Is there a Collatz-like sequence ending with $1,1,1,\ldots$?

I found a conjecture of some mathematics teacher. Is this known or solved problem? Is there a non-constant function $f$ such that for any $n$, iterating $f(n)$ end to the sequence $1,1,1,1,1,\ldots$. By iteration I mean the similar way as in Collatz problem we end to the sequence $2,1,2,1,\ldots$. For example, does the function $f(n)=\begin{cases}n/3\text{ if }n\equiv 0\pmod 3 \\(4n-1)/3\text{ if }n\equiv 1\pmod 3\\(5n-7)/3\text{ if} n\equiv 2\pmod 3\end{cases}$ end to the sequence $\ldots, 1,1,1,\ldots$ This is from the Finnish mathematical journal from college students called Solmu, http://matematiikkalehtisolmu.fi/2017/1/otaksuma.pdf

• You mean, like: $$f(n)=\begin{cases}n/2&n\text{ even}\\(n+1)/2&n\text{ odd}\end{cases}$$ – Thomas Andrews Mar 1 '17 at 17:44
• You'll probably want to state some more conditions on $f$ than you have here -- as currently written the constant function that always returns $1$ will satisfy your criteria. – Henning Makholm Mar 1 '17 at 17:45
• Can you understand the article you linked ? – A---B Mar 1 '17 at 17:49
• @HenningMakholm Hmm. It looks like the author and me forgot to check the easiest possible solution. – Jaakko Seppälä Mar 1 '17 at 18:52
• @user2219896: How about Thomas's example, then? Or $f(n) = \max(1,n-1)$? – Henning Makholm Mar 1 '17 at 18:58