# 斐波那契数

（重定向自斐波那契数列

• ${\displaystyle F_{0}=0}$
• ${\displaystyle F_{1}=1}$
• ${\displaystyle F_{n}=F_{n-1}+F_{n-2}}$（n≧2）

1123581321345589144233377610、 987……（OEIS数列A000045

## 起源

• 第一个月初有一对刚诞生的兔子
• 第二个月之后（第三个月初）它们可以生育
• 每月每对可生育的兔子会诞生下一对新兔子
• 兔子永不死去

## 表达式

### 初等代数解法

• ${\displaystyle a_{1}=1}$
• ${\displaystyle a_{2}=1}$
• ${\displaystyle a_{n}=a_{n-1}+a_{n-2}}$（n≥3）

#### 首先构建等比数列

${\displaystyle a_{n}+\alpha a_{n-1}=\beta (a_{n-1}+\alpha a_{n-2})}$

${\displaystyle a_{n}=(\beta -\alpha )a_{n-1}+\alpha \beta a_{n-2}}$

${\displaystyle {\begin{cases}\beta -\alpha =1\\\alpha \beta =1\end{cases}}}$

${\displaystyle {\begin{cases}\alpha ={\dfrac {{\sqrt {5}}-1}{2}}\\\beta ={\dfrac {{\sqrt {5}}+1}{2}}\end{cases}}}$

#### 求出数列{${\displaystyle a_{n}+\alpha a_{n-1}}$}

{\displaystyle {\begin{aligned}a_{n+1}+\alpha a_{n}&=(a_{2}+\alpha a_{1})\beta ^{n-1}\\&=(1+\alpha )\beta ^{n-1}\\&=\beta ^{n}\\\end{aligned}}}

#### 求数列{${\displaystyle b_{n}}$}进而得到{${\displaystyle a_{n}}$}

${\displaystyle b_{n+1}+{\frac {\alpha }{\beta }}b_{n}={\frac {1}{\beta }}}$
${\displaystyle b_{n+1}+\lambda =-{\frac {\alpha }{\beta }}(b_{n}+\lambda )}$，解得${\displaystyle \lambda =-{\frac {1}{\alpha +\beta }}}$。 故数列${\displaystyle \left\{b_{n}+\lambda \right\}}$为等比数列
${\displaystyle b_{n}+\lambda =\left(-{\frac {\alpha }{\beta }}\right)^{n-1}\left(b_{1}+\lambda \right)}$。而${\displaystyle b_{1}={\frac {a_{1}}{\beta }}={\frac {1}{\beta }}}$， 故有${\displaystyle b_{n}+\lambda =\left(-{\frac {\alpha }{\beta }}\right)^{n-1}\left({\frac {1}{\beta }}+\lambda \right)}$

${\displaystyle a_{n}={\frac {\sqrt {5}}{5}}\cdot \left[\left({\frac {1+{\sqrt {5}}}{2}}\right)^{n}-\left({\frac {1-{\sqrt {5}}}{2}}\right)^{n}\right]}$

#### 用数学归纳法证明表达式

• ${\displaystyle n}$为非负整数
${\displaystyle n=0}$时，${\displaystyle {\frac {1}{\sqrt {5}}}[\varphi ^{0}-(1-\varphi )^{0}]={\frac {1}{\sqrt {5}}}[1-1]=0=F_{0}}$，成立
${\displaystyle n=1}$时，${\displaystyle {\frac {1}{\sqrt {5}}}[\varphi ^{1}-(1-\varphi )^{1}]={\frac {1}{\sqrt {5}}}[\varphi -1+\varphi ]={\frac {1}{\sqrt {5}}}[2\varphi -1]={\frac {1}{\sqrt {5}}}\times {\sqrt {5}}=1=F_{1}}$，成立

${\displaystyle n=k+2}$
{\displaystyle {\begin{aligned}F_{k+2}&=F_{k+1}+F_{k}\\&={\frac {1}{\sqrt {5}}}[\varphi ^{k+1}-(1-\varphi )^{k+1}]+{\frac {1}{\sqrt {5}}}[\varphi ^{k}-(1-\varphi )^{k}]\\&={\frac {1}{\sqrt {5}}}[\varphi ^{k+1}+\varphi ^{k}-(1-\varphi )^{k+1}-(1-\varphi )^{k}]\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{k}({\color {brown}\varphi +1})-(1-\varphi )^{k}[{\color {green}(1-\varphi )+1}]\right\}\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{k}({\color {brown}\varphi ^{2}})-(1-\varphi )^{k}[{\color {green}(1-\varphi )^{2}}]\right\}\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{k+2}-(1-\varphi )^{k+2}\right\}\\\end{aligned}}}

