Sharkovsky ordering is a total ordering of the positive integers, one that appears very unusual on first acquantaintance.  It has the form: $3 ≺ 5 ≺ 7 ≺ ... 2 ⋅ 3 ≺ 2 ⋅ 5 ≺ 2 ⋅ 7 ... ≺ 2^{2} ⋅ 3 ≺ 2^{2} ⋅ 5 ≺ 2^{2} ⋅ 7 ≺ ... ,$ continuing in this pattern with increasing powers of 2 times the positive odd integers, and finishing with a final sequence of descending powers of 2.$$

Sharkovsky’s theorem is about continuous mappings $f$ of an interval  $I ⊂ \mathbb{R}$ into itself, and its iterations. Such a map $f$ is said to be periodic if there is an $n >0$ such that the points $x, f(x)$, ...., $f^{n−1}(x)$, are distinct, but $f^{n}(x) = x.$   In that case $n$ is called the (minimal) period of $f$. The set consisting of $x, f(x), f^{2}(x)$,  ..., $f^{n−1}(x)$  is called the the periodic orbit of $f$. $$

The first part of Sharkovsky’s theorem says that if f has a periodic point of least period $m$, and $m ≺ n$ in this the ordering, then $f$ also has a periodic point of least period $n$. The second part is that every possible initial segment of the Sharkovsky order is realized on a continuous interval map as the set of periods of its periodic points.  This has a couple of immediate consequences. If $f$  has a periodic orbit of period 3, then it has periodic orbits of all other periods. If $f$ has only finitely many periodic orbits, they must all have periods that are a power of 2.$$