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It is a well-known problem, known as Collatz problem, to determine whether iteratively applying the map $f(x)=\frac{x}{2}\text{ if $x$ is even and }f(x)=3x+1\text{ otherwise}$ on a positive integer $n$ always will give us $1$. One possibility for which this could not be true is if the sequence $(f^k(n))_{k\in\Bbb N}$ was diverging to infinity. There are no such $n$, and for what I know, it is believed there are none.

Looking at this Math SE question I was mildly surprised to find out that if we replace $3$ by $5$, there are (apparently) no known trajectories diverging to infinity.

This let me to the following question:

Is there an odd integer $a$ and an integer $n$ such that it is known that iterating the map $f(x)=\frac{x}{2}\text{ if $x$ is even and }f(x)=ax+1\text{ otherwise}$ gives a sequence diverging to infinity?

Note that allowing $a$ even would make the problem trivial. I was considering extending this question to more general functions, but I couldn't fina a formulation which would sieve out trivial cases. Let me know in the comments if you know of an example of a diverging sequence which (in your opinion) counts as non-trivial.

Thanks in advance.

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Wojowu
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  • Hmm, it is easy to see from examples with small numbers (for instance 5x+1, see http://math.stackexchange.com/a/717376/1714 and http://mathoverflow.net/a/200126/7710) , that in general there should be divergent sequences, however, as far as I know, no one example of this type of transformation is known to have divergent trajectories. – Gottfried Helms Oct 20 '15 at 18:45
  • @GottfriedHelms Yes, and actually, seeing some of these examples lead me to asking the question here. – Wojowu Oct 20 '15 at 19:33
  • Ah I see, I overlooked "reference-request" and misunderstand the focus of your question; sorry. Anyway, after J. Lagarias had stopped his engagement and his final bibliography did not contain such a result any such result must be very fresh and recently published at would be noticed quickly at least in wikipedia. So we can be pretty sure that such a definite knowledge does not yet exist. – Gottfried Helms Oct 21 '15 at 05:55
  • Let's say, hypothetically, that one were to prove that there exists a pair $(a,n)$ such that the sequence starting at $n$ with multiplier $a$ diverges, but that the proof did not exhibit a particular such pair. The way the question is posed, that would answer in the affirmative -- yet I strongly suspect that the real question is "has anybody produced a specific such pair and proven that it diverges?" – Mark Fischler Sep 08 '16 at 23:06
  • @MarkFischler You've guessed correctly that in mind I would like to see an explicit example, but if there were a nonconstructive proof (which I don't think you are claiming to have), I'd be very interested in hearing that as well. – Wojowu Sep 09 '16 at 07:48

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