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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 $\frac{1}{1 - px}$

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First error: In general, $\varphi(n)/\varphi(d) \not= \varphi(n/d)$.

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    @user9325, it's okay I found out that the name is generated by openid or whatever web software is used to register.2011-04-23