Here’s the simplest possible Egyptian fraction summing to 1:

1 = 1/2 + 1/3 + 1/6 = egypt(2,3,6)

But how many times does 1 = egypt()? Infinitely often, as is very easy to prove. Take this equation:

1/6 – 1/7 = 1/42

For any 1/n, 1/n – 1/(n+1) = 1/(n*(n+1)) = 1/(n^2 + n). In the case of 1/6, the formula means that you can re-write egypt(2,3,6) like this:

1 = egypt(2,3,7,42) = 1/2 + 1/3 + 1/7 + 1/42

Now try these equations:

1/6 – 1/8 = 1/24
1/6 – 1/9 = 1/18
1/6 – 1/10 = 1/15

Which lead to these re-writes of egypt(2,3,6):

1 = egypt(2,3,8,24)
1 = egypt(2,3,9,18)
1 = egypt(2,3,10,15)

Alternatively, you can expand the 1/3 of egypt(2,3,6):

1/3 – 1/4 = 1/12

Therefore:

1 = egypt(2,4,6,12)

And the 1/12 opens all these possibilities:

1/12 – 1/13 = 1/156 → 1 = egypt(2,4,6,13,156)
1/12 – 1/14 = 1/84 → 1 = egypt(2,4,6,14,84)
1/12 – 1/15 = 1/60 → 1 = egypt(2,4,6,15,60)
1/12 – 1/16 = 1/48 → 1 = egypt(2,4,6,16,48)
1/12 – 1/18 = 1/36 → 1 = egypt(2,4,6,18,36)
1/12 – 1/20 = 1/30 → 1 = egypt(2,4,6,20,30)
1/12 – 1/21 = 1/28 → 1 = egypt(2,4,6,21,28)

So you can expand an Egyptian fraction for ever. If you stick to expanding the 1/6 to 1/7 + 1/42, then the 1/42 to 1/43 + 1/1806 and so on, you get this:

1 = egypt(2,3,6)
1 = egypt(2,3,7,42)
1 = egypt(2,3,7,43,1806)
1 = egypt(2,3,7,43,1807,3263442)
1 = egypt(2,3,7,43,1807,3263443,10650056950806)
1 = egypt(2,3,7,43,1807,3263443,10650056950807,113423713055421844361000442)
1 = egypt(2,3,7,43,1807,3263443,10650056950807,113423713055421844361000443,12864938683278671740537145998360961546653259485195806)
1 = egypt(2,3,7,43,1807,3263443,10650056950807,113423713055421844361000443,12864938683278671740537145998360961546653259485195807,165506647324519964198468195444439180017513152706377497841851388766535868639572406808911988131737645185442)
[…]

Elsewhere Other-Accessible…

• A000058, Sylvester’s Sequence, at the Online Encyclopedia of Integer Sequences, with more details on the numbers above