prime) Code:
59^95 (161/154) 59^95 (158/156) 59^95 (157/149) 59^95 (149/145) 59^95 (137/122) I did some closer inspection of the entire set of what we have currently and we have completed six bases: Code:
251 263 293 306 14536 2147483647 We have four bases with only one sequence remaining: Code:
277^69: 166/120 9699690^24: 168/146 200560490130^14: 160/131 7420738134810^14: 182/161 OTOH, there were a lot more bases with two sequences left than I realized: Code:
37^105: 163/138 37^107: 155/137 42^98: 159/133 42^100: 155/147 55^93: 157/152 55^99: 172/126 241^67: 154/138 241^69: 160/119 257^65: 150/139 257^69: 161/128 281^67: 156/121 281^69: 148/148 288^64: 158/132 288^65: 158/137 385^61: 158/140 385^65: 161/155 552^58: 154/139 552^60: 165/157 660^56: 152/150 660^60: 170/140 1184^52: 160/152 1184^54: 166/154 2310^48: 161/140 2310^50: 168/140 12496^38: 156/145 12496^40: 164/158 14264^38: 157/145 14264^40: 167/159 14316^38: 159/129 14316^40: 167/152 131071^33: 161/138 131071^35: 172/171 510510^28: 145/143 510510^30: 172/151 I was going to ask if you wanted to run the c120 for 277^69, but I see you've signed off, so I'll see where it heads. It could easily be a pain with 3 and 7 sitting in a small factors list of five. 
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