[QUOTE=schickel;272376][SIZE="1"]PS. Clifford has mentioned the required form for a factorization to result in losing the 3, but I'm haven't been able to find that email yet....[/SIZE][/QUOTE]
I think it's something like this: All primes congruent to 1 mod 3 must be raised to [strike]an odd power[/strike] Edit: a power n with n != 2 mod 3. (since nearly all large factors will be ^1, this is the more important of the two; Edit: simplistically, in practice, this means you want the factors to be 1 mod 3) All primes congruent to 2 mod 3 must be raised to an even power [QUOTE=Batalov;272590]Don't forget  3 comes and goes... See [URL="http://www.factordb.com/sequences.php?se=1&eff=2&aq=611156&action=range&fr=4310&to=4350"]for example[/URL] 611156.[/QUOTE] Yes, it does. Still, it's more hopeful when it's gone. :smile: 
[QUOTE=MiniGeek;272588]Escape from 2^2*3 to 2^2 successful at 118 digits! :smile:
[url]http://factordb.com/sequences.php?se=1&aq=199710[/url] It's now on a slow downwardish slide, hopefully it will become the downdriver![/QUOTE]How about we delete the ones that manage to escape from the top post and move the sequence to the main reservation thread, unless indicated otherwise by the escapee? 
[QUOTE=MiniGeek;272592]I think it's something like this:
All primes congruent to 1 mod 3 must be raised to an odd power (since nearly all large factors will be ^1, this is the more important of the two) All primes congruent to 2 mod 3 must be raised to an even power.[/QUOTE]So does the exponent on the 3 matter? Also, does this change the sequences that get posted for work on this project? (Actually, at the very least, it lets out sequences with 5 raised to an odd power, correct? That takes 71 more out of play right off the bat....) 
Will take 183936, too.

[QUOTE=schickel;272594]How about we delete the ones that manage to escape from the top post and move the sequence to the main reservation thread, unless indicated otherwise by the escapee?[/QUOTE]
It'd be more work, and considering the 3 can come and go pretty easily, I don't really want to do it this way. I, for example, plan to run 199710 until it settles into a driver (currently driverless with 2^5*7^2*11) or grows too large for me to handle. I'd expect many others would do similar. I'd prefer to leave it in this thread until they say otherwise. (it's not like it prevents this subproject from being complete  we can mark sequences that are no longer 2^2*3 if that'd make things clearer) [QUOTE=schickel;272597]So does the exponent on the 3 matter? Also, does this change the sequences that get posted for work on this project? (Actually, at the very least, it lets out sequences with 5 raised to an odd power, correct? That takes 71 more out of play right off the bat....)[/QUOTE] AFAICT, no, the exponent on the 3 doesn't matter, because sigma(3^n) mod 3 is 1 for any n>=0. If you want to hit the most likely to break first, yes, you'd avoid sequences with 5 raised to an odd power, (because they can't lose it on the very next line) but depending on how hard it is to lose that 5 or change its power, it might be a very minor difference. A little more on my methods and why "sigma(3^n) mod 3 is 1" is important: see the [URL="http://www.mersennewiki.org/index.php/Aliquot_Sequences"]formula[/URL] for calculating the sigma of a number. If the current line is divisible by 3, that means it is 0 mod 3. Once we have its sigma, the next line is sigma  lastLine. Working mod 3, that's sigma  0, or sigma. So for the next line to not be divisible by 3, sigma != 0 mod 3 must be true. The sigma is the product of a series of numbers, which are the sigmas of the prime factors, e.g. sigma(2^2*3)=sigma(2^2)*sigma(3). If none of these numbers multiplied together are 0 mod 3, (i.e. all are 1 and 1, or 1 and 2 if you prefer) then the sigma will not be 0 mod 3, and so the next line will not be 0 mod 3. I'm sure I'm stating trivialities for mathematicians, but considering I'm the first in this thread to mention how to lose the 3, it might be of some use for learning for everyone. :smile: 
[QUOTE=MiniGeek;272592]All primes congruent to 1 mod 3 must be raised to an odd power (since nearly all large factors will be ^1, this is the more important of the two)[/QUOTE]
Correction: All primes congruent to 1 mod 3 must be raised to a power n with n != 2 mod 3. 
I'll take 243402.

[QUOTE=MiniGeek;272614]It'd be more work, and considering the 3 can come and go pretty easily, I don't really want to do it this way. I, for example, plan to run 199710 until it settles into a driver (currently driverless with 2^5*7^2*11) or grows too large for me to handle. I'd expect many others would do similar. I'd prefer to leave it in this thread until they say otherwise. (it's not like it prevents this subproject from being complete  we can mark sequences that are no longer 2^2*3 if that'd make things clearer)[/QUOTE]OK, whichever way you figure is easier (I'm all in favor of easier....)

Unreserving [URL="http://factordb.com/sequences.php?se=1&aq=199710"]199710[/URL], since it has a 2^5*7 guide and a c114 (full ECM done), size 120.
Reserving 247840. 
Just by way of encouragement.....
Just to help motivate everyone, here is what the status of bchaffin's latest termination was in my last pull:[code]734184 850. sz 111 2^2 * 3^3 * 83[/code]Yep, that's right, it was 2^2 * 3!!! :smile:
PS. Check out the [URL="http://factordb.com/aliquot.php?type=1&aq=734184&big=1"]slope[/URL] on that first downdriver run! 
[QUOTE=MiniGeek;272668]Reserving 247840.[/QUOTE]Hmmm....I left the file sorted by size, thinking it might make it easier for people to pick one out. Should we resort the list by sequence?

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