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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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up vote 2 down vote accepted

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

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What about in the case that $d|n$? – quanta Apr 23 '11 at 15:29
I see, even if $d|n$ it doesn't mean that $(d,n/d) = 1$. Thanks. – quanta Apr 23 '11 at 15:33
BTW, I did not mean to say that you made another error, I just wanted to make clear that I did not check the rest of your argument. – Phira Apr 23 '11 at 15:59
@user9325, it was confusing how you said "in general" since I had the condition $d|n$ but I see why my argument was wrong now. Thanks. BTW Why don't you have a name? You can be anonymous on this site but you don't have to have a random number as your identifier. – quanta Apr 23 '11 at 16:01
@user9325, how is this username generated? – quanta Apr 23 '11 at 16:14

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