Given a partition $\lambda$ and its Young diagram $\pmb{Y}_{\lambda}$, we say $\lambda$ is a $(t,s)$-core partition provided that neither $t$ nor $s$ is a hook length in $\pmb{Y}_{\lambda}$. We now recall a conjecture (now a theorem) of D. Armstrong which states "the total number $(s,t)$-core partitions is $\frac1{s+t}\binom{s+t}s$".
Focusing on the special case of $(s,s+1)$-core partitions, Stanley and Zanello proved the enumeration $C_s=\frac1{2s+1}\binom{2s+1}s$. Their argument utilizes a result of J. Anderson: $(s,s+1)$-cores correspond bijectively to the order ideals of
the poset $P(s,s+1)$ (that is, positive integers that are not contained in the numerical
semigroup generated by $s$ and $s+1$) by associating the lower ideal $\{a_1,\dots,a_j\}$, where $a_1>\cdots>a_j$, to the $(s,s+1)$-core partition $(a_1-(j-1), a_2-(j-2),\dots,a_{j-1}-1,a_j)$.
A few years ago, I considered core-partitions with distinct parts (see page 13) to the effect that the number of such $(s,s+1)$-core partitions equals the Fibonacci number $F_{s+1}$. This conjecture is now a theorem due to different people (see this paper and references therein).
QUESTION. Viewing $(s,s+1)$-cores of distinct partitions as embedded in the ordered ideals of the poset $P(s,s+1)$ (the above Catalan case), do they have any "interesting" structure?
Best Answer
The order ideal corresponding to a core partition with distinct parts cannot contain an element $x \geq s+2$, otherwise it must also contain $x-s, x-s-1$, resulting in two equal parts. For the same reason, it also can't contain two consecutive elements in $\{1,2,\dots, s-1\}$. Therefore the poset in questions is simply the poset of subsets of $\{1,2,\dots, s-1\}$ that do not contain consecutive elements, ordered by inclusion.
The underlying Hasse diagram of this poset is called the Fibonacci cube.