Is the exact value of $$ \sum_{d\mid n} \mu\left(\frac{n}{d}\right) \zeta^d $$ known? Here, $\mu$ denotes the Möbius function and $\zeta$ a root of unity. At first sight, it seems to me that this should be known, but I didn't manage to find the solution yet.
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5$\begingroup$ This is, almost by definition, $nM(\zeta,n)$, where $M$ is the necklace polynomial. I have no idea if this value has any more explicit description. $\endgroup$– WojowuCommented May 17, 2020 at 16:04
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$\begingroup$ I suppose that there is no trivial answer to my question yet. I was not aware of the necklace polynomial, but thanks to the above comment of @Wojowu, the paper by Trevor Hyde caught my attention: arxiv.org/abs/1811.08601. There, Hyde describes conditions under which $M(x,n)$ has factors of the form $x^m \pm 1$. So there are some criterions given under which $M(\zeta,n)$ vanishes, but this is far from computing $M(\zeta,n)$... $\endgroup$– LFMCommented May 17, 2020 at 19:26
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4$\begingroup$ The name of the paper by Trevor Hyde: Hyde - Cyclotomic factors of necklace polynomials. $\endgroup$– LSpiceCommented May 17, 2020 at 21:47
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In case of the root of unity is fixed one for all,
From the orthogonality of Dirichlet characters you can decompose the periodic sequence $n\to \zeta_{p^k}^n $ as a linear combination of the $\chi(\frac{n}{p^j}) 1_{p^j | n}, j\le k,\chi\bmod p^{k-j}$ (the coefficients of the linear combination are Gauss sums).
From $\zeta_N= \prod_{p^k \| N}\zeta_{p^k}^{a(p^k)}$ you get $$\zeta_N^n = \sum_{l|N}\sum_{\chi\bmod \frac{N}l} b(\chi) \chi(\frac{n}{l})1_{l| n}, \qquad b(\chi)=\frac1{\varphi(N/l)}\sum_{n=1}^{N/l} \overline{\chi(n)} \zeta_N^{nl}$$ from which the Dirichlet convolution $$\mu\star{\zeta_N} (n)=\sum_{d| n} \mu(n/d)\zeta_N^n $$ takes the explicit form $$= \sum_{l|N}\sum_{\chi\bmod \frac{N}l} b(\chi) (\chi\star\mu)(\frac{n}{l})1_{l| n}=\sum_{l|N}\sum_{\chi\bmod \frac{N}l} b(\chi)1_{l| n}\prod_{p^r\| \frac{n}{l}} (\chi(p^r)-\chi(p^{r-1}))$$
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$\begingroup$ If you don't tell what is unclear to you then I can't comment. The formula is messy but it is the formula, think to an arbitrary sum of multiplicative functions and the information you loose when not noticing it is. $\endgroup$– reunsCommented May 17, 2020 at 23:00
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$\begingroup$ It's less that it's unclear and more that it's not what I would call explicit, but, of course, that's a judgement call. $\endgroup$– LSpiceCommented May 17, 2020 at 23:46
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$\begingroup$ Well I don't agree, it is a finite sum of simple multiplicative functions, in what sense isn't that explicit. The original formula is shorter but it is not quite possible to guess its behavior for $n$ large. $\endgroup$– reunsCommented May 18, 2020 at 0:08