I was looking a A182137 in the OEIS. For n >= 0, define the Collatz iterate as c(n) = n/2 if n even; (3n+1)/2 if n odd. Note that this is the "normalized" iterate that rolls together the 3x+1 and subsequent x/2 steps for odd numbers. We say n shrinks with b steps if the trajectory of n under c reaches a value <= n within b steps. It is well known that if m == n (mod 2^b), then the first b iterations of c on m and n, respectively, make the same choices of n/2 or (3n+1)/2 on the first b steps. This means that if m == n (mod 2^b), m shrinks within b steps iff n shrinks within b steps. This implies that on any 2^b consecutive nonnegative integers, there is a fixed number a(b) that shrink within b steps. If my calculations are correct, a = A182137 (my computations agree up to a(22)), though the definition of that sequence seems somewhat dense to me. At any rate, if A182137 and my calculations are indeed correct, I have made interesting observation. We see that in general A182137(n+1) >= 2*A182137(n). But for a mysterious set of indices S = (2,5,8,10,13,16,18,21,24,27,29,32,35,36,..), we have equality A182137(n) == 2*A182137(n/2). I would think this means that there is no n for which the trajectory of n reaches a values <= n in exactly b steps for b in S. But this would seem to imply that there are no multiples of 3 in S, since the trajectory of 1 reaches a value <= 1 on steps that are a multiple of 3. Perhaps my calculations are wrong, or there is some subtle problem with my logic.