mersenneforum.org Aliquot sequences that start on the integer powers n^i
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 2020-08-21, 16:01 #474 chris2be8     Sep 2009 5×11×43 Posts 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
 2020-08-21, 16:34 #475 EdH     "Ed Hall" Dec 2009 Adirondack Mtns 22·3·17·23 Posts 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.
 2020-08-21, 16:59 #476 henryzz Just call me Henry     "David" Sep 2007 Liverpool (GMT/BST) 599210 Posts 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.
2020-08-21, 20:29   #477
RichD

Sep 2008
Kansas

3,637 Posts

Quote:
 Originally Posted by EdH ... but Aliquot sequences are only advanced by bumping them.
That's what I meant by a refresh - a bump.

2020-08-21, 21:45   #478
EdH

"Ed Hall"
Dec 2009

469210 Posts

Quote:
 Originally Posted by RichD That's what I meant by a refresh - a bump.
I understood that. I was replying to Chris.

 2020-08-22, 02:00 #479 RichD     Sep 2008 Kansas 3,637 Posts I was clarifying my original post for Chris.
2020-08-22, 11:03   #480
warachwe

Aug 2020

19 Posts

Quote:
 Originally Posted by garambois 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 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))

2020-08-22, 20:56   #481
henryzz
Just call me Henry

"David"
Sep 2007
Liverpool (GMT/BST)

23·7·107 Posts

Quote:
 Originally Posted by warachwe 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

 2020-08-22, 23:39 #482 RichD     Sep 2008 Kansas 3,637 Posts Table 18 All cells in n=18 should now be green. I'll start preliminary work on Table n=20.
 2020-08-23, 01:33 #483 EdH     "Ed Hall" Dec 2009 Adirondack Mtns 22×3×17×23 Posts 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.
 2020-08-23, 08:40 #484 garambois     "Garambois Jean-Luc" Oct 2011 France 2×463 Posts @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 !

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