Due to my not-so-advanced math skills, this question may take a few attempts to state clearly:
Consider the unordered pair (2-tuple) partitions of n (e.g. with n=4, we have {{4,0},{3,1},{2,2}}). Take the low value from each pair and add them together and call this total T1. Take the high value from each pair and add them together and call this total T2. Let r = T2 / T1. For what values of n will r be an integer? What can we conclude about the minimum and maximum values of r, integer or not, for n={1,2,3,...,infinity)?
Clearly the sequence of ratios r is infinite, as is the subset of integer ratios r. The minimum integer is either 0 or 1, depending upon whether or not we consider the null set equal to zero. As for the maximum, it appears that for all r, r < 4 (?), and I assume this has to do with things like modular arithmetic, congruences, and the elements of T2 >= n/2 and the elements of T1 <= n/2.
(The real question) Now restrict the pairs to just those composed of prime numbers (again, with n=4, we would toss out {{4,0},{3,1}} and end up with just {2,2} and 2/2=1. By restricting the pairs in this way and requiring r be an integer, we get what appears to be a finite set of values for n: {4, 6, 16, 18, 20, 32, 52, 72, 102, 180, 3212}. I say "appears" because after 3212, no more terms are found for n < 10^7. I suspect these are all the terms, because as n grows large, T2 will always contain elements that can't be canceled by elements in T1, thus the set is finite. Is this true? If not, is there a way to determine where the next term might be?
One more example with n=18:
Partitions are: {{9,9},{10,8},{11,7},{12,6},{13,5},{14,4},{15,3},{16,2},{17,1},{18,0}}.
{11,7} are both prime, and {13,5} are both prime, so T2 = 11+13 = 24, and T1 = 7 + 5 = 12. r = T2 / T1 = 24 / 12 = 2.
One other observation:
If we generalize and let the elements of T2 >= n/x and the elements of T1 <= n/x, and restrict x to prime number, the integer value of r will tend to be 2x-1 (i.e. in the case above, x=2, 2 is prime, and the integer value of r tends to be 3 = 2(2)-1.
Thanks
Mathematica:
okQ[n_] := Module[{p, q}, p = Select[Prime[Range[PrimePi[n]]], PrimeQ[2 n - #] &]; q = 2 n - p; Mod[Plus @@ q, Plus @@ p] == 0]; 2*Select[Range[2, 5000], okQ]