Exact deficit and the asymptotic size of the cone

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

$$\operatorname{Cone} =\operatorname{Cone}(\operatorname{Cone}_{2k+1} \cup\operatorname{Cone}_{2k+3}) \subseteq\mathcal P_n(*) .$$

The authors conjecture that

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

We prove the stronger exact identity

$$\boxed{ |\operatorname{Cone}|=2^{n-2}-3\binom{4k}{2k}.} \tag{1}$$

Thus the source's observed strict inequality $|\operatorname{Cone}|<2^{n-2}$ holds for every $k\ge2$, and the normalized deficit tends to zero.

The four source layers

For an integer $t$, write

$$B_t=\binom{m}{r+t}.$$

Since $m=2r$, binomial symmetry gives $B_{-t}=B_t$. The source's layer description gives

$$\begin{aligned} L_{r+1}&=B_1,\\ L_{r+3}&=7B_0+21B_1+7B_2+B_3,\\ L_{r+5}&=B_5+7B_4+21B_3+35B_2+35B_1+21B_0, \end{aligned} \tag{2}$$

and every odd layer of size at least $r+7$. Hence, if

$$T_7=\sum_{\substack{j\ge r+7\j\text{ odd}}}\binom nj,$$

then

$$|\operatorname{Cone}|=L_{r+1}+L_{r+3}+L_{r+5}+T_7. \tag{3}$$

The odd upper tail

Let

$$T_5=\sum_{\substack{j\ge r+5\j\text{ odd}}}\binom nj.$$

The integer $n=2r+7$ is odd. Complementation therefore shows that the full upper half, starting at $r+4$, has size $2^{n-1}$. The alternating-tail identity

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

with the even integer $s=r+4$ says that, within this upper half, the number of even sets minus the number of odd sets is $\binom{n-1}{r+3}$. Consequently

$$T_5=2^{n-2}-\frac12\binom{n-1}{r+3}, \qquad T_7=T_5-\binom n{r+5}. \tag{5}$$

Two Vandermonde collections

First, $n-1=m+6$, so Vandermonde's identity and $B_{-t}=B_t$ give

$$\frac12\binom{n-1}{r+3} =10B_0+15B_1+6B_2+B_3. \tag{6}$$

Second, $n=m+7$, and another Vandermonde expansion gives

$$\begin{aligned} \binom n{r+5} &=\sum_{i=0}^{7}\binom7i B_{5-i}\\ &=B_5+7B_4+21B_3+36B_2+42B_1+21B_0. \end{aligned} \tag{7}$$

Comparing (7) with the third line of (2),

$$L_{r+5}-\binom n{r+5}=-B_2-7B_1. \tag{8}$$

Now substitute (2), (5), (6), and (8) into (3). The difference from $2^{n-2}$ is

$$\begin{aligned} |\operatorname{Cone}|-2^{n-2} ={}&B_1+(7B_0+21B_1+7B_2+B_3)\\ &{}-B_2-7B_1-(10B_0+15B_1+6B_2+B_3)\\ =&-3B_0. \end{aligned}$$

This is exactly (1).

Taking the limit

Because $n-2=4k+5$, equation (1) gives

$$\frac{|\operatorname{Cone}|}{2^{n-2}} =1-\frac3{32}\frac{\binom{4k}{2k}}{2^{4k}}. \tag{9}$$

The remaining factor tends to zero elementarily. Indeed,

$$\frac{\binom{4k}{2k}}{2^{4k}} =\prod_{j=1}^{2k}\left(1-\frac1{2j}\right) \leq \exp\!\left(-\frac12\sum_{j=1}^{2k}\frac1j\right) \longrightarrow0.$$

Taking the limit in (9) proves the conjecture.