Number of Distinct Parenthesizations on Word

Theorem
Let $w_n$ denote an arbitrary word of $n$ elements.

The number of distinct parenthesizations of $w_n$ is the Catalan number $C_{n - 1}$:
 * $C_{n - 1} = \dfrac 1 n \dbinom {2 \paren {n - 1} } {n - 1}$

Proof
Let $w_n$ denote an arbitrary word of $n$ elements.

Let $a_n$ denote the number of ways $W_n$ elements may be parenthesized.

First note that we have:

and from Parenthesization of Word of $4$ Elements:
 * $a_4 = 5$

Consider a word $w_{n + 1}$ of $n + 1$ elements.

Then $w_{n + 1}$ can be formed as any one of:


 * $w_1$ concatenated with $w_n$
 * $w_2$ concatenated with $w_{n - 1}$
 * $\dotsc$ and so on until:
 * $w_n$ concatenated with $w_1$

Thus the $i$th row in the above sequence is the number of parenthesizations of $w_{n + 1}$ in which the two outermost parenthesizations contain $i$ and $n - i + 1$ terms respectively.

We have that:
 * there are $a_i$ parenthesizations of $w_i$
 * there are $a_{n - i + 1}$ parenthesizations of $w_{n - i + 1}$

Hence the total number of parenthesizations of $w_{n + 1}$ is the sum of all these parenthesizations for $1 \le i \le n$.

That is:
 * $(1): \quad a_{n + 1} = a_1 a_n + a_2 a_{n - 1} + \dotsb + a_n a_1$

Let us start with the generating function:
 * $\ds \map {G_A} z = \sum_{n \mathop = 1}^\infty a_n z^n$

Then:

Thus $\map {G_A} z$ satisfies the quadratic equation:
 * $\paren {\map {G_A} z}^2 - \map {G_A} z + z = 0$

By the Quadratic Formula, this gives:
 * $\map {G_A} z = \dfrac {1 \pm \sqrt {1 - 4 z} } 2$

Since $\map {G_A} 0 = 0$, we can eliminate the positive square root and arrive at:


 * $(2): \quad \map {G_A} z = \dfrac 1 2 - \dfrac {\sqrt {1 - 4 z} } 2$

Expanding $\sqrt {1 - 4 z}$ using the Binomial Theorem:


 * $\ds \map {G_A} z = \dfrac 1 2 - \dfrac 1 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \dbinom {\frac 1 2} n 4^n z^n$

where:
 * $\dbinom {\frac 1 2} 0 = 1$

and:
 * $\dbinom {\frac 1 2} n = \dfrac {\frac 1 2 \paren {\frac 1 2 - 1} \dotsm \paren {\frac 1 2 - n + 1} } {n!}$

As a result:
 * $\ds (3): \quad \map {G_A} z = -\dfrac 1 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \dbinom {\frac 1 2} n 4^n z^n$

We can expand $(3)$ as a Taylor series about $0$.

As such a series, when it exists, is unique, the coefficients must be $a_n$.

Hence: