ａ＝Ｆ2k-1，ｂ＝Ｆ2k+1，ｃ＝Ｆ2k-1Ｌ2kＦ2k+1のとき

　　（ａ＋ｂ＋ｃ）（１／ａ＋１／ｂ＋１／ｃ）を計算してみたい

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フィボナッチ数列

f(x)=(x)/(1-x-x^2)=(1/√5)/(1-αx)-(1/√5)/(1-βx)

α=(1+√5)/2、β=(1-√5)/2,αβ=-1,α^2+β^2=3、β=-1/α

an=1/√5・{α^n-β^n}

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F2n+1=Fn^2+Fn+1^2

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(a2n+1)^2=1/5・{α^4n+2-2(αβ)^2n+1+β^4n+2}=1/5・{α^4n+2+2+β^4n+2}

(a2n-1)^2=1/5・{α^4n-2-2(αβ)^2n-1+β^4n-2}=1/5・{α^4n-2+2+β^4n-2}

(ａ^2+ｂ^2+１）=1/5・{α^4n+2+2+β^4n+2+α^4n-2+2+β^4n-2+5}

(ａ^2+ｂ^2+１）=1/5・{α^4n+2+β^4n+2+α^4n-2+β^4n-2+9}

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a2n+1・a2n-1=1/5・{α^2n+1-β^2n+1}{α^2n-1-β^2n-1}

a2n+1・a2n-1=1/5・{α^4n+β^4n-α^2n+1β^2n-1-α^2n-1β^2n+1}

a2n+1・a2n-1=1/5・{α^4n+β^4n-α^2(αβ)^2n-1-(αβ)^2n-1β^2}

a2n+1・a2n-1=1/5・{α^4n+β^4n+α^2+β^2}

a2n+1・a2n-1=1/5・{α^4n+β^4n+3}

3a2n+1・a2n-1=1/5・{3α^4n+3β^4n+9}

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5(ａ^2+ｂ^2+１）-15ａｂ

={α^4n+2+β^4n+2+α^4n-2+β^4n-2+9-3α^4n-3β^4n-9}

={α^4n(α^2+α^-2)+β^4n(β^2+β^-2)-3α^4n-3β^4n}

={α^4n(α^2+α^-2-3)+β^4n(β^2+β^-2-3)}

={α^4n(α^2+β^2-3)+β^4n(α^2+β^2-3)}

＝0

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ａ＝Ｆ2k-1，ｂ＝Ｆ2k+1のとき、

(ａ^2+ｂ^2＋１）／ａｂ＝3

が成り立つ

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リュカ数列の漸化式は

　a0=2,a1=1

an=an-1+an-2 (n≧2)

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f(x)=Σanx^n=a0+a1x+Σanx^n (n≧2)

=a0+a1x+xΣan-1x^n-1+x^2Σan-2x^n-2 (n≧2)

=a0+a1x+x{f(x)-a0}+x^2f(x)

=2+x+x{f(x)-2}+x^2f(x)

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f(x)=(2-x)/(1-x-x^2)=(1)/(1-αx)+(1)/(1-βx)

α=(1+√5)/2、β=(1-√5)/2

an={α^n+β^n}

最も近い整数をとると

Ln〜[{α^n}+1/2]

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F2n=FnLn

L2n=(Ln)^2-2(-1)^n

F2n+1=(Fn)^2+(Fn+1)^2

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a=1/√5・{α^2n-1-β^2n-1}

b=1/√5・{α^2n+1-β^2n+1}

L={α^2n+β^2n}

c=abL

ab=1/5・{α^2n+1-β^2n+1}{α^2n-1-β^2n-1}

ab=1/5・{α^4n+β^4n-α^2n-1β^2n-1(α^2+β^2)}

αβ=-1,α^2+β^2=3

ab=1/5・{α^4n+β^4n+３}

a+b=1/√5・{α^2n-1（α^2+1)-β^2n-1(β^2+1)}

(a+b)L=1/√5・{α^2n-1（α^2+1)-β^2n-1(β^2+1)}{α^2n+β^2n}

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abL=1/5・{α^4n+β^4n+３}{α^2n+β^2n}

abL=1/5・{α^6n+β^6n+３{α^2n+β^2n}+α^4nβ^2n+α^2nβ^4}

c=abL=ab{α^2n+β^2n}

c=1/5・{α^4n+β^4n+３}{α^2n+β^2n}

ａ＋ｂ＋ｃ=(a+b)+abL

１／ａ＋１／ｂ＋１／ｃ=(a+b)/ab+1/abL={(a+b)L+1}/abL

(ａ＋ｂ＋ｃ)(１／ａ＋１／ｂ＋１／ｃ)={(a+b)/abL+1}{(a+b)L+1}

=(a+b)^2/ab+(a+b)/abL+　(a+b)L+1

(ａ^2+ｂ^2＋１）／ａｂ＝3より

(ａ^2+ｂ^2＋2ab+１）／ａｂ＝5

(a+b)^2/ab=5-1/ab

したがって、

(a+b)/abL-1/abが整数であればよい

＝(a+b-L)/abL

a+b=Lであればよいのであるが、これが成り立つので右辺は整数である。

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