Given a poset $P$, I am interested in the width (size of the maximal antichain) of $\mathcal{O}(P)$, i.e. the poset of downsets in $P$, ordered by inclusion.

As this is rather difficult, I'm starting with a simplification.

Consider a poset $P$ of $n$ distinct chains (i.e. presented as a product of chains), each of height $h$. Now, what is the size of the maximal antichain of subsets of $P$ that contain exactly one element from each chain? (The ordering we use is that a subset is greater than or equal to another subset if at each chain, it selects an element greater than or equal to that selected by the other poset).

By de Bruijn et al ("On the set of divisors of a number," 1952), I believe this question is the same as asking the maximal rank size of the product of the chains. However, that reformulation does not immediately provide a direct combinatorial formula for the calculation.

Here are some observations I have made thus far:

When $h = 1$, then this reduces to the width of antichains in the powerset of the chains, given by Sperner's theorem as $\binom{n}{\lfloor{n/2}\rfloor}$. Note that these are the central coefficients in the binomial triangle. (i.e. [1,2,3,6,10,20,...])

When $h = 2$, manual calculation of the first few terms leads to the sequence [1,3,7,19], which correspond to central trinomial coefficients.

When $h = 3$, manual calculation of the first few terms leads to the sequence [1,4,12], which seem derivable from higher multinomial formulae.

(some basic reference material is in https://link.springer.com/article/10.1007/BF00396270)