If you write a collatz-transformation this way $ \small a_{k+1} = (3 a_k + 1)/2^A $ where **A** is the number of all following even (divide-by-2) steps, then also all $\small a_k$ must be odd.      

With this write for one transformation $\small a_{k+1} = T(a_k;A) $ and for more $\small a_{k+m} = T(a_k;A_1,A_2,\ldots,A_m) $.    
For $\small A=1 $ these steps are increasing and if $\small A>1$ the steps are decreasing.     

A cycle occurs, if $\small a_m=T(a_0;A_0,A_1,\ldots,A_m) = a_0 $ thus $\small a_m = a_0 $.     

Now let's look at two different types of cycles - the "general" one, where the exponents $\small A_k $ have no a priori restriction, and the "primitive" one, where all but the last $ \small A_k = 1 $, thus $\small a_0 = T(a_0;1,1,1,1,\ldots,1,A) $ with say $\small N-1 $ times *1* and only one $\small A_{N} \gt 1$      

For the primitive cycle of the length *N* the smallest element $\small a_0 $ must have the form $\small a_0 = 2^{N-1} \cdot 2w - 1$ where *w* is some positive odd integer and that primitive cycle increases then up to $\small a_{N-1} = 3^{N-1}\cdot 2w-1 $ from where it must decrease by a consecutive set of *A* "even" transformations.    

The proof of Ray Steiner (and the subsequent proofs of J.Simons and B.de Weger) use that requirement of the "primitive cycle" (in their nomenclature "1-cycle") to show that such a cycle *cannot exist* except $\small a_0 = T(a_0;2) \qquad a_0=1 $ (where no exponents of value *1* occur - the degenerate case (also called "circuit").     


For the general cycle this is much more complicated and does definitely not have the properties which you sketch in your op.