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.

Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is:

$$
\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases}
$$

The first four rows of Clark's Triangle are given below.

![](https://cdn.mathpix.com/cropped/2023_12_21_c41919b703ea0b966244g-1.jpg?height=412&width=1152&top_left_y=1602&top_left_x=476)
Find and justify a formula for $\mathrm{Cl}(n, 3)$ in terms of $n$.

Solution:
Notice that $\mathrm{Cl}(3,3)=1=1^{3}$, and then for $n>3, \mathrm{Cl}(n, 3)=\mathrm{Cl}(n-1,2)+\mathrm{Cl}(n-1,3)$; replacing $\mathrm{Cl}(n-1,3)$ analogously on the right side yields the summation

$$
\mathrm{Cl}(n, 3)=\mathrm{Cl}(n-1,2)+\mathrm{Cl}(n-2,2)+\ldots+\mathrm{Cl}(3,2)+1
$$

Now we prove that $\mathrm{Cl}(n, 2)=(n-1)^{3}$. Use induction on $n$. First, rewrite $\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1=n^{3}-(n-1)^{3}$, and notice that $\mathrm{Cl}(2,2)=1=(2-1)^{3}$. Then if $\mathrm{Cl}(n, 2)=(n-1)^{3}$, using the recursive definition, $\mathrm{Cl}(n+1,2)=\mathrm{Cl}(n, 2)+\mathrm{Cl}(n, 1)=(n-1)^{3}+\left(n^{3}-(n-1)^{3}\right)=n^{3}$.

By the formula proved above, \mathrm{Cl}(n-1,2)=(n-2)^{3}$, so this formula is equivalent to

$$
\mathrm{Cl}(n, 3)=(n-2)^{3}+(n-3)^{3}+\ldots+2^{3}+1
$$

Use the identity $1^{3}+2^{3}+\cdots+m^{3}=\frac{m^{2}(m+1)^{2}}{4}$ and substitute $n-2$ for $m$ to obtain $\mathrm{Cl}(n, 3)=$ $\frac{(n-2)^{2}(n-1)^{2}}{4}$.

Final answer: Notice that $\mathrm{Cl}(3,3)=1=1^{3}$, and then for $n>3, \mathrm{Cl}(n, 3)=\mathrm{Cl}(n-1,2)+\mathrm{Cl}(n-1,3)$; replacing $\mathrm{Cl}(n-1,3)$ analogously on the right side yields the summation

$$
\mathrm{Cl}(n, 3)=\mathrm{Cl}(n-1,2)+\mathrm{Cl}(n-2,2)+\ldots+\mathrm{Cl}(3,2)+1
$$

Now we prove that $\mathrm{Cl}(n, 2)=(n-1)^{3}$. Use induction on $n$. First, rewrite $\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1=n^{3}-(n-1)^{3}$, and notice that $\mathrm{Cl}(2,2)=1=(2-1)^{3}$. Then if $\mathrm{Cl}(n, 2)=(n-1)^{3}$, using the recursive definition, $\mathrm{Cl}(n+1,2)=\mathrm{Cl}(n, 2)+\mathrm{Cl}(n, 1)=(n-1)^{3}+\left(n^{3}-(n-1)^{3}\right)=n^{3}$.

By the formula proved above, \mathrm{Cl}(n-1,2)=(n-2)^{3}$, so this formula is equivalent to

$$
\mathrm{Cl}(n, 3)=(n-2)^{3}+(n-3)^{3}+\ldots+2^{3}+1
$$

Use the identity $1^{3}+2^{3}+\cdots+m^{3}=\frac{m^{2}(m+1)^{2}}{4}$ and substitute $n-2$ for $m$ to obtain $\mathrm{Cl}(n, 3)=$ $\frac{(n-2)^{2}(n-1)^{2}}{4}$.