489
回編集
差分
:$$z^n=C_{n}+S_{n}z=-(S_{n-1})l+S_{n}\\$$
:$$\begin{array}{l}
z^1=&0+z\\
\end{array}$$
:$$z^n=\displaystyle-\left[\sum_{k=0}^{\lfloor (n-2)/2\rfloor}\binom{n-k-2}{k}(-1)^k(2\cos\theta)^{n-2k-2}\right]+\left[\sum_{k=0}^{\lfloor (n-1)/2\rfloor}\binom{n-k-1}{k}(-1)^k(2\cos\theta)^{n-2k-1}\right]z$$
==導出==