Every integer can be expressed as the sum of at most 3 triangular numbers. That is, the set of triangular numbers is an additive basis of order 3. The sum of the inverse triangular numbers is 2. (1/1 + 1/3 + 1/6 + 1/10 + ... = 2)
Is there a set of positive integers, such that every positive integer can be expressed as the sum of at most 3 members of the set, AND that the inverse sums of the members of the set is less than 2? If so, what set exists to minimise this sum of the inverses?
This problem can also be generalised in line with Fermat's polygonal theorem, which states that every positive integer can be expressed as the sum of at most n n-gonal numbers. Thus, our problem with n=4 becomes: is there a set of integers, such that every integer can be expressed as the sum of at most 4 members of the set, AND that the inverse sums of the members of the set is less than Pi^2 / 6 (the sum of the inverse squares). Similar questions can be asked with pentagonal, hexagonal numbers etc.
If you're really keen, can anyone prove my conjecture that that when n=2, no set has a sum of inverses which is finite. For instance, one set is the odd numbers (obviously, every positive integer is either odd or the sum of two odds). 1/1 + 1/3 + 1/5 + ... = infinity. But does a different set exist which has a finite sum of inverses?
Thus, the general question is about the relationship between n and the minimal sum of the inverse members of a set such that each positive integer can be expressed as the sum of at most n members of that set.
My thoughts:
One way to go about the problem for a particular additive basis is to simply choose each subsequent member of the set as the maximum it can be. For instance, with additive basis of 3, we start with 1. 4 is the next number we can't get to, so we include it. We can then get 5 = 4+1; 6 = 4+1+1; but can't get 7, so we include it. Thus, we get a series 1, 4, 7, 10 ... This is not helpful, as its inverses will sum to infinity.
Alternately, we can tweak the system to include the last number which can be reached with the previous members of the set. Thus, with an additive basis of 3, we start with 1. We can reach 3 but not 4, so we include 3 as the next set member. We can then reach 4 (3+1), 5 (3+1+1), 6 (3+3), 7 (3+3+1) but not 8; so we include 7 as the next member. We can now reach 8 (7+1), 9 (7+1+1), 10 (7+3), 11 (7+3+1) but not 12, so we include 11. This process yields the series 1, 3, 7, 11, 15, 19... Again, this is unhelpful, as it has become a linear series. However, the same approach with additive basis 4 looks like it may be fruitul. It yields a set beginning: 1, 4, 9, 22
A different approach would be to start with the Fermat polygonal theorem, and try to generate improvements. For instance, with an additive basis 3, consider the following series: 1, 4, 6, 11, 15, 22... (generated by adding 1 to every second triangular number) This series appears to work. Counter-examples, anyone?
However, none of this sets out a way to generate the optimal series. My conjecture is that for an additive basis of order n; there exists an optimal series in which each subsequent term can be generated through a polynomial of order n-1.