# Dirichlet Convolution of the Mobius Function with Itself

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.

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 $n$).

Note on the Previous Solution: Let $\operatorname{rad}(n)$ denote the radix, or squarefree part, of $n$, i.e., so that if $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ is the factorization of $n$ into powers of distinct primes then $\operatorname{rad}(n) = p_1 p_2 \cdots p_k$. Then we see that $\mu(\operatorname{rad}(n))$ is always non-zero and that $\mu^2(n / \operatorname{rad}(n)$ is the characteristic function of the cube-free integers. Also, if $\omega(n)$ counts the number of distinct prime factors of $n$ (the prime omega function) then the number of primes that divide $n$ exactly once is given by $$\#\{p \text{ prime } : p|n \text{ and } p^2 \nmid n\} = \omega(n) - \omega(n / \operatorname{rad}(n)).$$ This implies that the solution given by Essentially Lime in the previous response has the following exact closed-form expression for any $n \geq 1$: $$(\mu \ast \mu)(n) = (-2)^{\omega(n) - \omega(n / \operatorname{rad}(n))} \cdot \mu^2(\omega(n) - \omega(n / \operatorname{rad}(n)).$$ Curiously enough, parts of this formula for the solution can be re-expressed in terms of 1) the identity that $\sum_{d|n} \mu^2(d) = 2^{\omega(n)}$, or in other words the convolution $\mu^2 \ast 1 = 2^{\omega}$; and 2) the fact that $(-1)^{\omega(n)} = \mu(n)$ whenever $n$ is squarefree.

Alternate Solution Using Dirichlet Inverses: Another, perhaps shorter and more direct way to find this solution is to consider the Dirichlet inverse of the divisor function $d(n) := \sum_{d|n} 1 = (1 \ast 1)(n)$. Recall that the Dirichlet inverse $f^{-1}$ of any arithmetic function $f$ is the unique function (if one exists) satisfying $(f \ast f^{-1})(n) = \varepsilon = \delta_{n,1}$ where $\varepsilon$ is the multiplicative identity of Dirichlet convolutions. This inverse exists precisely when $f(1) \neq 0$.

Now consider the following phrasing of the problem: write $f := mu \ast \mu$. Then since $\mu \ast 1 = \varepsilon$ (an elementary fact which is seen here, for example, or proved by Dirichlet generating functions), we can invert the right-hand-side to obtain that $\varepsilon = f \ast 1 \ast 1 = f \ast d$. We can easily verify that $d(1) = 1 \neq 0$, so that the divisor function has a Dirichlet inverse. This implies that $f = d^{-1}$. The values of this inverse function are defined by $d(1) = 1$ and for $n > 1$ by the recurrence relation: $$d^{-1}(n) = -\sum_{\substack{d|n \\ d>1}} d(n) d^{-1}(n/d).$$ By computation, the first few values of this sum correspond to the sequence $\{d^{-1}(n)\}_{n \geq 1} = \{1,-2,-2,1,-2,4,-2,0,1,4,\ldots\}$ (A007427).

• There is also a related definition of so-called Moebius functions of order k defined in the exercises to Chapter 2 of Apostol's reference book on ANT. It constructs functions that satisfy $$\mu_k(n^k) = \mu(n), \forall n \geq 1.$$ These order-k functions satisfy a convolution identity of the following form: $$\mu_k(n) = \sum_{d^k|n} \mu_{k-1}\left(\frac{n}{d^k}\right) \mu_{k-1}\left(\frac{n}{d}\right)$$
– mds
Commented Dec 28, 2020 at 21:22