A pattern appearing in the powers of $\phi$

This can be seen from the following formula:

$$L_n = \varphi^n + \frac{1}{(-\varphi)^n}$$

Where $L_n$ are the Lucas numbers, which are integers. Because the term $\dfrac{1}{(-\varphi)^n}$ alternates between a tiny positive and negative value, we see that $\varphi^n$ must be just barely below or above an integer - hence the $.0$ and $.9$ pattern.

$\phi$ is the larger root of $$x^2-x-1=0$$

It's conjugate root is: $$\overline {\phi}=\frac {1-\sqrt 5}2\approx -0.61803$$

From the quadratic, we see that the sequence $$\{a_n\}=\{\phi^n +\bar {\phi}^n\}$$ satisfies the Fibonacci recursion: $$a_{n+1}=a_n+a_{n-1}\quad a_1=1\quad a_2=3$$

Of course $\bar {\phi}^n\to 0$ for large $n$ so we must have that $\phi^n$ is nearly an integer for large $n$. since $\bar {\phi}^n$ alternates sign we see the pattern you have noticed.

I cannot provide an explanation for why the $0$ and $9$ patterns appear, but I can give an explanation of why it seems like the powers converge to integers.

If you visit the following website: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phigits.html

Based on the special properties of $\phi$ (I won't prove it here),

$$\phi^2 -\phi^{-2} = 3,\\\phi^3 -\phi^{-3} = 4,\\\phi^4 -\phi^{-4} = 7,\\\phi^5 -\phi^{-5} = 11\\\cdots$$

It should be evident that as the exponent increases, the second term in each expression will converge to zero, and the entire expression converges to a number.