The Angry Statistician

Hofstadter mentions the following recursive relation in his great book "Gödel, Escher, Bach": $$\begin{align} g(0) &= 0;\\ g(n) &= n-g(g(n-1)). \end{align}$$ I claim that $g(n) = \left\lfloor \phi\cdot (n+1) \right\rfloor$, where $\phi = \frac{-1+\sqrt{5}}{2}$, and I'll show this using a technique that makes proving many identities of this type nearly automatic. Let $\phi\cdot n = \left\lfloor \phi\cdot n \right\rfloor + e$, where $0 < e < 1$ as $\phi$ is irrational, nor can $e = 1-\phi$, and note that $\phi$ satisfies ${\phi}^2 + \phi - 1 = 0$. Some algebra gives \[ \begin{align} n-\left\lfloor \left( \left\lfloor \phi\cdot n \right\rfloor + 1 \right) \cdot \phi \right\rfloor &= n-\left\lfloor \left( n\cdot \phi - e + 1 \right) \cdot \phi \right\rfloor \\ &= n-\left\lfloor n\cdot {\phi}^2 - e\cdot \phi + \phi \right\rfloor \\ &= n-\left\lfloor n\cdot \left(1-\phi\right) - e\cdot \phi + \phi \right\rfloor \\ ...

Here's a nice little problem from the 13th All Soviet Union Mathematics Olympiad. Given a set of real numbers $S$ containing 0 and 1 that's closed under finite means, show that it contains all rational numbers in the interval $\left[0,1\right]$. This isn't a difficult problem, but there's a particularly nice solution. First observe that if $x\in S$ then both $\frac{x}{4}$ and $\frac{3x}{4}$ are in $S$; average $\{0,x\}$ to get $\frac{x}{2}$, average $\{0, \frac{x}{2}\}$ to get $\frac{x}{4}$, average $\{\frac{x}{2}, x\}$ to get $\frac{3x}{4}$. We could show any rational number $\frac{m}{n}$ with $1\leq m < n$ is in $S$ if we had $n$ distinct elements from $S$ that summed to $m$. Let's exhibit one. Start with an array with $m$ 1s on the left, $n-m$ 0s on the right. Repeatedly replace adjacent $x,y$ values with $\frac{3(x+y)}{4}, \frac{(x+y)}{4}$, where either $x=1,y\neq1$ or $x\neq 0, y=0$, until there is one...