Starting any odd number $x$ with $3x + n$ and then dividing by $2$ until getting to the first odd, lets call it $y$.
For $3x + n$, when $n$ is not equal to $3^a$:
If $y$ ends in $1$ or $3$ or $7$ or $9$, it will have a sequence of $x$ made of numbers that end in a repeating pattern of 2 of the following: $1,3,5,7,9$ and every $y$ will have 2 of these patterns .
If $y$ ends in $5$ it will have an $x$ that has only 1 possible ending either: $1,3,5,7,9$
For example:
For $3x +1$ ($x$ on left $y$ on right):
___1,___5 → ___1
___7,___9 → ___1
___7,___9 → ___3
___5,___1 → ___3
___5 → ___3
___9,___7 → ___7
___1,___5 → ___7
___9,___7 → ___9
___5,___1 → ___9
For $3x + n$, when $n$ is equal to $3^a$:
If $y$ ends in $5$ it will have an $x$ that has only 1 possible ending either: $1,3,5,7,9$
If $y$ ends in $1$ or $3$ or $7$ or $9$, it will have one particular sequence made of all possible $1,3,5,7,9$ numbers expect if it is also the ending for $x$ if $y$ ends in $5$
(For every $n$ The order of the sequence is the same besides different beginnings)
For example:
For $3x +3$ ($x$ on left $y$ on right):
___3,___7,___5,___1 → ___1
___1,___3,___7,___5 → ___3
___9 ↦ ___5
___7,___5,___1,___3 → ___7
___5,___1,___3,___7 → ___9
When $n$ is not equal to $3^a$ the same pattern can lead to two different $y$, whereas when $n$ is equal to $3^a$ there is only one pattern for every possible $y$
Why is it happening? Can it be relevant to the conjecture? Any info about it will be appreciated.