Can you help with Syracuse $\frac{3x+1}{2}$ disjoint tree graph set-builder notation?

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These expressions are intended for my first paper, so I need a critique of my notation. Feel free to edit. Here is the Mathematica equivalent.

The function $\alpha$ is the $n^{th}$ integer that is $\not\equiv2\mod3 $, based on this OEIS sequence.
$\alpha(n) := 3 n - (5 + (-1)^n)/2$
The function $\beta$ is the maximum power of $2$ that divides $m$.
$\beta(m):=\max\left(n : 2^{n} | m\right)$
Here is the set-builder notation to define the creation of the Syracuse disjoint tree graph without any duplicates. $\textit{A}=\{\{x\to (x = \frac{3 x + 1}{2}):\beta((x = \alpha(j)) + 1) - 1\}\wedge \{x\to\frac{3 x + 1}{2^{\beta(3x+1)}}\}:1\leq j\},$
where $(\cdot)\to(\cdot)$ is a directed edge and $\wedge$ indicates a join.
Graph of 15 sequences: Fifteen sequences

Edit Here is my first attempt after studying all the comments: $$\beta(m):=\max\left(n : 2^{n} | m\right)$$ $$f((t, r, i)) := (2^{r - i} 3^{i - 1} (2 t - 1) - 1)/2^{\beta((2^{r - i} 3^{i - 1} (2 t - 1) - 1))}$$ $$g((t, r, i)) := (2^{r - i} 3^{i - 1} ((6 t + (-1)^{t} - 3)/2) - 1)/2^{\beta((2^{r - i} 3^{i - 1} ((6 t + (-1)^{t} - 3)/2) - 1))}$$ $$\{\left(f((t, r, i)), f((t, r, i + 1))\right)\ | \ \ 1\leq t\leq d\ \wedge \ 2\leq r\leq d\ \wedge \ 1\leq i\leq r-1\}$$ $$\{\left(g((t, r, i)), g((t, r, i + 1))\right)\ | \ \ 1\leq t\leq d\ \wedge \ 2\leq r\leq d\ \wedge \ 1\leq i\leq r-1\}$$ Where $d$ is the dimension of a $d\times d$ triangular prism, $t$ is the triangle, $r$ is the row, and $i$ is the row index.

Does this seem right?