WakeUp wrote: Let the function $f$ be defined on the set $\mathbb{N}$ such that $\text{(i)}\ \ \quad f(1)=1$ $\text{(ii)}\ \quad f(2n+1)=f(2n)+1$ $\text{(iii)}\quad f(2n)=3f(n)$ Determine the set of values taken $f$. Put $f(2)=:a$ and define the set S by all numbers that can be written with only digits $0$ and $1$ in base $3$: $S=\{0,1,3,4,9,10,12,13,27,...\}$ Then we have straightforward for $n\in\mathbb N$ \[f([2^n,2^{n+1}-1])=3^{n-1}a+S\cap[0,3^n-1],\] and $f(\mathbb N)$ is the (disjoint) union of this. I doubt there is a nicer way to write it See http://oeis.org/A005836/graph for the nice self-similar structure of $S$.

spanferkel wrote: I doubt there is a nicer way to write it "the number written in base 2 and then read in base 3" --> Its base three representation will only have 0's and 1's (no 2's). i.e. $f(2)=f(10_2)=10_3=3$ I seem to remember a Putnam problem concerning numbers written in one base and "read" in another.

flying2828 wrote: spanferkel wrote: I doubt there is a nicer way to write it "the number written in base 2 and then read in base 3" --> Its base three representation will only have 0's and 1's (no 2's). i.e. $f(2)=f(10_2)=10_3=3$ I seem to remember a Putnam problem concerning numbers written in one base and "read" in another. Of course... I missed $a=f(2)=3f(1)=3$