Collatz Conjecture, can the following skip a prime number?

Question

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 $3$)?

Answer 1

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 $3k$.