Given a Cayley graph of a group $G$ with finite generating set $A$ and exponential growth. Let $S_n$ be the elements whose word length is exactly $n$.
$\textbf{Question:}$ Is $f(n) = |S_{2n}|$ a log-concave function? i.e. $f(n)^2 \geq f(n+1)f(n-1)$.
Positive answer for all groups is possibly out of reach, so any extra hypothesis is welcome...
$\textbf{Motivation 1:}$ There are a few groups where the growth series is rational, so one can get a lot of numerical data. The above holds up to $n \geq 10^5$. In fact, in almost all cases [exceptions are described below], even $n \mapsto |S_n|$ is log-concave except for a few small $n$ (i.e. up to three values $\leq 15$).
I also tried a few linear groups and some semi-direct products (those that are easy to feed to a computer) and the same holds (except it can get hard to the values of $|S_n|$ for $n \geq 15$).
It could very well be that a counter-example has non-soluble word problem, so let me put a more precise version of the question for groups with rational growth function:
$\textbf{Question (restricted):}$ Is there a group with rational growth series so that there are 3 or more roots (of the same multiplicity) on the circle of convergence of the serie?
Indeed, for such groups, 3 or more roots seems to be a necessary condition for a negative answer to the question.
$\textbf{Motivation 2:}$ As a comparison, if $g(n)$ is the probability that a simple random walker returns to its initial point in $n$ steps. Then $g(n+1)/g(n)$ will never converge in bipartite graphs (in fact, there will be division by $0$). However, $n \mapsto -\ln g(2n)$ is concave.
[EDIT: The reason for the $-\ln $ and why this could have anything to do with growth is the following observation. Let $P_n$ be the distribution of a simple random walk starting at then identity after $n$ steps. Let $\| F\|_p^p = \sum_{g \in \mathrm{Supp} F} |F(g)|^p$. Let $\Phi_p(n):= \frac{\ln \| P_n\|^p_p}{ (1-p)}$. Then $\Phi_2(n) = \Phi_\infty(2n) = - \ln g(2n)$. Also, defining $\Phi_1 = \lim_{p \to 1} \Phi_p$, one has that $\Phi_1(n) = h(P_n)$ (the entropy of $P_n$). Lastly $\Phi_0(n) \in [ \ln |B_n|- \ln 2, \ln |B_n| ]$ (where $B_n$ is the ball of radius $n$). It turns out, $\Phi_1(n)$, $\Phi_2(n)$ and $\Phi_\infty(2n)$ are concave. So this also brings the question: how far is $\Phi_0(n)$ to be concave?..]
$\textbf{Motivation 3:}$ If one looks at self-avoiding walks in $\mathbb{Z}^d$ (say $c_n$ is the number of SA-walks of length $n$) then $c_{2n+2}/c_{2n}$ converges to $L^2$ (where $L = \lim c_n^{1/n} = \inf c_n^{1/n}$. If those self-avoiding walks could encode the geodesics of some group (lamplighter?), then it might support the fact that $|S_{2n+2}|/|S_{2n}|$ converges for these groups. [I don't think it is known whether $n \mapsto c_{2n}$ is log-concave or not.]
$\textbf{Important examples:}$ A [family of] example[s] due to Machi show that $|S_{n+1}|/|S_n|$ need not converge (these are the only examples I know where $n \mapsto |S_n|$ is not log-concave). These examples are given by free products $G_1 *G_2$ with $A = G_1 \setminus \{e_{G_1} \} \cup G_2 \setminus \{e_{G_2}\}$ (where $|G_1| \neq |G_2|$ and both are finite). In this example (like in the random walk), a bipartiteness phenomenon causes the trouble, but can one find a group where some sort "tri-partiteness" in $|S_n|$ happens?
$\textbf{Question (variations):}$ Let $\lim |S_{n}|^{1/n} = \ell$. Is it true that $\inf_n |S_{2n+2}|/|S_{2n}| \geq \ell^2$? that $\lim_n |S_{2n+2}|/|S_{2n}| = \ell^2$? Can one always find a $k$ so that $n \mapsto |S_{kn}|$ is log-concave? has ratio test converging to $\ell^k$?
This question is related to
On the size of balls in Cayley graphs
