Collatz Conjecture, why a rate of change of $*4$ in the following?

I prefer always the "Syracuse"-style of notation for the analysis of the Collatz-transformation: we would write, with odd $a$ and $b$ and where the exponent $A$ follows from the value $A=v_2(3a+1)$, the following one step: $$ b= { 3a+1 \over 2^A } \qquad \qquad a,b \text{ odd,} A \in \mathbb N^+ \tag 1$$ We can go backwards: $$ a= { 2^A b -1 \over 3 } \tag 2$$ But this way we see, that parameter $A$ is somehow free: having chosen some odd value $b$ we can choose some value $A_j$ either from $A_j \in \{2,4,6,8,...\}$ or from $A_j \in \{1,3,5,7,...\}$ (depending on $b$).

This gives for instance, that for $b=5$ all $a \in \{3,13,53,253,...\}$ can be found, which satisfy the eq $(1)$: $$ 5 = { 3a+1 \over 2^A } \qquad \qquad a \in \{3,13,53,...\} \text { and } A \in \{1,3,5,7,...\} \tag 3$$
We can see that there is a recursion in the set for $a$: $a_{j+1}=4 a_j+1$ and which can easily be explained by expanding $(2)$ while observing the modular requirements.

In another, but very similar way, one can find out the following table (with any $k \in \Bbb N$ and if we want, even with any $k \in \Bbb Z$) :

$$ \begin{array} {r|rl|rl|r|r|rl|rl|r|r} A & a_{A,k}&& b_{A,k} && &A & a_{A,k}&& b_{A,k} & \\ \hline 1 & 2 k \cdot 2^1 &+3 & 2 \cdot 3 k&+5& &2 & 2 k \cdot 2^2 &+1 & 2 \cdot 3 k&+1 \\ 3 & 2 k \cdot 2^3 &+13 & 2 \cdot 3 k&+5& &4 & 2 k \cdot 2^4 &+5 & 2 \cdot 3 k&+1 \\ 5 & 2 k \cdot 2^5 &+53 & 2 \cdot 3 k&+5& &6 & 2 k \cdot 2^6 &+21 & 2 \cdot 3 k&+1 \\ 7 & 2 k \cdot 2^7 &+253 & 2 \cdot 3 k&+5& &8 & 2 k \cdot 2^8 &+85 & 2 \cdot 3 k&+1 \\ \vdots &\vdots &\vdots &\vdots& &\vdots &\vdots &\vdots & \end{array} \tag 4$$ and for instance, for $A=5$ , and between $a_{5,k}-a_{5,k-1}$ with $k=1,2,3,\ldots$ we find differences of $\Delta_A = 2 \cdot 2^5$ while the $b_{A,k}$ have always the differences $6k$.

Moreover, we have already seen the sequence $\{3,13,53,253,...\}$ with its recursion rule (occuring here as residuals $r_A$ of the $a_{A,k}$). They can be written as $ r_A={5 \cdot 2^A-1\over 3} $ for odd $A$. That residuals transform to $r_A \to {3r_A+1 \over 2^A} =5$ .
For the other sequence of residuals (for the even $A$) I leave it as an exercise to explicate the analoguous properties... :-)

Firstly, note that the first number in the $+x$ sequence is always double the column C) number. So, in your first example, $13-3=10$ and $10=2\times5$. That also seems to work, so I would frase your question like this:

Let $n$ be an odd number, if the next odd number in it's Collatz sequence is $m$, then $\exists k\in\mathbb{N}$ such that the number $n+4^k2m$, also leads to $m$

Now, I don't know if there is a better phrasing for that, but that is what I've attempted to prove.

Say n leads to $n$, then $3n+1=2^lm$ for some $l$, so $2m=2^{-j}(3n+1)$, where $l=j+1$. Starting the sequence, $n+4^k2m$ would take you to $3(n+4^k2m)+1=(3n+1)+3·4^k2m=(3n+1)+3·4^k2^{-j}(3n+1)=(3n+1)(3·4^k2^{-j}+1)=$ $(3n+1)(3·2^{2k-j}+1)$, so if we take $k=j/2$, we get $(3n+1)(3·2^{2k-j}+1)=4(3n+1)=2^{l+2}m$.

Now, for this to work, we need $j$ to be even, so $l$ needs to be odd. I think you can prove that if $n$ is odd, $3n+1$ is an odd power of $2$ times the next odd number (unless the next odd number is $1$). Or, what is the same, that if $m$ is odd, $2^{2a}m-1$ is never a multiple of 3. I think you can prove this quite easily with modular congruences, but I'm a bit rusty on those.