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 $3$)?
Best Answer
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 $3k$.