I was studying the collatz conjecture and managed to show that the problem is equivalent the union of a set of conjectures (well uncountably many conjectures, one for each $f: \mathbb{C} \rightarrow \mathbb{C}$ ).
One such conjecture is, consider the function given by:
$$ f(x) = 2\left( 2x+ \frac{1}{5} \right) \sum_{k=1}^{\infty} \left[\frac{(-1)^{k+1}}{6^k} \frac{e^{-\frac{2\pi i}{5}\frac{6^{k}-1}{6^k}}e^{-\frac{4\pi ix}{5}\frac{6^{k}-1}{6^k}}}{1-e^{-\frac{2\pi i}{5}\frac{6^{k}-1}{6^k}}e^{-\frac{4\pi ix}{5}\frac{6^{k}-1}{6^k}}} \right] $$
The conjecture is that $$ f(n) = f(1) \ \forall n \in \mathbb{N} $$
This looks like a reasonably well behaved object, its some complex exponentials, in a geometric series, with nested 6^th roots. Is there a way to prove this without assuming Collatz?
A starting point, would be to break the conjecture into a statement for each natural number, and just trying to show that (in this case x = 1, x= 2)
$$ \frac{22}{5} \sum_{k=1}^{\infty} \left[\frac{(-1)^{k+1}}{6^k} \frac{e^{-\frac{6\pi i}{5}\frac{6^{k}-1}{6^k}}}{1-e^{-\frac{6\pi i}{5}\frac{6^{k}-1}{6^k}}} \right] = \frac{42}{5} \sum_{k=1}^{\infty} \left[\frac{(-1)^{k+1}}{6^k} \frac{e^{-2\pi i \frac{6^{k}-1}{6^k}}}{1-e^{-2 \pi i \frac{6^{k}-1}{6^k}}} \right] $$
Some findings from Wolfram Alpha, the RHS is equal to 0 from $k = 1 ... 155$ (I think that must be some kind of bug), and from $156$ onwards the sum isn't defined, (which seems very counterintuitive, and unlikely)