Problem of the Month
Hint for Problem 5: February 2023

1. Remember the condition that no prime number can divide every integer in a splendid sequence.

2. Think of an integer that is a divisor of every integer.

3. Try the positive integer $c=a_i+a_{i+1}$.

4. Prove that if the divisibility conditions (S1 and S2 in the problem statement) are satisfied by $(a_1,a_2,a_3,\dots ,a_n)$, then $a_1$ and $a_n$ must both divide every integer in the sequence.

5. Try to write down a few splendid sequences of length at least $5$. With the hint for (c) in mind, what do you notice about the largest integer in a splendid sequence? Show that, for each $n$, there is a largest possible value that an integer in any splendid sequence of length $n$ can take. For instance, it can be shown that in a splendid sequence of length $n\ge 2$, every integer must be less than $2^{n-2}$.

6. There is a famous sequence called the Catalan numbers where the $n$th Catalan number equal to ${\displaystyle \frac{1}{n+1}}{\displaystyle (\genfrac{}{}{0ex}{}{2n}{n})}$. The Catalan numbers arise in many interesting ways in mathematics. One such way is that the $n$th Catalan number is equal to the number of sequences $(b_1,b_2,\dots ,b_n)$ of length $n$ with the following properties.

   • Each $b_k$ is a positive integer.

   • $b_1=1$.

   • For each $k$ satisfying $1\le k\le n-1$, $b_{k+1}\le b_k+1$. That is, $b_{k+1}$ is no more than one more than $b_k$ (it is allowed to be less than or equal to $b_k$).

   In the solution, such sequences will be called tame sequences. One way to answer this question is to show that the number of tame sequences of length $n-1$ is equal to the number of splendid sequences of length $n$ and use the closed form for the number of tame sequences. To do this, we suggest trying to devise a way to use a tame sequence of length $n-1$ as "instructions" to construct a splendid sequence of length $n$. The idea from part (c) will probably be important.

   Note: In the solution, we will provide a proof that the number of tame sequences of length $n$ is the $n$th Catalan number. You might want to try to prove this yourself, but we recommend taking it for granted when trying to solve part (f).