Spring 2016, problem 13

Consider the binomial coefficients $\binom{n}{k}=\frac{n!}{k!(n-k)!}$, where $k\in\{1,2,\ldots, n-1\}$. Determine all positive integers $n$ for which $\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are all even numbers.

Comments

2 years ago

Hi, I'm Hubert from Paris ... Luca's theorem shows that $n=2^q$;

using the Wikipedia article notations with $p=2$ we want $N=\binom mn$ even for $1\leq n\leq m-1$,

  • if $m$ is not a power of $2$, then $m_k=m_j=1$ for some $j\lt k$ and the given formula says $\binom m{2^j} \equiv \binom 11 (mod\; 2)$ with $n_j=1,\; n_k=0, \; k\neq j$, odd;

  • now if $m=2^k,\; m_j=0,\; j\lt k$ and $n\lt m\Rightarrow~n_k=0$, and at least one of the $n_0,n_1,\cdots n_{k-1}$ is one, putting a $0=\binom 01$ in the given product, for each $n\in(0,m)$, making each $N$ even.

Note: this strange $\lt$ is the symbol "less than", can someone fix that ?

I need your help now ... :-(

The browser interprets the < as the start of an html tag or something like that. Use \lt and \leq inside $ signs instead.

Nelix 2 years ago
2 years ago

The number $n$ has the required property, if and only if $n = 2^k$ for some positive integer $k$.

I suspect, there's a nice proof for that, involving the binomial theorem and a little group theory, but all I could come up was this rather calculational Proof:

[The notation $\lfloor x \rfloor := floor(x)$ and $\lceil x \rceil := ceil(x)$ is used throughout.]

For every natural $a$, define $\nu(a) := b$, where $b$ is the largest integer, such that $2^b$ divides $a$. Or put another way: $\nu(a)$ is the numbert of twos in the prime factorization of $a$.

A particular case of a result known as Legendre's formula states, that for all $a \in \mathbb{N}$:

$\nu(a!) = \sum_{i=1}^{\infty} \lfloor \frac{a}{2^i} \rfloor$

Also, from the definition of the binomial coefficients, we have:

$\nu ( \binom{n}{k} ) = \nu(n!) -\nu(k!) - \nu([n -k]!)$ (*)

Now assume, that $n$ is not a power of 2. So $n$ lies strictly between two powers $2^j$ and $2^{j + 1}$ of two, i.e. there are $j,r \in \mathbb{N}$, with $0 \lt r \lt 2^j$, such that $n = 2^j + r$.

Then $\nu(n!) = \nu([2^j + r]!) = \sum_{i=1}^{\infty} \lfloor \frac{2^j + r}{2^i} \rfloor = \sum_{i=1}^{j} 2^{j-i }+\lfloor \frac{r}{2^i} \rfloor = \nu([2^j]!) + \nu(r!)$. Note, that the second manipulation only works, because we assume, that $r \lt 2^j$. Plugging this into the equation (*), we get:

$\nu ( \binom{n}{2^j} ) = \nu ( \binom{2^j + r}{2^j} ) = \nu([2^j]!) + \nu(r!) - \nu([2^j]!) - \nu(r!) = 0$

Therefore $\binom{n}{2^j}$ is odd so that $n$ does not have the property required in the problem statement.

This concludes the first half of the proof. We now assume, that $n = 2^j$ for some $j \in \mathbb{N}$ and prove it does indeed have said property.

Pick a $k \in \mathbb{N}$, such that $0 \lt k \lt 2^j$. Then:

$\nu ( \binom{n}{k} ) = \nu ( \binom{2^j}{k} ) = \nu([2^j]!) - \nu([2^j - k]!) - \nu(k!) = \sum_{i=1}^{j} \lceil \frac{k}{2^i} \rceil - \lfloor \frac{k}{2^i} \rfloor$. By the definition of the rounding functions, none of the summands is negative and - from $0 \lt k \lt 2^j$ - the last one is equal to 1. Therefore $\nu ( \binom{n}{k} ) >= 1$ and so $\binom{n}{k}$ is even. This completes the proof.

As an aside, using the general form of Legendre's formula, the proof takes little modification, to show that:

A prime number $p$ divides all of $\binom{n}{1},\binom{n}{2}, \ldots, \binom{n}{n-1}$ iff n is a power of $p$.

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