Is there a formula to calculate the sum of all proper divisors of a number?

Question

I don't need to list all proper divisors, I just want to get its sum. Because for a small number, checking all proper divisors and adding them up is not a big deal. However, for a large number, this would run extremely slow. Any idea?

Thanks,
Chan Nguyen

Answer 1

If the prime factorization of $n$ is $$n=\prod_k p_k^{a_k},$$ where the $p_k$ are the distinct prime factors and the $a_k$ are the positive integer exponents, the sum of all the positive integer factors is $$\prod_k\left(\sum_{i=0}^{a_k}p_k^i\right).$$

For example, the sum of all of the factors of $120=2^3\cdot3\cdot5$ is $$(1+2+2^2+2^3)(1+3)(1+5)=15\cdot4\cdot6=360.$$

For proper factors, subtract $n$ from this sum. This may or may not be faster, depending on the number and how you'd get the prime factorization, but this is the typical technique for high school contest problems of this sort.

Answer 2

Just because it is interesting:

There is actually a (less known) recursive formula for calculating $\sigma(n)$, the sum of the divisors of $n$.

$$\sigma(n) = \sigma(n-1) + \sigma(n-2) - \sigma(n-5) - \sigma(n-7) + \sigma(n-12) +\sigma(n-15) + ..$$ Here $1,2,5,7,...$ is the sequence of generalized pentagonal numbers $\frac{3n^2-n}{2}$ for $n = 1,-1,2,-2,...$ and the signs are repetitions of $1,1,-1,-1$. The summation continues until you try to calculate $\sigma$ of something negative. However, if $\sigma(0)$ occurs in the summation (this happens precisely when $n$ is a generalized pentagonal number), it should be replaced by $n$ itself. In other words $$ \sigma(n) = \sum_{i\in \mathbb Z_0} (-1)^{i+1}\left( \sigma(n - \tfrac{3i^2-i}{2}) + \delta(n,\tfrac{3i^2-i}{2})n \right), $$ where we set $\sigma(i) = 0$ for $i\leq 0$ and $\delta(\cdot,\cdot)$ is the Kronecker delta.

Note that calculating $\sigma(n)$ requires $\sigma(n-1)$ already, therefore its complexity is at least $\mathcal O(n)$, which makes it kind of useless for practical purposes. Note however the lack of reference to divisibility in this formula, which makes it a bit miraculous and therefore worth mentioning.

Here's a reference to the Euler's paper from 1751.

Answer 3

If $$n = a^p × b^q × c^r × \ldots$$ then total number of divisors $ = (p + 1)(q + 1)(r + 1)\ldots$

sum of divisors $\Large = [\frac{a^{(p+1)}-1}{(a \ – \ 1)} × \frac{b^{(q+1)}-1}{(b \ – \ 1)} × \frac{c^{(r+1)}-1}{(c \ – \ 1)}\ldots]$

for e.g. the divisors of $8064$ $$8064 = 2^7 × 3^2 × 7^1$$

total number of divisors $= (7+1)(2+1)(1+1) = 48$

sum of divisors $= [\frac{2^{(7+1)} –1]}{(2–1)} × \frac{3^{(2+1)} –1}{(3–1)} ×\frac{7^{(1+1)} –1}{(7–1)}]$

$= 255 × 7 × 8 = 26520$

P.S. Note that a divisor of an integer is also called a factor.

Answer 4

Here's a very simple formula:

$$\sum_{i=1}^n \; i\mathbin{\cdot}((\mathop{\text{sgn}}(n/i-\lfloor n/i\rfloor)+1)\mathbin{\text{mod}}2)$$

(for the sake of brevity, one can write $\mathop{\text{frac}}(n/i)$ instead of $n/i-\lfloor n/i\rfloor$).

This is a way to get the function $\text{sigma}(n)$, which generates OEIS's series A000203.

What you want is the function that generates A001065, whose formula is a slight modification of the one above (and with half its computational burden):

$$\sum_{i=1}^{n/2} \; i\mathbin{\cdot}((\mathop{\text{sgn}}(n/i-\lfloor n/i\rfloor)+1)\mathbin{\text{mod}}2)$$

That's it. Straight and easy.

Answer 5

If you want numerical values then the calculator at the site below will list all divisors of a given positive integer, the number of divisors and their sum. It also has links to calculators for other number theory functions such as Euler's totient function.

http://www.javascripter.net/math/calculators/divisorscalculator.htm

Answer 6

The other answers already talk about the basic formula, but there is a nice little trick if you're going into extremes:

say you have a very large exponent (that you normally wouldn't calculate by hand, but suppose even a computer struggles with it), like $2^x$ where x ~ $10^9$

You can actually reduce this to $log(n)$ operations.

For a shorter demonstration: sum of divisors of $x^7$. Normally you would calculate each value of $x^n$ and sum them. However, a faster way (not so noticeable for such a small exponent) is:

$x^0+x^1+x^2+x^3+x^4+x^5+x^6+x^7$ $=(x^0+x^1+x^2+x^3)*(1+x^4)$ $=(x^0+x^1)*(1+x^2)*(1+x^4)$

This works very nice for numbers such as 7, 15, etc. But it works for any other number too, as long as you simply add the part that you cannot include in polynomial factorization:

$x^0+x^1+...+x^{12}$ $=(x^0+x^1+x^2+x^3+x^4+x^5)*(1+x^6)+x^{12}$ $=(x^0+x^1+x^2)*(1+x^3)*(1+x^6)+x^{12}$

It's easy to write a computer program that does exactly this and it is able to sum to just about anything you can think of. $O(log(n))$ grows very slowly.

Answer 7

Here's a really simple formula:

$$ \sigma(x) = \sum_{n=1}^x \left| \text{sign}(x \text{ mod } n) - 1 \right| * n $$

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Short explanation:

$ \text{sign}(x \text{ mod } n) $ evaluates to $0$ if $n$ is a divisor of $x$ (or evaluates to $1$ if $n$ isn't a divisor of $x$), so we have to negate it by subtracting one from it ($0$ -> $-1$; $1$ -> $0$) and taking the absolute value of the result. We now have either $0$ or $1$ (depending on the divisibility), which can be multiplied by $n$ to get $0$ or $n$.

Answer 8

The typical brute foce approach in, say, C language:

 public int divisorSum(int n){
    int sum=0;
    for(int i=1; i<= n; i++){
        if(n % i == 0){
            sum +=i;
        }
    }
    return sum;
}