I've come across a family of posets which appear to have a couple of remarkable enumerative properties, and I'm wondering whether anyone has seen these before.


Take $n\geqslant3$, and let $\preccurlyeq$ be a partial order on the set $\{1,\dots,n\}$. Say that $\preccurlyeq$ has property M if:


1. There is a unique way to write $\{1,\dots,n\}$ as the union of two $\preccurlyeq$-chains, and
2. $\preccurlyeq$ is maximal with this property, i.e. any refinement of $\preccurlyeq$ destroys the uniqueness in (1).


For example, if $n=5$, there are four partial orders (up to isomorphism) with property M. Given as sets of covers, these are as follows:


$$\{(1,2),(2,3),(3,4)\},\{(1,2),(2,3),(2,5),(4,5)\},\{(1,3),(3,4),(2,3),(2,5)\},\{(1,3),(3,5),(2,4),(1,4),(2,5)\}$$


Now here are the (apparent) remarkable properties:


1. Up to isomorphism, there are exactly $2^{n-3}$ partial orders on $\{1,\dots,n\}$ with property M.
2. If $\preccurlyeq$ has property M, then the size of $\preccurlyeq$ (i.e. the number of pairs $i\preccurlyeq j$) is $\binom n2+1$.


I've checked these properties for $n\leqslant8$. Does anyone know whether they are true generally?


(I also have a conjectural construction of all partial orders with property M, coming from representation theory, but I can't prove anything!)



Has anyone seen these posets before?
I claim:


A poset with property M has precisely two maximal elements.


Precisely one maximal element is greater than every non-maximal element. Call this the supermaximal element.


Removing the supermaximal element (if $n>3$) leaves a poset with property M.


So there is a unique way to build any example from the $n=3$ example by repeatedly adding a new supermaximal element.


For an $n=k$ example, there are two ways to add a new supermaximal element (choose which of the existing maximal elements remains maximal), each of which increases the "size" by $k$.


Now the two properties you observed follow by induction.


