How can i find the sum of divisors of a given number in a faster way?

[jkmyoung]

Just to show the algorithm works on 24:
Start with 1 as a factor.
n = 24
i = 2! i divides n.
n => 12 (2^1)
n => 6 (2^2)
n => 3 (2^3)

This gives us (1, 2, 4, 8)
Using the formula, we go:
(2^4 -1)/(2-1) = 15/1 = 15
1 * 15 = 15. So the sum so far is 15. = 1 + 2 + 4 + 8. Check!

Next i = 3. 3 divides n.
n=> 1 (3^1)
Ok, so we have factors
(1, 2, 4, 8) * 1, and (1, 2, 4, 8) * 3
(3^2 - 1)/(3-1) = 8 /2 = 4
15 * 4 = 60
The sum of factors of 24 is 60.
Checking, we have factors
1, 2, 4, 8, 3, 6, 12, 24, rearrange to:
1, 2, 3, 4, 6, 8, 12, 24 => 60.

adv/disadv:
+ Don't have to check for duplicates.
- Complexity.