bijection between number of non-congruent triangles with perimeter 2n and integer sides and number of partitions of n into exactly three terms.

Question

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.

Answer 1

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.

Answer 2

Triangle with sides $x, y, z \in (0, +\infty), x\le y \le z$ can be constructed iff $z<x+y$. Let $n\ge 3$ be fixed. Take any $(a, b, c)\in \mathbb{N}^3:1\le a \le b \le c,\ a+b+c=n$. $$2a+2b+2c=(a+b)+(a+c)+(b+c)=2n$$ Define $k:=a+b, l:=a+c, m:=b+c$. Note, that $1\le k \le l \le m < k+l$, so there exist triangle with sides $k, l, m$. It could be shown, that $(a,b,c)\neq(d,e,f)$ with $d \le e \le f$ implies $(a+b\ne d+e \vee a+c\ne d+f \vee b+c\ne e+f)$ so triangles constructed on two different summations aren't congruent.

Conversely, take $(k, l, m)\in X=\{(p, q, r) \in \mathbb{N}^3:1\le p \le q \le r < p+q,\ p+q+r=2n\}$ - set of all noncongruent integer triangles represented by its sides lengths (in nondecreasing order for representation to be unique).

• assume in addition, that those numbers can be represented as $(a+b,a+c,b+c)$. Then from simple calculation $a=\frac{k+l-m}{2},b=\frac{k-l+m}{2},c=\frac{-k+l+m}{2},\ \ a+b+c=n$.
• now take any $(k',l',m')\in X$. $2|(k'+l'+m')$ implies that $2|(-k+l+m), 2|(k-l+m), 2|(k+l-m)$. Moreover $m < k+l$, so $k+l-m > 0$ and therefore numbers $a'=\frac{k'+l'-m'}{2},b'=\frac{k'-l'+m'}{2},c'=\frac{-k'+l'+m'}{2}$ are all positive integers, and $a'+b'+c'=n$.

Analogous to first part, if two triangles with sides $(k,l,m)$ and $(k',l',m')$ are non congruent, then $(a,b,c)\neq(a',b',c')$ (it is nontrivial and should be checked).

In conclusion $$(a,b,c)\mapsto (a+b,a+c,b+c)$$ is bijection indeed, with its inversion $$(k,l,m)\mapsto (\frac{k+l-m}{2},\frac{k-l+m}{2},\frac{-k+l+m}{2})$$