This article is one of those ‘half-finished thoughts’ put together late at night. Details are missing, and – in a spirit of collaboration – I’d be glad if you wanted to fill them in for me.

The estimable @onthisdayinmath (Pat in real life) recently posted about nearly-integers, and remarked that $\phi^n$ is very close to a whole number for moderately large values of $n$ (for instance, $\phi^{20} \approx 15\ 126.\ 999\ 887$). Aha!, I thought: I know why that is!

It’s all to do with binomial expansions: in particular, those of $\phi^n$ and $\phi^{-n}$, which are closely related ((I’m going to assume $n$ is an even number, so that the signs work out without too much jiggery-pokery; the almost-equivalent odd-$n$ version is left as an exercise.)):

$\phi^n = \left( \frac{\sqrt{5} + 1}{2}\right)^n=\left(\frac12\right)^n\left[1 + n \sqrt{5} + \frac{5n(n-1)}{2} + … \right]$

$\phi^{-n} = \left( \frac{\sqrt{5}-1}{2}\right)^n=\left(\frac12\right)^n\left[1 - n \sqrt{5} + \frac{5n(n-1)}{2} + … \right]$ (for even $n$, as assumed).

Adding these together gives $\left( \frac 12 \right)^n \left[ 2 + \left(2\times \frac52\right) n(n-1) + \left(2\times\frac{25}{24}\right)n(n-1)(n-2)(n-3) + … \right]$

… and it’s fairly easy to see that the big square bracket has to be an integer. It’s not so clear that it’s a multiple of $2^n$, although it is. ((Can anyone explain why? My induction proof doesn’t pan out.))

So, it turns out that $\phi^n + \phi^{-n}$ is, for even $n$, an integer. Moreover, since $\phi^{-1} < 1$, that means $\phi^{-n}$ tends to zero as $n$ increases. That means that $\phi^n$ gets progressively closer to an integer value as $n$ gets larger!

As a ballpark estimate of how close it gets, $\log_{10}\left(\phi^{-1}\right)\approx 0.21$, so dividing $n$ by 5 gives a reasonable guess for the number of 9s or 0s after the decimal point.

* Edited 2016-07-18 to correct an errant $\pi$. Thanks to @colinthemathmo for pointing it out!

* Edited 2016-07-31 to correct the previous correction. Hopefully tis one is better.