 Summatory bound for three to the prime omega

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 ${\tau }_3(n)=\#\{(a,b,c):abc=n\}$: