By 'bit-limited', I mean that we have a computer/calculator that handles binary numbers up to $n$ bits ($n>=1$), and any numbers greater than $2^n-1$ overflow by truncating the higher bits. Background information about the Collatz map and cycles can be found here.
For example, with 4 bits, we represent numbers from 0 to 15. Starting at 7, the map goes 7 -> 22 (overflow!) -> 6 -> 3 -> 10 -> 5 -> 16 (overflow!) -> 0 -> 0 -> 0... Using the same 4 bits, but starting from 11, we get 11-> 34 (overflow!) -> 2 -> 1 -> 4 -> 2 -> 1...
I have found that for $n = 9$, there is an additional cycle through 71: of the 512 possible numbers, the 71 cycle covers 124 of them. For $n=11$, there is a cycle through 55 covering 632/2048 numbers. I have tried higher numbers of bits, up to 22, and I haven't found any other cycles.
I'm trying to find a way to approach the question: are there higher values of $n$ that also result in cycles? or:
What values of m lead to cycles in a modified Collatz map
$$a_{i+1} = \begin{cases} (3\cdot a_i + 1) \mod{2^m} & a_i \mbox{ odd} \\ (\frac{a_i}{2}) \mod{2^m} & a_i \mbox{ even} \end{cases} $$
I found other treatments where the whole trajectory is found and then transformed modulo $2^m$, but that does not create cycles.