Is 993 the only example with such a property in the Collatz sequence?

Question

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:

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

Answer 1

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.