bijection between number of non-congruent triangles with perimeter 2n and integer sides and number of partitions of n into exactly three terms. Is there any a bijection between the following problems?
Let $f(n)$ be the number of non-congruent triangles with perimeter 2n
and integer sides. Let $h(n)$ be the number of partitions of n
into exactly three terms. 
I thought about it a lot, but I got nothing. I checked that for n=3,4,5 and 6 and it works.
Any help would be appreciated.

The number of non congruent triangles is given here ... https://oeis.org/A005044
The first few values are $0, 0, 0, \color{red}{1}, 0, \color{red}{1}, 1, \color{red}{2}, 1, \color{red}{3}, 2, \color{red}{4}, 3, \color{red}{5}, 4,\color{red}{7}, 5, \color{red}{8}, 7, \color{red}{10}, 8, \color{red}{12}, 10, \color{red}{14},\cdots$
The number of partitions into three parts is given here ...https://oeis.org/A001399
The first few values are $1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14, \cdots$
The generating function for partitions into three parts is given by 
\begin{eqnarray*}
\frac{x^3}{(1-x)(1-x^2)(1-x^3)}
\end{eqnarray*}
The generating function for the number of non congruent triangles is given
\begin{eqnarray*}
\frac{x^3}{(1-x^2)(1-x^3)(1-x^4)}= \frac{x^3+x^6}{(1-x^2)(1-x^6)(1-x^4)} = \\\frac{x^3}{(1-x^2)(1-x^4)(1-x^6)}+\color{red}{\frac{x^6}{(1-x^2)(1-x^4)(1-x^6)}}
\end{eqnarray*}
but we only want every second term in this, which is shown in red above ... & so clearly enumerates to the same as the number of partitions of $n$ into three parts.
