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I can not find any examples concerning this sum $\sum_{d|n}\frac{\Lambda \left({d}\right)}{d}$ where $\Lambda \left({n}\right)$ is the Von Mangoldt function. I suspect that there is no closed form solution in which case the real value that I need is the asymptotic expansion to the best possible order. The average over $N$ as $N \rightarrow \infty$ is about 0.566, that is $$\lim_{N \rightarrow \infty}\frac{1}{N} \sum_{n = 1}^{N} \sum_{d|n}\frac{\Lambda \left({d}\right)}{d} \rightarrow 0.566$$ so this suggests that $\sum_{d|n}\frac{\Lambda \left({d}\right)}{d} \sim 0.566$ + other terms.

I was summing over $d|2n$ instead of $d|n$ as stated.

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2 Answers 2

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$f(n)=\sum_{d|n}\Lambda(d)/d$ itself does not possess an asymptotic expansion since it can be very large and very small infinitely many times. When $n$ is a prime, we have $f(n)=\log n/n=o(1)$. However, when $n=\prod_{p\le x}p$, we have

$$ \log n=\sum_{p\le x}\log p\sim x $$

and

$$ f(n)=\sum_{p\le x}{\log p\over p}=\log x+O(1)\sim\log\log n. $$

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  • $\begingroup$ $f(n)\ne 2$ when $n=2^k$. $\endgroup$
    – jjagmath
    Jun 30, 2023 at 10:28
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We have \begin{align*} \sum_{n\le N} \sum_{d\mid n} \frac{\Lambda(d)}d &= \sum_{d\le N} \frac{\Lambda(d)}d \sum_{\substack{n\le N \\ d\mid n}} 1 \\ &= \sum_{d\le N} \frac{\Lambda(d)}d \biggl\lfloor \frac Nd \biggr\rfloor \\ &= \sum_{d\le N} \frac{\Lambda(d)}d \biggl( \frac Nd + O(1) \biggr) \\ &= N \sum_{d\le N} \frac{\Lambda(d)}{d^2} + O\biggl( \sum_{d\le N} \frac{\Lambda(d)}d \biggr) \\ &= N \sum_{d=1}^\infty \frac{\Lambda(d)}{d^2} + O\biggl( N \sum_{d>N} \frac{\Lambda(d)}{d^2} + \sum_{d\le N} \frac{\Lambda(d)}d \biggr) \\ &= N \sum_{d=1}^\infty \frac{\Lambda(d)}{d^2} + O(\log N). \end{align*} Therefore the limiting constant in question is $$ \sum_{d=1}^\infty \frac{\Lambda(d)}{d^2} = -\frac{\zeta'(2)}{\zeta(2)} \approx 0.569961, $$ which I've checked a couple of different ways (I'm not sure why the OP is getting around $0.8$).

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