Numbers in Row of Pascal's Triangle all Odd iff Row number 2^n - 1

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Then the non-zero elements of the $n$th row of Pascal's triangle are all odd if and only if:

$n = 2^m - 1$

for some $m \in \Z_{\ge 0}$.


$\begin{array}{r|rrrrrrrrrr} n & \binom n 0 & \binom n 1 & \binom n 2 & \binom n 3 & \binom n 4 & \binom n 5 & \binom n 6 & \binom n 7 & \binom n 8 & \binom n 9 & \binom n {10} & \binom n {11} & \binom n {12} \\ \hline 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 3 & 1 & 3 & 3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 4 & 1 & 4 & 6 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 5 & 1 & 5 & 10 & 10 & 5 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 6 & 1 & 6 & 15 & 20 & 15 & 6 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 7 & 1 & 7 & 21 & 35 & 35 & 21 & 7 & 1 & 0 & 0 & 0 & 0 & 0 \\ 8 & 1 & 8 & 28 & 56 & 70 & 56 & 28 & 8 & 1 & 0 & 0 & 0 & 0 \\ 9 & 1 & 9 & 36 & 84 & 126 & 126 & 84 & 36 & 9 & 1 & 0 & 0 & 0 \\ 10 & 1 & 10 & 45 & 120 & 210 & 252 & 210 & 120 & 45 & 10 & 1 & 0 & 0 \\ 11 & 1 & 11 & 55 & 165 & 330 & 462 & 462 & 330 & 165 & 55 & 11 & 1 & 0 \\ 12 & 1 & 12 & 66 & 220 & 495 & 792 & 924 & 792 & 495 & 220 & 66 & 12 & 1 \\ \end{array}$


As can be seen, the entries in rows $0, 1, 3, 7$ are all odd.


Proof

The statement:

$\dbinom n k$ is odd

is equivalent to:

$\dbinom n k \equiv 1 \pmod p$


The corollary to Lucas' Theorem gives:

$\displaystyle \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$

where:

$n, k \in \Z_{\ge 0}$ and $p$ is prime
the representations of $n$ and $k$ to the base $p$ are given by:
$n = a_r p^r + \cdots + a_1 p + a_0$
$k = b_r p^r + \cdots + b_1 p + b_0$


When $p = 2$, the digits $a_j, b_j$ are either $0$ or $1$.

In order for $\dbinom n k \equiv 1 \pmod p$, it is necessary and sufficient that $\dbinom {a_j} {b_j} \equiv 1 \pmod p$ for all $j \in \left\{ {0, 1, 2, \ldots, r}\right\}$.

In order for this to happen, $\dbinom {a_j} {b_j} = \dbinom 0 0$ or $\dbinom {a_j} {b_j} = \dbinom 1 0$ or $\dbinom {a_j} {b_j} = \dbinom 1 1$.

Suppose $a_i = 0$ for some $i \in \left\{ {0, 1, 2, \ldots, r}\right\}$.

Then if $b_i = 1$:

$\dbinom {a_j} {b_j} = \dbinom 0 1 = 0$

and so:

$\dbinom n k \equiv 0 \pmod p$

for whichever $k$ (and there will be at least one) has digit $b_i = 1$.


So the only way it can be assured that all $\dbinom {a_j} {b_j} \equiv 1 \pmod p$ for all $k \in \left\{ {0, 1, 2, \ldots, n}\right\}$ is for $a_j = 0$ for all $j \in \left\{ {0, 1, 2, \ldots, r}\right\}$.

That is, for $n = 2^m - 1$ for some $m \in \Z_{\ge 0}$.

$\blacksquare$


Sources