  A half-arc-transitive graph is a regular graph that is both vertex- and
edge-transitive, but is not arc-transitive. If such a graph has finite valency,
then its valency is even, and greater than $2$. In 1970, Bouwer proved that
there exists a half-arc-transitive graph of every even valency greater than 2,
by giving a construction for a family of graphs now known as $B(k,m,n)$,
defined for every triple $(k,m,n)$ of integers greater than $1$ with $2^m
\equiv 1 \mod n$. In each case, $B(k,m,n)$ is a $2k$-valent vertex- and
edge-transitive graph of order $mn^{k-1}$, and Bouwer showed that $B(k,6,9)$ is
half-arc-transitive for all $k > 1$.
  For almost 45 years the question of exactly which of Bouwer's graphs are
half-arc-transitive and which are arc-transitive has remained open, despite
many attempts to answer it. In this paper, we use a cycle-counting argument to
prove that almost all of the graphs constructed by Bouwer are
half-arc-transitive. In fact, we prove that $B(k,m,n)$ is arc-transitive only
when $n = 3$, or $(k,n) = (2,5)$, % and $m$ is a multiple of $4$, or $(k,m,n) =
(2,3,7)$ or $(2,6,7)$ or $(2,6,21)$. In particular, $B(k,m,n)$ is
half-arc-transitive whenever $m > 6$ and $n > 5$. This gives an easy way to
prove that there are infinitely many half-arc-transitive graphs of each even
valency $2k > 2$.
