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Looking for insights about this sequence:

  1. Jan 12, 2010 #1
    1 2 3 4 5 6
    1, 1, 2, 3/2, 4/(3/2), 5/(4/(3/2)), ...

    I suppose there are a lot of variations, but the general idea is the terms are defined by:

    f(n)*f(n+1)=n, where n is an integer. The top row is the term # (n), and the bottom one is the actual value of the term

    Particularly, I am looking for a way to find the partial sums of the terms.

    Any responses are appreciated. Sorry for the unclear formatting...
  2. jcsd
  3. Jan 12, 2010 #2


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    Here's a start; closed forms for the odd and even terms. However, I have started your series counting from 0. That is, where you write f(1) I have f(0). Otherwise it's the same.
    f(2n) & = 4^n n!^2 / (2n)! \\
    f(2n+1) & = (2n+1)! / n!^2 / 4^n
    Try it.

    Cheers -- sylas
    Last edited: Jan 12, 2010
  4. Jan 12, 2010 #3

    Wow! thank you.

    Could you maybe give me a hint as to how you found that?
  5. Jan 13, 2010 #4


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    Sure. Basically, I noted that the form of your fractions is as follows.
    To make life easy for myself, I started by looking only at even values of n. What I actually did was consider cases for 2n, so the equation becomes
    This looks a lot like a factorial, so I immediate thought about dividing everything by 2. Once I have a formula for the top line, I can see I will be able to get the bottom line by dividing (2n)! by the top line, so I just focus on the top line. There are n terms being multiplied, so the top line is
    [tex]2^n n(n-1)(n-2)(n-3)...1 = 2^n n![/tex]​
    The bottom line is therefore
    [tex]\frac{(2n)!}{2^n n!}[/tex]​
    So we divide these two, and obtain:
    [tex]\frac{(2^n n!)^2}{(2n)!} = \frac{4^n n!^2}{(2n)!}[/tex]​
    Given the way you numbered equations, this would actually be term number 2n+1. So I simply started numbering from 0. I'm also a pure mathematician, and for programming I like C better than Fortran... so I usually start counting from zero anyway. It often simplifies a problem like this.

    Finding the equation for the odd terms was a breeze, using f(n)*f(n-1) = n, which you was your defining relation (adjusted to start counting at zero).

    Cheers -- sylas
    Last edited: Jan 13, 2010
  6. Jan 13, 2010 #5
    Thanks, I read over it and then went through it myself and got the same thing :approve:
    Last edited: Jan 13, 2010
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