If you stare at a sufficiently large picture of Pascal's triangle the following entries in row $14$ might stand out to you:
$${14 \choose 4} = 1001, {14 \choose 5} = 2002, {14 \choose 6} = 3003.$$
We can ask more generally when three consecutive binomial coefficients are in arithmetic progression; this was asked and answered in 2014 here, and there is a fun connection to the central limit theorem I noticed while playing around with this: solutions must lie near the inflection points of the normal curve, since those are the only points near which the normal curve has a chance of being approximately straight.
The connection to the normal curve suggests that it might be interesting to generalize as follows:
Q1: When are there $m+1$ consecutive binomial coefficients ${n \choose k}, \dots {n \choose k+m}$ which can be interpolated by a polynomial of degree $m-1$?
The reason for the slightly funny indexing is that this corresponds to the $m^{th}$ finite difference being zero, which means it can be restated as
$$\sum_{i=0}^m (-1)^i {m \choose i} {n \choose k+i} = 0.$$
For example when $m = 2$ (the arithmetic progression case) this says
$${n \choose k} - 2 {n \choose k+1} + {n \choose k+2} = 0$$
and when $m = 3$ (four consecutive binomial coefficients which can be interpolated by a quadratic) this says
$${n \choose k} - 3 {n \choose k+1} + 3 {n \choose k+2} - {n \choose k+3} = 0.$$
The nicest thing that could happen here is that solutions exist and are near the points where the $m^{th}$ derivative $f^{(m)}$ of the normal curve is zero; these are given by the zeroes of the Hermite polynomials. However, the above condition is equivalent to the vanishing of a polynomial in $n, k$ of degree $m$ (which should be related to the $m$-th Hermite polynomial), which might not have integer solutions for large $m$.
I worked out the answer for $m = 3$; in this case something nice happens because even though you get a cubic polynomial there's a family of solutions $n = 2k+3$ by symmetry, corresponding to four points around the mean, and corresponding to the $x = 0$ root of the third Hermite polynomial $x^3 - 3x$; this happens for all odd $m$. So you can divide by $n - 2k-3$ and get a quadratic which can be solved, giving solutions corresponding to the other two roots $x = \pm \sqrt{3}$, which have the form
$$(n, k) = \left( \frac{s^2 - 7}{3}, \frac{n \pm s - 3}{2} \right)$$
where $3 \nmid s$ and $s \ge 7$. The smallest nontrivial solution at $s = 7$ is $(n, k) = (14, 2)$, corresponding to
$$\begin{align} {14 \choose 2} - 3 {14 \choose 3} + 3 {14 \choose 4} - {14 \choose 5} &= \\ 91 - 3 \cdot 364 + 3 \cdot 1001 - 2002 &= \\ 91(1 - 3 \cdot 4 + 3 \cdot 11 - 22) &= 0. \end{align}$$
But I don't know what happens for $m \ge 4$.
Q2: Is there a nice way to see that the above polynomial in $n, k$ of degree $m$ is related to the Hermite polynomials?
My guess as to what the relationship should be: if this polynomial is expressed in terms of $n$ and $s = n-2k-m$ then it should become a polynomial in $n$ and $s^2$, after dividing out by $s$ if $m$ is odd. The leading coefficients of this new polynomial should be the coefficients of the $m$-th Hermite polynomial.
I checked this using WolframAlpha for $m = 4$. In this case the polynomial in $n$ and $s^2$ works out to be
$$3n^2 - 6n(s^2 - 3) + (s^4 - 16s^2 + 24) = 0$$
and you can see that the coefficients of the leading terms $s^4 - 6ns^2 + 3n^2$ match the fourth Hermite polynomial. This polynomial rearranges to
$$6 \left( n - (s^2-3) \right)^2 = (2s^2 - 1)^2 + 5$$
which is an unexpected Pell's equation $x^2 - 6y^2 = -5$; we need solutions where $\frac{x+1}{2}$ is a square, which to my surprise happens multiple times, leading to the solutions
$$(n, k) = (13, 3), (62, 26), (1517, 711), (15039, 7472)$$
but no others that I can find.
