Ratio of (Collatz "path" bits) / (starting number bits) $\approx 2$ The "Path Records" page by Eric Roosendaal related to the Collatz conjecture (see here), reports a "striking" feature of the path records.

If we define $\text{MaxSeq}(M)$ (in Roosendaal it is $\text{Mx}(M)$) the maximum value reached starting the Collatz sequence from a positive integer $M$, a positive integer $N$ is called a path record if for all $M$ &lt; $N$ we have $\text{MaxSeq}(M) \lt \text{MaxSeq}(N)$.

The "striking" feature about path records is this:

$$\log_2{\text{MaxSeq}(N)}\approx 2\log_2{N} \tag{1}$$

meaning that the maximum number of bits to store the Collatz sequence values starting from a number with $k$ bits is about $2k$ and not much more than that.

So, now this is my comment, when we start from a path record $N$ we could think it has an "energy" of about $2$, because the number of its bits will be eventually doubled, but other bigger values in the sequence toward $\text{MaxSeq}(N)$ have a lower "energy", and lower and lower when they are more near to $\text{MaxSeq}(N)$. This $2$ factor is not so big after all.

My question is: someone can help giving some probabilistic/heuristic explanation for equation $(1)$?

I tried using the fact that each odd number in the sequence is on average $3/4$ of the previous one (see here), but with no success.
Try this and find out.
$$a,n,k,\ell \in \mathbb{N}$$ $$\forall\;n_0:=2^ka-1\;\exists\;\mathcal{C}^k(n_0):\;n_k=\frac{3^ka-1}{2^\ell}$$
Try this and find out.
$$a,n,k,\ell \in \mathbb{N}$$ $$\forall\;n_0:=2^ka-1\;\exists\;\mathcal{C}^k(n_0):\;n_k=\frac{3^ka-1}{2^\ell}$$