Number of integer triplets $(a,b,c)$ such that $aWhat is an equivalent combinatorial presentation for the problem? Can I use the stars and bars approach to find the number of integral solutions of $a+b+c=n$ where $a,b,c\geq 0$?
If in addition $a+b>c$, $b+c>a$, $a+c>b$ hold, then the problem can be seen as $a,b,c$ being the sides of a triangle with perimeter $n$. I would like a hint on how to do that as well.
 A: For the first part or your question:
Write $b=a+1+i$ and $c=(a+1+i)+1+j=a+2+i+j$ where $i,j\ge0$.  Then $$a+b+c=a+(a+1+i)+(a+2+i+j)=3a+2i+j+3,$$ and you can restate the question as the number of solutions to $$3a+2i+j=n-3$$ where $a,i,j\ge0$. This is the coefficient of $x^{n-3}$ in  $$\underbrace{(1+x^3+x^6+\cdots)}_{\text{contribution of } a}\cdot\underbrace{(1+x^2+x^4+\cdots)}_{\text{contribution of } i}\cdot\underbrace{(1+x^1+x^2+\cdots)}_{\text{contribution of } j}=\frac1{(1-x^3)(1-x^2)(1-x)},$$ or the coefficient of $x^n$ in $$\frac{x^3}{(1-x^3)(1-x^2)(1-x)}.$$
A: Let $Q(n,3)$ be the number of partitions of $n$ into unequal parts and $P(n,3)$ the number of partitions of $n$ for any $n$. If $a < b < c$ and $a+b+c = n$, then $a \leq b-1 \leq c-2$ and $(a,b-1, c-2)$ is a partition of $n-3$. Thus we need to calculate $P(n-3,3)$. For a $p-$partition of $n$, we have the recurrence ralation
\begin{equation*}
P(n,p) = P(n-p,p) + P(n-1, p-1)
\end{equation*}
Hence we have $P(n-3, 3) = P(n-6, 3) + P(n-4, 2)$. Noting that $P(k,2) = \left[\frac{k}{2}\right]$, we can calculate $P(n-3,3)$ recursively.
