Is it possible to find such a $l$,which that the known counter-example is only $l$, for any $k$, $f^k(n)≠1$? [Is there a modified Collatz rule 3n+$l$ where there is only one loop or "counter-example" and is not 3n+1?] $$f(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+l & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$
Here, $l$ is odd number and $l ≠1$
Because, for $n=l$ , $l$ is counterexample.
For example: $l=7$
The known counterexample are $(5,1),(7,7).$ [According to P..] [In other words, 3n+7 has two loops or two "counter-examples": 5-22-11-40-20-10-5... and 7-28-17-7...]
But, I'm looking for the known counterexample for function $f(n)$ is only $l$. I have tried many. But I did not succeed. Is there such a $l$ ?
If the question is unclear, please edit... I did everything I could.
There are two rules I am aware of that provide 1 "counterexample" for $l$:
$3n+3$ and $3n+9$
It is unproven if these are the only "counterexamples" for all numbers because we do not know if the Collatz Conjecture is true. However,
I checked both $3n+3$ and $3n+9$ for 1,000,000 starting numbers.
There was only 1 "counterexample" or loop for each:
$3n+3$: 12-6-3
$3n+9$: 36-18-9
Note: The files I uploaded are huge. You may need to download them.
I used some code I already wrote in Java that keeps track of the number of cycles and then dump the results into a .csv file. Since Excel can only hold so may numbers, I restricted my sample size to 1,000,000.
Gottfried Helms also worked on this problem and goes into greater detail here.