初等数论 1.4 Fibonacci数列

定义:Fibonacci数列: f 1 = 1 , f 2 = 1 f_1=1,f_2=1 f1=1,f2=1,且对 n ≥ 3 n\ge3 n3 f n = f n − 1 + f n − 2 f_n=f_{n-1}+f_{n-2} fn=fn1+fn2

约定: f 0 = 0 f_0=0 f0=0.

推论:Fibonacci数 f n = ( 1 + 5 2 ) n − ( 1 − 5 2 ) n 5 f_n=\dfrac{({\dfrac{1+\sqrt 5}{2})^{n}}-{(\dfrac{1-\sqrt 5}{2})^{n}}}{\sqrt 5} fn=5 (21+5 )n(215 )n
定理:Fibonacci数列的一些性质: f 2 n = f n 2 + 2 f n − 1 f n f_{2n}=f_{n}^{2}+2f_{n-1}f_n f2n=fn2+2fn1fn f n − 2 + f n + 2 = 3 f n f_{n-2}+f_{n+2}=3f_{n} fn2+fn+2=3fn f n + 1 f n − 1 − f n 2 = ( − 1 ) n f_{n+1}f_{n-1}-f_{n}^{2}=(-1)^{n} fn+1fn1fn2=(1)n f m + n = f m f n + 1 + f n f m − 1 f_{m+n}=f_{m}f_{n+1}+f_{n}f_{m-1} fm+n=fmfn+1+fnfm1 ∑ k = 1 n f k = f n + 2 − 1 \sum_{k=1}^{n}{f_{k}}=f_{n+2}-1 k=1nfk=fn+21 ∑ k = 1 n f 2 k − 1 = f 2 n − f 2 + f 1 \sum_{k=1}^{n}{f_{2k-1}}=f_{2n}-f_{2}+f_{1} k=1nf2k1=f2nf2+f1 ∑ k = 1 n f 2 n = f 2 n + 1 − f 1 \sum_{k=1}^{n}{f_{2n}}=f_{2n+1}-f_{1} k=1nf2n=f2n+1f1 ∑ k = 1 n f k 2 = f n f n + 1 \sum_{k=1}^{n}{f_{k}^{2}}=f_{n}f_{n+1} k=1nfk2=fnfn+1 ∑ k = 1 2 n − 1 f k f k + 1 = f 2 n 2 \sum_{k=1}^{2n-1}{f_{k}f_{k+1}=f_{2n}^{2}} k=12n1fkfk+1=f2n2
定义:黄金分割比(Golden ratio) φ : = lim ⁡ n → ∞ F n + 1 F n = 1 + 5 2 \varphi\coloneqq\lim\limits_{n\rightarrow\infty}{\dfrac{F_{n+1}}{F_n}}=\dfrac{1+\sqrt5}{2} φ:=nlimFnFn+1=21+5

定义:Lucas数列: L 1 = 1 , L 2 = 3 L_1=1,L_2=3 L1=1,L2=3,且对 n ≥ 3 n\ge3 n3 L n = L n − 1 + L n − 2 L_n=L_{n-1}+L_{n-2} Ln=Ln1+Ln2
L n = ( 1 + 5 2 ) n − ( 1 − 5 2 ) n L_n=(\dfrac{1+\sqrt 5}{2})^{n}-(\dfrac{1-\sqrt 5}{2})^{n} Ln=(21+5 )n(215 )n
Lucas数列也具有Fibonacci数列类似的性质.

定理:Zeckendorf表示:对于任意 n ∈ N n\in\N nN,存在唯一的一组Fibonacci子列,使得 n = ∑ i = 0 k F c i n=\displaystyle\sum_{i=0}^{k}{F_{c_i}} n=i=0kFci,其中 c i ≥ 2 , c i + 1 > c i + 1 c_i\ge2,c_{i+1}>c_i+1 ci2,ci+1>ci+1
定义:广义Fibonacci数列: g 1 = a , g 2 = b g_1=a,g_2=b g1=a,g2=b,且对 n ≥ 3 n\ge3 n3 g n = g n − 1 + g n − 2 g_n=g_{n-1}+g_{n-2} gn=gn1+gn2

g n = a f n − 2 + b f n − 1 n ≥ 3 g_n=af_{n-2}+bf_{n-1}\quad n\ge3 gn=afn2+bfn1n3

n n n为负整数时,递归定义Fibonacci数 f n − 2 : = f n − f n − 1 f_{n-2}\coloneqq f_{n}-f_{n-1} fn2:=fnfn1

f − n = ( − 1 ) n + 1 f n f_{-n}={(-1)}^{n+1}f_n fn=(1)n+1fn

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