Explicit Bijection between Triangles and Partitions Let $t(n)$ denote the number of nondegenerate triangles with integer side lengths and perimeter $n$ (where two triangles are considered the same if they have the same side lengths).
With some case work, it can be shown that
$$\sum_{n\ge 0} t(n)x^n = \frac{x^3}{(1-x^2)(1-x^3)(1-x^4)}.$$
However, is there a more direct proof of the above identity?
That is, is there an example of an explicit bijection between the set of the nondegenerate triangles with integer side lengths of perimeter $n$ and the set of partitions of $(n-3)$ with parts taken from $\{2,3,4\}$ ?
 A: Suppose that $n-3=2a+3b+4c$, where $a,b$, and $c$ are non-negative integers. We may think of this as a partition of $n-3$ into parts taken from $\{2,3,4\}$. If $p_1=p_2=a+b+c$, $p_3=b+c$, and $p_4=c$, the conjugate partition is $n-3=p_1+p_2+p_3+p_4$, with the proviso that $p_3$ and $p_4$ can be $0$.
Let 
$$\begin{align*}
r&=p_1+p_4+1=a+b+2c+1\;,\\
s&=p_2+1=a+b+c+1\;,\text{ and}\\
t&=p_3+1=b+c+1\;;
\end{align*}$$
clearly $r+s+t=n$, and $r\ge s\ge t\ge 1$. Moreover, $s+t-r=b+1>0$, so there is a non-degenerate triangle with sides $r,s$, and $t$ and perimeter $n$.
Now suppose that $r,s$, and $t$ are the sides of a non-degenerate triangle with perimeter $n$, where $r\ge s\ge t$. Let
$$\begin{align*}
p_1&=s-1\;,\\
p_2&=s-1\;,\\
p_3&=t-1\;,\text{ and}\\
p_4&=r-s\;.
\end{align*}$$
Clearly $p_1+p_2+p_3+p_4=r+s+t-3=n-3$, and $p_1=p_2\ge p_3$. Moreover, if $p_3<p_4$, then $t+s<r+1$ and hence $t+s\le r$, contrary to the assumption that the triangle is non-degenerate. Thus, $p_1\ge p_2\ge p_3\ge p_4$. Finally, let
$$\begin{align*}
c&=p_4\;,\\
b&=p_3-p_4\;,\text{ and}\\
a&=p_2-p_3\;;
\end{align*}$$
then
$$2a+3b+4c=2p_2+p_3+p_4=p_1+p_2+p_3+p_4=n-3\;.$$
This establishes the desired bijection.
By the way, this is OEIS A005044, with several interesting references; I especially liked this PDF and this one.
A: After some thinking, I've found a more visual correspondence between triangles with integer side lengths and the relevant type of partitions. This should be equivalent to Brian's answer, but I haven't checked.
The idea is as follows. Suppose we have a triangle with side lengths $a\ge b\ge c$ so that the triangle has perimeter $a+ b + c = n$. Draw three rows of dots such that the first row has $a$ dots, the second row has $b$ dots, and the third row has $c$ dots.
Since $a,b,c > 0,$ the first column in the resulting image will have three dots. Cut these dots off from the image.
If $a > b,$ the final few columns of the pictures will each only contain one dot. Move these singleton dots to columns with three dots, leading to columns with four dots. This is always possible since 
$$a - b \le c-1$$
by the triangle inequality.
Now read off the number of dots in each column to get a partition of $(n-3)$ with parts taken from the set $\{2,3,4\}.$
For example, below is an image showing how to start off with a triangle of perimeter $16$ with side lengths $7, 5,$ and $4,$ and then follow the above steps to get the partition $4 + 4 + 3 + 2$ of $13.$ 

To show that this procedure actually yields a bijection takes a bit more work, but since these steps are easy to reverse it is not too hard to show that it works out and that the triangle inequality ends up holding.
