Let there be a finite set of positive integers such that:
(a) no two members of the set are equal
(b) the sum of the inverse of each member of the set is equal to one
The smallest set (as defined by its number of members) is simply $$. The next is $[2, 3, 6]$.
There are (by my count) six different sets which have four members. However, all of them include the number '$2$'. This is self-evident, as, if they included only numbers greater than $2$, the greatest that four members could sum to is $1/3 + 1/4 + 1/5 + 1/6 = 19/20$.
Consequently, there is a way to have a five member set which doesn't use $2$. $[3, 4, 5, 6, 20]$.
What is the smallest set which uses neither $2$ nor $3$? It would seem possible to do it with $7$ members, as $1/4 + 1/5 + ... 1/10 = 1.095...$ However, I could not find a way to sum to exactly one with $7$ members. The smallest set appears to have $8$ members, such as $[4,5,6,8,10,12,20,40]$.
The general question, then, is this. What is the relationship between $n$ (minimal number permitted to be a member of the set) and $m$ (members of the minimal set). So far, we have $n=1,m=1$; $n=2,m=3$; $n=3,m=5$; $n=4,m=8$.
- Can anyone extend this series by considering $n=5$; or solve the general problem? Thanks