Working on a problem I encounter the following sum:


$\sum\_{r<R}\frac{\mu(r)^{2}\tau\_{k}(r)^{2}}{\varphi(r)}$ that I have to estimate on the above; here $\tau\_{k}(r)$ is the number of ways of writing $r$ as a product of $k$ natural integers. I now that $\sum\_{r<R}\tau\_{k}(r)=RP\_{k-1}(R)+E(R)$ with $P\_{k-1}(x)$ a polynomial of degree $k-1$ with leading coefficient $\frac{1}{k!}$ and $E(R)$ the error term.


Can anyone help me to obtain the desired estimate?



Particular sum on product partitions into k parts
* If you only need an upper bound of the right order of magnitude, I strongly recommend the use of Rankin's trick, which amounts to write
$$
\sum\_{r \leq R} f(r) \leq \prod\_{p \leq R} \left( \sum\_{\ell \geq 0} f(p^{\ell}) \right)
$$
for a nonnegative multiplicative function $f$. For $f(r) = \frac{\mu^2(r) \tau\_k(r)^2}{\phi(r)}$, this yields
$$
\sum\_{r \leq R} \frac{\mu^2(r) \tau\_k(r)^2}{\phi(r)} \leq \prod\_{p \leq R} \left( 1+ \frac{k^2}{p-1}\right) \leq \exp\left( k^2 \sum\_{p \leq R} \frac{1}{p-1} \right) \leq C^{k^2} \log(2R)^{k^2}, $$
for some constant $C > 1$.
* If you need an asymptotic estimate, then standard complex analytic methods, together with Vinogradov-Korobov estimates, yield
$$
\sum\_{r \leq R} \frac{\mu^2(r) \tau\_k(r)^2}{\phi(r)} = P\_k(\log(R)) + O(1),
$$
uniformly for $R \geq k^{ck^{\frac{2}{3}}}$, for some polynomial $P\_k$ of degree $k^2$, with leading coefficient
$$
\frac{1}{(k^2)!} \prod\_{p} \left( 1 - \frac{1}{p} \right)^{k^2} \left( 1 + \frac{k^2}{p-1} \right).
$$


