How to compute the sum with Mobious function

Question

Suppose $d = p_1p_2 \dots p_k$.

How to compute $\sum\limits_{lm = d} g(l)\tau(m)\log{m}$, where $g(l) = \sum\limits_{d~|~l} \mu(d)\mu\left(\dfrac{l}{d}\right)$

Any ideas or hints would be greatly appreciated.

Answer 1

Since $g$ is obtained from the Dirichlet-convolution $\,\mu * \mu\,$ and since the Dirichlet generating function for $\mu$ is $\dfrac 1{\zeta(s)}$ we deduce that $g(n)$ is generated by $\dfrac 1{\zeta(s)^2}$ : $$\tag{1}\frac 1{\zeta(s)^2}=\sum_{n=1}^\infty \frac {g(n)}{n^s}$$

Now you want : \begin{align} \tag{2}S(d)&:=\sum\limits_{lm = d} g(l)\tau(m)\log{m}=(g * (\tau\,\log))(d)\\ \end{align}

The generating function for $\tau$ (the number of divisors) is $\;\zeta(s)\,\zeta(s)$ and to get (for every term $n$) the product by the $\log(n)$ you only need to compute the derivative (since $\dfrac d{ds} n^{-s}=-\log(n)\,n^{-s}$) so that : $$\tag{3} \frac d{ds}\zeta(s)^2=-\sum_{n=1}^\infty \frac {\tau(n)\,\log(n)}{n^s}$$ One deduces that the generating function for $S$ will be given by the product of the generating function of $g$ and of $\tau\,\log$ that is : $$-\frac 1{\zeta(s)^2}\frac d{ds}\zeta(s)^2=\sum_{n=1}^\infty \frac {S(n)}{n^s}$$

But the von Mangoldt function $\Lambda(n)$ is generated by : $$-\frac {\zeta'(s)}{\zeta(s)}=\sum_{n=1}^\infty \frac {\Lambda(n)}{n^s}$$

so that the conclusion follows...