1993年，Hsu定义了Fibonacci立方体 ${\Gamma }_n$ [1] 作为新的内联网拓扑结构模型，斐波那契立方体可以作为一个等距离子图嵌入到布尔立方体(超立方体)中，它还包含一些其他特殊结构，如Lucas立方体作为它的子图，因此斐波那契立方体可以在容错计算中得到应用。Fibonacci立方体及其相关立方体的很多性质被大量的学者研究 [2] - [8] 。特殊地，Fibonacci立方体和Lucas立方体度序列多项式 [9] 也得到了研究。Zhang等 [10] [11] 在平面基本二部图的完美匹配集合上建立了一个有限分配格——匹配型分配格的概念。斐波那契立方体是平面六a角系统fibonaccenes的共振图，所以以有向斐波那契立方体作为Hasse图的分配格是匹配型分配格。Wang等 [12] 研究了有限分配格的凸扩张，给出Fibonacci立方体的一种更简单的画法。继而Wang等 [13] 提出了匹配型Lucas立方体的结构并且用一种统一的方式得到其度序列多项式。周玉玉等 [14] 提出了一类新的斐波那契相似立方体——匹配型分配格并研究了该类立方体的一些特殊性质。本文对周玉玉等 [14] 提出的斐波那契相似立方体给出了一个定向，并讨论了其与出度、入度相关的多项式，如度序列多项式、出度多项式及出入度多项式。

如果一个集合P有二元序关系 $\le$满足自反性、反对称性和传递性，那称P为偏序集。如果 $x<y$且 $x\le z<y$必有 $x=z$，则称偏序集P中y覆盖x。偏序关系与其覆盖关系相互唯一确定，偏序集的覆盖关系可以表示为以P中元素为顶点的有向图(通常画为Hasse图)：当且仅当y覆盖x时， $\left(y,x\right)$是一条弧。设Q是偏序集P的一个子集，若 $y\in Q$且 $y\le z$，则 $z\in Q$，则称Q为P的一个滤子， $ℱ\left(P\right)$表示一个偏序集P的所有滤子。P的所有滤子 $ℱ\left(P\right)$按照反包含关系( $Y^{\prime }\le Y$当且仅当 $Y^{\prime }\supseteq Y$)构成的偏序集是个分配格，称为滤子格，而且 $ℱ\left(P\right):=\left(ℱ\left(P\right),\supseteq \right)$是一个有限分配格。在有限分配格 $ℱ\left(P\right)$中， $\hat{1}$(或 $\hat{0}$)表示最大(或最小)元，从 $\hat{1}$到 $\hat{0}$按元素覆盖关系给分配格 $ℱ\left(P\right)$一个定向，这样最大元 $\hat{1}$只有出度，最大元 $\hat{1}$和最小元 $\hat{0}$之间的其他元素首先有入度然后有出度，而最小元 $\hat{0}$只有入度，这样的定向也是Z-变换图顶点间的定向，就可以定义并研究分配格的出度多项式及其出入度多项式。

引理2.1 [12] 滤子格 $ℱ\left(P\right)$可以由分配格的凸扩张“ $⊞$”得到，对任意的 $x\in P$，有：

$ℱ\left(P\right)\cong ℱ\left(P-x\right)⊞ℱ\left(P\ast x\right)$

其中， $P-x$和 $P\ast x$分别表示 $P\backslash \left\{x\right\}$和 $P\backslash \left\{y\in P|y\le x或x\le y\right\}$，均为P的子偏序集。

定义2.2 [14] “L-si-Fibonaccene”是在Fibonaccene的左端第二个六角形下面连接了一个额外六角形的Cata型六角系统，如下 图1 所示。本文中，我们将 $n\left(\ge 3\right)$个六角形的L-si-Fibonaccene记为 $Y_n$。

定义2.3 [14] 设 $n\ge 3$，称在 $Y_n$的内对偶图上的如 图2 所示的偏序集为 ${\psi }_n$，它是由“栅栏”(即“Zigzag”偏序)添加一个极大元得到的。

图2. 偏序 ${\psi }_n$

定理2.4 [11] 设 $n\ge 3$， $M\left(Y_n\right)\cong ℱ\left({\psi }_n\right)$是L-si-Fibonaccene $Y_n$的匹配型分配格，其Hasse图同构于 $Y_n$的有向Z-变换图。

以 ${\Psi }_3$为例，给出具体的 ${\Psi }_3$的构造过程。 ${\Psi }_3$对应的偏序如下 图3 所示。

其中， ${\psi }_3$的滤子有 $\{\varnothing \}$、 $\left\{x_1\right\}$、 $\left\{x_2\right\}$、 $\left\{x_1,x_2\right\}$、 $\left\{x_1,x_2,x_3\right\}$。然后按滤子的反包含关系构成一个滤子格，是匹配型分配格，覆盖关系为反包含关系，集合之间差一个元素。根据这样的构造就能得到 ${\Psi }_3$，如 图4 所示。

定理2.5 [14] 当 $n\ge 5$时，有：

${\Psi }_n\cong {\Psi }_{n-1}⊞{\Psi }_{n-2}\cong \left({\Psi }_{n-2}⊞{\Psi }_{n-3}\right)⊞{\Psi }_{n-2}$.

