# Thread: Lower integer value

1. ## 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)]

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

2. ## 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.

3. ## Re: Lower integer value

Direct computation also works for odd no induction is needed

4. ## Re: Lower integer value

Originally Posted by hedi
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

right