The Catalan number for n = 4 can be calculated using the formula C(n) = (2n)! / ((n+1)! * n!). Substituting n = 4, we get C(4) = (2*4)! / ((4+1)! * 4!) = 14.

The recursive formula for Catalan numbers is C(n) = C(n-1) + C(n-2), where C(0) = 1 and C(1) = 1.

The generating function for Catalan numbers is F(x) = 1 / (1 - x^2).

The number of ways to triangulate a convex n-gon is given by the Catalan number C(n-3).

There is a bijection between binary trees with n internal nodes and full binary trees with n+1 leaves. Therefore, the number of binary trees with n internal nodes is equal to the Catalan number C(n).

The asymptotic formula for Catalan numbers is C(n) ~ 4^n / (n^(3/2) * sqrt(pi)).

C(5) = (2*5)! / ((5+1)! * 5!) = 120 / (6 * 5) = 42.

The number of ways to arrange n distinct objects in a circle is given by the Catalan number C(n).

C(6) = (2*6)! / ((6+1)! * 6!) = 1440 / (7 * 6) = 93.

There is a bijection between Dyck paths of length 2n and lattice paths from (0, 0) to (n, n) that never go above the line y = x. Therefore, the number of Dyck paths of length 2n is equal to the Catalan number C(n).

C(7) = (2*7)! / ((7+1)! * 7!) = 2520 / (8 * 7) = 143.

The explicit formula for Catalan numbers is C(n) = (2n)! / (n+1)! * n!.

C(8) = (2*8)! / ((8+1)! * 8!) = 40320 / (9 * 8) = 204.

There is a bijection between Motzkin paths of length 2n and lattice paths from (0, 0) to (n, n) that never go below the line y = -x. Therefore, the number of Motzkin paths of length 2n is equal to the Catalan number C(n).

C(9) = (2*9)! / ((9+1)! * 9!) = 655360 / (10 * 9) = 286.