斐波那契数列

斐波那契数列性质

定义：

$$f_i=\{\begin{array}{ll}[i=1]& ,i\le 1\\ f_{i-1}+f_{i-2}& ,i\ge 2\end{array}$$

通项：

$$f_n=\frac{{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n}{\sqrt{5}}$$

性质：

$$\sum _{i=1}^n{f}_i=f_{n+2}-1$$

数学归纳法易证

$$\sum _{i=1}^n{f}_i^2=f_n{f}_{n+1}$$

数学归纳法易证

$$gcd(f_i,f_{i-1})=1$$

数学归纳法易证

$$f_n=f_m\times f_{n-m+1}+f_{m-1}\times f_{n-m}$$

证明：

显然

$$\left[\begin{array}{c}{f}_{n+1}\\ f_n\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]\times \left[\begin{array}{c}{f}_n\\ f_{n-1}\end{array}\right]$$

所以

$$\left[\begin{array}{c}{f}_{n+1}\\ f_n\end{array}\right]={\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]}^n\times \left[\begin{array}{c}1\\ 0\end{array}\right]={\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]}^{n-1}\times \left[\begin{array}{c}1\\ 1\end{array}\right]$$

由上式推得

$${\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]}^n=\left[\begin{array}{cc}{f}_{n+1}& f_n\\ f_n& f_{n-1}\end{array}\right]$$

于是

$$\begin{array}{rl}\left[\begin{array}{c}{f}_{n+1}\\ f_n\end{array}\right]& ={\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]}^m\times {\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]}^{n-m}\times \left[\begin{array}{c}1\\ 0\end{array}\right]\\ \\ & =\left[\begin{array}{cc}{f}_{m+1}& f_m\\ f_m& f_{m-1}\end{array}\right]\times \left[\begin{array}{c}{f}_{n-m+1}\\ f_{n-m}\end{array}\right]\end{array}$$

所以

$$f_n=f_m\times f_{n-m+1}+f_{m-1}\times f_{n-m}$$

$$gcd(f_n,f_m)=f_{gcd(n,m)}$$

证明：

$n=m$ 时显然

$n\ne m$ 时，不妨设 $n>m$

根据性质 4，

$$\begin{array}{rl}gcd(f_n,f_m)& =gcd(f_{m-1}\times f_{n-m}+f_m\times f_{n-m+1},f_m)\\ & =gcd(f_{m-1}\times f_{n-m},f_m)\end{array}$$

根据性质 3，

$$\begin{array}{rl}gcd(f_n,f_m)& =gcd(f_{n-m},f_m)\\ & =gcd(f_{nmodm},f_m)\end{array}$$

注意到这个和求 $gcd$ 的辗转相除是很相似的，其最终一定会得到 $gcd(f_n,f_m)=gcd(f_{gcd(n,m)},f_0)=f_{gcd(n,m)}$，原命题得证。