Eccoci.

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 $1,4,2,1, \dots$.
We shall define the sequence prefix $p(n)$ for the Collatz sequence starting with $a_1 = n$ as the sub-sequence of all numbers not a power of $2$ ($2^0=1$ 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 $\{a_1, a_2, \dots, a_m\}$ and $\{b_1, b_2, \dots, b_m\}$ in $S_m$ are said to belong to the same prefix family if $a_i \lt a_j$ if and only if $b_i \lt b_j$ for all $1 \le i,j \le m$.

For example, in $S_4$, $\{6, 3, 10, 5\}$ 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 $f(90)$.