Dirichlet Convolution of the Mobius Function with Itself

Question

I am attempting to find a formula for $$(\mu * \mu)(n)$$ where * represents the Dirichlet Convolution operator. I know this can be expressed as $$\sum_{d|n} \mu(d)\mu(\frac{n}{d})$$ but I'd like the formula to not include any sums over divisors. I know it will be necessary to include information about the factorization of n, but I'm not sure how. For reference, $$\mu(n)= \begin{cases}0,&\text{if $n$ has one or more repeated prime factors}\\1,&\text{if $n$=1}\\(-1)^k,&\text{if $n$ is a product of $k$ distinct primes}\end{cases}$$ Some initial thoughts: the Dirichlet Convolution of two multiplicative functions is multiplicative, and since $$\mu(n)$$ is multiplicative, then so is $$(\mu * \mu)(n)$$ Any information to point me in the right direction on this will be greatly appreciated.

Answer 1

As you say, this is multiplicative. This means that once you know how to calculate it for numbers of the form $p^a$ you can calculate it for arbitrary $n$ by writing $n$ as a product of powers of different primes $p^aq^b\cdots$, and then multiplying the corresponding values $(\mu*\mu)(n)=((\mu*\mu)(p^a))((\mu*\mu)(q^b))\cdots$.

If $a=1$, $\sum_{d\mid p}\mu(d)\mu(p/d)=-2$ (the two factors each contribute $-1$). If $a=2$, $\sum_{d\mid p^2}\mu(d)\mu(p^2/d)=1$, since the only way for $\mu(d)\mu(p^2/d)$ to be non-zero is if $d\leq p$ and $p^2/d\leq p$, which requires $d=p$. If $a>2$ then $\sum_{d\mid p^a}\mu(d)\mu(p^2/d)=0$, since for each term in the sum either $d$ or $p^a/d$ is divisible by $p^2$.

So your function is $0$ if $n$ is divisible by the cube of any prime. Otherwise it is $(-2)^k$, where $k$ is the number of primes that divide $n$ exactly once (i.e. their squares do not divide $n$).