I have the following formula of the Mobius inversion:
$$g(n) = \sum_{d|n}f(d) \iff f(n) = \sum_{d|n}g(\frac{n}{d})\mu(d)$$
The euler phi function has a divisor sum property: $\sum_{d|n}\phi(\frac{n}{d}) = \sum_{d|n}\phi(d)$
As I understand it:
In this case $f = g = \phi$ and $\mu(d) = 1 \,\text{for }\forall d$.
Are my assumptions correct(is the above understanding of the connection of Euler phi and Mobius correct)?
Why is $\mu(d) = 1$ if my assumptions are correct.
Best Answer
In the property for Euler's totient function we just use that the divisors $d$ are in bijection with the codivisors $n/d$, so that $$\sum_{d|n}\phi\left(\frac{n}{d}\right) = \sum_{d|n}\phi(d)=n$$ This is also called Sum Over Divisors Equals Sum Over Quotients.
However, the Moebius inversion formula, for $f(n)=\phi(n)$ gives $g(n)=\sum_{d\mid n}f(d)=n$, and not $g(n)=\phi(n)$. So this is impossible.