• ${\displaystyle n}$为非正整数
${\displaystyle n=0}$时，成立
${\displaystyle n=-1}$时，${\displaystyle {\frac {1}{\sqrt {5}}}[{\color {brown}\varphi ^{-1}}-{\color {green}(1-\varphi )^{-1}}]={\frac {1}{\sqrt {5}}}[({\color {brown}\varphi -1})-({\color {green}-\varphi })]={\frac {1}{\sqrt {5}}}[2\varphi -1]={\frac {1}{\sqrt {5}}}\times {\sqrt {5}}=1=F_{-1}}$，成立

${\displaystyle n=-k-2}$
{\displaystyle {\begin{aligned}F_{-k-2}&=F_{-k}-F_{-k-1}\\&={\frac {1}{\sqrt {5}}}[\varphi ^{-k}-(1-\varphi )^{-k}]-{\frac {1}{\sqrt {5}}}[\varphi ^{-k-1}-(1-\varphi )^{-k-1}]\\&={\frac {1}{\sqrt {5}}}[\varphi ^{-k}-\varphi ^{-k-1}-(1-\varphi )^{-k}+(1-\varphi )^{-k-1}]\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{-k-1}({\color {brown}\varphi -1})-(1-\varphi )^{-k-1}[{\color {green}(1-\varphi )-1}]\right\}\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{-k-1}({\color {brown}\varphi ^{-1}})-(1-\varphi )^{-k-1}[{\color {green}(1-\varphi )^{-1}}]\right\}\\&={\frac {1}{\sqrt {5}}}\left\{\varphi ^{-k-2}-(1-\varphi )^{-k-2}\right\}\\\end{aligned}}}

### 线性代数解法

${\displaystyle {\begin{pmatrix}F_{n+2}\\F_{n+1}\end{pmatrix}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}\cdot {\begin{pmatrix}F_{n+1}\\F_{n}\end{pmatrix}}}$

${\displaystyle {\begin{pmatrix}F_{n+2}&F_{n+1}\\F_{n+1}&F_{n}\end{pmatrix}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{n+1}}$

#### 构建一个矩阵方程

${\displaystyle {J_{n+1} \choose A_{n+1}}={\begin{pmatrix}0&1\\1&1\end{pmatrix}}\cdot {J_{n} \choose A_{n}},}$

#### 特征向量

${\displaystyle \left({\begin{pmatrix}0&1\\1&1\end{pmatrix}}-\lambda \cdot E\right)\cdot {\vec {x}}=0}$

${\displaystyle {\vec {x}}_{1}}$=${\displaystyle {\begin{pmatrix}1\\{\frac {1}{2}}(1+{\sqrt {5}})\end{pmatrix}}}$

${\displaystyle {\vec {x}}_{2}}$=${\displaystyle {\begin{pmatrix}1\\{\frac {1}{2}}(1-{\sqrt {5}})\end{pmatrix}}}$

#### 分解首向量

${\displaystyle {J_{1} \choose A_{1}}={\begin{pmatrix}0\\1\end{pmatrix}}}$

${\displaystyle {\begin{pmatrix}0\\1\end{pmatrix}}={\frac {1}{\sqrt {5}}}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1+{\sqrt {5}})\end{pmatrix}}-{\frac {1}{\sqrt {5}}}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1-{\sqrt {5}})\end{pmatrix}}}$ （4）

#### 用数学归纳法证明

${\displaystyle {J_{n+1} \choose A_{n+1}}={\begin{pmatrix}0&1\\1&1\end{pmatrix}}\cdot {J_{n} \choose A_{n}}}$=${\displaystyle \lambda \cdot {J_{n} \choose A_{n}}}$

${\displaystyle {J_{n+1} \choose A_{n+1}}={\begin{pmatrix}0&1\\1&1\end{pmatrix}}^{n}\cdot {J_{1} \choose A_{1}}=\lambda ^{n}\cdot {J_{1} \choose A_{1}}}$ （5）

