A student claims that f(n)=∑_d| ng(d) inverts g(n)=∑_d| nf(d). With g(n)=n, what happens for n=6?

False: the direct sum gives 12, not the value 2 from inversion

False: the direct sum gives 6, since it only counts n

True: for n=6 both values match φ(6)

True: summing g over divisors always inverts the relation

Exercise · ZW-MOBIUS-INVERSION-FORMULA-010 · ★★★ Advanced

A student claims that \(f(n)=\sum_{d\mid n}g(d)\) inverts \(g(n)=\sum_{d\mid n}f(d)\). With \(g(n)=n\), what happens for \(n=6\)?

Choose one option.

Show answer

Summing \(g(d)\) over \(d\mid 6\) gives \(1+2+3+6=12\), while \(\varphi(6)=2\). So the claim is false.

Formula that recovers an arithmetic function \(f\) from its sum over divisors, using the Möbius function \(\mu\).

\[f(n)=\sum_{d\mid n}\mu(d)\,g\!\left(\frac{n}{d}\right)\]