Given the sequence generator function: $$F_\lambda(n+2)=\lambda F_\lambda(n+1)+F_\lambda(n);\quad F_\lambda(0)=0, F_\lambda(1)=1$$ where $\lambda=1$ corresponds to the Fibonacci sequence, $\lambda=2$ corresponds to the Pell numbers, and so on, is it possible to generate a sequence such that $\lim_{n\to\infty} \varphi^c=\frac{a}{b}=\frac{a+b}{a}$ where $b=F_\lambda(n),\; a=F_\lambda(n+1),\,$ $\lambda \not\in \{1,4\}$, and $c \in \mathbb{N}$?
2026-03-25 03:03:07.1774407787
Are $\varphi$ and $\varphi^3$ the only powers of $\varphi$ that are also metallic ratios?
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After Claude's comment, I suspect you are asking for the values of $w$ in $$ w^2 - 5 v^2 = -4. $$ These are every other Lucas number, beginning with $w_1 = 1$ and $w_2 = 4,$ then continuing with $$w_{k+2} = 3 w_{k+1} - w_k$$ For example, with $w_3 = 11,$ the metallic number is $$ \frac{11 + \sqrt{125}}{2} = \frac{11 + 5 \sqrt{5}}{2} = \left( \frac{1 + \sqrt{5}}{2} \right)^5$$ $$ $$ With $w_4 = 29,$ the metallic number is $$ \frac{29 + \sqrt{845}}{2} = \frac{29 + 13 \sqrt{5}}{2} = \left( \frac{1 + \sqrt{5}}{2} \right)^7$$ $$ $$