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Homenumber of (nondistinct) prime factors function

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# number of (nondistinct) prime factors function

The number of (nondistinct) prime factors function $\Omega(n)$ counts with repetition how many prime factors a natural number $n$ has. If $\displaystyle n=\prod_{{j=1}}^{k}{p_{j}}^{{a_{j}}}$ where the $k$ primes $p_{j}$ are distinct and the $a_{j}$ are natural numbers, then $\displaystyle\Omega(n)=\sum_{{j=1}}^{k}a_{j}$.

Note that, if $n$ is a squarefree number, then $\omega(n)=\Omega(n)$, where $\omega(n)$ is the number of distinct prime factors function. Otherwise, $\omega(n)<\Omega(n)$.

Note also that $\Omega(n)$ is a completely additive function and thus can be exponentiated to define a completely multiplicative function. For example, the Liouville function can be defined as $\lambda(n)=(-1)^{{\Omega(n)}}$.

## Mathematics Subject Classification

11A25*no label found*

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