I want to prove that $$\frac{n}{\varphi(n)} = \sum_{d|n} \frac{\mu(d)^2}{\varphi(d)}.$$ First clear denominators to get $$n = \sum_{d|n} \mu(d)^2 \varphi(n/d).$$ Next I replaced $\mu(d)^2$ with $\lambda^{-1}(d)$ and rewrote as $$i * \lambda = \varphi$$ the computed then Bell series of both sides to get $$\frac{1}{1 - px} \cdot \frac{1}{1 + x} = \frac{1 - x}{1 - px}.$$ This is not an equality so I think I have a mistake somewhere?
- $\lambda$ is the Liouville function with Bell series $\frac{1}{1 + x}$.
- $\varphi$ is the Euler totient function with Bell series $\frac{1 - x}{1 - px}$.
- $i$ is the identity function with Bell series $\frac{1}{1 - px}$
First error: In general, $\varphi(n)/\varphi(d) \not= \varphi(n/d)$.