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

[hv at crypt.org]

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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