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Old 2020-07-10, 08:07   #1
garambois
 
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Oct 2011

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Default The infinite graph of the aliquots sequences

For a few years, I have also been working on the infinite graph of the aliquot sequences. (see the graph).
There are a lot of open questions about this graph, see here.
In particular, with Andrew R. Booker, from the University of Bristol, we wondered if there were all possible types of graphs (without meshes of course) in the main graph related to 1 and we asked ourselves the same question for the finite components of the graph (you can see an article here).

For my part, I reconsidered the problem of the infinite graph of the aliquot sequences, taking into account the length of the branches.
And I stumbled upon a problem.

Can branch lengths expressed as d(n) = s(n) - n = sigma(n) - 2n take all odd values ?
For a branch in the graph to have an odd length, the number must be a perfect square or double a perfect square.
For example, s(9) = sigma(9) - 9 = 13 - 9 = 4, so the branch length between 9 and 4 is 4 - 9 = sigma(9) - 2*9 = -5.
We tested all n = z^2 and n = 2 * z^2 for z from 1 to 20,000,000,000. There are only 25 values of n all smaller than n = 244036 = 494^2 such that -100 <= d(n) <= 100 excluding the value d(n) = 1 (which is obtained with all prime numbers).


The question :

Let z be an integer.
n=z^2 or n=2*z^2, so d(n) will be odd.
Is it worth letting the program run again beyond n=20,000,000,000 to find other values of n > 244036 such that -100 < d(n) < 100 or when n reaches a certain size, it is no longer possible for d(n) to be small ?
If someone could find a minor of d(n) as a function of n, it would save me from running a program for nothing.
We can't come up with such a proof !
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Old 2020-09-16, 11:35   #2
garambois
 
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OK.
If there's no answer here for more than 2 months, I'll restart the calculations beyond z=20,000,000,000. Because the answer to my question should not be trivial and it's worth letting the computer run to find other values of z such as -100 < d(n) < 100.
Thanks to all of you !
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