Counting cycles in an interval covering graph

If $A$ is the adjacency matrix of a directed covering graph, then $\mathrm{tr}(A^n)$ counts its pointed closed walks of length $n$. For a prime number $p$, subtracting the $\mathrm{tr}(A)$ constant walks and identifying the $p$ cyclic choices of starting point givesprimitive closed itineraries of length $p$.

For cycle points ordered $x_3<x_1<x_0<x_2<x_4$, successive gap vertices have arrows $0\to 3$, $1\to 1,2$, $2\to 0$, $3\to 0,1$. Their transition-matrix traces through length four are $1,3,1,7$, forcing periods one, two and four but not three. The affine interpolation through ordered values $(4,3,1,0,2)$ has no three-cycle: its only length-three closed interval itinerary is $111$, on which the map is $5-2x$ and its third iterate has only its fixed point. Two distinct length-four return words at vertex one give a horseshoe for the fourth iterate.

A five-cycle whose points are in orbit order along the real line gives four interval vertices with arrows $0\to 1\to 2\to 3$ and $3\to 0,1,2,3$. The transition-matrix traces for lengths one through four are $1,3,7,15$. Primitive closed words force at least one fixed point, one two-cycle, two three-cycles and three four-cycles. The second iterate has a two-branch horseshoe for an interval map on vertices two and three. The map itself need not have a one-step horseshoe: the unimodal affine example $F(x)=x+1$ for $x\le 3$, $F(x)=16-4x$ for $x\ge 3$ has precisely this five-cycle and has no two disjoint self-covering branches.