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The following arithmetic identity holds: \begin{align} \Lambda(n) = \sum_{d \mid n} \mu(d) \log \frac{n}{d} \end{align} where $\mu(n)$ is the Moebius function and $\Lambda(n)$ is the von Mangoldt function. Does the following related Dirichlet convolution simplify to known (or simpler) functions? \begin{align} n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d} \end{align}

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The only obvious thing is that the Dirichlet generating function for your convolution (sans the $n$ factor) is $-\dfrac{\zeta^\prime(s)}{\zeta(s+1)}$. I got nothing else... –  J. M. Jul 27 '12 at 17:45
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Well, the other obvious thing is that by Möbius inversion this function $f(n)$ satisfies $\sum_{d\mid n}f(d)=n\log n$. That doesn't look like a helpful property for a function to have, though. But as in your other question, you can use it to calculate $f(n)$ recursively without the Möbius function as $f(n)=n\log n-\sum_{d\mid n,d\lt n}f(d)$. –  joriki Jul 27 '12 at 18:33
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For the vast majority of the integers your dirichlet convolution is somewhat close to $\phi(n)/n \cdot n \log n$. The only (possible) exceptions are those integers with very many prime factors, say more than $(\log n)^{1 - \varepsilon}$. –  blabler Nov 30 '12 at 4:42
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up vote 1 down vote accepted

Yes,$$n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}=n \sum_{d \mid n} \frac{\mu(d)}{d} (\log(n)-\ln(d))$$ $$= \sum_{d \mid n} \frac{\mu(d)}{d} n\log(n)-n\frac{\mu(d)}{d}\ln(d)=n\ln(n) \sum_{d \mid n} \frac{\mu(d)}{d} -n\sum_{d\mid n}\frac{\mu(d)}{d}\ln(d)$$ $$=\ln(n)\phi(n)-n\sum_{d\mid n}\frac{\mu(d)}{d}\ln(d)=\ln(n)\phi(n)+\phi(n)\sum_{p\mid n}\frac{ln(p)}{p-1}$$ Thus, $$n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}=\phi(n)( \ln(n)+\sum_{p\mid n}\frac{ln(p)}{p-1})$$

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$\sum_{d|n} \mu(d)/d = \prod_{p|n} (1-1/p) = \phi(n)/n$ instead of $\phi(n)$. A similar mistake happens in your treatment of the second term - I believe that the final answer shouldn't have the $n$ factor in front. –  Sanchez Jan 30 '13 at 2:41
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My bad your right –  Ethan Jan 30 '13 at 2:54
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