TITLE: Updates on a least prime factor conjecture by Erdos
QUESTION [5 upvotes]: In the 1993 article "Estimates of the Least Prime Factor of a Binomial Coefficient," Erdos et al. conjectured that
$$\operatorname{lpf} {N \choose k} \leq \max(N/k,13)$$
With finitely many exceptional $(N,k)$. Here, $\operatorname{lpf}(x)$ denotes the smallest prime factor of $x$.
I am posting here to ask whether any progress has been made toward this conjecture.

REPLY [8 votes]: The authors of the paper you mention, Erdos, Lacampagne, and Selfridge, define $p(m)$ to be the least prime divisor of $m$ and concern themselves what can be said about $p(\binom{n}{k}).$ I suspect that Selfridge wrote the article. It has his style of saying a lot in a succinct way which is puzzling but solvable with some thought on the part of the reader. The conjecture stated in the abstract is $$p(\binom{n}{k}) \leq \max(\frac{n}{k},29).$$ That shouldn't be thought of as their big conjecture but rather a terse and amusing way of capturing some of the the main points.
The short answer to your question is that they did a lot of computation, made some observations that had theoretical backing and strong computational support. No-one, as far as I know,  has challenged or refuted them and perhaps it isn't especially attractive to try further computation. Or perhaps it is, but not to report "I didn't find anything else either."
Aside: They are perhaps more interested in the growth rate of $g(k),$ the minimal $n>k$ with $p(\binom{n}{k})>k.$ I feel compelled to quote a small stretch of the article:

That is a whole mess of conjectures, but not snappy enough for an abstract. That is the subject of section 1 of the paper. They and others explored $g(k)$ up to about $k=140$ and with more powerful computers the results were later extended to about $k=200.$ The current record lower bound is $$g(k) \geq exp(c(\log^3k/\log \log k)^{1/2}).$$
Getting back to the conjecture you ask about, the first puzzle is

The stated conjecture is clearly false. $\binom{n}{n-1}=n$ is prime when $n$ is. I can imagine Selfridge saying "Well of course we don't mean that." And if you read in further, the investigation is only for $k<n/2.$ The case $n=2k$ is a very small puzzle left to you.

Another puzzle is

Why $\frac{n}{k}?$ Is that best possible? Here is my take on that, read the article for a more elegant and general treatment: Suppose $p=q(k-1)!+1$ is prime. Then, for $n=pk,$ $$\binom{n}{k}=\prod_{i=0}^{k-1}\frac{n-i}{k-i}$$ where the $k$ factors are integers. If they all happen to be prime then $p(\binom{n}{k})=p=\frac{n}{k}.$ I can convince myself that for every $k$ we would expect that to happen infinitely often. Perhaps that is alluded to in the article or perhaps it is too obviously right (or wrong) to mention.

Another (small) puzzle is

How you could ever have $p(\binom{n}{k})>\frac{n}{k}?$ After all, there is a $0 \leq j <k$ with $\frac{n-j}{k}$ an integer, so $\frac{n}{k}$ seems a clear upper bound. And it is, for large enough $n.$
However, while $\binom{62}{6}$ is divisible by some divisor of
$\frac{60}{6},$ that divisor is $1$ !
$$\binom{62}{6}=\frac{62}{2}\cdot 61 \cdot \frac{60}{60}\cdot 59
   \cdot \frac{58}{2} \cdot \frac{57}{3}$$

They conjecture that, for $n \gt k^2$,  $p(\binom{n}{k}) \geq \frac{n}{k}$ with that one exception of $p(\binom{62}{6})=\frac{n-5}{3}=19$
They also conjecture that this one, along with $p(\binom{959}{6})=19,p(\binom{474}{66})=23$ and $p(\binom{284}{28})=29$ are the only cases with $p(\binom{n}{k}) \gt \min(\frac{n}{k},19).$
They find eight cases with $p(\binom{n}{k})=17>\frac{n}{k}.$
They say that they wrote a program to find all cases of $p(\binom{n}{k})=p>\frac{N}{k}$ with $p>5$ and $k \leq 12000.$ It must not have been entirely exhaustive because they say that there was only one output other than the twelve mentioned for $331 <k <625$ and that was $p(\binom{3574}{406})=13$. They continue "Thus, at this point in time, it is possible that $p(\binom{n}{k})\leq\max(\frac{n}{k},13).$" So that is short of making a conjecture, but I don't know that there are any exceptions known other than the thirten they mention.
One might wonder why they said $p>5.$ Anyone familiar with Pascal's Triangle $\mod 2$ will realize that for every $k>2$ there are lots of cases of $p(\binom{n}{k})=3$ with $2k<n<3k.$ The article gives a nice proof that there is always at least one case of $p(\binom{n}{k})=5$ with $3k<n<4k.$
There is much more in that article, but I will stop there.