mersenneforum.org  

Go Back   mersenneforum.org > Factoring Projects > Aliquot Sequences

Reply
 
Thread Tools
Old 2014-05-18, 03:53   #1
EdH
 
EdH's Avatar
 
"Ed Hall"
Dec 2009
Adirondack Mtns

2·1,607 Posts
Default Sociable Numbers

Everything I'm finding on line is several years old. Where would I locate anything current? One question is what the exhaustive search limit has reached for all orders. For order 4, it seems to be 5x1012, but that's of a few years ago.

Is anyone actively seeking them now?
EdH is offline   Reply With Quote
Old 2014-05-21, 09:37   #2
Drdmitry
 
Drdmitry's Avatar
 
Nov 2011

233 Posts
Default

The most recent list of known sociable cycles can be found here. The last time it was updated on March 2014.
About four years ago I did the search of all cycles such that
the minimal element is odd and is <=10^14;
the maximal element is <=10^16 and
the length is <=200.
About two years ago I extended the search to the cycles with minimal element divisible by 2 but not divisible neither by 4 nor by 3.
I do not think that anyone did the full search of sociable numbers beyond 5*10^12.
Drdmitry is offline   Reply With Quote
Old 2014-05-21, 13:11   #3
EdH
 
EdH's Avatar
 
"Ed Hall"
Dec 2009
Adirondack Mtns

2·1,607 Posts
Default

Thanks for the reply. As you may remember, I have "played around" with these before and am dabbling again. My previous ventures were in the 10^15 area, but quite spotty.

I had been reviewing the list you provided, but thank you for referencing it, in case I wasn't aware. Although the date shows 2014, the latest addition to the list seems to be from a long time previous. I guess I can consider this list complete, since the date is current?

I have renewed my interest and am playing with some more pari scripts, but have narrowed the parameters. I'm currently running a four-order only search right at 5x10^12, but am searching both odd and even. Since you've covered the odd pattern, I will see if I can gain improvement by narrowing my tests to even elements.

Not sure how long my attention span will extend this time, but for now, I'll run a "few" checks and see if I can rediscover the already known cycles.
EdH is offline   Reply With Quote
Old 2014-06-05, 15:15   #4
EdH
 
EdH's Avatar
 
"Ed Hall"
Dec 2009
Adirondack Mtns

2·1,607 Posts
Default

Quote:
Originally Posted by Drdmitry View Post
...
About two years ago I extended the search to the cycles with minimal element divisible by 2 but not divisible neither by 4 nor by 3.
I do not think that anyone did the full search of sociable numbers beyond 5*10^12.
Well, I've refined(?) a pari routine that seems to run well, but I'm wondering if the numbers you omitted in your search were because sociables would not exist, or the time involved would be outside your workable zone.

I've set up my machines to cover the numbers above 5*10^12 that you skipped, if I understood your description correctly, but I'm only set up to catch amicables and order four sequences. Am I wasting my time in this region? If so, is there a better region and search criteria?

Thanks.
EdH is offline   Reply With Quote
Old 2014-06-06, 09:59   #5
Drdmitry
 
Drdmitry's Avatar
 
Nov 2011

233 Posts
Default

Quote:
Originally Posted by EdH View Post
Well, I've refined(?) a pari routine that seems to run well, but I'm wondering if the numbers you omitted in your search were because sociables would not exist, or the time involved would be outside your workable zone.

I've set up my machines to cover the numbers above 5*10^12 that you skipped, if I understood your description correctly, but I'm only set up to catch amicables and order four sequences. Am I wasting my time in this region? If so, is there a better region and search criteria?

Thanks.
Well, the method I used works much faster if the Aliquot sequence goes down on average. So it was most optimal for odd numbers, however it also worked quite well for downdrivers. I thought about continuing the search for numbers starting with 2^2 times a prime number bigger than 7 but then I was distracted by other things.
One can modify the algorithm so it becomes more efficient not only for decaying sequences by starting with the number prior the largest number in the cycle. (This trick is explained in D. Moews, P. Moews 1993 paper).
Aliquot cycles containing drivers are theoretically possible but they are very unlikely (especially in such a "small" range as 5*10^12 -- 10^15). So I did not check them.
Finally it is useless to look for the amicable pairs in the range 5^10^12 -- 10^15 since the complete search of them has already been done (I do not remember by whom). On the other hand it makes sense to look for the cycles of length four. I guess there should be 2 -- 4 of them not discovered yet.
Drdmitry is offline   Reply With Quote
Old 2014-06-06, 14:25   #6
EdH
 
