Define the following two subsets of prime numbers
$$\Pi\_n:=\{p: \text{$p$ is prime, $p\leq n$}\}$$
and
$$B\_n:=\{p: \text{$p$ is prime, $p$ divides $\binom{n}k$ for some $0< k< n$}\}.$$
Denote their respective cardinalities by $\pi(n):=\vert\Pi\_n\vert$ and $\,b(n):=\vert B\_n\vert$.



> 
> **QUESTIONS.** Experimental evidence suggest that
> 
> 
> (a) $\pi(n)-b(n)=0$ or $1$.
> 
> 
> (b) If $\pi(n)-b(n)=1$, then the missing prime is the largest prime divisor of $n+1$.
> 
> 
> Are these true statements?
> 
> 
> 



Binomial coefficients and "missing primes"
Suppose that $p $ is a prime satisfying $p\le n$ and $p\nmid \binom{n}{k}$ for all $k=1,\ldots,n-1.$ Then from [Kummer's theorem](https://en.wikipedia.org/wiki/Kummer%27s_theorem) it follows that the base-$p$ representation of $n$ is $n=(a,p-1,\ldots,p-1)\_p$ with $1\le a\le p-1$. So $n=(a+1)p^r-1$, where $r$ is the number of base-$p$ digits in $n$. Then $n+1 = (a+1)p^r$ with $a+1 \leq p$, so any prime factor of $n+1$ besides $p$ is a prime factor of $a+1$, so it has to be less than $p$.


Here is a second solution, using [Lucas's theorem](https://en.wikipedia.org/wiki/Lucas's_theorem): writing $n = n\_0+n\_1p+\cdots +n\_rp^r$, where $0 \leq n\_i \leq p-1$ and $n\_r \geq 1$, 
each $k$ in $\{0,1,\ldots,n\}$ has base-$p$ representation $k\_0+k\_1p+\cdots +k\_rp^r$ where $0 \leq k\_i \leq p-1$ (maybe $k\_r = 0$ if $k$ is much smaller than $n$). Then Lucas' theorem says
$$
\binom{n}{k} \equiv \binom{n\_0}{k\_0}\binom{n\_1}{k\_1}\cdots \binom{n\_r}{k\_r} \bmod p.
$$
For digits $n\_i$ and $k\_i$, which are in $\{0,1,\ldots,p-1\}$, we have $\binom{n\_i}{k\_i} \equiv 0 \bmod p$ if and only if $n\_i < k\_i$ (otherwise the factors in $\binom{n\_i}{k\_i} = \frac{n\_i(n\_i-1)\cdots(n\_i-(k\_i-1))}{k\_i!}$ are all nonzero modulo $p$). Thus we have $\binom{n}{k} \not\equiv 0 \bmod p$ for *all* $k$ from $0$ to $n$ (let's allow $k=0$ and $k=n$ since those cases are not divisible by $p$) if and only if every $k \leq n$ has $k\_i \leq n\_i$ for $i < r$ and $k\_r \leq n\_r$, and that's equivalent to $n\_i = p-1$ for $i < r$ (if some $n\_i < p-1$ then at $k = (p-1)p^i$ we have $\binom{n}{k} \equiv 0 \bmod p$). Such $n$ have the form $(p-1)+(p-1)p\cdots+(p-1)p^{r-1} + n\_rp^r = (n\_r+1)p^r-1$. If $r = 0$ then $n = n\_r \leq p-1$. If $r \geq 1$ then $p \mid (n+1)$, and the first solution shows easily in this case why $p$ has to be the largest prime factor of $n+1$. In summary, every prime that is $\leq n$ divides some $\binom{n}{k}$ for $1 \leq k \leq n-1$ except when $n = mp^r-1$ where $r \geq 1$, $p$ is prime, and $2 \leq m \leq p$, in which case the prime $p$ divides no $\binom{n}{k}$.


