It's All Bases.

Let $\phi_{n} = \dfrac{\sqrt{n^2 + 4} + n}{2}$ .

$\phi_{n}$ is one solution to $x^2 - nx - 1 = 0$

$\implies \phi_{n}^2 = n\phi_{n} + 1 \implies \phi_{n} - 1 - n\phi_{n} \implies \dfrac{1}{\phi_{n}^2 - 1} = \dfrac{1}{n\phi_{n}} = \dfrac{(\dfrac{\sqrt{n^2 + 4} - n}{2})}{n}$ $= \dfrac{\phi_{n} - n}{n} \implies \dfrac{\phi_{n} - n}{n} = \dfrac{1}{\phi_{n}^2 - 1} = \dfrac{\dfrac{1}{\phi_{n}^2}}{1 - \dfrac{1}{\phi_{n}^2}} =$ $\sum_{m = 1}^{\infty} (\dfrac{1}{\phi_{n}^2})^{m} = \boxed{(0.010101010101 ...)_{\phi_{n}}}$ .