Collatz Conjecture, can the following skip a prime number? Following my previous questions at: Collatz Conjecture, why an increment of $+6$ in the following? and Collatz Conjecture, why a rate of change of $*4$ in the following?
Following the rules of the Collatz Conjecture, in this experiment I have created a list of all odd numbers until $33333$. The list includes 3 columns, such as in the following sample:




A) Starting Odd $(X)$
B) $(X * 3) +1$
C) $X/2$ repeat until odd




1
4
2, (1)


3
10
(5)


5
16
8, 4, 2, (1)


7
22
(11)


9
28
14, (7)


11
34
(17)


13
40
20, 10, (5)


15
46
(23).


17
52
26 (13)


19
58
(29)


21
64
32, 16, 8, 2. (1)


23
70
(35)


25
76
38, (19)




...
You will notice that all the final odd results in column C) represent a list of all the prime numbers. as denoted in the () in column C): 1, 5, 7, 11, 13, 17, 19, 23, 29.
Is it possible to skip a prime number in that list (with the exception of $3$)?

No, you should actually get every odd number which isn't a multiple of $3$ (and, in particular, every odd prime save $3$).
Note that any prime $p\ne2,3$ is either $1$ or $5$ mod $6$. Suppose $p=6k+5$. Then set $X=4k+3$, so that $3X+1=12k+10$ and dividing by $2$ gives $6k+5=p$. Now suppose $p=6k+1$. Then let $X=8k+1$, so that $3X+1=24k+4$. Dividing by $2$ twice gives $6k+1=p$, as desired.
In fact, since nothing here depends on $p$ being prime, this shows that any odd number which isn't a multiple of $3$ is a final number in your column (c).
Conversely, note that no multiple of $3$ can ever be a final number in (c). (In fact, no multiple of $3$ can ever be an element in that column!) After all, if $X=2k+1$, then the final number is of the form $2^{-n}(6k+4)$. Note that $6k+4$ isn't divisible by $3$, and so no number in the last column can be of the form $3k$.
