📂Number Theory

Sigma Functions in Number Theory

Theorem

For $\displaystyle \sigma (n) : = \sum_{d \mid n} d$, the following holds:

• [1]: For a prime number $p$, $$\sigma ( p^k ) = {{p^{k+1} - 1} \over {p-1}}$$
• [2]: If $\gcd (n , m ) = 1$, then $$\sigma (nm) = \sigma (n) \sigma (m)$$

Description

The sigma function, simply put, is the sum of divisors, for example, for $6$, it is $\sigma (6) = 1 + 2 + 3 + 6 = 12$. In analytic number theory, it is generalized as a divisor function.

Additionally, by mentioning the sigma function, it allows for a clean definition of Perfect Number. A perfect number is a number where the sum of its proper divisors (excluding itself) is equal to the number itself. Thus, a number $n$ that satisfies $\sigma (n) = 2n$ can be defined as a perfect number.

Proof

[1]

$$\sigma ( p^k ) = 1 + p + \cdots + p^{k} = {{p^{k+1} - 1} \over {p-1}}$$

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[2]

Let’s denote the divisors of $n$ as $1, d_{n1}, d_{n2}, \cdots, d_{nN}, n$, and the divisors of $m$ as $1, d_{m1}, d_{m2}, \cdots, d_{mM}, m$.

Since $\gcd(n,m) = 1$, $$\sum_{d \mid nm} d = 1 + d_{n1} + d_{m1} + d_{n1} d_{m1} + \cdots + nm$$, to sum up, $$\sum_{d \mid nm} d = (1 + d_{n1} + \cdots + n ) (1 + d_{m1} + \cdots + m) = \sum_{d | n} d_{n} \sum_{d | m} d_{m}$$

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