#### 化简矩阵方程

${\displaystyle {J_{n+1} \choose A_{n+1}}=\lambda ^{n}\cdot \left[{\frac {1}{\sqrt {5}}}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1+{\sqrt {5}})\end{pmatrix}}-{\frac {1}{\sqrt {5}}}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1-{\sqrt {5}})\end{pmatrix}}\right]}$

${\displaystyle {J_{n+1} \choose A_{n+1}}={\frac {1}{\sqrt {5}}}\cdot \lambda _{1}^{n}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1+{\sqrt {5}})\end{pmatrix}}-{\frac {1}{\sqrt {5}}}\cdot \lambda _{2}^{n}\cdot {\begin{pmatrix}1\\{\frac {1}{2}}(1-{\sqrt {5}})\end{pmatrix}}}$

#### 求A的表达式

${\displaystyle A_{n+1}={\frac {1}{\sqrt {5}}}\cdot \lambda _{1}^{n+1}-{\frac {1}{\sqrt {5}}}\cdot \lambda _{2}^{n+1}}$
${\displaystyle A_{n+1}={\frac {1}{\sqrt {5}}}\cdot (\lambda _{1}^{n+1}-\lambda _{2}^{n+1})}$
${\displaystyle A_{n+1}={\frac {1}{\sqrt {5}}}\cdot \left\{\left[{\frac {1}{2}}\left(1+{\sqrt {5}}\right)\right]^{n+1}-\left[{\frac {1}{2}}(1-{\sqrt {5}})\right]^{n+1}\right\}}$（7）

（7）即为An+1的表达式

### 组合数解法

${\displaystyle F_{n}=\sum _{i=0}^{\infty }{\binom {n-i}{i}}}$[1]

${\displaystyle F_{n-1}+F_{n}=\sum _{i=0}^{\infty }{\binom {n-1-i}{i}}+\sum _{i=0}^{\infty }{\binom {n-i}{i}}=1+\sum _{i=1}^{\infty }{\binom {n-i}{i-1}}+\sum _{i=1}^{\infty }{\binom {n-i}{i}}=1+\sum _{i=1}^{\infty }{\binom {n+1-i}{i}}=\sum _{i=0}^{\infty }{\binom {n+1-i}{i}}=F_{n+1}}$

### 黄金比例恒等式解法

${\displaystyle \varphi }$黄金比例${\displaystyle {\frac {1+{\sqrt {5}}}{2}}}$，则有恒等式${\displaystyle \varphi ^{n}=F_{n-1}+\varphi F_{n}}$${\displaystyle (1-\varphi )^{n}=F_{n+1}-\varphi F_{n}}$，其中${\displaystyle n}$为任意整数[注 1]，则

{\displaystyle {\begin{aligned}\varphi ^{n}-(1-\varphi )^{n}&=(F_{n-1}+\varphi F_{n})-(F_{n+1}-\varphi F_{n})\\&=(F_{n-1}-F_{n+1})+2\varphi F_{n}\\&=-F_{n}+2\varphi F_{n}\\&=F_{n}(2\varphi -1)\\&=F_{n}\times {\sqrt {5}}\\\end{aligned}}}

{\displaystyle {\begin{aligned}F_{n}&={\frac {1}{\sqrt {5}}}[\varphi ^{n}-(1-\varphi )^{n}]\\&={\frac {1}{\sqrt {5}}}\left[({\frac {1+{\sqrt {5}}}{2}})^{n}-({\frac {1-{\sqrt {5}}}{2}})^{n}\right]\\\end{aligned}}}

### 近似值

${\displaystyle n}$为足够大的正整数时

${\displaystyle F_{n}\approx {\frac {1}{\sqrt {5}}}\varphi ^{n}={\frac {1}{\sqrt {5}}}\cdot \left[{\frac {1}{2}}\left(1+{\sqrt {5}}\right)\right]^{n}\approx 0.4472135955\cdot 1.61803398875^{n}}$
${\displaystyle F_{-n}\approx -{\frac {1}{\sqrt {5}}}(1-\varphi )^{-n}=-{\frac {1}{\sqrt {5}}}\cdot \left[{\frac {1}{2}}\left(1-{\sqrt {5}}\right)\right]^{-n}\approx -0.4472135955\cdot (-0.61803398875)^{-n}}$

