用C#递归求第19位数

Question

有这样一列数 1 3 6 11 19 ....
求第十九位数 
请用C# 递归实现这是我面试时候公司出的笔试题

Answer 1

        private int getValue(int n)
        {
            if (1 == n)
            {
                return 1;
            }
            else if (2 == n)
            {
                return 3;
            }
            else
            {
                return getValue(n - 1) + getValue(n - 2) + 2;
            }
        }

Answer 2

       1
0 + 1= 3-2
1 + 3= 6-2
3 + 6=11-2
6 +11=19-2
11+19=??-2
        private static int GetData(int count)
        {
            if (count <= 1)
            {
                return 1;
            }
            else if (count == 2)
            {
                return 3;
            }
            else 
            {
                return GetData(count - 1) + GetData(count - 2) + 2;
            }
        }
比较笨的方法~

Answer 3

根据2楼的分析，我看到了公式f(n)=f(n-1)*2-f(n-3)
所以方法如下
public int GetNo(int n)
        {
            int resultNo = 0;
            if (n == 1)
            {
                resultNo = 1;
            }
            else
                if (n == 2)
                {
                    resultNo = 3;
                }
                else
                    if (n == 3)
                    {
                        resultNo = 6;
                    }
                    else
                    {
                        resultNo = GetNo(n - 1) * 2 - GetNo(n - 3);
                    }
            return resultNo;
        }

Answer 4

改下 private int getValue(int n)
        {
            if (1 == n)
            {
                return 1;
            }
            else if (2 == n)
            {
                return 3;
            }
            else if(0 >= n)
            {
                return 0;
            }
            {
                return getValue(n - 1) * 2 - getValue(n - 2);
            }
        }

Answer 5

f（n）=f(n-1)+f(n-2)+2 n > 2 不对吗？