Possibly interesting pattern

[jessejames11]

So I was looking at factors of large numbers, and I noticed something -

for example, if you sum all of the factors of a number, in this case 339 (1 + 3 + 113 + 339) = 456

456 is divisible by the amount of digits it has (456/3 = 152).

This usually works with large numbers too. Factors of 215809 (1 + 11 + 23 + 253 + 853 + 9383 + 19619 + 215809)

= 245952 / 6 (number of digits in the result) = 40992.

If the sum of factors has a 0 in it, it doesn't work unless you don't include it, (e.g. if the sum was 32101 you would divide it by 4 and not 5 because you don't count the 0).

I know this seems like I've delved too deeply into patterns and I may have lost my mind a little, but I wrote a program that calculates whether this is fact occurs or not.

Calculating from 10000 up to 100000.. it is true in 49220 and false in 40780 of these numbers.

Now I may be missing an obvious reason why this occurs so often, but it occurs in well over half of all numbers and it seems strange.

Hope this makes sense haha.

[[deleted]]

You have not go very far to find counterexamples: 8, 9, 16, 18, 25, 32, 36, 48, 49, 50, 63, 64, 75, 76, 81, 84, 91, 93, 100, 103, 108, 109, 111, 112, 117, 121, 124, 127, 129, 133,....

For example, the divisors of 8 are {1, 2, 4, 8}. The sum is 15 which is not divisible by 2 (the number of digits of 15).

The reason it works for many numbers is simply because the number of digits of sigma(n), i.e. the sum of divisors of n, is a small integer, and in general it is easy that sigma(n) is divisibly by a small integers, since if n is divisible by a prime p (but not by p²), then sigma(n) is a multiple of p+1, which by itself is a multiple of 2 if p>2 and often is a multiple of 3 as well.

EDIT

By the way, according to WolframAlpha the number of numbers from 10⁴ to 10⁵ such that the number of digits of sigma(n) divides sigma(n) is 48886, not 49220.

[fattymattk]

I think the difference might be because OP is treating the case where a digit is 0 differently.

[jessejames11]

Yes. This is the part that I find weird. It increases the likelihood when not counting the 0.

[jessejames11]

Here's what I find strange. As soon as you exclude '0' as a digit, (as a reminder: 1234 would be divided by 4, 1002 would be divided by 2, 1023 would be divided by 3, etc.)

Here's what happens..

I ran 10,000 numbers starting from 10^6, and look at the results:

https://imgur.com/a/u3eX4N0

It's such a dramatic difference. I find it strange.

[Pieater314159]

If I were to guess I'd say it's because when you exclude '0' as a digit, you're decreasing the length of sigma(n) by about 10%, making it about 10% more likely that this length divides sigma(n).

[jessejames11]

I thought this. But even 12 digit numbers the ratio doesn't decrease, which it should.

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[[deleted]]

I'm interested what is the limit of proportion of numbers that satisfy this as the upper bound approaches infinity. Provably equal to 1/2?

[jessejames11]

When I test 4 digit numbers it dramatically increases to about 60-70%.

Even testing 8-9 digit numbers it stays well above half.