Sharkovsky theorem

Order the positive integers by odd numbers from $3$ upward, then twice the odds, then four times the odds, and so on, followed by descending powers of two and finally $1$. If a continuous interval map has a cycle of one period, it has cycles of every period later in this order. In particular, period three forces every positive period.

A period-three orbit forces an interval-covering graph with adjacency matrixIts number of closed pointed length-$n$ itineraries is $\mathrm{tr}(A^n)=L_n$. Möbius removal of lower periods and division by $n$ give four primitive cyclic classes at $n=7$ and five at $n=8$.