Nowadays, we usually write noninteger numbers either as fractions (2/7) or decimals (0.285714). The floating point representation used in computers is another representation very similar to decimals. But the ancient Egyptians (as far as we can tell from the documents now surviving) used a number system based on unit fractions: fractions with one in the numerator. This idea let them represent numbers like 1/7 easily enough; other numbers such as 2/7 were represented as sums of unit fractions (e.g. 2/7 = 1/4 +1/28). Further, the same fraction could not be used twice (so 2/7 = 1/7 + 1/7 is not allowed). We call a formula representing a sum of distinct unit fractions an Egyptian fraction.
This notation is cumbersome and difficult to compute with, so the Egyptian scribes made large tables so they could look up the answers to arithmetic problems. However there is also some interesting mathematics associated with the problem of converting modern fraction notation into the Egyptian form. A number of famous mathematicians have looked at this problem, and invented different ways of doing this conversion process. Each of these methods has advantages and disadvantages in terms of the complexity of the Egyptian fraction representations it produces and in terms of the amount of time the conversion process takes. There are still some unsolved problems about whether some of these processes finish, or whether they get into an infinite loop.
To investigate some of these questions, I wrote a Mathematica notebook, now called "Algorithms for Egyptian Fractions", which as the title implies implements on the computer some of these computation methods. A version of this notebook was published as "Ten Algorithms for Egyptian Fractions" in Mathematica in Education and Research, vol. 4, no. 2, 1995, pp. 515, available online through MathSource.
Since then I have made several changes including improvements to the binary remainder method and two new sections on reverse greedy methods and brute force searches. I am making this updated version available here.
1  1  1  1  1  1  1  1  1  1  1  1  
1 = 

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6  7  8  9  10  14  15  18  20  24  28  30 
1  1  1  1  1  
k = 

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+ ··· + 

2  3  4  5  n 
(Hint: consider the largest power of two less than or equal to n.)
Numbertheoretic hacks, David Eppstein, Dept. Inf. & Comp. Sci., UC Irvine.
Last update: 23 Jun 2004, 22:43:43 PDT