------
## 最终结论
```python
def get_permutation(n, k):
    # 初始化可用数字列表
    numbers = list(range(1, n + 1))
    
    # 初始化阶乘列表
    factorials = [1] * n
    for i in range(1, n):
        factorials[i] = factorials[i - 1] * i
    
    # 调整k值以适应零索引
    k -= 1
    
    # 构建排列序列
    permutation = []
    
    for i in range(n):
        index = k // factorials[n - 1 - i]  # 确定当前位的索引
        permutation.append(str(numbers[index]))  # 添加当前位的数字到结果中
        numbers.pop(index)  # 从可用数字中移除已使用的数字
        k %= factorials[n - 1 - i]  # 更新k值以计算下一位
        
    return ''.join(permutation)

# 示例调用
n = 3
k = 3
print(get_permutation(n, k))  # 输出: "213"

n = 4
k = 9
print(get_permutation(n, k))  # 输出: "2314"
```