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:$$ \begin{align} 2+3+11+22+33 &= 2+5+8+12+20+24 \text{ and both total }71\\2+4+9+12+18 &= 2+5+6+12+20 \text{ and both total }45 \end{align} $$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.

Irrational Number Approximations by Egyptian Fractions

I've written about Egyptian fractions before in a post titled The Greedy Algorithm on the 2nd August 2020. However, this post deals with ways of representing proper fractions, like 5/7, as Egyptian fractions. How do we approximate an irrational number such as \(\pi\) by Egyptian fractions?

The question arose because the number associated with my diurnal age today, 26997, has a property that qualifies for admission in OEIS A132556:

A132556

Egyptian fraction representation for the cube root of 82.

In the case of \(82^{1/3}\) one obvious way of achieving this is to first get the initial digits of its digital representation. We find that:$$82^{1/3} \approx 4.3444814857686119017$$This means that the proper fraction to convert into an Egyptian fraction is:$$\frac{3444814857686119017}{10000000000000000000}$$Taking this fraction and using this algorithm, we find that:$$82^{1/3} \approx 4+\frac{1}{3}+ \frac{1}{90}+ \frac{1}{26997} +  \frac{1}{5832713646}+ \dots$$These initial fractions are in agreement with the OEIS output but subsequent ones are not. Of course, while the Egyptian fractions were ideally suited to the purposes for which the Egyptians put them, they were never intended as approximations for irrational numbers of whose existence they were oblivious.

Egyptian fraction representations are not unique and especially so when dealing with approximations of irrational numbers. The Engels expansion is an alternative and thus we have:$$82^{1/3} \approx 4+ \frac{1}{3}+ \frac{1}{90}+ \frac{1}{27000}+ \frac{1}{233280000} + \dots$$The best rational approximations for irrational numbers remain the progressive approximations afforded by the number's continued fraction. In case of the cube root of 82 we have the following progressive approximations:$$\frac{9}{2},\frac{ 13}{3}, \frac{126}{29}, \frac{391}{90}, \frac{ 1299}{299}, \frac{ 121198}{27897}, \frac{ 486091}{111887},\frac{ 2065562}{475445}$$Figure 1 shows an interesting and relevant response to the Quora question: Is there any pattern in the Egyptian fraction representation of Pi?

Figure 1