Ackerman

public func ack(_ m: Int, _ n: Int) -> Int {
    if m == 0 { return n + 1 }
    if n == 0 { return ack(m - 1, 1) }
    return ack(m - 1, ack(m, n - 1))
}

for i in (0...5) {
    for j in (0...5) {
        print("ack(\(i), \(j)) is \(ack(i, j))")
    }
}

/*
result:

ack(0, 0) is 1
ack(0, 1) is 2
ack(0, 2) is 3
ack(0, 3) is 4
ack(0, 4) is 5
ack(0, 5) is 6
ack(1, 0) is 2
ack(1, 1) is 3
ack(1, 2) is 4
ack(1, 3) is 5
ack(1, 4) is 6
ack(1, 5) is 7
ack(2, 0) is 3
ack(2, 1) is 5
ack(2, 2) is 7
ack(2, 3) is 9
ack(2, 4) is 11
ack(2, 5) is 13
ack(3, 0) is 5
ack(3, 1) is 13
ack(3, 2) is 29
ack(3, 3) is 61
ack(3, 4) is 125
ack(3, 5) is 253
ack(4, 0) is 13
ack(4, 1) is 65533

Program ended with exit code: 11 <- kaboom! ack(4, 2) is already too big
*/