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Mar 24, 2014 at 7:26 comment added მამუკა ჯიბლაძე Still slightly simpler expression for $b$: $$ b(m,n)=(m,\frac{n+1}4(8-\sin\frac{\pi n}2-5\,\sin^2\frac{\pi n}2)-1) $$
Mar 23, 2014 at 9:54 comment added მამუკა ჯიბლაძე As for $c$, we could replace it with $bcb$ which is $$ bcb(m,n)=(m,n+2\,\sin\frac{\pi n}2) $$
Mar 23, 2014 at 9:33 comment added მამუკა ჯიბლაძე Interesting idea! Btw your expression can be rewritten as $$ b(m,n)=(m,2n+1-\frac{n+1}4(5+\sin\frac{\pi n}2)\sin^2\frac{\pi n}2). $$ Not that it is much simpler but still...
Mar 11, 2014 at 20:46 comment added Per Alexandersson Ah, I see! Well, still, the Dragon fractal is where I'd start.
Mar 11, 2014 at 20:30 comment added Stefan Kohl As the group $G$ acts transitively on $\mathbb{Z}^2$, the orbit of $(0,0)$ under the action of $G$ is actually $\mathbb{Z}^2$, i.e. "everything" -- thus pictures of this would be pretty boring (everything black). But what my question considers are spheres (not balls!) of certain radii $r$ about $(0,0)$ -- these consist of the points which can be reached from $(0,0)$ by applying $r$ generators or inverses of generators, but no fewer. In particular, any sphere is finite.
Mar 11, 2014 at 19:57 history edited Per Alexandersson CC BY-SA 3.0
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Mar 11, 2014 at 19:26 history answered Per Alexandersson CC BY-SA 3.0