EdH's Avatar
 
"Ed Hall"
Dec 2009
Adirondack Mtns

2×1,607 Posts
Default

Quote:
Originally Posted by Drdmitry View Post
Well, the method I used works much faster if the Aliquot sequence goes down on average. So it was most optimal for odd numbers, however it also worked quite well for downdrivers. I thought about continuing the search for numbers starting with 2^2 times a prime number bigger than 7 but then I was distracted by other things.
One can modify the algorithm so it becomes more efficient not only for decaying sequences by starting with the number prior the largest number in the cycle. (This trick is explained in D. Moews, P. Moews 1993 paper).
Aliquot cycles containing drivers are theoretically possible but they are very unlikely (especially in such a "small" range as 5*10^12 -- 10^15). So I did not check them.
Finally it is useless to look for the amicable pairs in the range 5^10^12 -- 10^15 since the complete search of them has already been done (I do not remember by whom). On the other hand it makes sense to look for the cycles of length four. I guess there should be 2 -- 4 of them not discovered yet.
Thanks. I'm turning up amicables as a side-product on my way through four iterations. If I'm correct in my programming, which is intended to skip what you have searched, then I will be checking all the numbers you omitted. My current program checks only 0 mod 4 and even 0 mod 3 initial numbers and a test found 430324482433184, which is from the list of known sociables. My machines are currently spanning 5.5-5.7*10^12.

Thanks again for the info.
EdH is offline   Reply With Quote
Old 2014-08-14, 01:55   #7
EdH
 
EdH's Avatar
 
"Ed Hall"
Dec 2009
Adirondack Mtns

1100100011102 Posts
Default

After a long break (of a couple months), I have renewed my efforts and restarted many of my machines.

@Drdmitry: On Moews' page "A LIST OF ALIQUOT CYCLES OF LENGTH GREATER THAN 2" I can't find the credit for cycle number 56. Am I missing something right in front of me? More to my real question, did you discover this one, or do you know how it was found?

This cycle is the only one in the list which will meet my search criteria any time relatively soon and I hope it proves my script valid when I reach it. (Actually, I hope even more to find a new one...)
EdH is offline   Reply With Quote
Old 2014-09-22, 15:02   #8
MichelMarcus
 
Nov 2011
Saint Maur, France

2×52 Posts
Default

Cycle number 56 was discovered by Andre Needham.
See this line in http://djm.cc/sociable.txt
42,46,48-51,53-57: A. Needham
MichelMarcus is offline   Reply With Quote
Old 2014-09-23, 13:27   #9
EdH
 
EdH's Avatar
 
"Ed Hall"
Dec 2009
Adirondack Mtns

2×1,607 Posts
Default

Quote:
Originally Posted by MichelMarcus View Post
Cycle number 56 was discovered by Andre Needham.
See this line in http://djm.cc/sociable.txt
42,46,48-51,53-57: A. Needham
Thanks - I overlooked it somehow and a search for "56" was obviously ineffective...
EdH is offline   Reply With Quote
Old 2014-11-19, 19:27   #10
EdH
 
EdH's Avatar
 
"Ed Hall"
Dec 2009
Adirondack Mtns

2·1,607 Posts
Default

I have found one - an aliquot cycle of order four - that I have not found listed anywhere, as of yet. Is it possible this is a new discovery? The cycle is:
Code:
14592614233912 = 2^3 * 11 * 13 * 263 * 48501071
17674178946248 = 2^3 * 11 * 2417 * 8783 * 9461
18500448943672 = 2^3 * 71 * 60527 * 538127
16677107567048 = 2^3 * 220511 * 9453671
A Thank You to Dana Jacobsen (danaj) for pointing me toward his ntheory module for Perl, which has allowed me to discover this cycle much sooner than I would have running the previous PARI version of my search routine.
EdH is offline   Reply With Quote
Old 2014-11-19, 19:31   #11
Batalov
 
Batalov's Avatar
 
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3

5·1,811 Posts
Thumbs up

Nice one! Congrats!

Looking at djm.cc list, "2E3.11." and "2E3.71." might be good seeds for a focused search for similar ones.

EDIT: Is it of a form given by Borho ? No, it is not.

Last fiddled with by Batalov on 2014-11-19 at 19:56
Batalov is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Carmichael numbers and Devaraj numbers devarajkandadai Number Theory Discussion Group 0 2017-07-09 05:07
6 digit numbers and the mersenne numbers henryzz Math 2 2008-04-29 02:05
LLT numbers, linkd with Mersenne and Fermat numbers T.Rex Math 4 2005-05-07 08:25

All times are UTC. The time now is 17:24.

Fri Jul 3 17:24:03 UTC 2020 up 100 days, 14:57, 2 users, load averages: 1.78, 1.70, 1.59

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.