Consider sequence $x_n$, such $x_1=1, x_2=2x_0=1, x_1=1, x_3=3$ x_2=3$ and $x_{n+2}=x_{n+1}(5x_n^2+2)$ for all n>1n>0. How it was done: it is well known, that $F_{2n}=F_n L_n$, where $L_n$ - n-th Lucas number. But we now, that $$ L_n=\phi^n+(-1/\phi)^n $$ Using Binet's formula: $$ F_n^2=\frac{1}{5}(\phi^{2n}+(-1/\phi)^{2n}-2(-1)^n)\to L_{2n}=5F_n^2+2(-1)^n $$ So we have: $$ F_{4n}=F_{2n}(5F_n^2+2(-1)^n) $$
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Consider sequence $x_n$, such $x_1=1, x_2=2, x_3=3$ and $x_{n+2}=x_{n+1}(5x_n^2+2)$.x_{n+2}=x_{n+1}(5x_n^2+2)$ for all n>1. How it was done: it is well known, that $F_{2n}=F_n L_n$, where $L_n$ - n-th Lucas number. But we now, that $$ L_n=\phi^n+(-1/\phi)^n $$ Using Binet's formula: $$ F_n^2=\frac{1}{5}(\phi^{2n}+(-1/\phi)^{2n}-2(-1)^n)\to L_{2n}=5F_n^2+2(-1)^n $$ So we have: $$ F_{4n}=F_{2n}(5F_n^2+2(-1)^n) $$ |
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Consider sequence $x_n$, such $x_1=1, x_2=2, x_3=3$ and $x_{n+2}=x_{n+1}(5x_n^2+2)$. |
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