Odd and (almost) even rows in the Pascal's triangle

Question

I came across the following proposition:

$\binom{n}{k}$ is even for all $1 ≤ k ≤ n-1$ iff $n=2^m$ for some $m \in \mathbb N$

$\binom{n}{k}$ is odd for all $0 ≤ k ≤ n$ iff $n=2^m-1$ for some $m \in \mathbb N_0$

I guess one could prove it using the Lucas' theorem or the neat fact that

$$ \mbox{(#Odd entries in } n^{th} \mbox{ row of Pascal's triangle)} = 2^{\mbox{(#1's in binary representation of $n$)}} $$

Yet it seems to be an overkill in this case (and the book I use doesn't introduce the theorem). I wonder if there is a way to prove the proposition at hand directly.

Any help is appreciated.

Answer 1

It's easy to prove that $(1+x)^{2^m} = 1+x^{2^m}$ in $\mathbb Z_2[x]$ for all $m$ by induction.

Now you need to prove it is not true for $n$ not a power of $2$.

If $n=2^m q$ where $q>1$ is odd and $m\geq 0$ then $(1+x)^{2^mq}= (1+x^{2^m})^q = 1+qx^{2^m} +\dots$. Since $q$ is odd, that means that $\binom{n}{2^m}$ is odd.

This result generalizes to all primes $p$. If $\binom n k$ is divisible by $p$ for $1\leq k\leq n-1$ the $n=p^m$ for some $m$. Again you can prove by induction that $(1+x)^{p^m}=1+x^{p^m}$ in $\mathbb Z_p[x]$, again you prove that $\binom{p^mq}{p^m}\equiv q\pmod p$.

Answer 2

HINT: Note first that if $\binom{2^m-1}k$ is odd for $0\le k\le 2^m-1$, then automatically $\binom{2^m}k$ is even for $1\le k\le 2^m-1$, since $\binom{2^m}k=\binom{2^m-1}{k-1}+\binom{2^m-1}k$.

Now use the generalized Pascal’s identity,

$$\binom{n}k=\sum_{i=0}^r\binom{r}i\binom{n-r}{k-i}$$

with $n=2^{m+1}-1$ and $r=2^m$:

$$\binom{2^{m+1}-1}k=\sum_{i=0}^{2^m}\binom{2^m}i\binom{2^m-1}{k-i}\;.$$

From your induction hypothesis you know the parity of each binomial coefficient in summation on the righthand side.

(If you’ve not seen the generalized Pascal’s identity before, it can easily be proved by induction on $r$.)

Answer 3

Actually, you can see in the triangle itself that there is a diagonal of odd numbers starting with the 1 in row $2^k$ and reaching row $2^{k+1} -1.$ So every row in between has such an odd entry, this being by simple propagation. Starting with row 4, diagonal odd entries: 1,5,15,35. Or, starting with row 8: 1,9,45, 165,495,1287,3003,6435.

On the same note, there is a diagonal of even numbers from row $2^k$ to $2^{k+1}-2.$ From row 4: 4,10,20. From row 8: 8,36,120,3309,792,1716,3432.

I guess you need to work out what happens at row $2^k$ and row $2^k - 1$ on your own. Life is a never ending quest for renewal. However, you can certainly do it by Legendre's theorem for $\mbox{ord}_2 (n!)$ using a sum of floor functions, in the case that $n = 2^k.$ Then the fact that row $2^k -1$ are all odd is just induction.

EDIT: I see, @Thomas gives a very simple answer for row $2^m,$ so there you go.

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