# [seqfan] Re: Destinies of sums of proper divisors.

David Wilson dwilson at gambitcomm.com
Mon Jul 6 23:59:40 CEST 2009

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franktaw at netscape.net wrote:
> Actually, if n is large with many prime divisors, and is divisible by
> larger power of 2, the next term is quite likely to continue to be
> divisible by that power of 2;  it is perhaps as likely to to increase
> this exponent as it is to decrease it.

Not sure I agree (I don't have time to mull it over too deeply right
now, so I may embarrass myself here, but...)

Let s(n) = sigma(n), f(n) = sigma(n) - n (sum of aliquot parts).

Let n = 2^a*j (j odd) and let s(j) = 2^b*k (k odd).

Then
f(n) = s(2^a*j) - 2^a*j
= odd*s(j) - 2^a*j
= odd*2^b*k - 2^a*j
= 2^b*odd - 2^a*odd

Thus, if a != b, the power of 2 in f(n) is min(a, b) <= a.

In other words, for the power of 2 in n = 2^a*j to grow with application
of f, the power of 2 in s(j) must be precisely 2^a. My gut feeling is
that such numbers would be fairly rare. This would mean than mean that
repeated application of f would exert downward pressure on a (the power
of 2 in n). And as a decreases and j increases, the likelihood that the
power of 2 in s(j) is 2^a shrinks yet further.

I checked the trajectory of 276 for about 80 iterations, and after the
first few iterations, the power of 2 settled at 2^2 and stayed there
(and 7 became an omnipresent divisor). Does that pattern continue, or
does the power of 2 in n start in increase at some point?

> So these trajectories do not,
> in general, get stuck in this particular pattern.
>
> On the other hand, almost everybody who has worked with these things
> believes that there are, indeed, divergent trajectories, and that 276
> is very likely one of them.
>
>
>
> -----Original Message-----
> From: David Wilson <dwilson at gambitcomm.com>
>
> I also wanted to make this observation on aliquot sequences:
>
> Let s(n) be the sum of divisors of n, and f(n) = s(n)-n be the sum of
> aliquot divisors. We are discussing the trajectory of f(n).
>
> Note that if n is a large number of the form 28k with k odd, then f(n)
> is likely to be a yet larger number of this form. If a sufficiently
> large number of this form occurs in the trajectory of n, the subsequent
> trajectory is likely retain that form and grow. Once it has grown
> sufficiently large, in the unlikely event it loses the form 28k, it is
> likely to recover that form after a few iterations
> and resume its growth.
>
> For this reason, I conjecture that f(n) is very likely to include
> divergent trajectories, and that n = 276 is probably one of them.
>
>
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>

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