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Let $\chi_G(q)$ be the chromatic polynomial of a graph $G$ with $n$ vertices. (More generally, $\chi_{\mathcal{A}}(q)$ can be the characteristic polynomial of a finite hyperplane arrangement $\mathcal{A}$.) Write $\widetilde{\chi}_G(q)=(-1)^n\chi_G(-q)=\sum a_kq^k$, which is obtained by replacing each coefficient of $\chi_G(q)$ by its absolute value. We can think of $\widetilde{\chi}_G(q)$ as defining a probability distribution $p$ on $\{1,\dots,n\}$, with $p(k)=a_k/\widetilde{\chi}_G(1)$. (We always have $a_0=0$, except for the empty graph.) The mean $\mu$ of $p$ is given by $\mu=\widetilde{\chi}_G^\prime(1)/\widetilde{\chi}_G(1)$, and the mode $m$ by $a_m = \max_k a_k$. (Possibly $m$ is multivalued.)

I am wondering whether anything can be said in general about $m$. If all the zeros of $\widetilde{\chi}_G(q)$ are real, then by a theorem of Darroch (see for instance Exercise 1.48(a) of Enumerative Combinatorics, vol. 1, second ed.) $|\mu-m|<1$. To what extent does this result hold "in general"? Is it approximately true (where we need to define "approximately" precisely) for all graphs, or at least for "generic" graphs?

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    $\begingroup$ Of course you, Richard, are aware of this, but for the general audience it might be worth mentioning: it was proved by June Huh in "Milnor numbers of projective hypersurfaces and the chromatic polynomial of graphs" (arxiv.org/abs/1008.4749) that the coefficients $a_1,\ldots,a_n$ form a unimodal sequence. I don't see how this is directly related to your question, but it feels at least spiritually related. EDIT: In fact, Huh showed that the coefficients $a_1,\ldots,a_n$ form a log-concave sequence (which is a stronger property than unimodality). $\endgroup$ Commented Mar 2, 2023 at 21:39
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    $\begingroup$ I looked at the characteristic polynomial $\chi_{L_n}(q)$ of the Linial arrangement $L_n$, since none of the zeros except $q=0$ are real. The mode is always 2. By mathoverflow.net/questions/441994, the mean approaches $2.278\cdots$ as $n\to\infty$, so Darroch's formula $|\mu-m|<1$ continues to hold. $\endgroup$ Commented Mar 4, 2023 at 22:37

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