Divisor Sum of Square-Free Integer/Examples/66/Proof 2

Proof
From Sigma of Integer:
 * $\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.

We have that:
 * $66 = 2 \times 3 \times 11$

Hence: