Three to the distinct-prime-factor count ($3^{\omega (n)}$)

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.