[seqfan] Re: Question about abundant numbers (A5101)

[Allan Wechsler]

This line of thinking leads me to wonder whether anyone has investigated
the number of distinct values that partial sums of the divisors of N can
take on.

This starts out looking a lot like https://oeis.org/A107748, but then
A107748(10) = 14, where the sequence I'm thinking of ought to have 16.
Either I made an arithmetic error or we haven't collected this butterfly
yet.

On Mon, Dec 12, 2022 at 9:09 PM <hv at crypt.org> wrote:

> This is a minor variation on weird numbers (A006037). A quick check shows
> that the first few exceptions go:
>
> 9272 9730 10430 10570 10990 11410 11690 12110 12530 12670 13370 13510 13736
> 13790 13930 14770 15610 15890 16030 16310 16730 16870 17570 17990 18410
> 18830
> 18970 19390 19670 19810
>
> Take 9272 as an example: 9272 = 2^3 * 19 * 61, sigma(9272) = 2 * 9272 + 56.
> So finding a solution is equivalent to finding a subset of factors that
> sum to 55. Since 55 == 17 (mod 19), and sigma(2^3) = 15, clearly there
> can be no such solution.
>
> It is slightly intriguing that such a high proportion of the exceptions
> are multiples of 10.
>
> Hugo
>
> Alonso Del Arte <alonso.delarte at gmail.com> wrote:
> :Given an abundant number *n*, is it always the case that there are at
> least
> :two subsets of *n*'s divisors that add up to *n* + 1?
> :
> :e.g., *n* = 70, we see that obviously 1 + 70 = 71 but also 35 + 14 + 10 +
> :7 + 5 = 71.
> :
> :Al
> :
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