$Erd\ddot{o}s\ Straus\ Homework$
Brandon Burkholder
10/5/2014
The first function here looks at $\frac{4}{n}$ for each n up to the input integer x. It then attempts to break that function into as many triplets of unit fractions as possible in the form $\frac{4}{n} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ where $a \leq b \leq c$. This is done by first considering the stipulation $a \leq b \leq c$ such that we know the smallest $\frac{1}{a}$ can be is $\frac{1}{3}$ of the total sum of unit fractions. This sets an upper limit on what $a$ can be, because as the indexes get higher $\frac{1}{a}$ gets smaller until it reaches the smallest it can be at $a = \frac{3 \times n}{4}$. Then for each iteration that we try for $a$, we know that the smallest which $\frac{1}{b}$ can be is $\frac{1}{2}$ of what is remaining after we subtract the value $\frac{1}{a}$ from $\frac{4}{n}$. So once again we have an upper limit for what $b$ can be. We then use this nested loop to test subtract every combination of $\frac{1}{a} + \frac{1}{b}$ from $\frac{4}{n}$ and if this result is a unit fraction then we have found a sequence of three Egyptian fractions which satisfie the criteria of the conjecture.

Next we plot the number of solutions for each value of n up to 1000

Now plotting only the prime values of n up to 1000 we can see that the 43rd and 52nd prime have relatively few solutions to the conjecture. These correspond to values of $n$ of 139 and 191 respectively. These numbers do not seem to have anything in common or hold any special significance, seemingly a consequence of the random nature of the number of solutions for a given $n$.

