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I've seen that the cake numbers give the largest number of regions that can be created by cutting 3-D space by N planes. I have a variation on that question. How many regions can be created by N planes that all pass through the origin (subspaces)? In 2-D, using lines it looks like 2N regions can be created. For 3D and planes I'm not sure, but I imagine it will be less that the cake numbers.

Gene
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    Yep... $2N$. Proof by induction: If $N=1$: two regions. Each additional plane adds $2$ regions. – David G. Stork Sep 19 '18 at 18:37
  • Given that the three coordinate planes cut $\mathbb R^3$ into eight regions (orthants), $2N$ is certainly not the right answer in three dimensions. – Greg Martin Sep 19 '18 at 18:44
  • Thanks David, but I'm looking for the number of regions in 3D cut by planes. I know for 3 planes the answer is 8. It probably grows as a polynomial in N, but at a slower rate than the cake sequence. I need to compare it to the number of vertices in the unit cube which is 2^N. – Gene Sep 19 '18 at 18:45

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Let's first plane $p_0$ be $z=0$. Consider the plane $p'_0$ $z=1$. All partition of space with $p_0$, $p_1$,...$p_n$ will create a pattern of lines $a_1$,... $a_n$ on a plane $p'_0$ and vice versa.

For every flat piece on plane $p'_0$ there are two subspaces: one above $p_0$ and one below. So your number is twice the $(n-1)$-th central polygonal number: $$ k = (n-1)^2+(n-1)+2 = n^2-n+2. $$

Vasily Mitch
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  • I think I see this. Every possible set of cuts by N planes sets up a set of polygonal cones with their vertex at the origin. You have just cut all the cones facing in the z > 0 direction. The most cones is the CPN in N-1 dimensional space. Each cone is reflected through the origin and so we double count. Very elegant. – Gene Sep 19 '18 at 19:19