Remember the condition that no prime number can divide every integer in a splendid sequence.
Think of an integer that is a divisor of every integer.
Try the positive integer \(c=a_i+a_{i+1}\).
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.
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 \(n\geq 2\), every integer must be less than \(2^{n-2}\).
There is a famous sequence called the Catalan numbers where the \(n\)th Catalan number equal to \(\dfrac{1}{n+1}\dbinom{2n}{n}\). 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 \(1\leq k\leq n-1\), \(b_{k+1}\leq b_k+1\). 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).