1. Sequence

For an increasing sequence of natural numbers $a_1,\ a_2,\ a_3,\ \ldots$ it holds that $a_{(a_k)}=3k$, where $k$ is also natural number.
Find out value of $a_{100}$ and $a_{2010}$.

I've really no idea how to solve this. Thanks in advance.

2. $a_n=3^n$ perhaps?

3. Originally Posted by fhactor
For an increasing sequence of natural numbers $a_1,\ a_2,\ a_3,\ \ldots$ it holds that $a_{(a_k)}=3k$, where $k$ is also natural number.
Find out value of $a_{100}$ and $a_{2010}$.

I've really no idea how to solve this. Thanks in advance.
[I'm sure we have had this problem here previously, but I can't locate it.]

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

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

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

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

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

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

5. I believe I got it.
$a_9=18$, $a_{27}=54$, $a_{81}=162$, so $a_{100}$ should be $181$.
$a_{729}=1458$, $a_{1458}=2187$, so $a_{2010}=2187+(2010-1458) \cdot 3=3843$.

True or false?