Is it possible to find closed form of the Collatz conjecture?

Question

So recently I was thinking about converting the recurrent definition of Collatz conjecture into a closed-form expression, which would map any $n,\space n\in\mathbb{N}$ to the $n^{th}$ iteration of the sequence with a given seed $m$. Something similar can be found in this comment: $$C(n)=\frac{n}{2}\cos^2\left(\frac{\pi n}{2}\right)+(3n+1) \sin^2\left(\frac{\pi n}{2}\right)$$ Although, this is not exactly what I'm looking for. The reason for this is that $C(n)$ gives the next member of sequence which has $n$ as initial seed, when the formula I'm looking for has arbitrary initial seed independent of $n$. The recursive formula looks like this: $$C_{n}=\frac{C_{n-1}}{2}\cos^2\left(\frac{\pi C_{n-1}}{2}\right)+(3C_{n-1}+1) \sin^2\left(\frac{\pi C_{n-1}}{2}\right)\\C_{1}=m=\text{arbitrary initial integer seed}$$ But, as far as I know, it is not possible to mathematically analyze expressions like this - I need to find a closed form of it. Is there any way to do it?

Answer 1

There's no known formula for the $n$th iterate of the Collatz function. The long-term behavior of iterated functions can be very complicated even if the function being iterated is simple (see e.g. logistic map or Mandelbrot set) - that's why the Collatz problem is hard.