How to find out the sum involving floor function [closed]

I came up with an equation while solving a question. The question is - suppose we have n numbers, from 1 up to n. How many groups of 3 numbers (repetition allowed) can be formed whose sum will be n.

The equation I came up with is something like this:

$\sum_{i=1}^{\left \lfloor \frac{n}{3} \right \rfloor} \left \lfloor \frac{n-i}{2} \right \rfloor-i+1$

This is giving correct result I hope. For example, for n=7, we can form 4 groups, (1,1,5), (1,2,4), (1,3,3), (2,2,3). My question is, how to get a more compact form of this equation? I am stuck since I don't find a way of finding the sum involving floor function.

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closed as off topic by Ryan Budney, Pietro Majer, Alain Valette, Loop Space, Gerry MyersonNov 17 '11 at 11:22

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This would be more appropriate for math.stackexchange.com. – Ryan Budney Nov 17 '11 at 8:19
Actually, it has been done to death at m.se, a little searching there should turn up the tools to solve the problem. – Gerry Myerson Nov 17 '11 at 11:24

Your sum is related to OEIS A001399 Number of partitions of n into at most 3 parts which is close to your formula a(n)=sum{k=0..floor(n/3), floor((n-3k+2)/2)}
a(n)=Nearest integer to $\frac{n^2}{12}$
actually the OP wants exactly 3 parts, so the OEIS' sequence has to be shifted by 3, and the answer should be the closest integer to $n^2/12$ – Pietro Majer Nov 17 '11 at 8:35
Also note that since the gf of $a(n)$ is $\frac{x^3}{(1−x)(1−x^2)(1−x^3)}$, with poles at sixth roots of unity, which are all simple but $x\neq 1$, $a(n)−n^2/12$ has periodicity $6$. – Pietro Majer Nov 17 '11 at 13:56