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garambois 2020-07-10 08:07

The infinite graph of the aliquots sequences
For a few years, I have also been working on the infinite graph of the aliquot sequences. ([URL=""]see the graph[/URL]).
There are a lot of open questions about this graph, [URL=""]see here[/URL].
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 ([URL=""]you can see an article here[/URL]).

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).

[B]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 !

garambois 2020-09-16 11:35

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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