TITLE: Length of the longest chain in dominance order
QUESTION [6 upvotes]: If $\Pi_n$ is the set of partitions of $n$, then for $\lambda, \mu\in \Pi_n$ we say $\mu$ dominates $\lambda$ if $\sum\limits_{i=1}^k \lambda_i \leq \sum\limits_{i=1}^k \mu_i$ for all $k$. This gives a partial order on $\Pi_n$ which is not a total order if $n>5$. For example, $(4,1,1)$ and $(3,3)$ are not comparable in $\Pi_6$. This is closely related to the usual ordering on dominant weights in Lie theory. 
Question 1: What is the length of a maximal chain in the dominance order on $\Pi_n$? Given the Young diagram for $\lambda$, we get something smaller in the dominance order by moving one square down and to the left in such a way that we get another Young diagram (i.e. applying a simple lowering operator). So this is asking for a longest such "path" from the horizontal Young diagram $(n,0,\ldots,0)\in \Pi_n$ to the vertical Young diagram $(1,1,\ldots, 1,1)\in \Pi_n$. 
Question 2: Is there a formula for the number of maximal chains?

REPLY [4 votes]: This is Exercise 7.2(f) in Stanley "Enumerative Combinatorics II": The length of the longest chain in dominance order is
$$\frac{1}{3}m(m^2+3r-1)$$
where $n = \binom{m+1}{2} + r$ with $0 \leq r \leq m$.