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Does anyone ever write the Collatz conjecture as a single algebraic, recursive sequence? For example, a crude version might be:

$$ g(n+1)=\delta _{1,g(n)}+(1-\delta _{1,g(n)})*\left(\left(\frac{cos(\pi*g(n))+1}{2}\right)*\frac{g(n)}{2}+\left(1-\left(\frac{cos(\pi*g(n))+1}{2}\right)\right)*(3g(n)+1)\right) $$ Where:

  • $\delta_{1,g(n)}$ is the Kronecker delta function with $i=1$; and

  • The $\frac{cos(\pi*g(n))+1}{2}$ term is always evaluates to 0 at odd numbers and 1 at even ones and so handles the toggling between $n/2$ and $3n+1$ (Perhaps there is a better way to achieve this?)

So that if you want Collatz(10), start with $n=0$ and $g(0)=10$ and the recursive equation gives the sequence:

5, 16, 8, 4, 2, 1, 1, 1....

Given a recursion like this, are there ways to investigate its convergence properties? Proving that the sequence converges to 1 irrespective of $g(0)$ (assuming $g(0) > 0$ and integral), or that the sub-sequence that omits the first Kronecker delta function converges to 0 would prove the conjecture would it not? Are there other means of writing the conjecture in a combined algebraic form that might be easier to deal with? Wolfram provides several alternate forms but they don't seem to be much or any easier to deal with. Is there a way to do away with the use of the Kronecker delta function all-together? Has this sort of thing already been tried and deemed a dead end?

Perhaps convergence such as this cannot be proved for the same reasons that sequence analysis of the conjecture has not resulted in a proof...

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  • $\begingroup$ Is the smooth map $f(x)=\frac12x\cos^2\left(\dfrac{\pi\cdot x}2\right)+\frac{3x+1}2\sin^2\left(\dfrac{\pi\cdot x}2\right)$ the kind of thing you're thinking of? It's tricky from there because it turns out the function has infinitely many fixed points in-between the integers. In fact the answer to my one remaining question on here proves it's impossible because on $\Bbb R$ as a metric space, a projection necessary to resolve the conjecture necessarily has the trivial pseudometric. $\endgroup$ Commented Aug 25, 2022 at 16:10
  • $\begingroup$ The form you give is certainly more concise than mine. Since I posted, I have come across that equation as well. What about this idea of writing it as a recursive function and using the delta function to cause convergence to 1 (rather than the 421 sequence) and to see if it could be shown to converge? $\endgroup$ Commented Aug 26, 2022 at 23:43
  • $\begingroup$ Not so sure about that. I did some work on differences and ratios between terms at one point which led to a dynamical system which terminates for all rational numbers, but I was unable to retrace my steps and see what it told us about the Collatz conjecture. $\endgroup$ Commented Aug 30, 2022 at 6:41

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