## 和黄金分割的关系

${\displaystyle {\frac {f_{n+1}}{f_{n}}}\approx a={\frac {1}{2}}(1+{\sqrt {5}})=\varphi \approx 1{.}618{...}}$

${\displaystyle {\frac {1}{1}}=1\qquad {\frac {2}{1}}=1+{\frac {1}{1}}\qquad {\frac {3}{2}}=1+{\frac {1}{1+{\frac {1}{1}}}}\qquad {\frac {5}{3}}=1+{\frac {1}{1+{\frac {1}{1+{\frac {1}{1}}}}}}\qquad {\frac {8}{5}}=1+{\frac {1}{1+{\frac {1}{1+{\frac {1}{1+{\frac {1}{1}}}}}}}}}$

${\displaystyle F_{n}={\frac {1}{\sqrt {5}}}\left[\left({\frac {1+{\sqrt {5}}}{2}}\right)^{n}-\left({\frac {1-{\sqrt {5}}}{2}}\right)^{n}\right]={\varphi ^{n} \over {\sqrt {5}}}-{(1-\varphi )^{n} \over {\sqrt {5}}}}$

${\displaystyle \varphi =1+{\frac {1}{1+{\frac {1}{1+{\frac {1}{1+{\frac {1}{1+...}}}}}}}}}$

${\displaystyle \varphi ={\sqrt {1+{\sqrt {1+{\sqrt {1+{\sqrt {1+...}}}}}}}}}$

## 恒等式

• ${\displaystyle F_{n}}$可以表示用多个1和多个2相加令其和等于${\displaystyle n}$的方法的数目。

• ${\displaystyle F_{0}+F_{1}+F_{2}+F_{3}+...+F_{n}=F_{n+2}-1}$

1. 若第1个被加数是2，有${\displaystyle F_{n}}$种方法来计算加至${\displaystyle n-1}$的方法的数目；
2. 若第2个被加数是2、第1个被加数是1，有${\displaystyle F_{n-1}}$种方法来计算加至${\displaystyle n-2}$的方法的数目。
3. 重复以上动作。
4. 若第${\displaystyle n+1}$个被加数为2，它之前的被加数均为1，就有${\displaystyle F_{0}}$种方法来计算加至0的数目。

• ${\displaystyle F_{1}+2F_{2}+3F_{3}+...+nF_{n}=nF_{n+2}-F_{n+3}+2}$
• ${\displaystyle F_{1}+F_{3}+F_{5}+...+F_{2n-1}=F_{2n}}$
• ${\displaystyle F_{2}+F_{4}+F_{6}+...+F_{2n}=F_{2n+1}-1}$
• ${\displaystyle {F_{1}}^{2}+{F_{2}}^{2}+{F_{3}}^{2}+...+{F_{n}}^{2}=F_{n}F_{n+1}}$
• ${\displaystyle F_{n}F_{m-k}-F_{m}F_{n-k}=(-1)^{n-k}F_{m-n}F_{k}}$，其中${\displaystyle m,n,k}$${\displaystyle F}$的序数皆不限于正整数。[注 2]
• 特别地，当${\displaystyle n=m-k}$时，${\displaystyle {F_{n}}^{2}-F_{n+k}F_{n-k}=(-1)^{n-k}{F_{k}}^{2}}$
• 更特别地，当${\displaystyle k=1}$${\displaystyle k=-1}$时，对于数列连续三项，有${\displaystyle {F_{n}}^{2}-F_{n-1}F_{n+1}=(-1)^{n-1}}$
• 另一方面，当${\displaystyle (m,n,k)=(n+1,n,-2)}$时，对于数列连续四项，有${\displaystyle F_{n}F_{n+3}-F_{n+1}F_{n+2}=(-1)^{n+1}}$[注 3]
• ${\displaystyle {F_{m}}{F_{n}}+{F_{m-1}}{F_{n-1}}=F_{m+n-1}}$[11]（以${\displaystyle 1}$代入${\displaystyle k}$${\displaystyle 1-n}$代入${\displaystyle n}$
• ${\displaystyle F_{m}F_{n+1}+F_{m-1}F_{n}=F_{m+n}}$[11]

