Unity as a Sum of Egyptian Fractions

Continuing with Achmad Damar's "104 Number Theory", I found this little problem interesting. The author asks: let \(k\) be an even number. Is it possible to write 1 as the sum of the reciprocals of \(k\) odd integers? He approaches the problem by assuming that:$$ 1=\frac{1}{n_1}+ \cdots + \frac{1}{n_k}$$for odd integers \(n_1, \dots, n_k \). Removing denominators produces:$$n_1 \cdots n_k=s_1+ \cdots + s_k $$where all \(s_i\) are odd because numbers that are the products of odd numbers are themselves odd. This is impossible because the LHS is odd and the RHS is even because there are an even number (\(k\)) of integers and the sum of each pair of odd  integers is even. Thus it is not possible to represent 1 as the sum of the reciprocals of an even number of odd integers. However, if \(k\) is odd, then it is possible. The example is given of unity expressed as the sums of reciprocals of nine odd integers. See Figure 1.

Figure 1

This got me thinking about how this result was obtained. There are 114 odd integers between 3 and 231 inclusive and 7,032,112,662,630 ways to sample 9 reciprocals at a time (that's over seven trillion ways). Nonetheless, when running this program in SageMathCell, the above combination of fractions was quickly spat out and after that the program timed out. Running the same program in my jupyter notebook (to avert the program timing out), no further combinations were generated. Is this combination of reciprocals unique? I'm not sure.

Certainly we can state that there are infinitely many disjoint sets of positive integers in which the sum of those reciprocals is equal to unity (source). I have a program that generates Egyptian fractions for rational numbers \(p/q\) where \(p<q\). Using this program, I can determine that: $$ \begin{align} \frac{5}{7} &= \frac{1}{2}+\frac{1}{5}+\frac{1}{70}\\ \frac{2}{7} &= \frac{1}{4}+\frac{1}{48}\\1 &= \frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{48}+\frac{1}{70} \end{align}$$Alternatively, I could take another pair of fractions that add to 1 such as 7/11 and 4/11. This gives:$$ \begin{align} \frac{7}{11} &= \frac{1}{2} + \frac{1}{8} + \frac{1}{88} \\ \frac{4}{11} &= \frac{1}{3} + \frac{1}{33} \\ 1 &= \frac{1}{2}+ \frac{1}{3}+ \frac{1}{8} + \frac{1}{33} + \frac{1}{88} \end{align} $$Obviously we could continue this process indefinitely. This Mathematics Stack Exchange source states that:

R. L. Graham showed that \( a(n)>0 \) for \( n>77 \), where \( a(n) \) is the number of ways to express 1 as the sum of distinct unit fractions such that the sum of the denominators is \( n \). This implies that for values of \( n \leq 77 \) such representations may not be possible. In the two examples shown earlier the sum of denominators is greater than 77. However, representations where \(n \leq 77 \) are certainly possible. For example from the same Stack Exchange source we see that:$$ 2+3+11+22+33 = 2+5+8+12+20+24 \text{ and both total }71\\2+4+9+12+18 =2+5+6+12+20 \text{ and both total }45$$and the sum of reciprocals of LHS's and RHS's all total 1.

The simplest representation of 1 using Egyptian fractions is:$$1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$$but once we impose special conditions such as all denominators must be odd or prime or whatever, then things get more complicated. This site offers some useful insights into a problem that can well be explored in far more depth than I've attempted here.