TITLE: Divisibility of an expression involving the Möbius function
QUESTION [4 upvotes]: Let $d$ and $n$ be integers, with $n\geq 2$, and define $$R(n,d) := \sum_{k\mid n} \mu( n/k)d^k,$$ where $\mu$ is the Möbius function. It can be shown (see below for an arithmetic proof) that $n\mid R(n,d)$. My question is whether anyone knows of a combinatorial proof of this fact? Feel free to make $d\geq 1$ if this makes the question have a more combinatorial flavor. 
Sketch: Suppose $n$ has prime decomposition $n = p_1^{e_1}\cdots p_r^{e_r}$. We prove that $n\mid R(n,d)$ by induction on $r\geq 1$. When $r = 1$, so $n = p^e$, one computes $$R(p^e, d) = d^{p^e} - d^{p^{e-1}} = d^{p^{e-1}}\left(d^{p^{e-1}(p-1)} - 1\right).$$ If $p\mid d$, it's easy to see $n = p^e\mid d^{p^{e-1}}$, so $n\mid R(n,d)$. If $p\nmid d$, then Fermat's little theorem gives that $d^{p-1} = 1 + pm$ for some $m\in \mathbb{Z}$, so $$d^{p^{e-1}(p-1)} - 1 = (1 + pm)^{p^{e-1}} - 1 = (1 + p^em') - 1 = p^em' = nm'$$ for some $m'\in \mathbb{Z}$. This proves the base case of the induction. 
To do the induction step, assume $n = ap^e$, where $p\nmid a$, and you know that $a\mid R(a,d)$ for all integers $d$. One now computes $$R(n,d) = R(a, d^{p^e}) - R(a, d^{p^{e-1}}),$$ and hence by induction $a\mid R(n,d)$. On the other hand, $$R(a,d^{p^e}) - R(a,d^{p^{e-1}}) = \sum_{k\mid a}\mu(a/k)\left(d^{kp^e} - d^{kp^{e-1}}\right),$$ and every term on the right hand side is divisible by $d^{p^e} - d^{p^{e-1}}$, which in turn is divisible by $p^e$ from our work in the previous paragraph. Thus $p^e\mid R(n,d)$. Since $(a,p) = 1$, it follows that $n = ap^e\mid R(n,d)$, completing the proof. 
Context: When $d\geq 2$, the expression  $R(n,d)$ is the degree of the $n$-th dynatomic polynomial of any polynomial map $f\colon \mathbb{C}\to\mathbb{C}$ of degree $d$, so these expressions have some dynamical significance. For many (but not all) such $f$, the  $n$th dynatomic polynomial of $f$ is the polynomial whose roots are exactly the $f$-periodic points with exact period $n$. Since these points break up into cycles of length $n$ under the dynamics of $f$, it therefore follows that $R(n,d)$ is divisible by $n$.

REPLY [1 votes]: As zxy has pointed out in the comments, the expression
$$\frac{1}{n} \sum_{d \mid n} \mu(n/d) k^{d}$$
counts the number of necklaces on $n$ beads with $k$ colors (I inverted the role of $d$ and $k$ in your notation because this seems more natural). In particular, it is an integer.
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