特别地，当m = n时，{\displaystyle {\begin{aligned}F_{2n-1}&=F_{n}^{2}+F_{n-1}^{2}\\F_{2n}&=(F_{n-1}+F_{n+1})F_{n}\\&=(2F_{n-1}+F_{n})F_{n}\end{aligned}}}

• ${\displaystyle \varphi ^{n}=F_{n-1}+\varphi F_{n}}$${\displaystyle (1-\varphi )^{n}=F_{n+1}-\varphi F_{n}}$，其中${\displaystyle \varphi }$黄金比例${\displaystyle {\frac {1+{\sqrt {5}}}{2}}}$${\displaystyle n}$为任意整数[注 1]

## 数论性质

### 公约数和整除关系

• ${\displaystyle F_{n}}$整除${\displaystyle F_{m}}$，当且仅当n整除m，其中n≧3。
• ${\displaystyle \gcd(F_{m},F_{n})=F_{\gcd(m,n)}}$
• 任意连续三个菲波那契数两两互素，亦即，对于每一个n
gcd(Fn, Fn+1) = gcd(Fn, Fn+2) = gcd(Fn+1, Fn+2) = 1

### 斐波那契素数

§ 公约数和整除关系所述，${\displaystyle F_{kn}}$总能被${\displaystyle F_{n}}$整除，故除${\displaystyle F_{4}=3}$之外，任何斐氏素数的下标必同为素数。由于存在任意长英语Arbitrarily large的一列连续合数，斐氏数列中亦能找到连续任意多项全为合数。

### 与其他数列的交集

1、3、21、55为仅有的斐氏三角形数曾猜想此结论，后来由罗明证明。[19]

## 推广

### 和卢卡斯数列的关系

${\displaystyle F_{n}L_{n}=F_{2n}}$

### 反费波那西数列

${\displaystyle G_{n+2}=G_{n}-G_{n+1}}$

#### 证明关系式

${\displaystyle \varphi }$表示黄金分割数${\displaystyle {\frac {1+{\sqrt {5}}}{2}}}$，则有${\displaystyle \varphi (1-\varphi )=-1}$
${\displaystyle (-1)^{m}=[\varphi (1-\varphi )]^{m}=\varphi ^{m}(1-\varphi )^{m}}$，因此
{\displaystyle {\begin{aligned}(-1)^{m+1}F_{-m}&=(-1)^{m+1}\times {\frac {1}{\sqrt {5}}}[\varphi ^{-m}-(1-\varphi )^{-m}]\\&=(-1)\times {\color {brown}(-1)^{m}}\times {\frac {1}{\sqrt {5}}}[\varphi ^{-m}-(1-\varphi )^{-m}]\\&=(-1)\times {\color {brown}\varphi ^{m}(1-\varphi )^{m}}\times {\frac {1}{\sqrt {5}}}[\varphi ^{-m}-(1-\varphi )^{-m}]\\&=(-1)\times {\frac {1}{\sqrt {5}}}[\varphi ^{-m+m}(1-\varphi )^{m}-(1-\varphi )^{-m+m}\varphi ^{m}]\\&=(-1)\times {\frac {1}{\sqrt {5}}}[(1-\varphi )^{m}-\varphi ^{m}]\\&={\frac {1}{\sqrt {5}}}[\varphi ^{m}-(1-\varphi )^{m}]\\&=F_{m}\\\end{aligned}}}

## 应用

1970年，尤里·马季亚谢维奇指出了偶角标的斐波那契函数

${\displaystyle y=F_{2x}}$

## 程序参考

function fib(n) {
var fib_n = function(curr, next, n) {
if (n == 0) {
return curr;
}
else {
return fib_n(next, curr+next, n-1);
}
}
return fib_n(0, 1, n);
}

