Prime omega function ($\omega (n)$)

The prime omega function $\omega (n)$ counts the distinct prime factors of a positive integer $n$, so it is the cardinality of the prime support of an integer $P(n)$. For a prime factorization with distinct prime numbers $p_1,\dots ,p_r$ and positive exponents,At $n=1$ there are no prime factors, so the empty sum gives $\omega (1)=0$. The total number of prime factors instead gives $\Omega (n)={\sum }_i{a}_i$. ^[ref]

For $72=2^3\cdot 3^2$, the prime support of an integer is $\{2,3\}$ and $\omega (72)=2$. The comparison examples give $\omega (30)=3$, $\omega (32)=1$ and $\omega (36)=2$. Raising an existing prime factor to a higher power does not affect the prime omega function: $\omega (p^k)=1$ for a prime number $p$ and $k\ge 1$. Thus the prime omega function is a strongly additive arithmetic function, withThe reason is that coprime integers have disjoint prime supports of an integer. For arbitrary positive integers, their prime support of an integer combines by set union, and the shared prime factors are exactly those of the greatest common divisor. Counting this set union gives the more general identityFor $a=12$, $b=18$, both prime supports of an integer are $\{2,3\}$ and $\gcd (a,b)=6$, so $\omega (216)=2+2-2=2$. In particular $\omega (4)=1\ne 2=\omega (2)+\omega (2)$: the prime omega function is not a completely additive arithmetic function.

The prime omega function is useful for counting choices involving distinct prime factors. Every squarefree divisor of $n$ chooses a subset of its prime support of an integer, so there are $2^{\omega (n)}$ squarefree divisors. For $72$, this gives $2^2=4$, namely $1,2,3,6$. Also $\omega (\mathrm{rad}(n))=\omega (n)$, because the radical of an integer keeps exactly that prime support of an integer. The total number of prime factors records what this count discards, and the two counts agree exactly on squarefree integers.

The multiplicative arithmetic function $f(n)=3^{\omega (n)}$ counts assignments of three labels to each distinct prime factor. It also counts pairs of squarefree divisors whose least common multiple is the radical of an integer $n$.

The Euler product and Mertens second theorem give ${\sum }_{n\le D}{3}^{\omega (n)}/n\ll {\log }^{3}D$. The log-weighted convolution bound for three to the prime omega, with the Chebyshev estimate $\psi (y)\ll y$, then gives ${\sum }_{\sqrt{D}\le n\le D}{3}^{\omega (n)}\ll D{\log }^{2}D$. A direct bound for the full range follows from $3^{\omega (n)}\le {\tau }_3(n)$, where ${\tau }_3(n)=\#\{(a,b,c):abc=n\}$:

For $f(n)=3^{\omega (n)}$,If $p^a\Vert n$, its contribution to $(f\ast \Lambda )(n)$ is $f(n)(a-1+1/3)\log p$. Since $3(a-1+1/3)\ge a$, summing over the prime factors proves the bound. Here $\ast$ is Dirichlet convolution and $\Lambda$ is the Von Mangoldt function.