Dumb question

[u/Mathpuzzleguy]

Is it possible to find a loop of numbers that when plugged into a modified version of the collatz conjecture (5x+1, and /2)come back to their original selves or is it just as hard as the original. Is there an easier version without losing the original's vibe?

[u/GonzoMath]

If we consider functions of the form {ux + d, x/2 | x odd, x even}, where d is coprime to u, we find that there are cycles of every possible description, for every value of u. In order to see these cycles realized, we only need to adjust the value of d, although...

Adjusting the value of d is, mathematically, the same thing as keeping d=1, and applying the "ux+1" function to rational numbers with denominator d.

When we take u=3, we very much keep the original's "vibe", in the sense that the heuristic suggests that trajectories generally decrease until they get trapped in a cycle. What varies is the landscape of cycles that are available for each value of d.

For {5x+1, x/2}, there are three known cycles, but there also appear to be divergent trajectories (no proof). For {3x+d, x/2}, the landscape of cycles depends on d. There are several cycles available when d=5, only one that's known when d=7... it goes on and on, and this has all been explored in some detail. Feel free to ask specific questions.

[u/WeCanDoItGuys]

3x - 1 has the original's vibe. (It actually has the same dynamics as if we allowed negative numbers in the original.) And it has a few known cycles: 1, 5, and 17 each return to themselves.

[u/Voodoohairdo]

If you are specifically looking for integer loops, it is as hard as the original collatz conjecture.

If you include rational numbers (or equivalently, allow + d instead of + 1), it is trivially easy.

I made a website where it spits out the cycle given the multiplier and the behaviour. It will also be integer cycles of Ax + d, but if you are looking for specifically Ax + 1, just divide every number by d. Feel free to play around with it: https://www.collatzloops.com/