Catalan numbers

func f(i) { i==0 ? 1 : (i * f(i-1)) }
func c(n) { f(2*n) / f(n) / f(n+1) }

With memoization:

func c(n) is cached {
    n == 0 ? 1 : (c(n-1) * (4 * n - 2) / (n + 1))
}

Calling the function:

15.times { |i|
    say "#{i}\t#{c(i)}"
}

Output:

0       1
1       1
2       2
3       5
4       14
5       42
6       132
7       429
8       1430
9       4862
10      16796
11      58786
12      208012
13      742900
14      2674440