*Also posted on the Math Stackexchange: When is the number of areas obtained by cutting a circle with $n$ chords a power of $2$?*

# Introduction

Recently, a friend told me about the following interesting fact:

Place $n$ points on a circle and draw a line between every pair of points. Suppose that no three lines intersect at one point. Then the number of regions which are separated by the lines is equal to the sum of the first five numbers in the $n-1$st row of Pascal's triangle!

See this image image (from Wikipedia). Here, $n$ is the number of points, $c$ is the number of lines and $r_G$ is the number of regions:

Here is a great video by 3Blue1Brown on this subject: Circle Division Solution. The series is A000127 in the OEIS.

# Preliminary results

The following is known (see again Wikipedia for instance):

For $n$ points, the number of resulting regions is $$1+\binom n2+\binom n4 = \sum_{i=0}^4 \binom{n-1}i=\text{sum of first } 5 \text{ numbers in $n$th row of Pascal's triang.}=\frac{1}{24}n(n^3-6n^2+23n-18)+1.$$

In particular, for $n\in\{1,2,3,4,5,10\}$, the number of areas is a power of $2$.

# My question

**Is it true that, for any other $n$, the number of areas is not a power of two?**

# Some attempts

First off, we can simply check that for $n\in\{6,7,8,9\}$, the number of areas is not a power of two. So the question is equivalent to: *Is it true that, for any $n\geq 11$, the number of areas is not a power of $2$?*

The following Proposition is easy to prove:

**Proposition.** For $n> 5$, we have that $f(n)< 2^{n-1}$, where $f(n)$ denotes the number of regions.

*Proof.* For $n>5$ we have $$f(n)=\sum_{i=0}^{n-1} \binom{n-1}i-\sum_{i=5}^{n-1}\binom{n-1}i = 2^{n-1}-\sum_{i=5}^{n-1}\binom{n-1}i<2^{n-1}.\square$$

However, this only proves that $f(n)\neq 2^{n-1}$ for any $n>6$. There could still be some $m\in\mathbb N$ with $m<n$ such that $f(n)=2^m$.

User Pazzaz at MSE said that he checked all cases up to $n=10^{10}$ and none of them were powers of $2$.