2

Let $r>0$, $k\ge 2r$ be integers and let $P=\binom{k}{r}\left(\frac{r}{k}\right)^r\left(\frac{k-r}{k}\right)^{k-r}$ be the probability that a Binomial hits its expectation.

Question: Is it true that $1/P\in\mathbb Z$ if and only if $r=1$, $k=2$?

Background: I'm studying satifiability thresholds for random regular constraint satisfaction problems. In this context, it's of interest if the threshold is integer or not, since the behavior at the threshold is tricky. For one particular case, this question came up.

Progress: So far, I have nothing better than the naive approach, which is to completely expand $$\frac{1}{P}=\frac{(r-1)!k^{k-1}}{r^{r-1}(k-r)^{k-r}(k-1)^{\underline{r-1}}},$$ where $a^{\underline b}=a!/(a-b)!$ is the falling factorial. The second ingredient is that $\gcd(k,k-1)=1$. For $r=1$ we have $P^{-1}=(k/(k-1))^{k-1}$, which is $2$ for $k=2$ and otherwise not in $\mathbb Z$ since $a^b\not\in\mathbb Z$ for $a\in\mathbb Q\setminus\mathbb Z$ and $b\in\mathbb Z_{>0}$. For $r=2$ we have $P^{-1}=k^{k-1}/(2(k-2)^{k-2}(k-1))\not\in\mathbb Z$ because $k-1$ in the denominator does not cancel out. For $r=3$ we have $k\ge 6$, so $k-1\ge 5$ and thus the $(r-1)!=2$ in the numerator does not suffice to eliminate $k-1$, and still, $k^{k-1}$ does not help at all. For $r=4$ we have $3!/4^3=3/32$ and $k\ge 8$, so the $3$ in the numerator does not suffice, again. For $r=5$ we're left with $4!$ and $k\ge 10$, which further complicates things. Computing $10\le k\le 25$ by hand gives $P^{-1}\not\in\mathbb Z$, then we can argue that $4!<k-1$.

Matija
  • 3,526

1 Answers1

1

The required key result is Sylvester's Theorem. I ran into it here when I researched the special case $k=2r$.

Since we covered the base case already, we may assume $r>1$. Notice that $(k-1)^{\underline{r-1}}=(k-1)\cdots(k-r+1)$ is a product of $r-1$ consecutive integers greater than $r-1$, so there exists $\Delta\in\{1,\dots,r-1\}$ and a prime $p\in\mathbb Z_{>r-1}$ such that $(k-\Delta)\in p\mathbb Z_{>0}$. On the other hand, for the greatest common divisor $g=\gcd(k,k-\Delta)$ we have $\Delta\in g\mathbb Z_{>0}$ and thereby $g\le\Delta\le r-1$. This shows that $(r-1)!k^{k-1}\not\in p\mathbb Z_{>0}$ and thereby $P_{r,k}^{-1}\not\in\mathbb Z_{>0}$ for all $r>1$ and $k\ge 2r$.

Matija
  • 3,526