mersenneforum.org  

Go Back   mersenneforum.org > Factoring Projects > Aliquot Sequences

Reply
 
Thread Tools
Old 2020-08-21, 16:01   #474
chris2be8
 
chris2be8's Avatar
 
Sep 2009

35748 Posts
Default

Factordb itself factors everything below 70 digits. At present I'm factoring the range from 70 to 80 digits. So I've probably contributed to several sequences.

But there's been a lot of what looks like junk added to factordb over the past year of so. Numbers like 24681280*46##+251 obviously won't be part of a sequence. Does anyone know where they are coming from?

Is there a way to check if a given number is from a sequence (eg clicking More information)?

Chris
chris2be8 is offline   Reply With Quote
Old 2020-08-21, 16:34   #475
EdH
 
EdH's Avatar
 
"Ed Hall"
Dec 2009
Adirondack Mtns

D3C16 Posts
Default

factordb does factor <71 digit composites (and larger via helper elves), but Aliquot sequences are only advanced by bumping them. The bump may advance a few lines at a time, but if it stalls, it stays there even after the stalled composite is factored, until another bump. I have been doing all my sequence runs off line with Aliqueit to lessen the elves' load (having served as an elf myself).

Alas, I have found no way to tie a number to a particular sequence within the db. I suppose that's not feasible because of how many sequences would be listed for smaller numbers.
EdH is offline   Reply With Quote
Old 2020-08-21, 16:59   #476
henryzz
Just call me Henry
 
henryzz's Avatar
 
"David"
Sep 2007
Cambridge (GMT/BST)

131478 Posts
Default

Conjecture 2 is bugging me. It is fairly easy to show that the 5 maintains the 3 unless the sequence begins 2^(4*5^(2n-1)*k).
If the sequence begins with 2^(4*5^n*k) then we know 2^20-1 = 3*5^2*11*31*41 is a divisor of s(2^(4*5^n*k)).
This means we have to account for 11 and 41. In the same way as 5, 11 maintains the 3 unless the sequence begins 2^(4*5^(2n-1)*11^(2m-1)*k). Accounting for 41 gives a requirement of beginning with 2^(4*5^(2n-1)*11^(2m-1)*41^(2o-1)*k).
s(2^(4*5^(2n-1)*11^(2m-1)*41^(2o-1)*k)) is always divisible by 2^9020-1 which contains yet more primes that preserve the 3.
Expanding the lower bound like this is easy but I can't quite think how to turn it into a proof. Maybe there is something based on proving there will always be more and more prime factors that are 2 mod 3.
henryzz is online now   Reply With Quote
Old 2020-08-21, 20:29   #477
RichD
 
RichD's Avatar
 
Sep 2008
Kansas

61028 Posts
Default

Quote:
Originally Posted by EdH View Post
... but Aliquot sequences are only advanced by bumping them.
That's what I meant by a refresh - a bump.
RichD is offline   Reply With Quote
Old 2020-08-21, 21:45   #478
EdH
 
EdH's Avatar
 
"Ed Hall"
Dec 2009
Adirondack Mtns

22×7×112 Posts
Default

Quote:
Originally Posted by RichD View Post
That's what I meant by a refresh - a bump.
I understood that. I was replying to Chris.
EdH is offline   Reply With Quote
Old 2020-08-22, 02:00   #479
RichD
 
RichD's Avatar
 
Sep 2008
Kansas

2·3·523 Posts
Default

I was clarifying my original post for Chris.
RichD is offline   Reply With Quote
Old 2020-08-22, 11:03   #480
warachwe
 
Aug 2020

2·3 Posts
Default

Quote:
Originally Posted by garambois View Post
But, do you also think that it is no longer worthwhile to formulate conjectures like those of post #447, or do you think that this kind of conjecture can still be of interest ?
Personally, I think these conjectures are interesting in a way that the prove seem somewhat doable, but I agree that we should look more into other phenomena.

Quote:
Originally Posted by henryzz View Post
Conjecture 2 is bugging me. It is fairly easy to show that the 5 maintains the 3 unless the sequence begins 2^(4*5^(2n-1)*k).
If the sequence begins with 2^(4*5^n*k) then we know 2^20-1 = 3*5^2*11*31*41 is a divisor of s(2^(4*5^n*k)).
This means we have to account for 11 and 41. In the same way as 5, 11 maintains the 3 unless the sequence begins 2^(4*5^(2n-1)*11^(2m-1)*k). Accounting for 41 gives a requirement of beginning with 2^(4*5^(2n-1)*11^(2m-1)*41^(2o-1)*k).
s(2^(4*5^(2n-1)*11^(2m-1)*41^(2o-1)*k)) is always divisible by 2^9020-1 which contains yet more primes that preserve the 3.
Expanding the lower bound like this is easy but I can't quite think how to turn it into a proof. Maybe there is something based on proving there will always be more and more prime factors that are 2 mod 3.
I think I found a way.
S(24k) = 24k-1=(22k-1)(22k+1).

