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Thread: Sequence

  1. #1
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    Sequence

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

    I've really no idea how to solve this. Thanks in advance.
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  2. #2
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    $\displaystyle a_n=3^n$ perhaps?
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  3. #3
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    Quote Originally Posted by fhactor View Post
    For an increasing sequence of natural numbers $\displaystyle a_1,\ a_2,\ a_3,\ \ldots$ it holds that $\displaystyle a_{(a_k)}=3k$, where $\displaystyle k$ is also natural number.
    Find out value of $\displaystyle a_{100}$ and $\displaystyle a_{2010}$.

    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 $\displaystyle a_1=n$. 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 $\displaystyle a_{2010}.$
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    Quote Originally Posted by Opalg View Post
    [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 $\displaystyle a_{2010}.$
    Cool, thanks for help, I am on it.
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  5. #5
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    I believe I got it.
    $\displaystyle a_9=18$, $\displaystyle a_{27}=54$, $\displaystyle a_{81}=162$, so $\displaystyle a_{100}$ should be $\displaystyle 181$.
    $\displaystyle a_{729}=1458$, $\displaystyle a_{1458}=2187$, so $\displaystyle a_{2010}=2187+(2010-1458) \cdot 3=3843$.

    True or false?
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