TITLE: A pattern appearing in the powers of $\phi$
QUESTION [13 upvotes]: \begin{align}
\phi^5 &= 11,\underline{0}901699\cdots\\
\phi^6 &= 17,\underline{9}44271\cdots\\
\phi^7 &= 29,\underline{6}34441\cdots\\
\phi^8 &= 46,\underline{9}7871\cdots\\
\phi^9 &= 76,\underline{0}1315 \cdots\\
\phi^{10} &= 122,\underline{99}18\cdots\\
\phi^{11} &= 199,\underline{00}502\cdots\\
\phi^{12} &= 321,\underline{99}6894\cdots\\
\phi^{13} &= 521,\underline{00}191\cdots\\
\phi^{14} &= 842,\underline{99}881\cdots\\
\phi^{15} &= 1364,\underline{000}73\cdots\\
\phi^{16} &= 2206,\underline{999}54\cdots\\
\end{align}
Why there is a $0$ $9$ patterns in the powers of the golden ratio

REPLY [11 votes]: $\phi$ 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.