# 費波那契數

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

1123581321345589144233377610、 987……（OEIS數列A000045

## 起源

• 第一個月初有一對剛誕生的兔子
• 第二個月之後（第三個月初）牠們可以生育
• 每月每對可生育的兔子會誕生下一對新兔子
• 兔子永不死去

## 表達式

### 初等代數解法

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

#### 首先構建等比數列

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

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

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

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

#### 求出數列${\displaystyle \left\{a_{n}+\alpha a_{n-1}\right\}}$

{\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}}}

#### 求數列${\displaystyle \left\{{b_{n}}\right\}}$ 進而得到${\displaystyle \left\{a_{n}\right\}}$

${\displaystyle b_{n+1}+{\frac {\alpha }{\beta }}b_{n}={\frac {1}{\beta }}}$
${\displaystyle b_{n+1}+\lambda =-{\frac {\alpha }{\beta }}(b_{n}+\lambda )}$ ，解得${\displaystyle \lambda =-{\frac {1}{\alpha +\beta }}}$ 。 故數列${\displaystyle \left\{b_{n}+\lambda \right\}}$ 為等比數列
${\displaystyle b_{n}+\lambda =\left(-{\frac {\alpha }{\beta }}\right)^{n-1}\left(b_{1}+\lambda \right)}$ 。而${\displaystyle b_{1}={\frac {a_{1}}{\beta }}={\frac {1}{\beta }}}$ ， 故有${\displaystyle b_{n}+\lambda =\left(-{\frac {\alpha }{\beta }}\right)^{n-1}\left({\frac {1}{\beta }}+\lambda \right)}$

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

#### 用數學歸納法證明表達式

• ${\displaystyle n}$ 為非負整數
${\displaystyle n=0}$ 時，${\displaystyle {\frac {1}{\sqrt {5}}}[\varphi ^{0}-(1-\varphi )^{0}]={\frac {1}{\sqrt {5}}}[1-1]=0=F_{0}}$ ，成立
${\displaystyle n=1}$ 時，${\displaystyle {\frac {1}{\sqrt {5}}}[\varphi ^{1}-(1-\varphi )^{1}]={\frac {1}{\sqrt {5}}}[\varphi -1+\varphi ]={\frac {1}{\sqrt {5}}}[2\varphi -1]={\frac {1}{\sqrt {5}}}\times {\sqrt {5}}=1=F_{1}}$ ，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}}}

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

### 線性代數解法

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

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

#### 構建一個矩陣方程式

${\displaystyle J_{n}}$ 為第${\displaystyle n}$ 個月有生育能力的兔子數量，${\displaystyle A_{n}}$ 為這一月份的兔子數量。

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

#### 特徵向量

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

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

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

#### 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）

#### 用數學歸納法證明

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

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

#### 化簡矩陣方程式

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

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

#### 求A的表達式

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

（7）即為${\displaystyle A_{n+1}}$ 的表達式

### 組合數解法

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

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

### 黃金比例恆等式解法

${\displaystyle \varphi }$ 黃金比例${\displaystyle {\frac {1+{\sqrt {5}}}{2}}}$ ，則有恆等式${\displaystyle \varphi ^{n}=F_{n-1}+\varphi F_{n}}$ ${\displaystyle (1-\varphi )^{n}=F_{n+1}-\varphi F_{n}}$ ，其中${\displaystyle n}$ 為任意整數[註 1]，則

{\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

${\displaystyle n}$ 為足夠大的正整數時，則

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

## 和黃金分割的關係

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

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

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

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

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

## 恆等式

• ${\displaystyle F_{n}}$ 可以表示用多個1和多個2相加令其和等於${\displaystyle n}$ 的方法的數目。

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

1. 若第1個被加數是2，有${\displaystyle F_{n}}$ 種方法來計算加至${\displaystyle n-1}$ 的方法的數目；
2. 若第2個被加數是2、第1個被加數是1，有${\displaystyle F_{n-1}}$ 種方法來計算加至${\displaystyle n-2}$ 的方法的數目。
3. 重複以上動作。
4. 若第${\displaystyle n+1}$ 個被加數為2，它之前的被加數均為1，就有${\displaystyle F_{0}}$ 種方法來計算加至0的數目。

