Does the $5x + 1$ sequence for 7 reach a power of 2 or does it get stuck in a period?

Question

This is much like the $3x + 1$ iteration, except that if $x$ is odd, you do $5x + 1$ [and $\frac{x}{2}$ if $x$ is even]. If $x = 7$, then we have 7, 36, 18, 9, 46, 23, 116, 58, 29, 146, 73, 366, 183, 916, 458, 229, ...

I've iterated this twenty thousand times and found no power of 2. It's also possible that there is a period but it's so large that I'm not spotting it. And it's also possible that I have made a mistake somewhere along the way.

Surely someone else has also calculated this, even though $5x + 1$ is not as famous as $3x + 1$. I have tried other starting $x$ and seen that they quickly reach a period. Has anyone determined what happens with $x = 7$?

Answer 1

In Lagarias' paper from 1985, he mentions (on the bottom of p. 12) this (5x+1) variation. He says that stochastic models predict that almost all orbits escape to infinity. However, this is a heuristic only; it was not proved (in 1985) that even a single orbit escapes to infinity. He calls this open problem (C3), on p. 22. This appears to remain open circa 2006, for here is a paper by Volkov, behind a paywall unfortunately, that continues study of this problem. Volkov gives three cycles, and computational (and heuristic) evidence that all other orbits diverge.

Answer 2

This is giving me some deja vu from centuries ago, but I don't know if it's because I actually found the answer and then forgot it. So I'm not sure if it reaches a period that has evaded detection by you and your colleages. I can tell you this much: it sure doesn't hit a power of $2$. Consider the sequence modulo $8$:

$$7, 4, 2, 1, 6, 7, 4, 2, 5, 2, 1, 6, 7, 4, 2, 1, \ldots$$

In order for this to hit a power of $2$, it would have to hit $4$ itself.