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Thread: Help me with this recurrence equation

  1. #1
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    Unhappy Help me with this recurrence equation

    $\displaystyle a_1 = 3$
    $\displaystyle 2a_{n+1} = a_n^2 + 2a_n$
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  2. #2
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    Quote Originally Posted by OliverQ View Post
    $\displaystyle a_1 = 3$
    $\displaystyle 2a_{n+1} = a_n^2 + 2a_n$
    So what would you like to know?

    $\displaystyle \frac{a_n}{2} \cdot \left(a_n + 2\right)$

    As it is the equation says that the next term is half the sum of the previous term squared and twice the previous term

    For example

    $\displaystyle a_2 = \frac{a_1}{2}\left(a_1+2\right) = \frac{3}{2}\left(3+2\right) = \frac{15}{2}$
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  3. #3
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    Hello, OliverQ!

    Note: Everything here is, of course, my opinion.

    I got tired of writing "I think ... ", "It seems that ... ", "The pattern might be ..."


    $\displaystyle a_1 \:=\: 3, \quad 2a_{n+1} \:=\: a_n^2 + 2a_n$

    If you want a function for the $\displaystyle n^{th}$ term,
    . . I've made only a little progress.


    I found some patterns among consecutive terms, but not all of them.


    . . $\displaystyle \begin{array}{ccccccc}
    a_1 &=& 3 &=& \quad\dfrac{(2-1)(2+1)}{1} \\ \\
    a_2 &=& \dfrac{15}{2} &=& \quad\dfrac{(4-1)(4+1)}{2} \\ \\
    a_3 &=& \dfrac{285}{8} &=& \quad\dfrac{(17-2)(17+2)}{2^3} \\ \\
    a_4 &=& \quad\;\;\dfrac{85,\!785}{128}\qquad &=& \qquad\dfrac{(293-8)(293+8)}{2^7} \end{array}$

    . . $\displaystyle \begin{array}{ccccccc}
    a_5 &=& \dfrac{7,\!390,\!046,\!705}{32,\!768} &=& \dfrac{(85,\!849 - 64)(85,\!849 + 64)}{2^{15}}\end{array}$



    The numerators have the sum and difference of two integers:

    . . $\displaystyle \begin{array}{ccccc}1 & 2\pm1 &=& ? \\ 2 & 4\pm1 &=&4 \pm 2^0\\ 3 & 17\pm2 &=& 17\pm2^1\\ 4 & 293\pm8&=& 293 \pm 2^3 \\ 5 & 85,\!849\pm64 &=& 85,\!849 \pm 2^6\end{array}$

    The exponents seem to be triangular numbers: 0, 1, 3, 6, . . .

    $\displaystyle \text{Hence, the numerators are of the form: }\;C \pm 2^{\frac{(n-2)(n-1)}{2}} $



    The denominators are powers-of-2:

    . . $\displaystyle \begin{array}{ccccccc}
    1 & 1 &=& 2^{1-1} \\
    2 & 2 &=& 2^{2-1} \\
    3 & 2^3 &=& 2^{4-1} \\
    4 & 2^7 &=& 2^{8-1} \\
    5 & 2^{15} &=& 2^{16-1}
    \end{array}$

    The exponents are of the form: .$\displaystyle 2^{n-1}-1$

    $\displaystyle \text{Hence, the denominators are of the form: }\:2^{2^{n-1}-1} $



    $\displaystyle \text{The general term is: }\;a_n \;=\;\frac{\left(C - 2^{\frac{(n-2)(n-1)}{2}}\right)\left(C + 2^{\frac{(n-2)(n-1)}{2}}\right)} {2^{2^{n-1}-1} } $



    Now what is generating those $\displaystyle C$-values? . $\displaystyle 2,\;4,\;17,\;293,\;85,\!849\;\hdots $

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