Subfield: Combinatorics


Question: The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.

Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.

PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.

![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473)

As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power 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 and prove a closed formula (that is, a formula with a fixed number of terms and no "...") for $\operatorname{Le}(n, k)$ in terms of $n, k$, and the $\mathrm{Pa}$ function.

Solution:
The formula is $\operatorname{Le}(n, k)=\frac{1}{(n+1) \cdot \operatorname{Pa}(n, k)}$, or equivalently, $\frac{1}{(k+1) \cdot \operatorname{Pa}(n+1, k+1)}$. Because $\operatorname{Pa}(n, 0)=$ $\operatorname{Pa}(n, n)=1$, when $k=0$ or $k=n$, the formula is equivalent to the definition of $\operatorname{Le}(n, 0)=$ $\operatorname{Le}(n, n)=\frac{1}{n+1}$.

To prove the formula for $1 \leq k \leq n-1$, use induction on $k$. The base case $k=0$ was proved above. If the formula holds for a particular value of $k<n$, then it can be extended to the case $k+1$ using two identities:



$$
\begin{aligned}
\operatorname{Le}(n+1, k+1) & =\operatorname{Le}(n, k)-\operatorname{Le}(n+1, k) \text { and } \\
(n+1) \operatorname{Pa}(n, k) & =\frac{(n+1) !}{k !(n-k) !}
\end{aligned}
$$

Using the first identity and the inductive hypothesis yields

$$
\begin{aligned}
\operatorname{Le}(n+1, k+1) & =\operatorname{Le}(n, k)-\operatorname{Le}(n+1, k) \\
& =\frac{1}{(n+1) \cdot \operatorname{Pa}(n, k)}-\frac{1}{(n+2) \cdot \operatorname{Pa}(n+1, k)}
\end{aligned}
$$

Applying the second identity to the right side of the equation yields:

$$
\begin{aligned}
\operatorname{Le}(n+1, k+1) & =\frac{k !(n-k) !}{(n+1) !}-\frac{k !(n+1-k) !}{(n+2) !} \\
& =\frac{k !(n-k) !}{(n+2) !}[(n+2)-(n+1-k)] \\
& =\frac{k !(n-k) !}{(n+2) !}(k+1) \\
& =\frac{(k+1) !((n+1)-(k+1)) !}{(n+1) !(n+2)} \\
& =\frac{1}{(n+2) \operatorname{Pa}(n+1, k+1)}
\end{aligned}
$$

Final answer: The formula is $\operatorname{Le}(n, k)=\frac{1}{(n+1) \cdot \operatorname{Pa}(n, k)}$, or equivalently, $\frac{1}{(k+1) \cdot \operatorname{Pa}(n+1, k+1)}$. Because $\operatorname{Pa}(n, 0)=$ $\operatorname{Pa}(n, n)=1$, when $k=0$ or $k=n$, the formula is equivalent to the definition of $\operatorname{Le}(n, 0)=$ $\operatorname{Le}(n, n)=\frac{1}{n+1}$.

To prove the formula for $1 \leq k \leq n-1$, use induction on $k$. The base case $k=0$ was proved above. If the formula holds for a particular value of $k<n$, then it can be extended to the case $k+1$ using two identities:



$$
\begin{aligned}
\operatorname{Le}(n+1, k+1) & =\operatorname{Le}(n, k)-\operatorname{Le}(n+1, k) \text { and } \\
(n+1) \operatorname{Pa}(n, k) & =\frac{(n+1) !}{k !(n-k) !}
\end{aligned}
$$

Using the first identity and the inductive hypothesis yields

$$
\begin{aligned}
\operatorname{Le}(n+1, k+1) & =\operatorname{Le}(n, k)-\operatorname{Le}(n+1, k) \\
& =\frac{1}{(n+1) \cdot \operatorname{Pa}(n, k)}-\frac{1}{(n+2) \cdot \operatorname{Pa}(n+1, k)}
\end{aligned}
$$

Applying the second identity to the right side of the equation yields:

$$
\begin{aligned}
\operatorname{Le}(n+1, k+1) & =\frac{k !(n-k) !}{(n+1) !}-\frac{k !(n+1-k) !}{(n+2) !} \\
& =\frac{k !(n-k) !}{(n+2) !}[(n+2)-(n+1-k)] \\
& =\frac{k !(n-k) !}{(n+2) !}(k+1) \\
& =\frac{(k+1) !((n+1)-(k+1)) !}{(n+1) !(n+2)} \\
& =\frac{1}{(n+2) \operatorname{Pa}(n+1, k+1)}
\end{aligned}
$$