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Thread: Lower integer value

  1. #1
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    Lower integer value

    Hi,
    Let g(n) denote the lower integer value (Liv) of (n^2)/4 for n>=4.Prove the following recursion:

    g(n)=Liv[n*g(n-1)/(n-2)]
    Thank's in advance

    I tried proving by induction with no much success.
    I will be happy to get help.

    https://en.wikipedia.org/wiki/Floor_...ling_functions

    the floor here
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  2. #2
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    Re: Lower integer value

    Case 1: $n = 2k$ for some positive integer $k\ge 2$.
    Then $g(n) = \text{Liv}\left(\dfrac{(2k)^2}{4}\right) = k^2$.
    Test: $g(n) \stackrel{?}{=} \text{Liv}\left(\dfrac{n g(n-1)}{n-2} \right)$
    $g(n-1) = \text{Liv}\left(\dfrac{(2k-1)^2}{4}\right) = \text{Liv}\left(\dfrac{4k^2-4k+1}{4} \right) = \text{Liv}\left(k^2-k+\dfrac{1}{4}\right) = k(k-1)$
    so
    $$\text{Liv}\left(\dfrac{n g(n-1)}{n-2} \right) = \text{Liv}\left( \dfrac{2k\cdot k(k-1)}{2k-2} \right) = k^2$$

    Try for the case when $n = 2k+1$ for some integer $k\ge 2$.

    Then try the base cases for $k=2$. Then turn it into an induction argument.
    Last edited by SlipEternal; Oct 29th 2018 at 06:07 AM.
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  3. #3
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    Re: Lower integer value

    Direct computation also works for odd no induction is needed
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  4. #4
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    Re: Lower integer value

    Quote Originally Posted by hedi View Post
    Direct computation also works for odd no induction is needed
    There is a much simpler recursion formula:

    $\displaystyle g(2) = 1,\quad g(n + 1) = g(n) + \left\lceil {\dfrac{n}{2}} \right\rceil $ SEE HERE
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  5. #5
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    Re: Lower integer value

    right
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