<s>[INST] Parametrizing the square spiral Related to this question concerning number spirals I have another one, more specific.
While it is rather easy to arrange the natural numbers along an Archimedean spiral by
$$x(n) = \sqrt{n}\cos(2\pi\sqrt{n})$$
$$y(n) = \sqrt{n}\sin(2\pi\sqrt{n})$$
it's much more difficult to arrange them along a square spiral by a closed formula. It's very easy to construct the square spiral algorithmically ("go along straight lines and always turn to the right if possible") but I'm totally stuck how the functions $x(n),y(n)$ would look like as formulaic expressions. The only thing I'm quite sure about is that they will make use of the square root function - but how are the "radii" and the turns coded?
This is how the two spirals look like (the Archimedean spiral being appropriately scaled and rotated to align the square numbers):

And here you can see them being morphed into each other.
Can anyone give me a hint (or the solution)?
 [/INST] [This answer is inspired by user SMM's answer. Thanks for it.]

Consider "piecewise linear approximations" of the sine and cosine function, periodically defined on the unit interval, i.e. $x \in [0,1]$. 
Let
$$\boxed{\cos_\bigcirc(x) = \cos(2\pi x)\\\sin_\bigcirc(x) = \sin(2\pi x)}$$
and compare this to
$$\boxed{\cos_\square(x) = \begin{cases}
+1 & \text{ for } \frac{0}{8} \leq x \leq \frac{1}{8} \\
+2 - 8x & \text{ for } \frac{1}{8} \leq x \leq \frac{3}{8} \\
-1  & \text{ for } \frac{3}{8} \leq x \leq \frac{5}{8} \\
-6 + 8x & \text{ for } \frac{5}{8} \leq x \leq \frac{7}{8} \\
+1 & \text{ for } \frac{7}{8} \leq x \leq \frac{8}{8} \\
\end{cases}
\\ \\\sin_\square(x) = \begin{cases}
+0 + 8x & \text{ for } \frac{0}{8} \leq x \leq \frac{1}{8} \\
+1 & \text{ for } \frac{1}{8} \leq x \leq \frac{3}{8} \\
+4 - 8x & \text{ for } \frac{3}{8} \leq x \leq \frac{5}{8} \\
-1 & \text{ for } \frac{5}{8} \leq x \leq \frac{7}{8} \\
-8 + 8x & \text{ for } \frac{7}{8} \leq x \leq \frac{8}{8} \\
\end{cases}}$$
or written more readable:
$$\cos_\square(x) = \begin{cases}
+1 & \text{ for } 0 \leq 8x \leq 1 \\
+2 - 8x & \text{ for } 1 \leq 8x \leq 3 \\
-1  & \text{ for } 3 \leq 8x \leq 5 \\
-6 + 8x & \text{ for } 5 \leq 8x \leq7 \\
+1 & \text{ for }7 \leq 8x \leq 8 \\
\end{cases}
\\ \\\sin_\square(x) = \begin{cases}
+0 + 8x & \text{ for } 0 \leq 8x \leq 1 \\
+1 & \text{ for } 1 \leq 8x \leq 3 \\
+4 - 8x & \text{ for } 3 \leq 8x \leq 5 \\
-1 & \text{ for } 5 \leq 8x \leq7 \\
-8 + 8x & \text{ for }7 \leq 8x \leq 8 \\
\end{cases}$$
These are the plots:

$\cos_\square$ and $\sin_\square$ are especially well suited to arrange numbers on a square with integer coordinates around the origin with uniform distance $1$ along the square.
This only works for multiples of $8$ with $8n = (2n+1)^2 - (2n-1)^2$. In this case the positions of the $8n$ numbers $k = 0, 1, \dots, 8n-1$ are given by
$$\boxed{x^{(n)}_\square(k) = n\cos_\square(\frac{k}{8n})\\ \\y^{(n)}_\square(k) = n\sin_\square(\frac{k}{8n})}$$
Compare this to the positions of $8n$ numbers on a circle around the origin with uniform distance $\frac{2\pi}{8}$ along the circle:
$$\boxed{x^{(n)}_\bigcirc(k) = n\cos_\bigcirc(\frac{k}{8n})\\
y^{(n)}_\bigcirc(k) = n\sin_\bigcirc(\frac{k}{8n})}$$


For the (circular) Archimedean spiral we have 
$$x_\bigcirc(k) = -\frac{\sqrt{k}}{2}\cos_\bigcirc(\frac{\sqrt{k}}{2}-\frac{1}{8}) $$
$$y_\bigcirc(k) = -\frac{\sqrt{k}}{2}\sin_\bigcirc(\frac{\sqrt{k}}{2}-\frac{1}{8})$$
Written for the sake of comparison with the square spiral:
$$\boxed{x_\bigcirc(k) = - x_\bigcirc^{(\sqrt{k}/2)}(2k-\frac{1}{8})\\
y_\bigcirc(k) = -y_\bigcirc^{(\sqrt{k}/2)}(2k-\frac{1}{8})}$$
Note that the factors $-1$, $\frac{1}{2}$ and the phase $\frac{1}{8}$ (which corresponds to $\frac{\pi}{4}$) where chosen to align the Archimedean with the square spiral, especially the square numbers. 
The formulas
$$x_\bigcirc(k) = \sqrt{k}\cos_\bigcirc(\sqrt{k})$$
$$y_\bigcirc(k) = \sqrt{k}\sin_\bigcirc(\sqrt{k})$$
would give an Archimedean spiral as well.

This is for the square spiral. Let $k'$ be the greatest odd perfect square smaller than $k$. Let $\hat{k} = (\sqrt{k'}-1)/2$. Let $x_\square(0) = 0 $ and $y_\square(0) = 0 $ and for $k > 0$
$$\boxed{x_\square(k) = x_\square^{(\hat k)}(k - k' - \hat k + 1) \\
y_\square(k) = y_\square^{(\hat k)}(k - k' - \hat k + 1)} $$
Note that $k - k' - \hat k + 1$ being negative doesn't pose a problem since $\cos_\square$ and $\sin_\square$ are periodic in both directions.

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