For any natural number $n\ge 1$ with the factorization $n=\prod_{k=1}^r p_k^{e_k},$ the sum of the Möbius function over the divisors $d\mid n$ and divided by those divisors $d$ equals
$$\sum_{d\mid n}\frac{\mu(d)}d=\prod_{k=1}^r\left(1-\frac 1{p_k}\right).$$
Proofs: 1
Proofs: 1
