OEIS [A109388][1] $\{a_n\}_{n\ge1}$ 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 

> $\{a_n\}_{n\ge1}$ is the size of the largest antichain in 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 $b_n(\lfloor\frac{2n-1}{3}\rfloor+1)$ is the **largest** antichain, i.e. the claim in OEIS.  
  
[1]: https://oeis.org/A109388