what is so special about the number $3$ in the $3n+1$ conjecture?

Question

Recently, I was quite intrigued by the $3n+1$ conjecture and it left me wondering what is so special about the number $3$? With the same rules apply, does there exist another positive integer other than $1$ such that regardless of the starting point, you seem to end hitting a certain number?

So for instance, let us take $k$ a positive integer, we start the algorithm by choosing any positive integer, say $n$, if this number is odd then multiply by $k$ and $+1$ again. If it is even then divide by $2$ as usual. Then we repeat the algorithm depending on the parity of the resulting number. So my question is that does there exist $k\geq 5$ such that given any starting point, we suspect to hit a certain number (usually $1$?) after a certain number of iterations of this algorithm?

I can see that for $k$ even, any starting point will be sent to infinity, but other than that I couldn't make any other observations.

It is just a curious thought of mine and hopefully I have explained my thought sufficiently clear. Thank you so much for help in advance!

Answer 1

It's odd, and it's less than $4$.

Odd --> Thus for each odd $n$, $3n+1$ will be even and we can do divisions by $2$.

Less than $4$ --> This is more subtle. When we generate an even number, on average among all even numbers two such divisions are possible. Thus we expect statistically, each odd number tends to be followed by a smaller odd number because $3<4$, favoring the conjecture that we always go down to $1$. Higher odd coefficients, greater than $4$, are not as cooperative.

Answer 2

It's prime. That means the mappings it produces aren't related to scalings of a primes variant. It's also odd, and that is needed to hit even numbers. It being small, has the advantage of not needing a complex conditions for it to decrease easily. $k=5$ has $kj+1$ needing $j\equiv -(5^{-1})\pmod {2^m}$ to decrease enough that division by $2^m$ or better occurs. This makes division by $8$ or higher power of $2$ a $1$ in $4$ or less occurence.