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An Egyptian fraction is a finite sum of distinct unit fractions, such as $$\frac{43}{48} = \frac{1}{2}+\frac{1}{3}+\frac{1}{16}.$$

Does there exist a number in the range $(0.5, 1)$ that when written as an Egyptian fraction, none of the representations with the minimum length include $\frac{1}{2}$?

For example, I proved computationally (by checking all examples in my other question) that if you can write a number in the range $(0.5, 1)$ as $\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}$ you can also write it as $\frac{1}{2}+\frac{1}{s_1}+\frac{1}{s_2}$. I'm wondering if it is a general rule for any minimum length.

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  • $\begingroup$ Is $1/5+1/7+1/9+1/11$ of necessary form $1/2+1/a+1/b+1/c$? $\endgroup$
    – Fedor Petrov
    Commented Oct 30, 2023 at 14:07
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    $\begingroup$ This was very hard to find but yes. 1/2+1/23+1/715+1/2072070 $\endgroup$
    – Peyman
    Commented Oct 30, 2023 at 14:37
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    $\begingroup$ $1/2+1/30+1/90+1/2310$ has smaller maximal denominator. $\endgroup$
    – Peter Mueller
    Commented Oct 30, 2023 at 16:06
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    $\begingroup$ $\ldots$ or even smaller: $1/2+1/33+1/105+1/198$. $\endgroup$
    – Peter Mueller
    Commented Oct 30, 2023 at 19:32
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    $\begingroup$ @Peyman If I want to write the reduced fraction $n/d$ as an Egyptian fraction with $\le r$ summands, I do the following: For integers $k=1,2,\ldots$ I consider binary variables $a[i]$, where $i$ runs through the divisors of $kd$, and try to solve the binary linear program $\sum a[i]\le r$ and $\sum ia[i]=kn$. $\endgroup$
    – Peter Mueller
    Commented Oct 31, 2023 at 7:04

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Suppose this is known for all Egyptian fractions with minimal representation in $k$ fractions. Then if there's a counterexample in $k+1$ fractions $\frac{1}{n_1} + \frac{1}{n_2} + \cdots \frac{1}{n_{k+1}}$ we require $\frac{1}{n_1} + \frac{1}{n_2} + \cdots \frac{1}{n_k} < \frac12$. This allows fast enumeration of potential counterexamples: your claim for sums of 3 fractions only requires 75 sums to be tested, and for sums of 4 fractions there are 11122 sums to be tested. By brute force, there are two counterexamples among sums of four fractions:

$$\frac{1}{3} + \frac{1}{7} + \frac{1}{43} + \frac{1}{47} = \frac{1}{2} + \frac{1759}{84882} \\ \frac{1}{3} + \frac{1}{7} + \frac{1}{43} + \frac{1}{137} = \frac{1}{2} + \frac{1669}{247422}$$ where $\frac{1759}{84882}$ and $\frac{1669}{247422}$ cannot be expressed as sums of three or fewer unit fractions.

The main tool for efficient calculation is that if $n_1 > n_2 > \cdots > n_k$ then $\frac{1}{n_1} + \frac{1}{n_2} + \cdots \frac{1}{n_k} < \frac{k}{n_k}$. The other optimisation which makes a complete search of the space take seconds rather than hours is efficient determination of two-part Egyptian fractions. If $\frac{n}{d} = \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$ where wlog $\gcd(n,d) = 1$ then there is some $c$ for which $a+b = cn$ and $ab = cd$. Then $$\{a,b\} = \frac{cn \pm \sqrt{c^2 n^2 - 4cd}}{2}$$ Let $c = u^2 v$ where $v$ is squarefree. Then $v(u^2 vn^2-4d) \in \square$ so $v \mid 4d$. Let $w = \frac{4d}v$. Now we have $$\{a,b\} = \frac{u^2 vn \pm uv\sqrt{u^2 n^2 - w}}{2}$$ If the square root evaluates to $s$ then $u^2 n^2 - w = s^2$ so $w = (un-s)(un+s)$, and by factoring $4d$ we can quickly enumerate all candidates for $v$ and $u$.

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  • $\begingroup$ Thanks. How can we show that mentioned fraction doesn't have a representation of three unit fractions? $\endgroup$
    – Peyman
    Commented Nov 2, 2023 at 15:08
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    $\begingroup$ @Peyman If $a=1/x+1/y+1/z$ for $x<y<z$, then $x<3/a$, leaving finitely many $x$ to check. Given $x$, we have $1/y > (a-1/x)/2$, so finitely many $y$'s to consider. Now solve for the $z$'s and check if the numerator is $1$. $\endgroup$
    – Peter Mueller
    Commented Nov 2, 2023 at 15:43
  • $\begingroup$ A bit simpler argument: $\frac{n}d = \frac{a+b}{ab}$ is equivalent to $(na-d)(nb-d) = d^2$, which reduces the problem of finding $a,b$ to factoring of $d^2$. $\endgroup$
    – Max Alekseyev
    Commented Nov 3, 2023 at 22:31

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