■フィボナッチ数・リュカ数の問題(その30)
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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b=1/√5・{α^2n+1-β^2n+1}
a=1/√5・{α^2n-1-β^2n-1}
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)}=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
簡単にならないので、ここで打ち切り
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