Greatest common divisor of some binomial coefficients Note that this is now cross-posted on mathoverflow.
While making some computation, I stumbled upon a curious relation among some binomial coefficients.
Consider the sequence of binomial coefficients $a(k,n)$, $0 \le k \le n-1$, $n \ge 2$ defined as follows:
$$a(k,n) = \binom{\frac{n(n-1)}{2}+3+k}{\lfloor\frac{n-1}{2}\rfloor+1+k}$$
and the sequence ($n \ge 2$):
$$b_n=\gcd(a(0,n),a(1,n),\ldots,a(n-2,n),a(n-1,n))$$
I conjecture that:
$$b_n \gt 1 \space\space\space\space \forall n \ge 2$$
The first few values of $b_n$ starting from $b_2$ are $2,5,6,143,102,253,899,703$. I didn't check for $n \gt 9$.
Any hint for proving or disproving the conjecture?
Also, can we say something about the values of $b_n$?
 A: I believe your conjecture is true. Here is an attempted proof that's pretty close to working but ends up relying on a famous number theory conjecture.
First note: If $p$ is prime, then $p | n!$ if and only if $n \ge p$.
Let $r(n,k) = \frac{n(n-1)}{2} + 3 + k$ and $s(n,k) = \lfloor \frac{n-1}{2} \rfloor + 1 + k$, so that $a(k,n) = \binom{r(n,k)}{s(n,k)}$. Among the coefficients being GCD'd to compute $b_n$, the smallest $r$ will be $r(n,0) = \frac{n(n-1)}{2} + 3$; the largest $s$ is $s(n,n-1) = \lfloor \frac{n-1}{2} \rfloor + n$; and $r-s$ will be the same across all of the binomial coefficients in question. This means the conjecture will be proved if we can show there's some prime $p$ satisfying
$$\begin{align}
\left\lfloor \frac{n-1}{2} \right\rfloor + n &< p \\
\frac{n(n-1)}{2} - \left\lfloor \frac{n-1}{2} \right\rfloor + 2 &< p \\
p &\le \frac{n(n-1)}{2} + 3 \\
\end{align}$$
We actually have $\left\lfloor \frac{n-1}{2} \right\rfloor + n \le \frac{n(n-1)}{2} - \left\lfloor \frac{n-1}{2} \right\rfloor + 2$ for all $n \ge 1$, so we can ignore the first inequality above.
Let $x(n) = \frac{n(n-1)}2$. Then $\left\lfloor \frac{n-1}{2} \right\rfloor \le \sqrt{x/2}$, and for my proof to work I need to find $p$ with $x-\sqrt{x/2}+2 < p \le x+3$.
This last statement is true for all the $n$ I could quickly check. Also, the statement being true for sufficiently large $x$ (i.e. sufficiently large $n$) is basically Legendre's conjecture but a little stronger due to the fixed constant. See also Wikipedia's sections on upper bounds for prime gaps and conjectures related to prime gaps for more reading material. Basically, people in the field think the conjecture is probably true, but cutting edge number theory methods have been sloooowly chipping away at the problem by proving slightly better bounds over time.
Even if the conjecture I need is true, that would only prove the statement for sufficiently large $n$. We'd still need to check some number of smallish cases by hand. I've already checked at least up through $n=10,000$ though, and it seems fine at least that far.
Good luck getting a full answer! I think it's totally possible that this question may be answerable by some much simpler means. This is just the first approach I thought about.
