Subfield: Combinatorics


Question: Leibniz's Harmonic Triangle: Consider the triangle formed by the rule

$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$

This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.

![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470)

For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
Find three distinct sets of positive integers $\{a, b, c, d\}$ with $a<b<c<d$ such that

$$
\frac{1}{3}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}
$$

Solution:
Note that $\frac{1}{3}=\operatorname{Le}(2,0)=\operatorname{Le}(3,0)+\operatorname{Le}(3,1)$. Also $\operatorname{Le}(3,0)=\operatorname{Le}(4,0)+\operatorname{Le}(4,1)$ and $\operatorname{Le}(4,0)=$ $\operatorname{Le}(5,0)+\operatorname{Le}(5,1)$. So

$$
\begin{aligned}
\frac{1}{3} & =\operatorname{Le}(5,0)+\operatorname{Le}(5,1)+\operatorname{Le}(4,1)+\operatorname{Le}(3,1) \\
& =\frac{1}{6}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}
\end{aligned}
$$

Therefore $a=6, b=12, c=20, d=30$ is a solution.

On the other hand, a similar analysis yields

$$
\begin{aligned}
\frac{1}{3} & =\operatorname{Le}(4,0)+\operatorname{Le}(5,1)+\operatorname{Le}(5,2)+\operatorname{Le}(3,1) \\
& =\frac{1}{5}+\frac{1}{30}+\frac{1}{60}+\frac{1}{12}
\end{aligned}
$$

so $a=5, b=12, c=30, d=60$ is another such solution.

Finally, notice that $\operatorname{Le}(3,1)=\frac{1}{12}=\operatorname{Le}(11,0)$, so that $\frac{1}{12}$ can be rewritten as $\operatorname{Le}(12,0)+$ $\operatorname{Le}(12,1)=\frac{1}{13}+\frac{1}{156}$. Then

$$
\begin{aligned}
\frac{1}{3} & =\operatorname{Le}(4,0)+\operatorname{Le}(4,1)+\operatorname{Le}(3,1) \\
& =\operatorname{Le}(4,0)+\operatorname{Le}(4,1)+\operatorname{Le}(12,0)+\operatorname{Le}(12,1) \\
& =\frac{1}{5}+\frac{1}{20}+\frac{1}{13}+\frac{1}{156},
\end{aligned}
$$

yields $a=5, b=13, c=20, d=156$.

In general, once a triple $a, b, c$ has been found, rewriting (for example) $\frac{1}{a}=\operatorname{Le}(a-1,0)=$ $\operatorname{Le}(a, 0)+\operatorname{Le}(a, 1)=\frac{1}{a+1}+\frac{1}{a(a+1)}$ creates an appropriate quadruple, although on occasion these values duplicate another value in the quadruple.

Final answer: Note that $\frac{1}{3}=\operatorname{Le}(2,0)=\operatorname{Le}(3,0)+\operatorname{Le}(3,1)$. Also $\operatorname{Le}(3,0)=\operatorname{Le}(4,0)+\operatorname{Le}(4,1)$ and $\operatorname{Le}(4,0)=$ $\operatorname{Le}(5,0)+\operatorname{Le}(5,1)$. So

$$
\begin{aligned}
\frac{1}{3} & =\operatorname{Le}(5,0)+\operatorname{Le}(5,1)+\operatorname{Le}(4,1)+\operatorname{Le}(3,1) \\
& =\frac{1}{6}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}
\end{aligned}
$$

Therefore $a=6, b=12, c=20, d=30$ is a solution.

On the other hand, a similar analysis yields

$$
\begin{aligned}
\frac{1}{3} & =\operatorname{Le}(4,0)+\operatorname{Le}(5,1)+\operatorname{Le}(5,2)+\operatorname{Le}(3,1) \\
& =\frac{1}{5}+\frac{1}{30}+\frac{1}{60}+\frac{1}{12}
\end{aligned}
$$

so $a=5, b=12, c=30, d=60$ is another such solution.

Finally, notice that $\operatorname{Le}(3,1)=\frac{1}{12}=\operatorname{Le}(11,0)$, so that $\frac{1}{12}$ can be rewritten as $\operatorname{Le}(12,0)+$ $\operatorname{Le}(12,1)=\frac{1}{13}+\frac{1}{156}$. Then

$$
\begin{aligned}
\frac{1}{3} & =\operatorname{Le}(4,0)+\operatorname{Le}(4,1)+\operatorname{Le}(3,1) \\
& =\operatorname{Le}(4,0)+\operatorname{Le}(4,1)+\operatorname{Le}(12,0)+\operatorname{Le}(12,1) \\
& =\frac{1}{5}+\frac{1}{20}+\frac{1}{13}+\frac{1}{156},
\end{aligned}
$$

yields $a=5, b=13, c=20, d=156$.

In general, once a triple $a, b, c$ has been found, rewriting (for example) $\frac{1}{a}=\operatorname{Le}(a-1,0)=$ $\operatorname{Le}(a, 0)+\operatorname{Le}(a, 1)=\frac{1}{a+1}+\frac{1}{a(a+1)}$ creates an appropriate quadruple, although on occasion these values duplicate another value in the quadruple.