Variations of the Collatz problem


$$ f(n) = \begin{cases} 3n+K & \text {$n$ odd} \\ \frac{n}{2} & \text {$n$ even} \end{cases} $$ 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 $k$.

  • $\begingroup$ 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? $\endgroup$
    – Shuri2060
    Jul 19, 2017 at 9:52
  • $\begingroup$ @Shuri all $±K$ variations $\endgroup$ Jul 19, 2017 at 9:54
  • $\begingroup$ 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. $\endgroup$
    – Shuri2060
    Jul 19, 2017 at 9:57
  • $\begingroup$ @Shuri I am new,İf you understand my question ,can you edit for me? $\endgroup$ Jul 19, 2017 at 10:00
  • 1
    $\begingroup$ 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. $\endgroup$
    – user451844
    Jul 19, 2017 at 10:27

1 Answer 1


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.

  • $\begingroup$ Sorry I didn't understand anything. $\endgroup$ Jul 19, 2017 at 10:36
  • 1
    $\begingroup$ $f(f(f(K)))=f(f(4K))=f(2K)=K$ so the sequence never goes to 1 if it starts at $K$! $\endgroup$
    – didgogns
    Jul 19, 2017 at 10:39
  • $\begingroup$ Now I understood... $\endgroup$ Jul 19, 2017 at 10:56
  • $\begingroup$ Thanks for answer.✓ $\endgroup$ Jul 19, 2017 at 11:40

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