Divisor Counting Function from Prime Decomposition

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Theorem

Let $n$ be an integer such that $n \ge 2$.

Let the prime decomposition of $n$ be:

$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$

Let $\map {\sigma_0} n$ be the divisor counting function of $n$.


Then:

$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$


Proof 1

We have:

$d \divides n \implies \forall i: 1 \le i \le r: d = p_1^{l_1} p_2^{l_2} \ldots p_1^{l_1}, 0 \le l_i \le k_i$


For each $i$, there are $k_i + 1$ choices for $l_i$, making $\paren {k_1 + 1} \paren {k_2 + 1} \cdots \paren {k_r + 1}$ choices in all.

By the Fundamental Theorem of Arithmetic and hence the uniqueness of prime decomposition, each of these choices results in a different number, therefore a distinct divisor.

$\blacksquare$


Proof 2

From Divisor Counting Function of Power of Prime we have:

$\forall j \in \closedint 1 r: \map {\sigma_0} {p_j^{k_j} } = k_j + 1$

The result follows immediately from Divisor Counting Function is Multiplicativeā€Ž.

$\blacksquare$


Sources