For what natural numbers $n$ and $k$ does the sequence

$$
\binom{n}{k-1}, \quad\binom{n}{k}, \quad\binom{n}{k+1}
$$

form an arithmetic progression?
According to the task requirement, the difference between adjacent binomial coefficient pairs must be equal, that is, the difference of differences is 0:

$$
\binom{n}{k-1}-2\binom{n}{k}+\binom{n}{k+1}=0
$$

Here we assume that $k-1 \geq 0$ and $k+1 \leq n$, that is,

$$
1 \leq k \leq n-1
$$

If equation (1) holds - and only then - the three binomial coefficients form an arithmetic sequence.

Multiply by $(k+1)!(n-k+1)!/ n!$. This exists and is positive due to (2), so (1) holds if and only if

$$
\begin{gathered}
k(k+1)-2(k+1)(n-k+1)+(n-k)(n-k+1)=0 \\
n^{2}-4 n k+4 k^{2}-n-2=0
\end{gathered}
$$

Accordingly,

$$
n=(n-2 k)^{2}-2
$$

is 2 less than a perfect square:

$$
n=u^{2}-2
$$

where $u$ is a natural number and here $u=n-2 k$ or $u=2 k-n$, that is,

$$
k=k_{1}=\frac{n-u}{2}=\frac{u^{2}-u}{2}-1=\binom{u}{2}-1
$$

or

$$
k=k_{2}=\frac{n+u}{2}=\binom{u+1}{2}-1
$$

From the last form, it is clear that we get integer values for $k$.

Here $u \geq 2$ must hold so that $n$ is a positive integer. For $u=2$, however, the first and second inequalities in (2) do not hold for the two $k$ values.

If $u \geq 3$, then

$$
k_{1}=\binom{u}{2}-1 \geq 1 \quad \text { and } \quad k_{1}=\frac{n-u}{2}<n
$$

Since $k_{1}+k_{2}=n$ and $k_{1}<k_{2}$, both $k$ values satisfy (2).

We started from the necessary and sufficient condition for the task requirements and performed equivalent transformations, so the pairs of $n, k$ numbers that meet the condition are those for which there is some integer $u$ greater than 2 such that

$$
n=u^{2}-2 \quad \text { and } \quad k=\binom{u}{2}-1 \quad \text { or } \quad k=\binom{u+1}{2}-1
$$

Remarks. 1. Even the $k=0$ and $k=2$ values corresponding to $u=2$ are acceptable if we consider $\binom{n}{k}$ to be 0 when $k$ is negative or greater than $n$. In this case, we get the arithmetic sequences 0, 1, 2 or 2, 1, 0. (One contestant also noted that with this agreement, any positive integer $n$ and $k \leq -2$ or $k \geq n+2$ also meet the condition.)

2. The easily verifiable

$$
\binom{n}{k-1}+2\binom{n}{k}+\binom{n}{k+1}=\binom{n+2}{k+1}
$$

relation, when subtracting the corresponding sides of (1), shows that the task requirement is satisfied if and only if

$$
4\binom{n}{k}=\binom{n+2}{k+1}
$$

3. Multiplying equation (3) by 4 and completing the square with respect to $n$,

$$
8 k+9=(2 n-4 k-1)^{2}
$$

results, which is the square of an odd number. Denoting this by $2 v+1$, we get

$$
k=\frac{(2 v+1)^{2}-9}{8}=\frac{v^{2}+v-2}{2}=\binom{v+1}{2}-1
$$

$$
2 n-4 k-1=2 v+1 \quad \text { or } \quad 4 k+1-2 n=2 v+1
$$

From this,

$$
n=2 k+v+1=v^{2}+2 v-1=(v+1)^{2}-2
$$

or

$$
n=2 k-v=v^{2}-2
$$

4. It can be seen from the two $k$ values in the solution that the larger $k$ value corresponding to any allowed $n$ value is the same as the smaller $k$ value corresponding to the next $n$ value.
5. If we ask whether more than three consecutive binomial coefficients can form an arithmetic sequence, the answer is no. In this case, the first, second, and third elements of the sequence, as well as the second, third, and fourth elements, would form a three-term arithmetic sequence. However, in the solution of the problem, it was found that the two $k$ values corresponding to a given $n$ value correspond to symmetrically placed binomial coefficients, so one of the two resulting three-element sequences is increasing, the other is decreasing, and they cannot be consecutive elements of the same sequence.
6. Arithmetic sequences are sequences in which the differences between consecutive elements are all the same. Similarly, we can examine sequences - so-called second-order arithmetic sequences - in which these differences form an arithmetic sequence, and higher-order arithmetic sequences can be defined similarly. The question arises: can four consecutive binomial coefficients form a second-order arithmetic sequence? This problem leads to a cubic equation in $n$ and $k$, similar to the above. However, an infinite number of solutions can be found with the following simple consideration: If $n$ is odd, of the form $2 u+1$, then the sequence of binomial coefficients corresponding to $n$ has two equal binomial coefficients in the middle, and the binomial coefficients before and after these are also equal. The differences of the numbers

$$
\binom{2 u+1}{u-1}, \quad\binom{2 u+1}{u}, \quad\binom{2 u+1}{u+1}, \quad\binom{2 u+1}{u+2}
$$

are thus of the form $a, 0, -a$, and this is an arithmetic sequence. This also leads to the conclusion that the mentioned cubic equation can be easily solved: the resulting polynomial can be factored into a first and a second-degree product, and thus the other solutions can be easily found.