Continuous function with a point of periodicity

I have trouble in following question.

Suppose $f : [a, b] \to [a, b]$ is continuous and $x_0$ is a point having period $3$. How many points of period $5$ are there? How many orbits of period $5$? How many points of period $29$ are there?

I think there should be a specific function so that we can check the periodicity, but apparently there's anything wrong in this wording of this problem according to class. Could anyone give me the clue? Might be bifurcation?

• By euler.genepeer.com/period-three your keywords are "Sharkovsky's ordering" and "Period 3 implies chaos" by Li and Yorke. – LutzL Dec 2 '17 at 11:25

By possibly replacing $f(x)$ with $-f(-x)$ we can assume that the 3-periodic point satisfies $$x_0<x_1=f(x_0)<x_2=f(x_1).$$ Then following the very nice article of Uhland Burkart (1982): "Interval mapping graphs and periodic points of continuous functions" consider the intervals $I_1=[x_0,x_1]$ and $I_2=[x_1,x_2]$. The dynamic of $f$ imposes a dynamic on the intervals by $f(I_1)\supset I_2$ and $f(I_2)\supset I_1\cup I_2$, which gives rise to the directed graph with edges $(1,2),(2,2),(2,1)$ Each cyclic path of length 5 gives rise to a different fixed point of $f^5$. Starting at $2$, each times the path visits the left cycle there is a $(2,1)$ grouping in the node sequence. Counting those visits, there has to be at least one and at most $(5-1)/2=2$. In a hole-and-peg pattern, one has to distribute $k$ left-cycle-$(2,1)$ pegs into $5-k$ holes, where the remaining empty holes correspond to right cycles. Each of the thus obtained combinations corresponds to a path not starting in $1$. In each 5 cycle there are $5-k$ such paths, thus the number of cycles is $$\frac1{5-1}\binom{5-1}{1}+\frac1{5-2}\binom{5-2}2=2$$

k=1, cycle: [2,2,2,(2,1)]
k=2, cycle: [2,(2,1),(2,1)]


In consequence there are at least 2 orbits of 5 cycles of $f$.

Counting orbits of period 29 by the same formula gives at least $$\sum_{k=1}^{14}\frac1{29-k}\binom{29-k}{k}=39650$$ orbits, each with 29 fixed points of $f^{29}$.

This method only works for prime numbers as cycle length, for composite numbers the cycles of smaller period will also be counted in the binomials, invalidating this sum formula.