Difference between lines dividing planes and planes dividing space

Let a(n) represent the number of regions that the plane R2 is broken into by n lines (no 2 of which are parallel, and no 3 of which intersect in a single point).

Let b(n) represent the number of regions that the 3-dimensional space R3 is broken into by n planes (no 2 of which are parallel, and no 3 of which intersect in a shared line).

Find either a recursive formula or a closed-form solution for a(n) and b(n).

So I know that a(n)=C(n+1,2)+1. Why should b(n) be any different? I simply cannot visualize it.

$a(n)$ is known as the lazy caterer's sequence, while $b(n)$ is known as the cake number. They are different starting from $n=3$, notice placing the three "canonical planes" split space into $8$ quadrants, while $a(n)=7$

• Thats useful! Thanks! – Blessoul Jan 28 '15 at 3:59
• It would be useful to show why the cake number is cubic in $n$ while the lazy caterer's number is quadratic. – Ross Millikan Jan 28 '15 at 4:18
• @RossMillikan: It took me over 4 years to come across this question, but I think my answer addresses your comment. – robjohn Sep 29 '19 at 7:27

$$n$$ points divides $$\mathbb{R}^1$$ into at most $$\binom{n}{0}+\binom{n}{1}$$ pieces.

When dividing $$\mathbb{R}^2$$, line $$n$$ meets the $$n-1$$ previous lines at $$n-1$$ points, dividing line $$n$$ into $$\binom{n-1}{0}+\binom{n-1}{1}$$ pieces. Each of those pieces of line $$n$$ divides a piece of $$\mathbb{R}^2$$ in two, adding $$\binom{n-1}{0}+\binom{n-1}{1}$$ pieces of $$\mathbb{R}^2$$. We start with $$1=\binom{n}{0}$$ piece of $$\mathbb{R}^2$$, so after adding $$n$$ lines, we get \begin{align} a(n) &=\binom{n}{0}+\sum_{k=1}^n\left[\binom{k-1}{0}+\binom{k-1}{1}\right]\\ &=\binom{n}{0}+\binom{n}{1}+\binom{n}{2} \end{align} pieces of $$\mathbb{R}^2$$. We used the Hockey Stick Identity to evaluate the summations above.

When dividing $$\mathbb{R}^3$$, plane $$n$$ meets the $$n-1$$ previous planes at $$n-1$$ lines, dividing plane $$n$$ into $$\binom{n-1}{0}+\binom{n-1}{1}+\binom{n-1}{2}$$ pieces. Each of those pieces of plane $$n$$ divides a piece of $$\mathbb{R}^3$$ in two, adding $$\binom{n-1}{0}+\binom{n-1}{1}+\binom{n-1}{2}$$ pieces of $$\mathbb{R}^3$$. We start with $$1=\binom{n}{0}$$ piece of $$\mathbb{R}^3$$, so after adding $$n$$ planes, we get \begin{align} b(n) &=\binom{n}{0}+\sum_{k=1}^n\left[\binom{k-1}{0}+\binom{k-1}{1}+\binom{k-1}{2}\right]\\ &=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3} \end{align} pieces of $$\mathbb{R}^3$$.

Thus, $$b(n)-a(n)=\binom{n}{3}$$.

• Well done!.......... – Ross Millikan Sep 29 '19 at 14:18