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

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

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.

• @Issac: Thank you! In fact, I thought of prime factorization, but the algorithm for factorization is not fast too. – Chan Feb 18 '11 at 23:10
• The sum of divisors can also be written using $\sum_{i=0}^{a_k}p_k^i = (p_k^{a_k + 1})/(p_k - 1)$ for the individual factors, as may be seen from the PlanetMath article: planetmath.org/encyclopedia/FormulaForSumOfDivisors.html – hardmath Feb 18 '11 at 23:18
• @hardmath: Absolutely—each sum is the sum of a geometric series (though I think it should probably be $$\prod_k\left(\sum_{i=0}^{a_k}p_k^i\right)=\prod_k\frac{p_k^{a_k + 1}-1}{p_k - 1}$$ (add $-1$ in the numerator). – Isaac Feb 18 '11 at 23:41
• can anyone explain why this works? – akashchandrakar May 31 '16 at 11:42
• @aksam: Take the example of 120, as in the answer. A positive integer factor is the product of 0, 1, 2, or 3 factors of 2, 0 or 1 factor of 3, and 0 or 1 factor of 5. Expanding $(1+2+2^2+2^3)(1+3)(1+5)$ gives the sum of all possible such products. – Isaac May 31 '16 at 18:42

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.

• Many thanks for a great information. Although I don't understand it completely now, I will go back to it when I'm ready. – Chan Feb 19 '11 at 4:58
• Is the formula correct? I get a negative sign for i=1 in your sum, and $\sigma(n-\frac{3 1^2 - 1}{2})$ has a positive sign in your first equation. Most likely, I made a mistake... (I tried it by hand using n=6). – Unapiedra Oct 11 '13 at 17:25
• "it should be replaced by $n$ itself". So do that: $\delta(...) n$, also I find that it should be $(-1)^{i+1}$. Doing this gives me correct result for all my test cases. – Unapiedra Oct 11 '13 at 23:08
• Thanks for pointing out this little gem! And it's far from useless. For certain purposes - like the SPOJ DIVSUM challenge where the sigma function needs to be computed in bulk - all lower sigma values are available via memoisation (caching), so that the computation of any one value needs nothing more than addition and table lookups. I coded it for SPOJ DIVSUM and it passed with flying colours (0.5 s for computing half a million sigmas, compared to one minute for one two-digit sigma w/o memoisation). – DarthGizka Feb 20 '16 at 0:31

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 one 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.

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.

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

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;
}

• It is a stack exchange for mathematics not for c-programming.So please write it in terms of mathematics language. – Ripan Saha Jun 27 '15 at 14:24
• Also, this method was already proposed by the OP. – wythagoras Jun 27 '15 at 14:32