User:Bernhard Oemer/Catalan Numbers Proof

Fifth proof
This proof is based on the Dyck words interpretation of the Catalan numbers, so Cn is the number of ways to correctly match n pairs of brackets. We denote a (possibly empty) correct string with c and its inverse (where [ and ] are exchanged) with c+. Since any c can be uniquely decomposed into c = [ c1 ] c2, summing over the possible spots to place the closing bracket immediately gives the recursive definition
 * $$C_0 = 1 \quad \text{and} \quad C_{n+1} = \sum_{i=0}^n C_i\,C_{n-i}\quad\text{for }n\ge 0.$$

Now let b stand for a balanced string of length 2n containing an equal number of [ and ] and $$\textstyle B_n = {2n\choose n} = d_n C_n$$ with some factor dn ≥ 1. As above, any balanced string can be uniquely decomposed into either [ c ] b or ] c+[ b, so
 * $$B_{n+1} = 2 \sum_{i=0}^n B_i C_{n-i}.$$

Also, any incorrect balanced string starts with c ], so
 * $$B_{n+1} - C_{n+1} = \sum_{i=0}^n {2i+1 \choose i} C_{n-i} = \sum_{i=0}^n \frac{2i+1}{i+1} B_i C_{n-i}.$$

Subtracting the above equations and using Bi = di Ci gives
 * $$C_{n+1} = 2 \sum_{i=0}^n d_i C_i C_{n-i} - \sum_{i=0}^n \frac{2i+1}{i+1} d_i C_i C_{n-i} = \sum_{i=0}^n \frac{d_i}{i+1} C_i C_{n-i}.$$

Comparing coefficients with the original recursion formula for Cn gives di = i + 1, so
 * $$C_n = \frac{1}{n+1}{2n\choose n}.$$