Originally Posted by

**Opalg** [I'm sure we have had this problem here previously, but I can't locate it.]

To start you off, let $\displaystyle a_1=n$. Then $\displaystyle a_n = a_{a_1}=3\times1 = 3$. But the sequence is increasing, and $\displaystyle 1<n$, so it follows that $\displaystyle a_1<a_n$. In other words, $\displaystyle n<3$. Therefore the only possibility is that $\displaystyle n=2$.

Thus $\displaystyle a_1=2$ and $\displaystyle a_2=3$. Therefore $\displaystyle a_3 = 3\times2=6$, and $\displaystyle a_6 = 3\times3=9$. But if $\displaystyle a_3=6$ and $\displaystyle a_6 = 9$ then $\displaystyle a_4$ and $\displaystyle a_5$ have to fit in between 6 and 9 (because the sequence is increasing). The only way that can happen is if $\displaystyle a_4=7$ and $\displaystyle a_5=8$.

Continue exploring in that way, using the facts that $\displaystyle a_{(a_k)}=3k$ and that the sequence is increasing. You will find that $\displaystyle a_k$ is uniquely determined for each $\displaystyle k$. After a while, a pattern should emerge, which will enable you to predict $\displaystyle a_{100}$ and $\displaystyle a_{2010}.$