Tangled Knot Function I am seeking a function $f: \mathbb{R}^3 \to \mathbb{R}^3$
that has these properties:
(1) When iterated $n$ times starting from some $p$,
connecting the points in order 
with segments and closing last to first,
$$(p, f(p), f^2(p), \ldots, f^n(p), p)$$
results in a simple (non-self-intersecting) closed polygonal
cycle $K$.
(2) When $K$ is viewed as a knot, it is highly tangled,
e.g., it has large crossing number, or large unknotting number.
The tangledness, however defined, should increase with $n$, the faster the better.
(3) These properties should hold for infinitely many $n$.
Expressed differently, I would like a way to generate
an infinite variety of increasingly tangled stick knots
via a simple function iteration.
My requirements are a bit loose, as I just want to 
simply generate knotty examples.
Likely some weaving is known to accomplish this...?
          
(This question is related to an earlier question,
"Complexity of random knot with vertices on sphere.")
 A: There is a body of work starting with Birman and Williams, and continued by Ghrist, Holmes, and Sullivan, concerning knottedness of closed orbits of flows. The flows they consider are certain Axiom A flows on $S^3$ with interesting basic sets, for example the Lorentz attractor. Williams' paper "Lorentz knots are prime" shows that all the closed orbits of the Lorentz attractor are prime knots. And, being an Axiom A basic set, closed orbits are dense in the Lorentz attractor. So if $f$ is the time $\epsilon$ map of a flow on $S^3$ having the Lorentz attractor as a basic set, then you'll get lots of examples by choosing $p$ on longer and longer closed orbits, as long as you are careful to connect $f^k(p)$ to $f^{k+1}(p)$ by a flow segment. I am guessing that these knots will be more and more complicated as you choose $p$ to have longer and longer orbit.
A: If you parametrize a torus in $\mathbb{R}^3$ as $(x(u,v),y(u,v),z(u,v))$, $0\le u,v<1$, you can easily generate the torus knot $(3,q)$ (with crossing number $2q$) for $q$ large enough and not divisible by three by letting $f(u,v) = (u+3/q^2 \mod 1,v+1/q \mod 1)$ and $n=q^2$. So you just have to embed a continuum of these tori with $q$ varying continuously, and you'll have a function $f$ that generates every knot $(3,q)$. You can probably also improve on how $n$ grows with $q$.