• ${\displaystyle F_{1}+2F_{2}+3F_{3}+...+nF_{n}=nF_{n+2}-F_{n+3}+2}$
• ${\displaystyle F_{1}+F_{3}+F_{5}+...+F_{2n-1}=F_{2n}}$
• ${\displaystyle F_{2}+F_{4}+F_{6}+...+F_{2n}=F_{2n+1}-1}$
• ${\displaystyle {F_{1}}^{2}+{F_{2}}^{2}+{F_{3}}^{2}+...+{F_{n}}^{2}=F_{n}F_{n+1}}$
• ${\displaystyle F_{n}F_{m-k}-F_{m}F_{n-k}=(-1)^{n-k}F_{m-n}F_{k}}$ ，其中${\displaystyle m,n,k}$ ${\displaystyle F}$ 的序數皆不限於正整數。[註 2]
• 特別地，當${\displaystyle n=m-k}$ 時，${\displaystyle {F_{n}}^{2}-F_{n+k}F_{n-k}=(-1)^{n-k}{F_{k}}^{2}}$
• 更特別地，當${\displaystyle k=1}$ ${\displaystyle k=-1}$ 時，對於數列連續三項，有${\displaystyle {F_{n}}^{2}-F_{n-1}F_{n+1}=(-1)^{n-1}}$
• 另一方面，當${\displaystyle (m,n,k)=(n+1,n,-2)}$ 時，對於數列連續四項，有${\displaystyle F_{n}F_{n+3}-F_{n+1}F_{n+2}=(-1)^{n+1}}$ [註 3]
• ${\displaystyle \varphi ^{n}=F_{n-1}+\varphi F_{n}}$ ${\displaystyle (1-\varphi )^{n}=F_{n+1}-\varphi F_{n}}$ ，其中${\displaystyle \varphi }$ 黃金比例${\displaystyle {\frac {1+{\sqrt {5}}}{2}}}$ ${\displaystyle n}$ 為任意整數[註 1]

• ${\displaystyle {F_{m}}{F_{n}}+{F_{m-1}}{F_{n-1}}=F_{m+n-1}}$ [11]
• ${\displaystyle F_{m}F_{n+1}+F_{m-1}F_{n}=F_{m+n}}$ [11]

特別地，當${\displaystyle m=n}$ 時，{\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}}}

## 數論性質

### 公因數和整除關係

• ${\displaystyle F_{n}}$ 整除${\displaystyle F_{m}}$ ，若且唯若${\displaystyle n}$ 整除${\displaystyle m}$ ，其中${\displaystyle n\geqq 3}$
• ${\displaystyle \gcd(F_{m},F_{n})=F_{\gcd(m,n)}}$
• 任意連續三個菲波那契數兩兩互質，亦即，對於每一個${\displaystyle n}$
${\displaystyle \mathrm {gcd} (F_{n},F_{n+1})=\mathrm {gcd} (F_{n},F_{n+2})=\mathrm {gcd} (F_{n+1},F_{n+2})=1}$

### 費波那契質數

§ 公因數和整除關係所述，${\displaystyle F_{kn}}$ 總能被${\displaystyle F_{n}}$ 整除，故除${\displaystyle F_{4}=3}$ 之外，任何斐氏質數的下標必同為質數。由於存在任意長英語Arbitrarily large的一列連續合數，斐氏數列中亦能找到連續任意多項全為合數。

### 與其他數列的交集

1、3、21、55為僅有的斐氏三角形數曾猜想此結論，後來由羅明證明。[19]

## 推廣

### 和盧卡斯數列的關係

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

### 反費波那西數列

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

#### 證明關係式

${\displaystyle \varphi }$ 表示黃金分割數${\displaystyle {\frac {1+{\sqrt {5}}}{2}}}$ ，則有${\displaystyle \varphi (1-\varphi )=-1}$
${\displaystyle (-1)^{m}=[\varphi (1-\varphi )]^{m}=\varphi ^{m}(1-\varphi )^{m}}$ ，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}}}

## 應用

1970年，尤裏·馬季亞謝維奇指出了偶角標的費波那契函數

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

## 延伸閱讀

• 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.)
• 克裏福德A皮科夫.數學之戀.湖南科技出版社.

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## 註釋

1. 這可以透過${\displaystyle \varphi ^{2}=1+\varphi }$ ${\displaystyle {\frac {1}{\varphi }}=\varphi -1}$ ${\displaystyle {\frac {1}{1-\varphi }}=-\varphi }$ 此三個等式，以及費氏數列的遞迴定義，以數學歸納法證明。
2. ^ 例如當${\displaystyle (m,n,k)=(4,-8,6)}$ 時，${\displaystyle F_{-8}F_{-2}-F_{4}F_{-14}=(-21)\times (-1)-3\times (-377)=(-1)^{-14}F_{12}F_{6}=1\times 144\times 8=1152}$
3. ^ 亦即「頭尾兩項乘積」與「中間兩項乘積」恆相差1
4. ^ 利用指數律${\displaystyle \varphi ^{m+n}=\varphi ^{m}\times \varphi ^{n}}$ 、性質${\displaystyle \varphi ^{2}=1+\varphi }$ ，以及「若${\displaystyle a,b,c,d}$ 是有理數，${\displaystyle x}$ 是無理數，且滿足${\displaystyle a+bx=c+dx}$ ，則${\displaystyle a=c,b=d}$ 」證明。