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We can place $n\in\mathbb{N}$ points on the surface of a sphere in a configuration so as to maximize the answer. A plane is defined by $3$ points. We create all $\binom{n}{3}$ planes from the $n$ points on the surface of the sphere. What is a formula or way to compute $f(n)$: the maximum number of regions the sphere can be cut into by the planes?

For example, I hand-counted for the first few $n$. $f(1)=1$, $f(2)=1$, $f(3)=2$, and $f(4)=11$. For $f(5)$, I have explicitly counted $26$ regions for a configuration but according to my lower bound conjecture, I should have $f(5)\geq 41$.

I have conjectures for both the lower and upper bound on $f(n)$.

My conjecture for a lower bound on $f(n)$: $f(n)\geq 1+\binom{n}{3}+6\binom{n}{4}$. This is my conjecture because, first, there is the original region ($1$). Then, as each plane enters the sphere, it adds another region (splitting a previous region) ($\binom{n}{3}$). Note that, for each subset of $4$ points, the $\binom{4}{3}=4$ planes formed using them have $\binom{4}{2}=6$ intersections. Each time a plane intersects withe another plane, it adds $1$ region (by splitting a region into $2$). There are $\binom{n}{4}$ such subsets of $4$ points from the $n$. I am certain that the maximal number of regions cannot be LESS than shown via this count. But I'm not sure whether I should be incorporating the counts from planes formed from larger subsets of the $n$ points.

My conjecture for an upper bound on $f(n)$: $f(n)\leq 1+\binom{n}{3}+\binom{\binom{n}{3}}{2}$. This is my conjecture because, first, there is the original region ($1$). Then, as each plane enters the sphere, it adds another region (splitting a previous region) ($\binom{n}{3}$). Then, as each plane intersects with another plane, it adds another region (splitting a previous region) ($\binom{\binom{n}{3}}{2}$). I am certain that we cannot have MORE regions than shown via this count. But given the additional constraints on the $\binom{n}{3}$ planes (that they all come from just $n$ points on the sphere's surface), $f(n)$ may be less than this count for $n>4$.

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    $\begingroup$ oeis.org/A144841 Maybe this sequence may help you. $\endgroup$ – Meow Mix Nov 25 '16 at 19:46
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    $\begingroup$ One thing to notice is that, by considering stereographic projection, this question is equivalent to: Given $n$ points in the plane, no three colinear, if we draw all circles containing three of these points, into how many regions is the plane partitioned? $\endgroup$ – munchhausen Jun 27 '18 at 13:00

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