First easily have $f(0)=0$ Plug $n=1$: $f(f(1))=3$ If $f(1)>3$, then let $f(1)=a$ We have $f(a)=3<f(1)$, contradition So $f(1)\le 3$, but $f$ is strictly increase, so $f(1)=2$ Hence, $f(2)=3$ Now plug $n=2$: $f(3)=6$ Then I stuck here Someone pls continue my approach

parmenides51 wrote: Let $f : N \to N$ be a strictly increasing function such that $f(f(n))= 3n$, for all $n \in N$. Find $f(2010)$. Note: $N = \{0,1,2,...\}$ Posted (and solved) many many times (at least 2008, 2010, 2011, 2015, 2016, 2018, 2020, 2021. Dont hesitate to use the search function (see here ). Set (for example copy/paste) in the "search term" field the exact following string : +"increasing" +"f(f(n))=3n" You'll get in the ten first results (excluded your own post and this post itself) all the help you are requesting for.

The result is 3843 I think