Let $r$ be a rational number with $\frac{1}{2}<r<1$
Let $(a_1,\cdots,a_k)$ be a solution of $r=\frac{1}{a_1}+\cdots +\frac{1}{a_k}$ with distinct positive integers $a_1,\cdots , a_k$ , in other words , a representation as a sum of dinstinct positive egyptian fractions with length $k$. WLOG, we can assume $a_1<\cdots <a_k$.
Let us call a solution optimal, if the length $k$ is as small as possible.
Prove or disprove the following conjecture : One of the optimal solutions has denominator $2$. In other words, if we want to find one of the shortest egyptian representation of a number $r$ with $\frac{1}{2}<r<1$ , we can always begin with $\frac{1}{2}$.
The greedy algorithm (always use the smallest possible denominator) can lead to catastrophical results , for example for $\frac{11}{199}$ we get length $11$ whereas the optimal length is $3$. The conjecture asks , if "greedy" can be used at least for the first denominator for the case that it is $2$.
- Brute force check reveals that the conjecture is true for optimal length $2$ or $3$.
- If an optimal solution contains $3$ as the smallest denominator and $4-6$ for the second-smallest, we can easily show that the conjecture is true. In some cases we can make the length even smaller by using denominator $2$. If desired, I can further work this out.
Any idea or reference ? I thought about induction over $k$ , but the induction step does not work if the given representation has the property $a_1+\cdots a_{k-1}<\frac{1}{2}$.
Final question : Can this be extended to every rational $r$ with $0<r<1$ concerning the choice of the first denominator ? Can we always begin with the smallest possible denominator ?