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

an=1/√5・{α^n-β^n}

間引いたフィボナッチ数列{F2^n}、すなわち、１，１，３，２１，９８７，・・・

α=(1+√5)/2、β=(1-√5)/2

F2^n=1/√5・{α^2^n-β^2^n}

では

　　Σ1/F2^n=(7-√5)/2

が成り立つ

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

F(2^n-1)L(2^n)=F(2^n+1-1)+1

ΠL2^i=F(2^n+1)

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

P=(1+1/2)(1+1/13)(1+1/610)・・・=Π(1+1/F2^n+1-1)

=Π(1+F(2^n+1-1)/(F2^n+1-1)

=Π(F(2^n-1)L(2^n))/(F2^n+1-1)

=Π(F(2^n-1)/(F2^n+1-1)・F2(2^m+1)

=F(2^m+1)/F(2^m+1-1)→φ

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