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 2021-10-29, 09:49 #1 Drdmitry     Nov 2011 271 Posts Numerical and Statistical Analysis of Aliquot Sequences I want to share a paper, where the authors did a great job in investigating a behaviour of aliquot sequences. Well, it was published in Experimental Mathematics in 2020, but only appeared on arxiv several days ago. The main objective of the paper is to find an evidence in favour of one of the following hypotheses:All (or almost all) aliquot sequences converge to 1 or to a cycle. Many (or almost all) even aliquot sequences diverge to infinity. The authors do that by investigating the geometric mean of an amplification factor $s_k(n)/s_{k-1}(n)$ for all n up to $2^{40}$ and k between 1 and 10. The idea is that if that value is smaller than 1 then the aliquot sequence decreases on average and that would support Conjecture 1. Otherwise, we have an opposite situation and Conjecture 2 would become more favorable. The computations show that the geometric mean $\nu_k(N)$ of $s_k(n)/s_{k-1}(n)$ where n is even and changes from 1 to N grows with k but decays with N. For example, for $N=2^{40}$ the values of $\nu_k(N)$ are smaller than 1 for k up to 5 and bigger than 1 for bigger values of k. Then the authors tried some smarter ways of computing the geometric mean. For example, they add weightings to terms $s_k(n)/s_{k-1}(n)$ according to the number of aliquot preimages of n. Hence, untouchable numbers are not counted in the mean but if a number has ten preimages it is counted ten times. After such weightings are implemented, the geometric mean becomes even smaller. The conclusion of this part of the paper is that most probably the values $\nu_k(N)$ and their weighted analogues become smaller than 1 as N becomes large enough. That is a good sign for everyone who wants to terminate all aliquot sequences. But in the second part of the paper, the authors take into account the guides and drivers, i.e. the fact that aliquot sequences tend to stick with certain factorisation patterns and are very hard to escape from them. The authors considered 8000 aliquot sequences and extended them until they terminate or become larger than $2^{288}$. The data coming from those sequences indicate that the downdriver is the most frequent driver, but the number of all the other (up)drivers is way bigger than that of downdrivers. And also the geometric mean value of $s_k(n)/s_{k-1}(n)$ for the terms of those sequences is around 1.133. That is a rather strong evidence towards Conjecture (b). An interesting observation from the paper's data: the downdriver appears to be the easiest driver to go away from. The average length of the downdriver is 30.488. It is followed by $2^5\cdot 3\cdot 7$ with the average length 33.139, then $2^3\cdot 3\cdot 5$, $2^2\cdot 7$, $2^4\cdot 31$, $2\cdot 3$ and the stickiest driver is $2^6\cdot 127$ with the average length 188.307.
 2021-10-30, 17:09 #2 Stargate38     "Daniel Jackson" May 2011 14285714285714285714 2A816 Posts Which sequences did they run? Any chance of getting a txt file with all the sequence starters in it?
2021-10-30, 21:21   #3
EdH

"Ed Hall"
Dec 2009

47·89 Posts

Quote:
 Devitt [1976] used the average order of consecutive terms in a sequence, s(n)/n, taken over even values of n, namely 5π 2 /24 − 1 = 1.0562 as evidence that most aliquot sequences diverge, seemingly providing evidence in favor of Guy and Self- ridge.
My emphasis. I think this is wrong, but am ready for correction. Half of the sequences are odd and all odd terminate. Some of the even terminate, which means more than half terminate. Where am I mistaken?

2021-10-31, 02:26   #4
axn

Jun 2003

34×5×13 Posts

Quote:
 taken over even values of n
My emphasis. Looks like only even values are being considered, so the "most" applies to only the even values?

2021-10-31, 04:00   #5
Drdmitry

Nov 2011

10F16 Posts

Quote:
 Originally Posted by Stargate38 Which sequences did they run? Any chance of getting a txt file with all the sequence starters in it?
This is what paper says: "We collected data from 8000 aliquot sequences: eight sets of 1000 sequences, with
each sequence starting at $2^{16+32n} + 2k$, where 0 ≤ n ≤ 7 and 0 ≤ k < 1000.
"

I would suggest to try to contact the authors and ask them for data.

Last fiddled with by Drdmitry on 2021-10-31 at 04:53

2021-10-31, 13:51   #6
EdH

"Ed Hall"
Dec 2009

101278 Posts

Quote:
 Originally Posted by axn My emphasis. Looks like only even values are being considered, so the "most" applies to only the even values?
I agree that that is what they meant, but it isn't what they wrote, if I read the words displayed. (All emphasis below by me.):
Quote:
 On the other hand, Guy and Selfridge [1975] conjectured that, starting with an even value of n, many, perhaps almost all, such sequences diverge.
They have clearly shown their meaning. But, this has not:
Quote:
 Devitt [1976] used the average order of consecutive terms in a sequence, s(n)/n, taken over even values of n, namely 5π 2 /24 − 1 = 1.0562 as evidence that most aliquot sequences diverge, seemingly providing evidence in favor of Guy and Self- ridge.
This says I studied evens and decided most of all sequences diverge. To me, this needs to say:
Code:
as evidence that most even aliquot sequences diverge
which is what Guy and Selfridge conjectured.

But, it's obviouslyly just me. . .

2021-10-31, 17:56   #7
garambois

"Garambois Jean-Luc"
Oct 2011
France

26×11 Posts

Quote:
 Originally Posted by EdH My emphasis. I think this is wrong, but am ready for correction. Half of the sequences are odd and all odd terminate. Some of the even terminate, which means more than half terminate. Where am I mistaken?

"all odd terminate"
This is not proven !
I would not dare to formulate the conjecture that there are sequences that have only odd terms and are Open-End.
But I would gladly formulate (unofficially of course) the following conjecture :
There exist sequences with all odd terms that are increasing for k iterations with k as large as one wants.
I looked for such sequences using the "regina" file and the record sequence I found is the one starting with 2551185 and whose terms are all odd and which is increasing until index 6.
And this starting odd number has only 7 digits.
I can easily assume that other odd numbers of millions of digits could be the start of increasing odd-term sequences for thousands or millions of iterations.

For the even sequences, I side with Guy and Selfridge.
A few years ago, with some friends, we wrote two pdf documents and even today, I have three computer programs constantly running to update these pdf documents.
But it will take a few years before the calculations are accurate enough.
And these works make me intuitively lean towards the existence of divergent sequences.
http://www.aliquotes.com/vitesse_croissance.pdf
http://www.aliquotes.com/infirmer_catalan.pdf

2021-10-31, 20:07   #8
EdH

"Ed Hall"
Dec 2009

418310 Posts

Quote:
 Originally Posted by garambois "all odd terminate" This is not proven !
Point taken! Thanks!

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