Exact excess and the asymptotic size of $\mathcal D$

Statement

For $k\ge 2$, put

$$n=4k+7,\qquad m=4k,\qquad r=2k.$$

Section 4 of Bovdi and Leung's construction gives a maximal commutative algebraic system $\mathcal D\subseteq\mathcal P_n(*)$ with

$$\begin{aligned} |\mathcal D_{r+3}| &=7\binom mr+21\binom m{r+1} +7\binom m{r+2}+\binom m{r+3},\\ |\mathcal D_j|&=\binom nj \qquad(j\ge r+5\text{ odd}), \end{aligned} \tag{1}$$

and no other layers. They conjecture that

$$\lim_{k\to\infty}\frac{|\mathcal D|}{2^{n-2}}=1.$$

We prove the stronger exact identity

$$\boxed{ |\mathcal D|-2^{n-2} =\frac{2(4k+3)(2k-1)}{(2k+1)(2k+2)} \binom{4k}{2k}.} \tag{2}$$

In particular, the excess is positive for every $k\ge2$, and its ratio to $2^{n-2}$ tends to zero. This proves the conjecture.

The odd upper tail

Let

$$T=\sum_{\substack{j\ge r+5\j\text{ odd}}}\binom nj.$$

Because $n=2r+7$ is odd, complement symmetry gives

$$\sum_{j=r+4}^{n}\binom nj=2^{n-1}. \tag{3}$$

We also use the elementary alternating-tail identity

$$\sum_{j=s}^{n}(-1)^j\binom nj =(-1)^s\binom{n-1}{s-1}. \tag{4}$$

Here $s=r+4=2k+4$ is even. Thus the difference between the even and odd terms in the upper half is

$$\sum_{j=r+4}^{n}(-1)^j\binom nj=\binom{n-1}{r+3}.$$

The odd terms in that upper half begin at $r+5=2k+5$, exactly the terms defining $T$. Combining this observation with (3) gives

$$T=2^{n-2}-\frac12\binom{n-1}{r+3}. \tag{5}$$

The exceptional layer

Write

$$B_j=\binom m{r+j}.$$

The exceptional layer in (1) is

$$F=7B_0+21B_1+7B_2+B_3. \tag{6}$$

Since $n-1=m+6$, Vandermonde's identity and $B_{-j}=B_j$ give

$$\begin{aligned} \frac12\binom{n-1}{r+3} &=\frac12\binom{m+6}{r+3}\\ &=\frac12\sum_{i=0}^{6}\binom6i B_{3-i}\\ &=10B_0+15B_1+6B_2+B_3. \tag{7} \end{aligned}$$

Equations (5)--(7) therefore imply

$$|\mathcal D|-2^{n-2}=F-\frac12\binom{n-1}{r+3} =-3B_0+6B_1+B_2. \tag{8}$$

The adjacent-binomial ratios are

$$\frac{B_1}{B_0}=\frac{2k}{2k+1},\qquad \frac{B_2}{B_0}=\frac{2k(2k-1)}{(2k+1)(2k+2)}.$$

Substitution into (8) yields

$$\begin{aligned} |\mathcal D|-2^{n-2} &=\left( -3+\frac{12k}{2k+1} +\frac{2k(2k-1)}{(2k+1)(2k+2)} \right)\binom{4k}{2k}\\ &=\frac{2(4k+3)(2k-1)}{(2k+1)(2k+2)} \binom{4k}{2k}, \end{aligned}$$

which is (2).

Taking the limit

The rational prefactor in (2) tends to $4$, while the standard central-binomial estimate gives

$$\frac{\binom{4k}{2k}}{2^{4k}}=O(k^{-1/2})\longrightarrow0.$$

Since $2^{n-2}=2^{4k+5}=32\,2^{4k}$, division of (2) by $2^{n-2}$ proves

$$\frac{|\mathcal D|}{2^{n-2}} =1+O(k^{-1/2})\longrightarrow1.$$

This resolves the asymptotic-size conjecture.