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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?

Ignatij
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    If there were a simple closed expression found for the iterates, most likely the conjecture would be easy to solve. So, it seems unlikely that one exists, given how much effort has gone into solving it. – Jair Taylor Jun 08 '22 at 15:05
  • The collatz sequence is not just formed by a recurrence relation. The map is so dynamic that the existence of a general closed form would be a miracle. And as other mentioned, the existence of such a closed form would likely solve Collatz. – Peter Jun 08 '22 at 15:41

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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.

Karl
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