Möbius inversion formula

Statement

Let $f$ and $g$ be arithmetic functions. Then $g(n) = \sum_{d \mid n} f(d)$ for all positive integers $n$ if and only if $$f(n) = \sum_{d \mid n} \mu(d) \, g\!\left(\frac{n}{d}\right)$$ for all positive integers $n$. Here $\mu$ is the Möbius function.

Remarks

In terms of the Dirichlet convolution, the statement is: $g = f * \mathbf{1}$ if and only if $f = g * \mu$. The proof uses $\mu * \mathbf{1} = \varepsilon$ and the associativity of $*$. For example, the sum of the totient over the divisors gives $\sum_{d \mid n} \varphi(d) = n$. So $\varphi(n) = \sum_{d \mid n} \mu(d) \, n/d$.

History

August Ferdinand Möbius introduced the formula into number theory in 1832.

Explanations

L1 human

If you know all the divisor sums of $f$, the formula recovers $f$ itself. It undoes the operation "sum over the divisors", as subtraction undoes addition.