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Distributing the natural numbers as circles evenly along the Archimedean spiral yields surprising patterns when changing the radius of the circles: they cover more and more of the plane, finally covering it completely. But shortly before this happens, more or less obvious and intricate patterns shortly show up, and I'd like to understand them.

The numbers are arranged by these formulas with $\hat k = \frac{\sqrt{k}}{2}$

$$x_\alpha(k) = -\hat k\cos(\alpha\cdot 2 \pi\cdot \hat k)$$ $$y_\alpha(k) = -\hat k\sin(\alpha\cdot 2 \pi\cdot \hat k)$$

– the distance between consecutive numbers along the spiral being controlled by the parameter $\alpha$.

This is how the number spirals look like for $\alpha = \frac{\sqrt{2}}{2}, 1, \sqrt{2}, 2,2 \sqrt{2},4, \ldots = \sqrt{2}^k$:

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[Click image to enlarge.]

When the radius of the circles is enlarged until they almost cover the plane, one observes different spot patterns: a spiral, a cross with $8$-fold rotation symmetry, another cross with $4$-fold rotation symmetry and a series of horizontal stripes.

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[Click image to enlarge.]

When we compare $\alpha = 0.99, 1, 1.01$ we see that the straight cross for $\alpha = 1$ is the limit of two bundles of eight spirals running in opposite directions.

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[Click image to enlarge.]

Here is my question:

How can the cross with the 8-fold rotational symmetry (for $\alpha=1$) be explained? ("For $\alpha=1$ there is a cross with a 8-fold rotational symmetry, because ....")

Note that for $\alpha=1$ the square numbers are aligned along the horizontal axis:

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One final remark: The observed patterns come as a surprise (somehow) because "from the distance" the spiral looks almost like a set of concentric circles thus having seemingly total rotational symmetry. But in fact the spiral has no rotational symmetry at all (only for 360°). And the ever changing patterns are due to this fact: the concentric circles do show only patterns reflecting the density of numbers along them.

In this example the density is $10$ per $2\pi$. Note how the triangle numbers (compared to the square numbers in the case of the spiral) are arranged along the horizontal line. Let $\triangle(k) = \frac{k(k+1)}{2}$ take the values $0,1,3,6,10,15,\dots$ We observe along the horizontal line $n(k) = 10\triangle(k) + 1$ (to the right) and $m(k) = n(k) + 5k$ (to the left):

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To see how the spiral patterns continue here for $\alpha = 4\sqrt{2}, 8, 8\sqrt{2}$:

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There's a little Youtube video where you can see the number spiral continuously change for $\alpha$ going from 1 to 4.

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Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.

In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.

Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).

This also explains the other symmetric patterns. We can work out that if $\alpha = \sqrt 2^c$ for $c\le 3$, then the circles that lie just before the positive $x$-axis will be $k=2^{-c}\cdot((2n+1)^2-1) = 2^{2-c}\cdot n\cdot(n+1)$, which implies that there will be $2^{3-c}\cdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^{3-c}$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 \le c \le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.

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  • $\begingroup$ Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument. $\endgroup$ – Hans-Peter Stricker Mar 28 at 11:15
  • $\begingroup$ You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $\alpha$ going from 1 to 8. $\endgroup$ – Hans-Peter Stricker Mar 28 at 16:33
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This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $\alpha=2$, which corresponds to the function $z_k:=\sqrt{k}e^{2\pi i\sqrt{k}}$ (it is different from OP's function by a half). Take different cases of $k$.

(i) $k=n^2+m$. $$z_n=x_n+iy_n=\sqrt{n^2+m}\,e^{2\pi i\sqrt{n^2+m}}=\sqrt{n^2+m}\,e^{2\pi i n(1+\frac{m}{n^2})^{1/2}}\approx ne^{\pi im/n}$$ so $x_n\approx n$, $y_n\approx n(\frac{\pi m}{n}-\frac{m^3\pi^3}{6n^3})\approx\pi m-\frac{m^3\pi^3}{6n^2}$, that is $$y_n\approx \pi m-\frac{A}{x_n^2}\to \pi m$$ This explains the right-hand lines in OP's diagrams with $m=\ldots,-1,0,1,2,\ldots$.

(ii) $k=n^2+n+m$. $$z_n=\sqrt{n^2+n+m}\,e^{2\pi i\sqrt{n^2+n+m}}\approx (n+\tfrac{1}{2})e^{2\pi in}e^{\pi i}e^{\pi i m/n}\approx-(n+\tfrac{1}{2})e^{\pi im/n}$$ This is similar to case (i) but reflected, so the lines appear on the left side.

Now consider $\alpha=2\sqrt{2}$ which corresponds to $z_k:=\sqrt{k}e^{2\pi i\sqrt{2k}}$. The same type of analysis gives $$z_n\approx n\,e^{2\pi im/n}$$

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  • $\begingroup$ You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $\alpha$ going from 1 to 8. $\endgroup$ – Hans-Peter Stricker Mar 28 at 16:34
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To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,\ldots$ with $a_k = 8\triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,\ldots$ with $b_k = a_k +1$, $k=0,1,2,\ldots$.

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For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:

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What needs to be shown analytically is that for example for the numbers $a_{k},b_k, a_{k+1}, b_{k+1}$ we have

$$|p(a_k) - p(b_k)| \approx |p(a_{k+1}) - p(b_{k+1})|$$

and

$$|p(a_k) - p(a_{k+1})| \approx |p(b_k) - p(b_{k+1})|$$

with $p(k) = \langle x(k), y(k)\rangle$ and $x(k) = -\hat k\cos(2 \pi\cdot \hat k)$, $y(k) = -\hat k\sin(2 \pi\cdot \hat k)$ and $\hat k = \frac{\sqrt{k}}{2}$.

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  • $\begingroup$ Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_{k+1})$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_{k+1})$), since the line between $p(a_k)$ and $p(b_{k})$ is perpendicular to centre. $\endgroup$ – Milten Mar 28 at 11:42
  • $\begingroup$ I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $\frac{\pi}{8n}$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though. $\endgroup$ – Milten Mar 28 at 11:51
  • $\begingroup$ Can you give an analytical expression for this requirement, please? I would add it to my answer. $\endgroup$ – Hans-Peter Stricker Mar 28 at 11:52
  • $\begingroup$ Did it. But still there's the ugly gap between 23 and 9. $\endgroup$ – Hans-Peter Stricker Mar 28 at 12:08
  • $\begingroup$ Hmm, we could use the dot product: $(p(a_{k+1}-p(a_k))\cdot(p(a_k)-p(b_k)) \approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_{k+1}) \parallel p(a_k)$, which means they have the same angle. So we could write it as $2\pi \hat a_{k+1} \approx 2\pi (\hat a_{k}+1) \iff \sqrt{a_{k+1}} \approx \sqrt{a_k} + 2$. (+1 since we go once around the spiral). $\endgroup$ – Milten Mar 28 at 12:16

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