<s>[INST] $N \equiv 3 (\textrm{mod } 4)$ and Collatz conjecture Can the Collatz conjecture also be interpreted as behaviour, transformation of number of form $N\equiv 3(\textrm{mod }4)$ to the form of  $N\equiv 1(\textrm{mod }4)$ 
Because integers of the form  $N\equiv 3(\textrm{mod }4)$  can be later always only ones be divided by $2$. Number of form  $N\equiv 1(\textrm{mod }4)$  have much more diversity.
$3,7,11,15,19,23,29,35, \ldots, N \equiv 3(\textrm{mod }4)  $(distance $4$ between numbers)
using $(3n+1)/2$ or applying $+2,+4,+6,\ldots$ turns into
$5,11,17,23,29,35,41,47,\ldots, N \equiv 1 (\textrm{mod }4)$ or $3$ (50%) (distance $6$ between numbers)
again using $(3n+1)/2$ or applying $+3,+6,+9$ turns into
$8,17,26,35,44,53,62,71,\ldots$ (distance 9 between numbers), $N(\textrm{mod }4)$ is various.
 [/INST] To add to @coffemath approach: we can separate the class of odd numbers a,b in the following way such that $b=(3a+1)/2^A$ where $A$ is the value which makes b an odd integer:
$$ \begin{array} {r|c|l||r|c|l}
    & k \ge 0& & && k\gt 0\\
  A & a=& b= & & A & a= & b= \\
\hline \\
 0 & 2^0 \cdot 2  k +0 & k      & & 1 & 2^1 \cdot 2  k -1  & 6 k - 1 \\
 2 & 2^2 \cdot 2  k +1 & 6 k + 1 & &3 & 2^3 \cdot 2  k -3  & 6 k - 1 \\
 4 & 2^4 \cdot 2  k +5 & 6 k + 1 & &5 & 2^5 \cdot 2  k -11 & 6 k - 1 \\
 6 & 2^6 \cdot 2  k +21 & 6 k + 1 & &7 & 2^7 \cdot 2  k -43 & 6 k - 1 \\
 &\cdots & \cdots & && \cdots & \cdots \\
 A & 2^A \cdot 2  k + r & 6 k + 1 & &A & 2^A \cdot 2  k -s & 6 k - 1 \\
 \end{array} $$
where the residues in the columns below the a are dependent on A by
$$r = { 2^A -1  \over 4-1} \qquad s ={ 2^A +1  \over 4-1}  $$
Of course, the entry at $A=0$ should not occur in this table, if we look only at odd a but leave it there for completeness/smoothness of the table.
(That this table covers all odd positive integers is a nice exercise, left to the reader...) 

[added:]
The known "trivial" cycle is that where a=1 in the category A=2 where k=0 such that $$ 2^2 \cdot 2  k + 1 \to 6 k + 1  \quad||_{k=0}\quad   2^2 \cdot 2 \cdot 0+1 \to 1 $$
If we look also at the negative odd integers we have the cycle $-1 \to -1 $ at A=1, k=0 such that 
$$ 2^1 \cdot 2  k - 1 \to 6 k - 1  \quad||_{k=0}\quad   2^1 \cdot 2 \cdot 0 -1 \to -1 $$
and the cycle $-5 \to -7 \to -5 $ by $A_1=1$, $k_1=-1$ and $A_2=2$, $k_2=-1$ :
$$ 2^1 \cdot 2  k - 1 \to 6 k - 1 \quad ||_{k=-1} \quad -5 \to -7 \\
 2^2 \cdot 2  k + 1 \to 6 k + 1 \quad||_{k=-1}\quad  -7 \to -5 \\ $$
The other known cycle in the negative integers is longer and not shown here.
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