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Let be N an integer.

$T^{(k)} (N) $ is the 3x+1 function with k the number of iterations.

Starting from N=993, after 65 steps, you reach 130, which is $65\cdot 2$.

Do you believe that there are infinitely many N such that:

$T^{(k)} (N) =2\cdot k$?

Jyrki Lahtonen
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Enzo Creti
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    Why this obsession with $993$? You have posted several very similar questions about it...is there any reason for this? At a minimum, you should link to all the parallel questions you have been asking. – lulu Jul 26 '20 at 11:26

1 Answers1

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Pick any $k$ and go backwards $k$ steps from $2k$ to find a suitable $N$ for this $k$. In fact, quite often you have two choices of ging backwards, so you may find many $N$ for your $k$. What always works is $N=2^k\cdot 2k$, of course.

Hagen von Eitzen
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