Source: Philippe Fondanaiche, citing Tangente n° 65/66 L. Pianaro, with his original extensions.
I think the key here is weeding out fractions with large factors in their denominators that can't possibly add to exactly 1 because there aren't enough other fractions sharing that factor. At least, it's a key first step. The fraction 1/7, plus anything, is not going to equal 1 if none of the other fractions have 7 in their denominators, but if they do, you can make 1/7 + 1/14 + 1/21 + 1/42 = 1/6 and then 1/2 + 1/3 + 1/7 + 1/14 + 1/21 + 1/42 = 1. The factors to consider here are not only primes, but also powers of primes. (i.e., 1/9 + any number of 1/3, 1/6, 1/12, 1/15, 1/21, etc. are not going to make 1 if some of the other fractions don't have a factor of 9 in the denominator). 1. We ignore 1/7 and 1/9 because no sum of the other numbers is going to give 7ths or 9ths in the denominator. Also, 1/5 and 1/10 are not enough to add up to anything (1/10, 2/10, and 3/10 are going to leave tenths left over no matter what else you add). What's left is 1/2, 1/3, 1/4, 1/6, 1/8. The sum of all these is 11/8, so we need all of them except 3/8, which means 1/2 + 1/3 + 1/6, the only way to do this with egyptian fractions whose denominators are 10 or less. 2. Ignore 1/13, 1/16, 1/17, 1/19, 1/23, 1/25 immediately. Ignore 1/11 & 1/22 for the same reason we ignored 5ths in part 1. Similarly, 1/7, 1/14, and 1/21 are going to leave us stuck with some number of 42nds from 1 to 6, so ignore those too. 1/9 + 1/18 = 3/18 = 1/6 so this can go in our solution (but only together) 1/8 + 1/24 = 4/24 = 1/6 so this can go in our solution (but only together) 1/5, 1/10, 1/15, 1/20 = (12, 6, 4, 3)/60, which allows us to see that 1/5 + 1/20 = 5/20 = 1/4 and 1/10 + 1/15 = 10/60 = 1/6, so the 5ths are usable (in these pairs). So here's what usable numbers we have: 1/12 1/9 + 1/18 = 1/6 1/8 + 1/24 = 1/6 1/10 + 1/15 = 1/6 1/6 1/5 + 1/20 = 1/4 1/4 1/3 1/2 Now, if we simply take all the smallest usable fractions, we get 1/12 + 1/6 + 1/6 + 1/6 + 1/6 + 1/4 = 1, or, explicitly, 1/24 + 1/20 + 1/18 + 1/15 + 1/12 + 1/10 + 1/9 + 1/8 + 1/6 + 1/5 = 1 It would be nice if it was always this simple, but I expect it will not be, as the set of usable fractions gets larger. I may solve the N<=50 and 100 problems later. 5. This procedure, starting with the largest prime or prime-power factors present in the set, and determining what combinations the fractions with those powers can possibly appear in the solution, seems to be a general approach. You clearly need to try to use smaller fractions, and this process generally takes you through the smaller fractions first. I don't necessarily know that the best answer will be the first one you can make from fractions considered in this approach, though.
EGYPTIAN FRACTIONS
N is the maximum value of the denominators.
m(N) is the corresponding maximum number of different egyptian fractions
such that their sum is equal to 1.
The series described hereafter are of the form S = [a,b,c,d,e...]such as:
a < b < c < d < e......<= 100
1/a + 1/b + 1/c + 1/d +.... = 1
1) N <=10
m(10) = 3
S = [2,3,6]
2) N<=25
m(25) = 10
S = [5,6,8,9,10,12,15,18,20,24]
3) N<=50
m(50) = 19
S = [7,10,11,14,15,16,18,20,21,22,24,28,30,33,35,36,40,42,48]
4) N<=100
m(100) = 43
S=[17,18,21,22,24,25,26,27,28,30,32,33,34,38,39,40,44,45,48,50,52,54,55,56,60,
63,66,70,72,75,76,77,78,80,84,85,88,90,91,95,96,99,100]
How to proceed:
For any integer a, it is natural to consider all the possible ways allowing
to split 1/a according to the following partitions:
1/a = 1/b + 1/c with c>b>a (1)
or
1/a = 1/b + 1/c +1/d with d>c>b>a (2)
or
1/a = 1/b + 1/c +1/d +1/e with e>d>c>b>a (3)
etc..
Some partitions are well known, such as:
1/a = 1/(a+1) + 1/[a*(a+1)] whatever a>0
or
1/[a*(a*b+a*c+b*c)] +1/[b*(a*b+a*c+b*c)] +1/[c*(a*b+a*c+b*c)] = 1/[a*b*c]
whatever a,b,c>0
But a simple computer program allows to determine the comprehensive list of
the possible representations:
Partitions of category (1):
a b c
2 3 6
3 4 12
4 5 20
6 12
5 6 30
6 7 42
8 24
9 18
10 15
7 8 56
8 9 72
10 40
12 24
9 10 90
12 36
10 12 60
14 35
15 30
12 14 84
15 60
16 48
18 36
20 30
21 28
14 18 63
21 42
15 18 90
20 60
24 40
16 20 80
24 48
18 22 99
24 72
27 54
30 45
20 28 70
30 60
36 45
21 28 84
30 70
22 33 66
24 32 96
33 88
36 72
40 60
42 56
26 39 78
28 42 84
44 77
30 45 90
48 80
50 75
55 66
32 48 96
35 60 84
36 60 90
63 84
40 72 90
42 78 91
Partitions of the category (2)
a b c d
2 3 7 42
3 8 24
3 9 18
3 10 15
4 5 20
4 6 12
3 4 14 84
4 15 60
4 16 48
4 18 36
4 20 30
4 21 28
5 9 45
5 10 30
5 12 20
6 7 42
6 8 24
6 9 18
6 10 15
etc...up to
28 78 84 91
All the integers <=100 are in these partitions except:
¤ the prime numbers > or = 23 and their multiple,
¤ the multiples 51 and 68 of the prime number 17,
¤ the squares 49,64,81 and the multiple 98 of 49.
Globally there are 66 integers appearing in the partitions.
Starting from 1/2 + 1/3 + 1/6 = 1, I have created an arborescence by
replacing each fraction
by a partition of the category (1) or of the category (2).
Due to the fact that some fractions rarely appear
in the possible partitions such as:
1/76 and 1/95 in 1/38 + 1/76 + 1/95 = 1/20
or
1/17 and 1/34 in 1/10 = 1/17 + 1/34 + 1/85,
it is relatively easy to get quickly a representation with 43 fractions.
Is it possible to get 44 fractions or more ? I hand on the torch to the
readers of your POTW...
PS There are many sites on the Web about Egyptian fractions (Stan Wagon, David
Eppstein, Eric Weissteins's treasure, K.S. Brown math pages and so on .....)