Summatory Function $F(n) = 1 $ for all $n$ odd, and $F(n) = 2$ for all n even

62 Views Asked by At

So, I have this summatory function $$ F(n)=\sum_{d\mid n}f(n)$$ that goes $F(n) = 1$ for $2\nmid n$, and $F(n)=2$ for $2\mid n$. This summatory function is multiplicative. I need to describe the function $f$ by using the Möbius inversion formula.

So I got $$f(n)=\sum_{d\mid n} \mu(d)\times F\left(\frac{n}{d}\right),$$ where $\mu(n)$ is the Möbius function. Now, since $f(n)$ is multiplicative, it is enough to describe $f(p^a)$ for $p$ prime, correct? If I do so, I get something like $f(p^a) = 1\times2+(-1)^1 \times 2 = 0$ if $p=2$, and $f(p^a)=1\times1 + (-1)^1 \times 1 = 0$ if $p$ is odd.

Is this the correct way of doing it?