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.

<img_3774>

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.
Consider the parity of each entry: define

$$
\operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases}
$$

If $j$ is an integer such that $2^{j} \leq n<2^{j+1}$, and $k<2^{j}$, prove that

$$
\operatorname{PaP}(n, k)=\operatorname{PaP}\left(n-2^{j}, k\right) .
$$

Solution:
Proceed by induction on $j$. The claim is that $\mathrm{Pa}(n, k) \equiv \mathrm{Pa}\left(n-2^{j}, k-2^{j}\right)+\mathrm{Pa}\left(n-2^{j}, k\right) \bmod 2$. If $j=0$, so that



$2^{j}=1$, then this congruence is exactly the recursive definition of $\mathrm{Pa}(n, k)$ when $0<k<n$. For $k=0$ and $k=n$, the left-hand side is 1 and the right-hand side is either $0+1$ or $1+0$. For other values of $k$, all three terms are zero.

Now, assume $\mathrm{Pa}(n, k) \equiv \operatorname{Pa}\left(n-2^{j}, k-2^{j}\right)+\mathrm{Pa}\left(n-2^{j}, k\right) \bmod 2$ for some $j \geq 0$, and let $n \geq 2^{j+1}$. For any $k$, apply the inductive hypothesis three times:

$$
\begin{aligned}
\operatorname{Pa}(n, k) \equiv & \operatorname{Pa}\left(n-2^{j}, k-2^{j}\right)+\operatorname{Pa}\left(n-2^{j}, k\right) \bmod 2 \\
\equiv & \operatorname{Pa}\left(n-2^{j}-2^{j}, k-2^{j}-2^{j}\right)+\operatorname{Pa}\left(n-2^{j}-2^{j}, k-2^{j}\right) \\
& \quad+\operatorname{Pa}\left(n-2^{j}-2^{j}, k-2^{j}\right)+\operatorname{Pa}\left(n-2^{j}-2^{j}, k\right)
\end{aligned}
$$

But $u+u \equiv 0 \bmod 2$ for any integer $u$. Thus the two occurrences of $\mathrm{Pa}\left(n-2^{j}-2^{j}, k-2^{j}\right)$ cancel each other out, and

$$
\begin{aligned}
\operatorname{Pa}(n, k) & \equiv \operatorname{Pa}\left(n-2^{j}-2^{j}, k-2^{j}-2^{j}\right)+\operatorname{Pa}\left(n-2^{j}-2^{j}, k\right) \\
& \equiv \operatorname{Pa}\left(n-2^{j+1}, k-2^{j+1}\right)+\operatorname{Pa}\left(n-2^{j+1}, k\right) \bmod 2 .
\end{aligned}
$$

$\operatorname{Pa}(n, k) \equiv \operatorname{Pa}\left(n-2^{j}, k-2^{j}\right)+\operatorname{Pa}\left(n-2^{j}, k\right) \bmod 2$ when $j \geq 0$ and $n \geq 2^{j}$. If $2^{j} \leq n<2^{j+1}$ and $0 \leq k<2^{j}$, then $\mathrm{Pa}\left(n-2^{j}, k-2^{j}\right)=0$, so

$$
\mathrm{Pa}(n, k) \equiv \operatorname{Pa}\left(n-2^{j}, k-2^{j}\right)+\mathrm{Pa}\left(n-2^{j}, k\right) \equiv \operatorname{Pa}\left(n-2^{j}, k\right) \bmod 2 .
$$

Thus $\operatorname{PaP}(n, k)=\operatorname{PaP}\left(n-2^{j}, k\right)$.
Notice that

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

Examining the right hand side in the case where $n=2^{j}$ and $0<k<n$, the second factor, $\mathrm{Pa}(n-1, k-1)$ is an integer, and the first factor has an even numerator, so $\mathrm{Pa}(n, k)$ is even. Therefore $\operatorname{PaP}(n, k)=0$.

Now when $n=2^{j}$. For $2^{j}<n<2^{j+1}$, proceed by induction on $n$. Then $\operatorname{PaP}(n, k) \equiv \operatorname{PaP}(n-1, k-1)+\operatorname{PaP}(n-1, k) \bmod 2$, while $\operatorname{PaP}(n-1, k-1)=\operatorname{PaP}\left(n-1-2^{j}, k-1\right)$ and $\operatorname{PaP}(n-1, k)=\operatorname{PaP}\left(n-1-2^{j}, k\right)$. But $\operatorname{PaP}\left(n-1-2^{j}, k-1\right)+\mathrm{PaP}\left(n-1-2^{j}, k\right) \equiv \operatorname{PaP}\left(n-2^{j}, k\right) \bmod 2$, establishing the statement.