Sequence

**fhactor** · January 10th 2011, 12:14 PM

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.

**Also sprach Zarathustra** · January 10th 2011, 12:28 PM

$a_n=3^n$ perhaps?

**Opalg** · January 10th 2011, 01:06 PM

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 $1<n$ , so it follows that $a_1<a_n$ . 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 $a_{2010}.$

**fhactor** · January 10th 2011, 01:36 PM

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 $1<n$ , so it follows that $a_1<a_n$ . 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.

**fhactor** · January 10th 2011, 02:21 PM

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?

Thread: Sequence

LinkBack

Thread Tools

Display

Sequence

Similar Math Help Forum Discussions

Arithmetic Sequence problem and sumac sequence

Randomized sequence of blocked 1-0-sequence and assignment of random integers

Recursive sequence (Fibonacci Sequence) & Limits

sequence membership and sequence builder operators

set, geometric sequence,fraction and arithmetic sequence