# Possible variation of the $"3n+1"$ problem as $"3n+K"$

Variations of the Collatz problem

Let

$$f(n) = \begin{cases} 3n+K & \text {n odd} \\ \frac{n}{2} & \text {n even} \end{cases}$$ with $$K\in\mathbb Z^*$$

I am looking for $K\neq1$ such that $(\forall n\in\mathbb{N})\,\,\, f^k(n)=1$ for large enough $k$.

• Just checking, but $+1$ is just $1$? I also find your question slightly confusing - I'm failing to interpret '( Variations of the "3n+1""3n+1" problem) "any" ...' and 'I mean from the "any n∈Nn∈N", we don't know counterexample for now, which that'. Could we have a link/explanation to this problem? Jul 19, 2017 at 9:52
• @Shuri all $±K$ variations Jul 19, 2017 at 9:54
• I don't quite understand the emphasis placed on $\pm$ signs here. $K\in\mathbb Z^*$ and $K\neq1$ is what normally would be written. Jul 19, 2017 at 9:57
• @Shuri I am new,İf you understand my question ,can you edit for me? Jul 19, 2017 at 10:00
• if k is a multiple of 3 all odd numbers hit multiples of 3. k=3 would lead to $3 \mapsto 12\mapsto 6\mapsto 3$ for example and k=6 leads to $3\mapsto 15\mapsto 51\ldots$ k=9 leads to $3\mapsto 18 \mapsto 9\mapsto 36\mapsto 18\ldots$ etc.
– user451844
Jul 19, 2017 at 10:27

If $K\neq 1$ and it's odd, your sequence will always contain a cycle: $$K, 4K, 2K, K$$
If $K$ is even, one can see the sequence increases infinitely once it hits odd number since it never comes back into even number.
• $f(f(f(K)))=f(f(4K))=f(2K)=K$ so the sequence never goes to 1 if it starts at $K$! Jul 19, 2017 at 10:39