k-Sperner family

A $k$-Sperner family contains no inclusion chain of length $k+1$. Every maximal chain in a Boolean lattice therefore meets it in at most $k$ members, so its Lubell mass is at most $k$.

For positive weights $w(i)$ depending only on rank, put $u_i=\left(\genfrac{}{}{0ex}{}{n}{i}\right)w(i)$. The largest weight of a k-Sperner family in $\mathcal{P}([n])$ is the sum of the $k$ largest $u_i$. Every maximizing family is a union of $k$ full rank levels. To see the equality assertion, its Lubell mass must be $k$; decomposing it into $k$ antichains by chain height and using equality in the LYM inequality forces each part to be a full level.

If $t$ is the $k$th largest rank weight $u_i$, exactly $h$ levels have weight greater than $t$, and exactly $s$ have weight $t$, then there are $\left(\genfrac{}{}{0ex}{}{s}{k-h}\right)$ maximizing families. For fixed $n,k$, the possible counts are exactly $\left(\genfrac{}{}{0ex}{}{s}{j}\right)$ with $1\le j\le k$ and $j\le s\le n+1-k+j$.

Among $k$-Sperner subfamilies of $\mathcal{P}([n])$, the maximum cardinality is the sum of the $k$ largest binomial coefficients $\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. The union of the corresponding levels attains the bound. For the upper bound, the Lubell mass is at most $k$, and assigning its available mass to the levels with largest binomial coefficients maximizes cardinality.