The Sigma Function in Number Theory
Theorem
For $\displaystyle \sigma (n) : = \sum_{d \mid n} d$, the following hold.
- [1]: For a prime $p$, $$\sigma ( p^k ) = {{p^{k+1} - 1} \over {p-1}}$$
- [2]: If $\gcd (n , m ) = 1$, then $$\sigma (nm) = \sigma (n) \sigma (m)$$
Explanation
The sigma function is, simply put, the sum of divisors; taking $6$ as an example, $\sigma (6) = 1 + 2 + 3 + 6 = 12$. In analytic number theory, it is generalized to the divisor function.
Meanwhile, by referring to the sigma function, one can neatly define a perfect number. A perfect number is a number whose sum of divisors excluding itself equals itself. Thus, one can define a perfect number as an $n$ satisfying $\sigma (n) = 2n$.
Proof
[1]
$$ \sigma ( p^k ) = 1 + p + \cdots + p^{k} = {{p^{k+1} - 1} \over {p-1}} $$
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[2]
Let the divisors of $n$ be $1, d_{n1}, d_{n2}, \cdots, d_{nN}, n$ and the divisors of $m$ be $1, d_{m1}, d_{m2}, \cdots, d_{mM}, m$.
Since $\gcd(n,m) = 1$, $$ \sum_{d \mid nm} d = 1 + d_{n1} + d_{m1} + d_{n1} d_{m1} + \cdots + nm $$ rearranging, $$ \sum_{d \mid nm} d = (1 + d_{n1} + \cdots + n ) (1 + d_{m1} + \cdots + m) = \sum_{d | n} d_{n} \sum_{d | m} d_{m} $$
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