give a closed form to the sum $ \sum_{(a, b, c) \in D \cap \mathbb{N}^3} \sum_{k=0}^{\min(a,b,c)} \binom{n}{3k-3\min(a,b,c)+a+b+c} $ One of the questions that I have seen so far was asking to give a closed form to the finite sum
$$ \displaystyle\sum_{(a, b, c) \in D \cap \mathbb{N}^3} 
\sum_{k=0}^{\min(a,b,c)} \binom{n}{3k-3\min(a,b,c)+a+b+c} $$
where n is a strictly positive integer and $D$ represents the full triangle in $\mathbb{R}^3$ of the edges $(n, 0, 0);(0, n, 0)$ and $(0, 0, n)$.
I tried to solve this problem by using the trick of change of variables in order to simplify the given formula but without getting a nice results, do you have any ideas?
 A: The full triangle $D$ is the portion of the plane passing by $(n,0,0),(0,n,0),(0,0,n)$ contained in the first octant.
The plane equation is $x+y+z=n$, therefore every $(a,b,c) \in D \cap \mathbb{N}^3$, satisfies $a+b+c=n$ and $a,b,c$ are non-negative integers.
Hence we can rewrite the sum as:
$$ \sum_{(a, b, c) \in D \cap \mathbb{N}^3} 
\sum_{k=0}^{\min(a,b,c)} \binom{n}{3k-3\min(a,b,c)+n} \\= \sum_{(a, b, c) \in D \cap \mathbb{N}^3} 
\sum_{k=0}^{\min(a,b,c)} \binom{n}{3(\min(a,b,c)-k)} \\= \sum_{(a, b, c) \in D \cap \mathbb{N}^3} 
\sum_{k=0}^{\min(a,b,c)} \binom{n}{3k} $$
Then we need to count how many triples are there such that $\min(a,b,c) \ge k$.
Let $a'=a-k \ge 0$, $b'=b-k \ge 0$, $c'=c-k \ge 0$; we can rewrite $a+b+c=n$ as $a'+b'+c'=n-3k$.
From Stars and Bars method, the number of non-negative integer solutions of $a'+b'+c'=n-3k$, for $n \ge 3k$, is ${n-3k+3-1 \choose 3-1}={n-3k+2 \choose 2}$.
We have also that $\min(a,b,c) \le \lfloor \frac{n}{3} \rfloor$.
Therefore, changing the order of the double sum, we can rewrite it as:
$$\sum_{(a, b, c) \in D \cap \mathbb{N}^3} 
\sum_{k=0}^{\min(a,b,c)} \binom{n}{3k} = \sum_{k=0}^{\lfloor \frac{n}{3} \rfloor} \sum_{(a, b, c) \in D \cap \mathbb{N}^3, \\ \min(a,b,c) \ge k} 
\binom{n}{3k} =
\sum_{k=0}^{\lfloor \frac{n}{3} \rfloor} {n-3k+2 \choose 2}{n \choose 3k}$$
and I don't think it can be furtherly simplified.
All the above assumes that $0 \in \mathbb{N}$, otherwise it must be slightly modified for the first addendum of the last formula.
