I have constructed a tree that looks something like this: $$ 1,2,4,8,16, \begin{cases} 32,64,\begin{cases} 128, 256, \begin{cases} 512...\\ 85... \end{cases}\\ 21, 42, 84... \end{cases}\\ 5, 10, \begin{cases} 20, 40, \begin{cases} 80...\\ 13... \end{cases}\\ 3, 6, 12... \end{cases} \end{cases} $$ to visualize all Collatz sequences which arrive at $1$ within $10$ steps. I observed that, as soon as the tree splits into two paths, dividing a "top" number in any brace by the number directly below it always seems to yield a value between 6 and 7.
For example, $$ \frac{32}{5}=6.4, \frac{20}{3}=7, \frac{512}{85}\approx6.0235 $$ . This also holds when I divide a number on the tree by the number occupying the same position in the brace below the one it occupies, e.g.: $$ \frac{128}{20}=6.4,\frac{85}{13}\approx6.54 $$ . It's pretty obvious why $32/5=64/10=128/20=256/42=512/80=6.4$. But if we let $S_n$ be the sequence on the tree: $$32, 64, 21, 42, 84...$$ always taking the low path, and we let $T_n$ be the similar sequence: $$5, 10, 3, 6, 12...$$ , can it be proved that $$6\leq\frac{S_n}{T_n}\leq7$$ for all real $n$ with what is currently known about the Collatz conjecture?