mersenneforum.org Subproject #11: 2^2 * 3 - Lose the '3'
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2011-09-24, 03:40   #12
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17×251 Posts

Quote:
 Originally Posted by schickel 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....
I think it's something like this:
All primes congruent to 1 mod 3 must be raised to an odd power 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:
 Originally Posted by Batalov Don't forget - 3 comes and goes... See for example 611156.
Yes, it does. Still, it's more hopeful when it's gone.

Last fiddled with by Mini-Geek on 2011-12-31 at 03:15

2011-09-24, 03:53   #13
schickel

"Frank <^>"
Dec 2004
CDP Janesville

2×1,061 Posts

Quote:
 Originally Posted by Mini-Geek Escape from 2^2*3 to 2^2 successful at 118 digits! http://factordb.com/sequences.php?se=1&aq=199710 It's now on a slow downward-ish slide, hopefully it will become the downdriver!
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?

2011-09-24, 04:06   #14
schickel

"Frank <^>"
Dec 2004
CDP Janesville

212210 Posts

Quote:
 Originally Posted by Mini-Geek 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.
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....)

 2011-09-24, 07:08 #15 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100011101001102 Posts Will take 183936, too.
2011-09-24, 13:17   #16
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17·251 Posts

Quote:
 Originally Posted by schickel 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?
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:
 Originally Posted by schickel 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....)
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 formula 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.

Last fiddled with by Mini-Geek on 2011-09-24 at 13:26

2011-09-24, 13:55   #17
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

426710 Posts

Quote:
 Originally Posted by Mini-Geek 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)
Correction:
All primes congruent to 1 mod 3 must be raised to a power n with n != 2 mod 3.

 2011-09-24, 14:23 #18 bchaffin   Sep 2010 Portland, OR 7·53 Posts I'll take 243402.
2011-09-24, 15:20   #19
schickel

"Frank <^>"
Dec 2004
CDP Janesville

84A16 Posts

Quote:
 Originally Posted by Mini-Geek 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)
OK, whichever way you figure is easier (I'm all in favor of easier....)

 2011-09-25, 02:40 #20 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 17×251 Posts Unreserving 199710, since it has a 2^5*7 guide and a c114 (full ECM done), size 120. Reserving 247840.
 2011-09-25, 04:59 #21 schickel     "Frank <^>" Dec 2004 CDP Janesville 2·1,061 Posts 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 Yep, that's right, it was 2^2 * 3!!! PS. Check out the slope on that first downdriver run! Last fiddled with by schickel on 2011-09-25 at 05:01 Reason: Add PS.
2011-09-25, 05:04   #22
schickel

"Frank <^>"
Dec 2004
CDP Janesville

2·1,061 Posts

Quote:
 Originally Posted by Mini-Geek Reserving 247840.
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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