The number of distinct prime factors function $\omega(n)$ counts how many distinct prime factors $n$ has. Expressing $n$ as $$n = \prod_{i = 1}^k {p_i}^{a_i},$$ where the $p_i$ are distinct primes, there being $k$ of them, and the $a_i$ are positive integers (not necessarily distinct), then $\omega(n) = k$

Obviously for a prime $p$ it follows that $\omega(p) = 1$ When $n$ is a squarefree number, then $\Omega(n) = \omega(n)$ where $\Omega(n)$ is the number of (nondistinct) prime factors function. Otherwise, $\Omega(n) > \omega(n)$

$\omega(n)$ is an additive function, and it can be used to define a multiplicative function like the Möbius function $\mu(n) = (-1)^{\omega(n)}$ (as long as $n$ is squarefree).