Dirichlet sum over number of divisors is
$$\sum_{k=1}^n{\tau(k)} = 2\sum_{k=1}^{\lfloor{\sqrt n}\rfloor}\lfloor{\frac{n}{k}}\rfloor \ - {\lfloor\sqrt n\rfloor}^2$$
What would be this formula for the sum
$$\sum_{k=1}^n{\tau(ak)}$$
if $a$ is a constant? I know I can't use the multiplicity property for $\tau(ak)$ unless $a$ and $k$ are relatively primes.