#include <stdio.h>
#include <math.h>

int main()
{
int n;
double constant_a = (1 + sqrt(5)) / 2;
double constant_b = (1 - sqrt(5)) / 2;
double constant_c = sqrt(5) / 5;
double value_1 = 0;
int value_2 = 0;
scanf("%d", &n);
if(n > 0)
{
for (int i = 0; i < n; i++)
{
value_1 = constant_c * (pow(constant_a, i) - pow(constant_b, i));
value_2 = (int)value_1;
printf("%d\n", value_2);
}
return 0;
}
else
{
return -1;
}
}


c++二变量求某项版

#include <iostream>
using namespace std;
int main () {
int x,y,n;
x=1;y=0;
cin >> n;
for (int i=0;i<n;i=i+1) {
x=x+y;
y=x-y;
}
cout << y;
return 0;
}


c++通项公式版

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
unsigned long long n;
double ca = (1 + sqrt(5)) / 2;
double cb = (1 - sqrt(5)) / 2;
double cc = sqrt(5) / 5;
double v1 = 0;
double v2 = 0;
cout <<" ";
cin>>n;
if(n > 0)
{
for (unsigned long long i = 0; i < n; i++)
{
v1 = cc * (pow(ca, i) - pow(cb, i));
v2 = (int)v1;
cout <<v2<<endl;
}
return 0;
}
else
{
return -1;
}
cout <<'/b';
}

# Fibonacci numbers module

def fib(n):    # write Fibonacci series up to n
a, b = 0, 1
while b < n:
print(b, end=' ')
a, b = b, a+b
print()

def fib2(n):   # return Fibonacci series up to n
result = []
a, b = 0, 1
while b < n:
result.append(b)
a, b = b, a+b
return result

fibs = [0, 1]
numZS = int(input('How many Fibonacci numbers do you want? '))
for i in range(numZS-2):
fibs.append(fibs[-2] + fibs[-1])
print fibs


Common Lisp

(defun fibs (x)
(cond ((equal x 0) 1)
((equal x 1) 1)
(t (+ (fibs (- x 1))
(fibs (- x 2))))))

(defun fibs (x)
(do ((n 0 (+ n 1))
(i 1 j)
(j 1 (+ i j)))
((equal n x) i)))


func fibonacci(n int) int {
if n < 2 {
return n
}

return fibonacci(n-2) + fibonacci(n-1)
}


func fibonacci(n int) int {
a, b := n%2, 1

for i := 0; i < n/2; i++ {
a += b
b += a
}

return a
}


Java语言通项公式版：

public int fibonacci(int n){
if(n<2){
return n;
}else {
return fibonacci(n-1)+fibonacci(n-2);
}
}


Java语言快捷版：

public int fibonacci(int n){
if(n<2){
return n;
}else {
int[] ans = new int[n];
ans[0] = 1;
ans[1] = 2;
for(int i=2;i<n;i++) {
ans[i]=ans[i-1]+ans[i-2];
}
return ans[n-1];
}
}


C语言阵列版：

#include <stdio.h>
#include <stdlib.h>
int main()
{
int n,s,L;
printf("輸入長度");
scanf("%d",&L);
while(L<0)
{
printf("錯誤");
return 0;
}
int a[L];
int x=1,y=2;
a[0]=x;
a[1]=x;
a[2]=y;
for(n=3;n<L;n++)
{
a[n]=a[n-1]+a[n-2];
}
for(n=0;n<L;n++)
{
printf("%d ",a[n]);
}
system("pause");
return 0;
}


Python Lambda 递归版:

fib = lambda n: n if n<2 else fib(n-1) + fib(n-2)


## 延伸阅读

• KNUTH, D. E. 1997. The Art of Computer ProgrammingArt of Computer Programming, Volume 1: Fundamental Algorithms, Third Edition. Addison-Wesley. Chapter 1.2.8.
• Arakelian, Hrant (2014). Mathematics and History of the Golden Section. Logos, 404 p. ISBN 978-5-98704-663-0, (rus.)
• 克里福德A皮科夫.数学之恋.湖南科技出版社.

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## 注释

1. 这可以透过${\displaystyle \varphi ^{2}=1+\varphi }$${\displaystyle {\frac {1}{\varphi }}=\varphi -1}$${\displaystyle {\frac {1}{1-\varphi }}=-\varphi }$此三个等式，以及斐波那契数列的的递归定义，以数学归纳法证明。
2. ^ 例如当${\displaystyle (m,n,k)=(4,-8,6)}$时，${\displaystyle F_{-8}F_{-2}-F_{4}F_{-14}=(-21)\times (-1)-3\times (-377)=(-1)^{-14}F_{12}F_{6}=1\times 144\times 8=1152}$
3. ^ 亦即“头尾两项乘积”与“中间两项乘积”恒相差1