Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 32A / b

Write the period-three points in increasing order and take the two intervals between consecutive points. For either possible cyclic ordering, the interval-covering transition graph is, after interchanging its vertices,The loop at the first vertex gives a fixed point. For every $n\ge 2$, the closed itineraryhas least period $n$. The interval-covering periodic-orbit theorem supplies a point with that itinerary and therefore a cycle of least period $n$. Thus $F$ has periodic orbits of every positive period. This is the period-three case of the Sharkovsky theorem.

For the Fibonacci transition matrix above,where $L_n$ is the $n$th Lucas number. It counts closed symbolic itineraries of length $n$. Since $7$ is prime, the itineraries fixed by the seventh shift have least period either $1$ or $7$. Hence the number of cyclic classes of primitive length-seven itineraries isDistinct primitive cyclic itineraries give distinct cycles, so $F$ has at least four distinct $7$-cycles. This is the fibonacci transition-graph cycle count.

An itinerary fixed by the eighth shift has least period dividing $8$. Those with period dividing $4$ are exactly the itineraries fixed by the fourth shift. Thus the primitive length-eight itineraries formpointed words. Each cycle has eight cyclic shifts, so they givedistinct $8$-cycles. Therefore at least five such cycles are forced.