Merge pull request #1263 from 24OI/TrisolarisHD-patch-1

该递推关系的解为：

$$

H_n=\frac{C_{2n}^{n}}{n+1}(n=1,2,3,\cdots)

H_n = \frac{\binom{2n}{n}}{n+1}(n \geq 2, n \in \mathbf{N_{+}})

$$

或者是

关于 Calalan 数的常见公式：

$$

H_n=H_0*H_(n-1)+H_1*H_(n-2)+...+H_(n-1)*H_0, (n>=2), H_0=H_1=1

H_n = \begin{cases}

    \sum_{i=1}^{n} H_{i-1} H_{n-i} & n \geq 2, n \in \mathbf{N_{+}}\\

    1 & n = 0, 1

\end{cases}

$$

或者是

$$

H_n=H_(n-1)*(4*n-2)/(n+1)

H_n = \frac{H_{n-1} (4n-2)}{n+1}

$$

或者是

$$

H_n=C_{2n}^{n}-C_{2n}^{n-1}

H_n = \binom{2n}{n} - \binom{2n}{n-1}

$$