OEIS [A109388](https://oeis.org/A109388) $\{a\_n\}\_{n\ge1}$ is an integer sequence with $a\_n=\binom{n}{\lfloor \frac{n}{3} \rfloor}\times 2^{n-\lfloor\frac{n}{3}\rfloor}$, I noticed that OEIS says



> 
> $a\_n$ is the **size** of the largest antichain in the partial ordering
> $(0,1,a)^n$ where $0$ and $1$ are less than $a$.
> 
> 
> 


According to the binomial theorem, $(1+2)^n=\sum\_{i=0}^{n} \binom{n}{i}\times2^{i}$. Let $b\_n(i):=\binom{n}{i}\times2^{i}$ denote the number of elements in $i$-th level in $(0,1,a)^n$.
It is obvious that $b\_n(\lfloor\frac{2n-1}{3}\rfloor+1)$ is maximal in $\{b\_n(i)\}\_{0\leq i \leq n}$, and $b\_n(\lfloor\frac{2n-1}{3}\rfloor+1)=\binom{n}{\lfloor \frac{n}{3} \rfloor}\times 2^{n-\lfloor\frac{n}{3}\rfloor}$.


I know every $b\_n(i)$ elements in $i$-th level form an antichain in $(0,1,a)^n$. My question is how to prove that $b\_n(\lfloor\frac{2n-1}{3}\rfloor+1)=a\_n$ is the **size of largest antichain**, i.e. the claim in OEIS.



Largest antichain in partial ordering in OEIS
The poset $(0,1,a)^n$ is isomorphic to the face lattice of an $n$-dimensional cube, with the empty face removed. This lattice can be proved to be Sperner by the same argument Lubell used to prove the Spernicity of the boolean algebra. A reference is the slides [here](https://math.mit.edu/~rstan/transparencies/garsiafest.pdf). Spernicity is also a consequence of the theorem of Greene and Kleitman, On the structure of Sperner $k$-families, *J. Combin. Theory Series A* **20** (1976), 41-68, that for any finite poset $P$, some maximum size antichain is a union of orbits of the automorphism group of $P$. For the face lattice of an $n$-cube, all the elements of a given rank form an orbit.


