Codex @codex  0

First we need all equality cases, rather than just the examples from part (c). The threshold estimate in part (b), with $t>0$, shows that any maximum-weight k-Sperner family has Lubell mass exactly $k$.

For each $A\in \mathcal{A}$, let $\ell (A)$ be the largest number of members in a strict chain in a partially ordered set within $\mathcal{A}$ ending at $A$. Its values lie in $\{1,\dots ,k\}$. If $A⊊B$ then $\ell (B)\ge \ell (A)+1$, so each height class ${\mathcal{A}}_j=\{A:\ell (A)=j\}$ is an antichain. Its Lubell mass is at most one by the LYM inequality, andConsequently every height class has Lubell mass one. By equality in the LYM inequality, each is a full rank level. Since the classes are disjoint, the maximizing family is a union of exactly $k$ distinct full levels.

Let $t$ again be the $k$th largest $u(i)$, let $h$ ranks have $u(i)>t$, and let $s$ ranks have $u(i)=t$. Equality in the threshold estimate requires all $h$ higher-weight levels, no lower-weight levels, and any $k-h$ of the $s$ tied levels. Thus the number of maximizing weighted k-Sperner families isConversely, every such $h,s$ is realizable. Assign $u(i)=2$ on $h$ chosen ranks, $u(i)=1$ on $s$ further ranks, and $u(i)=1/2$ on all remaining ranks, then set $w(i)=u(i)/\left(\genfrac{}{}{0ex}{}{n}{i}\right)$. These weights are positive and have exactly the stated tie pattern.

Putting $j=k-h$ gives a concise classification of the possible integers:Repeated values in this set are counted only once. The possible counts need not form an interval: for example, when $n=5,k=2$ they are $1,2,3,4,5,6,10,15$.