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I have already understand the basic proof of Catalan numbers through monotonic lattice paths. However, I am know figuring out the specific case: if the two paths are symmetric about a diagonal line (from lower-left to upper-right), they should be counted only once.

For example, the two matrices (the boundary between 0 and 1 in upper triangular matrix is the monotonic path), \begin{bmatrix}1&0&0\\&1&1\\&&1\end{bmatrix}\begin{bmatrix}1&1&0\\&1&0\\&&1\end{bmatrix} They are isomorphic by my definition, and should be counted only once. Thus, how to get the number of equivalent classes in this case? Thanks!

Erus Izumi
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1 Answers1

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Let $C_n$ be the $n^\text{th}$ Catalan number, which is the total number of paths without regard to the reflection condition.

Let $R_n$ be the number of Catalan paths which are equal to their own reflection through the skew-diagonal.

Using Burnside's lemma, the number of non-equivalent up to reflection paths is equal to $$ \frac12(C_n+R_n). $$ There is a well-known formula for $C_n$, so you just need to find a formula for $R_n$. Note that a path counted by $R_n$ is completely determined by its first $n$ steps, because the remaining $n$ steps are the reverse-complement of the first $n$ steps. These first $n$ steps must be a sequence of $n$ symbols, each of which is either $\to$, for a right step, or $\downarrow$, for a downward step. This sequence is valid if and only if, for each initial segment, there are at least as many $\to$'s as $\downarrow$'s. Using this answer by Marc van Leeuwen, or my alternate solution, we conclude that $$ R_n=\binom{n}{\lfloor n/2\rfloor}. $$ Therefore, the number of paths up to reflection symmetry is $$ \frac12\left[\frac1{n+1}\binom{2n}n+\binom{n}{\lfloor n/2\rfloor}\right]. $$

Mike Earnest
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  • I've seen that the requirement for not crossing the diagonal stands for right step is strictly more than downward step through out the count for the first half of palindrome sequence. So is this R_n equivalent to Bertrand's ballot theorem? – Erus Izumi Mar 21 '23 at 07:35
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    They are closely related. When counting $R_n$, one method is to break it up into several sub-problems, where the $k^\text{th}$ sub-problem counts the number of sequences of $k$ downsteps and $n-k$ right-steps which never go below the diagonal, for each $k\in {0,1,\dots,\lfloor n/r\rfloor}$. This sub-problem is exactly the ties-allowed variant of Bertrand's ballot problem. – Mike Earnest Mar 21 '23 at 15:20
  • I meant to say $\lfloor n/2\rfloor$ in my last comment. – Mike Earnest Mar 22 '23 at 03:57