22k+1 is 2 (mod 3) , so there exist a prime p such that p is 2 (mod 3) and p2m-1 divide 22k+1 ,but p2m not divide 22k+1

gcd(22k-1,22k+1)=gcd(2,22k+1)=1, so no other factor of p from 22k-1.
Hence p2m-1 preserved 3 for s(24k), so 3 divide s(s(24k))
warachwe is offline   Reply With Quote
Old 2020-08-22, 20:56   #481
henryzz
Just call me Henry
 
henryzz's Avatar
 
"David"
Sep 2007
Cambridge (GMT/BST)

10110011001112 Posts
Default

Quote:
Originally Posted by warachwe View Post
Personally, I think these conjectures are interesting in a way that the prove seem somewhat doable, but I agree that we should look more into other phenomena.



I think I found a way.
S(24k) = 24k-1=(22k-1)(22k+1).

22k+1 is 2 (mod 3) , so there exist a prime p such that p is 2 (mod 3) and p2m-1 divide 22k+1 ,but p2m not divide 22k+1

gcd(22k-1,22k+1)=gcd(2,22k+1)=1, so no other factor of p from 22k-1.
Hence p2m-1 preserved 3 for s(24k), so 3 divide s(s(24k))
Brilliant. Now to think about conjecture 3
henryzz is online now   Reply With Quote
Old 2020-08-22, 23:39   #482
RichD
 
RichD's Avatar
 
Sep 2008
Kansas

2·3·523 Posts
Default Table 18

All cells in n=18 should now be green.

I'll start preliminary work on Table n=20.
RichD is offline   Reply With Quote
Old 2020-08-23, 01:33   #483
EdH
 
EdH's Avatar
 
"Ed Hall"
Dec 2009
Adirondack Mtns

64748 Posts
Default

The preliminary work on base 79 has been finished through i=79.

Here is a list of all the prime terminations encountered:
Code:
7 (1)
11 (1)
13 (1)
19 (1)
23 (1)
37 (1)
41 (3)
43 (1)
79 (1)
109 (1)
277 (1)
373 (1)
647 (1)
2053 (1)
3209 (1)
56417 (1)
274201 (1)
1404653 (1)
39449441 (1)
3726897833 (1)
15149890507 (1)
7003404108271 (1)
1750258119649999 (1)
50030448965430187 (1)
582160055705443697 (1)
737287429537747247 (1)
11218513019153099281 (1)
51295583920298014697 (1)
126184888328246734874859217 (1)
281050352710245559603460861942264287 (1)
612334801072206651528322236763533211 (1)
144410932495341043816499009968224193057 (1)
5215871181653976380322743907608410822343857033 (1)
10098441202754878623864731377586611856943613601 (1)
58715257934974315365589544756228858137685070659 (1)
4514080124904825140989752683500717658279017890401 (1)
144247534946592562812896661356728761240274391235398539509968323519 (1)
1543397657597549030942319129765250282733366278065161881086980020577 (1)
21210775475881297038103544006601558250238621656859058445205548774369 (1)
969294053696923120396927330592354921021401157551137054187134211303939583373802685159736925954726601223772976624910063809731121765969 (1)
As shown, only the prime 41 terminated more than one sequence.
EdH is offline   Reply With Quote
Old 2020-08-23, 08:40   #484
garambois
 
garambois's Avatar
 
Oct 2011

25×11 Posts
Default

@warachwe :
Thank you for your informed response (post #480) !

@ Everyone :
For the moment, I'm not making any more conjectures like the ones in post #447. I'm trying to spot other different things, especially concerning the occurrences of prime numbers that end sequences. And also, for n=2^i, for n=18^i, for n=m^i, with m and i in the same parity, I know that s(n) is odd and therefore the sequence that starts with n most likely ends. But, I don't understand why all these sequences without exception end. I'm looking forward to the first Open-End sequence for n=2^i, for n=18^i, for n=m^i, with m and i in the same parity ! But we'll have to wait until the amount of data we have increases. But with the computing power available to us at the moment, I estimate that in one week, we'll make more progress than in several years at the rate of a few months ago !
That said, I have no idea what "enough data" means. Is it going to take several months, a year, or more ?

@warachwe & henryzz :
If I understood correctly, the conjecture (2) is demonstrated !
Wow! Thanks to you !
Regarding conjecture (3), I didn't put a star. Because I doubt that the prime number 3 is kept indefinitely during the first 7 iterations for all the sequences starting with 2^(36*k). The calculations have not really been pushed beyond k=15 yet... But you never know !

@RichD and EdH :
Thanks a lot for your help !


garambois is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Broken aliquot sequences fivemack FactorDB 45 2020-05-16 15:22
Broken aliquot sequences schickel FactorDB 18 2013-06-12 16:09
A new theorem about aliquot sequences garambois Aliquot Sequences 34 2012-06-10 21:53
poaching aliquot sequences... Andi47 FactorDB 21 2011-12-29 21:11
New article on aliquot sequences schickel mersennewiki 0 2008-12-30 07:07

All times are UTC. The time now is 19:45.

Fri Oct 30 19:45:17 UTC 2020 up 50 days, 16:56, 1 user, load averages: 2.27, 1.76, 1.73

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.