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Under the Collatz rules:

  • $n\to 837n+1$ if $n$ is odd
  • $n\to n/2$ if $n$ is even.

What is the simplest argument/proof to show that there is a Collatz sequence with a starting number that goes off to infinity?

$837$ is just an arbitrary, large odd number. If for some reason $837$ is inconvenient to make such a "simple" argument, then you are welcome to use a different multiplier.

I am interested in different proofs/arguments. Would something akin to Will Jaggy's proof to my previous question work here? If so, how?

Or is this open for all multipliers (the $a$ in $an+1$) and all starting values?

Thanks

Adam Rubinson
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    This is open , for example , for $5n+1$ and start value $7$. – Peter Jul 16 '24 at 16:09
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    @Peter, Is it open for all $an+1,$ and for all starting values? – Adam Rubinson Jul 16 '24 at 16:11
  • I think that we can only in exceptional cases be sure that the sequence is diverging. In particular , should there be such a diverging sequence in the usual Collatz conjecture , we probably would not be able to prove it. – Peter Jul 17 '24 at 08:33
  • I've not yet perfectly understood the argument that R. Crandall used in his 1978 article about $1093n+1$, arguing from the detail, that $1093$ is a Wieferich-prime. Perhaps there is something in this case useful for your question... – Gottfried Helms Jul 17 '24 at 08:50
  • My first comment pointing to Crandall's paper might be misformulated. The article is about the 3x+1 problem, and the $1093x+1$ is just a short paragraph in the generalization chapter. Better to look for it is to take some token from my (locally) saved filename: "Crandall 1978 Onthe3x+1Problem S0025-5718-1978-0480321-3.pdf" for the google input field. – Gottfried Helms Jul 17 '24 at 19:55
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    @GottfriedHelms I didn't know what you meant but now I do. Oops indeed from me. This is the link I meant: https://mathoverflow.net/questions/462321/5n1-sequence-starting-at-7 – Adam Rubinson Jul 18 '24 at 14:39

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