#include <iostream>

using namespace std;
const int N = 30 + 3;
long long h[N];


//catalan数代码实现

//1. 基于公式  H(n) = (4*n - 2)/(n + 1) H(n-1);
void catalan1(){
    h[0] = 1;
    for( int i = 1; i <= N; i++ ){
        h[i] = h[i-1] * (4 * i - 2) / ( i + 1 );
    }
}

//2. 基于公式 H(n) = ∑ H(i)*H(n-i-1)
void catalan2(){
    h[0] = 1;
    for(int i = 1; i <= N; i++){
        for(int j = 0; j <= i; j++){
            h[i] += h[j] * h[i-j-1];
        }
    }
}

//3. 基于公式 H(n) = C(2n, n) / n + 1
     // 首先基于递推公式求  C(2n, n)
    long long C(long long m, long long n){
        if(n == 0 || n == m) return 1;
        return C(m-1, n) + C(m - 1, n - 1);
    }

void catalan3(){
    h[0] = 1;
    for( int i = 1; i <= N; i++){
        //cout<<"C("<< 2*i <<"," << i << ")"<<" "<<C(2*i,i)<<endl;
        h[i] = C(2* i, i) / (i + 1);
    }

}

//4. 基于公式 H(n) = C(2n, n) - C(2n, n - 1)
void catalan4(){
    h[0] = 1;
    for( int i = 1; i <= N; i++){
        h[i] = C(2* i, i) - C(2 * i, i -1);
    }

}


int main(){
    //catalan1();  
    //catalan2();  
    //catalan3();  
    catalan4();
    for(int i = 0; i <= N; i++){
        cout<<i<<" "<<h[i]<<endl;
    }

    return 0;
}