This is an exhibit of the fact, but it isn't really a proof - it doesn't explain why those two functions sum to 1, just shows (arguably, just claims) that they do. You could replace the curve with any function $f$ with $f(\pi/2)=1$.

This isn't a Fibonacci identity per se; for all $a$ and $b$, $(a+b)^3 = a^3+b^3+3ab(a+b)$.

This is true of any even $r$ - the value decreases as long as the iterates are even, and as soon as one iterate is odd it enters the short cycle here (which might be of length 1, if the series goes to 0).

The first of these can be attained much more easily: $(1+\sqrt{x})^{\sqrt{x}} = \sqrt{x}^{\sqrt{x}}\cdot(1+\frac1{\sqrt{x}})^{\sqrt{x}}$, and the latter clearly goes to $e$ as $x\to\infty$.

The combination of this and the other answer make me wonder whether there's any known correlation between the 'Khinchin-ness' of a number and its irrationality measure...

I seem to remember this cropping up in one of the Games Of No Chance collections, but I'll have to dig around for it a bit. AFAIK the value of the initial position is unknown and no position is known to have value $\geq \omega+\omega$ (there are positions known to have value $\omega$ and I believe there are some with value $\omega+n$ for various $n$), but don't hold me to either of those.

I think you want to not just 'replace 1 by -1' but just 'multiply each entry $\equiv 0\pmod p$ by -1'; your code itself gets it right, but the description suggests that once a value has been set to -1 it can't be flipped back.

@EmilJeřábek In recursively enumerable fashion, maybe, but that construction wouldn't be recursive (in some abstract sense) without specifying at least a bound on the size of the largest of those groups.

Gorgeous! May I ask how this was made?

Since there is a symmetry you could try doing a higher-order fit - something like $y^2=p(x)$ with a higher-degree $p()$ - with precise numerical data for a handful of points, and see what your coefficients look like...

@JoeSilverman, David: Those are both excellent explanations - it's great to get both the group-theoretic and geometric versions; thank you!

For my own understanding, why does the latter statement hold? At least at first glance, it's not clear to me that addition 'respects' components; is it just a continuity argument ($P+0$ is, so $P+(1/n)P$ is, so...)?

One nice one, per journals.cambridge.org/… : if all sets of reals are Lebesgue measurable, then the additive groups of $\mathbb{R}$ and $\mathbb{C}$ are (no longer) isomorphic.

A relevant paper for question 1 is arxiv.org/abs/0811.3057 .

Also, for completeness-sake, there is no square-free word over an alphabet of size 2: suppose the alphabet is $\{a,b\}$ and WLOG that the first element of the sequence is $a$. Then the next must be $b$ (else $aa$ is a square), and the next after that must be again $a$ (else $bb$ is a square). But now there's no choice for the fourth element; $a$ will yield $a^2$, while $b$ will yield $(ab)^2$.

@NoamD.Elkies Indeed. ($x$ and $n$ are canonically interchangeable anyway, right? :-) )