Inspired by Priyanshu Mishra 2

The full explanation is way too long so I will give the summary here (proofs omitted as they are tedius):

Lemma 1: The following is equivalent to n-Mishra Interesting

• (1) $a_1 = 1$
• (2) $a_{i-1} \leq a_i \leq i \textrm{ or } a_i = n + 1, \; \forall i \in \{ 2, \ldots, n\}$

This is a much easier definition to work with, as we can actually count these. First consider the sequences that have no entries that are $n + 1$ . These correspond exactly to the Catalan numbers, specifically the interpretation of lattice paths below the diagonal, with the values of the sequence being the height of the columns. See here on Wikipedia to learn more about them. That means the number of n-Mishra interesting sequences of length n with no entries that have value $n+1$ is exactly $\frac{1}{n+1} \binom{2n}{n}$ Now note that if $a_j = n + 1$ then all subsequent entries in the sequence are $n+1$ . Thus we count all of them that have $a_2 = n + 1$ , then $a_3 = n + 1$ (and $a_2 \neq n + 1$ ), etc. all the way to only $a_n = n + 1$ . This is exactly the same as the cases with no values of $n + 1$ but for shorter sequences (lengths 1, 2, ... n-1). So to count all of them, we sum them all of them (and the number without $n + 1$ in the sequence) to get our final count: $N_n = \sum_{k=1}^{n} \frac{1}{k+1} \binom{2k}{k}$ .

Now we need to use Stirling's approximation to simplify $\frac{1}{k+1} \binom{2k}{k} \approx \frac{1}{k+1} \frac{4^k}{\sqrt{\pi k}}$ , which is true as $k$ becomes large. So we get: $\lim_{n \rightarrow \infty} N_n 4^{-n} n^{3/2} = \lim_{n \rightarrow \infty} \left( \sum_{k=1}^{n} \frac{1}{k+1} \binom{2k}{k} \right) 4^{-n} n^{3/2} = \lim_{n \rightarrow \infty} \left( \sum_{k=1}^{n} \frac{1}{k+1} \frac{4^k}{\sqrt{\pi k}} \right) 4^{-n} n^{3/2}$

$= \frac{1}{\pi^{1/2}} \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{n^{3/2}}{k^{3/2}+k^{1/2}} \frac{1}{4^{n-k}} = \frac{1}{\pi^{1/2}} \lim_{n \rightarrow \infty} \sum_{k=0}^{n-1} \frac{n^{3/2}}{(n-k)^{3/2}+(n-k)^{1/2}} \frac{1}{4^k}$

To evaluate that last sum, you can prove that as the fractional term in front of the $1 / 4^{k}$ can be ignored ( this is intuitive I actually had a hard time doing this rigorously). So that simplifies to: $\frac{1}{\pi^{1/2}} \lim_{n \rightarrow \infty} \sum_{k=0}^{n-1} \frac{1}{4^k} = \frac{4}{\pi^{1/2} 3}$ . Thus we have $a = 4$ , $b = 3$ , and $c = 2$ , thus $a + b + c = 9$ , our final answer.