32

Is there any way of determining if $\binom{n}{k} \equiv 0\pmod{n}$. Note that I am aware of the case when $n =p$ a prime. Other than that there does not seem to be any sort of pattern (I checked up to $n=50$). Are there any known special cases where the problem becomes easier? As a place to start I was thinking of using $e_p(n!)$ defined as:

$$e_p(n!) = \sum_{k=1}^{\infty}\left \lfloor\frac{n}{p^k}\right \rfloor$$

Which counts the exponent of $p$ in $n!$ (Legendre's theorem I believe?)

Then knowing the prime factorization of $n$ perhaps we can determine if these primes appear more times in the numerator of $\binom{n}{k}$ than the denominator.

Essentially I am looking to see if this method has any traction to it and what other types of research have been done on this problem (along with any proofs of results) before. Thanks!

Martin Sleziak
  • 53,687
Patrick
  • 2,106

3 Answers3

31

Below is a picture of the situation. Red dots are the points of the first 256 rows of Pascal's triangle where $n\mid {n \choose k}$.

enter image description here

It appears that "most" values fit the bill.

One can prove the following:

Proposition: Whenever $(k, n)=1$, we have $n \mid {n\choose k}$.

This follows from the case where $n$ is a prime power (considering the various prime powers dividing $n$).

However, it happens quite often that $(k,n) \neq 1$ but $n \mid {n\choose k}$ still. For instance, $10 \mid {10\choose 4}=210$, but $(10,4) \neq 1$ (this is the smallest example). I do not think that there is a simple criterion.

In fact, it is interesting to consider separately the solutions $(n,k)$ into those which are relatively prime (which I'll call the trivial solutions) and those which are not. It appears that the non-trivial solutions are completely responsible for the Sierpinski pattern in the triangle above. Indeed, here are only the trivial solutions:

enter image description here

and here are the non-trivial solutions:

enter image description here

Let $f(n)$ be the number of $k$'s between $0$ and $n$ where $n \mid {n\choose k}$. By the proposition we have $\varphi(n)< f(n) < n$.

Question: is $$\text{lim sup } \frac{f(n)-\varphi(n)}{n}=1?$$

Here is a list plot of $\frac{f(n)-\varphi(n)}{n}$. The max value reached for $1<n<2000$ is about $0.64980544$. The blue dots at the bottom are the $n$'s such that $f(n) = \varphi(n)$.

enter image description here

Community
  • 1
Bruno Joyal
  • 54,711
  • 4
    It is nice to see Sierpinski's triangle appear. – Pedro Oct 31 '13 at 06:06
  • @PedroTamaroff Indeed! – Bruno Joyal Oct 31 '13 at 06:07
  • There is a sort of study of this type of thing; the first sequence is such $n$ that $f(n) - \phi(n)$ is a new world record. A subsequence is when $\frac{f(n) - \phi(n)}{n}$ is a new world record. These resemble Ramanujan/ Alaoglu-Erdos highly abundant and superabundant. If things go well, you may be able to identify a sequence corresponding to colossally abundant numbers as well. http://en.wikipedia.org/wiki/Colossally_abundant_number Not sure in the final case, there are requirements... – Will Jagy Oct 31 '13 at 06:53
  • Marie, there is some reason to consider high values of $$ \frac{f(n)}{\phi(n)} $$ instead, likely to factor well. – Will Jagy Oct 31 '13 at 07:43
  • 1
    Very helpful stuff! I too got excited when I saw Sierpinski's triangle. One of my favorite things in math is seeing familiar structures when exploring new problems. Really cool. – Patrick Oct 31 '13 at 11:46
  • Dear @WillJagy: I will check $f/\varphi$ and post the results. – Bruno Joyal Oct 31 '13 at 18:15
  • I'm still a little lost on why you chose the sequence: $\frac{f(n)-\phi(n)}{n}$ in your second part. Asking if this is bounded above by $1$ as $n \to \infty$ is saying that the excess of $f(n)$ over $\phi(n)$ is asymptotically equal to $n$ correct? This seems strange since both functions are $\lt n$. Can you elaborate please? – Patrick Oct 31 '13 at 18:18
  • @Patrick, it is a matter of judgement to take a number theoretic function and decide what comparisons might be revealing. That was a first attempt, i think there are other combinations that will have more explanatory power; but all interesting in some way. – Will Jagy Oct 31 '13 at 18:24
  • Fair point. Also, are you answering on behalf of Marie (aka Bruno)? Or did I make a mistake in assuming $\varphi(n) = \phi(n)$ (the totient function) and so you thought I had a question about your comment? – Patrick Oct 31 '13 at 18:35
  • @Patrick, on behalf of Bruno, and $\phi = \varphi$ – Will Jagy Oct 31 '13 at 21:39
  • I would love to know the mean asymptotic density, i.e. $\lim_{n\to\infty} \frac{2}{n^2}\sum_{k=1}^n f(k)$. It clearly has positive density (since the proportion of pairs of relatively prime integers is $\frac{6}{\pi^2}$), but is the density $1$ or does it approach some other limit? – Steven Stadnicki Nov 01 '13 at 01:24
  • 1
    Thank you for adding those additional two triangle images...really paints an even clearer picture wow. What software did you use to generate these? – Patrick Nov 01 '13 at 12:52
  • 1
    @Patrick You are welcome, it was fun. I used SAGE. – Bruno Joyal Nov 01 '13 at 13:21
  • @BrunoJoyal Loved the post and the pics! I was thinking $$ \binom{n}{k} = \frac{n}{k} \times \binom{n-1}{k-1} $$ and so if $\gcd(k,n)=1$, the proposition above follows. More generally, one wonders when $k$ is forced to divide $\binom{n-1}{k-1}$ there.

    So, for instance, when $\gcd(n,k)=2$ and $4\not \mid k$, we are good iff the base-2 representation of $k-1$ has an on bit that is not on in that of $n-1$ (this would be (n-1).__and__(k-1) != k - 1 in SAGE). These cases seem to generate a Sierpinski triangle too, still don't see why though.

     – Pablo Rotondo Dec 08 '13 at 17:07
  • @BrunoJoyal: You are missing more trivial solutions than just those where $\gcd(n,k)=1$. Whenever $n$ satisfies the property that $p^j||n\Rightarrow p^j\nmid k$, we have that $n|\binom{n}{k}$. If you take this into account I believe it will make the picture clearer. – Eric Naslund Nov 11 '15 at 18:32
  • You might want to clarify what (k,n) = 1 means? – Nike Dattani Nov 03 '20 at 22:24
4

Well, $n = p^2$ when $k$ is not divisible by $p.$ Also $n=2p$ for $k$ not divisible by $2,p.$ Also $n=3p$ for $k$ not divisible by $3,p.$

enter image description here

Will Jagy
  • 139,541
1

It is not too hard to see that if $(n,k)=p$ where $p$ is a prime then $n\mid{n\choose k}$ exactly when $$p\mid \frac{{n/p\choose k/p}}{n/p}.$$ From this we can produce a large family of nontrivial examples. For $\ell\geq 2$, $p$ a prime, $n=p(\ell!p+1)$, and $k=\ell p$ then $n\mid{n\choose k}$. The first nontrivial example of $10\mid {10\choose 4}$ corresponds to $\ell=p=2$.

Matthew Just
  • 11