Collatz prefix families (noch nicht übersetzt)

The Collatz sequence is defined as: $a_{i+1} = \left\{ \large{\frac {a_i} 2 \atop 3 a_i+1} {\text{if }a_i\text{ is even} \atop \text{if }a_i\text{ is odd}} \right.$.

The Collatz conjecture states that starting from any positive integer, the sequence eventually reaches the cycle 1,4,2,1....
We shall define the sequence prefix p(n) for the Collatz sequence starting with a₁ = n as the sub-sequence of all numbers not a power of 2 (2⁰=1 is considered a power of 2 for this problem). For example:
p(13) = {13, 40, 20, 10, 5}
p(8) = {}
Any number invalidating the conjecture would have an infinite length sequence prefix.

Let S_m be the set of all sequence prefixes of length m. Two sequences {a₁, a₂, ..., a_m} and {b₁, b₂, ..., b_m} in S_m are said to belong to the same prefix family if a_i < a_j if and only if b_i < b_j for all 1 ≤ i,j ≤ m.

For example, in S₄, {6, 3, 10, 5} is in the same family as {454, 227, 682, 341}, but not {113, 340, 170, 85}.
Let f(m) be the number of distinct prefix families in S_m.
You are given f(5) = 5, f(10) = 55, f(20) = 6771.

Find f(90).