number of distinct prime factors functionNumberOfDistinctPrimeFactorsFunction2006-07-27 18:17:482006-07-28 15:19:10Definition% this is the default PlanetMath preamble. as your knowledge
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The {\em 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 \PMlinkname{number of (nondistinct) prime factors function}{NumberOfNondistinctPrimeFactorsFunction}. Otherwise, $\Omega(n) > \omega(n)$.
$\omega(n)$ is an additive function, and it can be used to define a multiplicative function like the M\"obius function $\mu(n) = (-1)^{\omega(n)}$ (as long as $n$ is squarefree).