Number of prime factors counted with multiplicity

Definition

Let $n$ be a positive integer with prime factorization $n = p_1^{e_1} \cdots p_k^{e_k}$, with distinct primes $p_i$. Then $\Omega(n) = e_1 + e_2 + \dots + e_k$. So $\Omega(n)$ is the number of prime factors of $n$, counted with multiplicity.

Remarks

For example, $\Omega(1) = 0$, $\Omega(12) = 3$ and $\Omega(360) = 6$. The function $\Omega$ is completely additive: $\Omega(mn) = \Omega(m) + \Omega(n)$ for all positive $m$ and $n$. The Liouville function is $\lambda(n) = (-1)^{\Omega(n)}$.

Explanations

L1 human

The function $\Omega(n)$ counts all the prime factors of $n$, with repeats. For $12 = 2 \cdot 2 \cdot 3$, there are three factors, so $\Omega(12) = 3$, while $\omega(12) = 2$.