OEIS 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

$\{a_n\}_{n\ge1}$ 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.

OEIS 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.

OEIS 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

$\{a_n\}_{n\ge1}$ is 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 the $b_n(\lfloor\frac{2n-1}{3}\rfloor+1)$ is the largest antichain, i.e. the claim in OEIS.

OEIS 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

$\{a_n\}_{n\ge1}$ 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.