根据定理2.5就可以得到如 图5 所示的 ${\Psi }_n$的递归结构图。

图5. ${\Psi }_n$的递归结构图

前七个有向斐波那契相似立方体 ${{\Psi }^{\prime }}_0$、 ${{\Psi }^{\prime }}_1$、 ${{\Psi }^{\prime }}_2$、 ${{\Psi }^{\prime }}_3$、 ${{\Psi }^{\prime }}_4$、 ${{\Psi }^{\prime }}_5$、 ${{\Psi }^{\prime }}_6$，如 图6 所示，其中 ${{\Psi }^{\prime }}_0$表示只有一个顶点的平凡格或图。

在画上述有向立方体的过程中，若y覆盖x，则将y放在x上方，当忽略掉所有弧的方向就得到偏序集的Hasse图。由此前七个斐波那契相似立方体，也就是其Hasse图， ${{\Psi }^{\prime }}_0$、 ${{\Psi }^{\prime }}_1$、 ${{\Psi }^{\prime }}_2$、 ${{\Psi }^{\prime }}_3$、 ${{\Psi }^{\prime }}_4$、 ${{\Psi }^{\prime }}_5$、 ${\Psi }_6$，如 图7 所示。

1) 度序列多项式

设 $d_{n,k}:=d_k\left({\Psi }_n\right)$表示 ${\Psi }_n$中度为k的顶点数目，也可表示为 $d_{n,k}=\left|\left\{v\in V\left({\Psi }_n\right)|{\mathrm{deg}}_{{\Psi }_n}\left(v\right)=k\right\}\right|$，则称如下计数多项式

$D_n\left(x\right)={\displaystyle \underset{k\ge 0}{\sum }{d}_{n,k}}{x}^k$

为 ${\Psi }_n$的度序列多项式。部分 $D_n\left(x\right)$如下：

$\begin{array}{l}{D}_0\left(x\right)=1,\\ D_1\left(x\right)=2x,\\ D_2\left(x\right)=4x^2,\\ D_3\left(x\right)=x+3x^2+x^3,\\ D_4\left(x\right)=x+7x^3+x^4,\\ D_5\left(x\right)=2x^2+7x^3+4x^4+x^5,\\ D_6\left(x\right)=2x^2+4x^3+12x^4+4x^5+x^6.\end{array}$

通过在 ${\Psi }_n$的递归结构中虚添一个 ${\Psi }_{n-3}$及部分边，如 图8 所示。可得 $d_{n,k}$的递归结构。

命题3.1 当 $n\ge 5$时，

$d_{n,k}=d_{n-1,k-1}+d_{n-2,k-1}-d_{n-3,k-2}+d_{n-3,k-1}.$

由命题3.1可得如下推论。

命题3.2 当 $n\ge 5$时，

$D_n\left(x\right)=x{D}_{n-1}\left(x\right)+x{D}_{n-2}\left(x\right)+\left(x-x^2\right)D_{n-3}\left(x\right).$

因此，可得 $D_n\left(x\right)$的生成函数。

定理3.3 $D_n\left(x\right)$的生成函数为：

${\displaystyle \underset{n\ge 0}{\sum }{D}_n\left(x\right)y^n}=\frac{1+xy+\left(-x+2x^2\right)y^2+\left(2x^2-3x^3\right)y^3+\left(x-3x^2+2x^3\right)y^4}{1-xy-x{y}^2-\left(x-x^2\right)y^3}.$

证明 由命题3.2得：

$\begin{array}{c}{\displaystyle \underset{n\ge 0}{\sum }{D}_n}\left(x\right)y^n={\displaystyle \underset{n\ge 5}{\sum }{D}_n}\left(x\right)y^n+{\displaystyle \underset{n=0}{\overset{4}{\sum }}{D}_n}\left(x\right)y^n\\ ={\displaystyle \underset{n\ge 5}{\sum }\left(x{D}_{n-1}\left(x\right)+x{D}_{n-2}\left(x\right)+\left(x-x^2\right)D_{n-3}\left(x\right)\right)y^n}+{\displaystyle \underset{n=0}{\overset{4}{\sum }}{D}_n}\left(x\right)y^n\\ =x{\displaystyle \underset{n\ge 5}{\sum }{D}_{n-1}}\left(x\right)y^n+x{\displaystyle \underset{n\ge 5}{\sum }{D}_{n-2}}{y}^n+x{\displaystyle \underset{n\ge 5}{\sum }{D}_{n-3}}\left(x\right)y^n-x^2{\displaystyle \underset{n\ge 5}{\sum }{D}_{n-3}}\left(x\right)y^n+{\displaystyle \underset{n=0}{\overset{4}{\sum }}{D}_n}\left(x\right)y^n\\ =xy{\displaystyle \underset{n\ge 0}{\sum }{D}_n}\left(x\right)y^n-xy\left(D_0\left(x\right)+D_1\left(x\right)y+D_2\left(x\right)y^2+D_3\left(x\right)y^3\right)+x{y}^2{\displaystyle \underset{n\ge 0}{\sum }{D}_n}\left(x\right)y^n\\ -x{y}^2\left(D_0\left(x\right)+D_1\left(x\right)y+D_2\left(x\right)y^2\right)+x{y}^3{\displaystyle \underset{n\ge 0}{\sum }{D}_n}\left(x\right)y^n-x{y}^3\left(D_0\left(x\right)+D_1\left(x\right)y\right)\\ -x^2{y}^3{\displaystyle \underset{n\ge 0}{\sum }{D}_n}\left(x\right)y^n+x^2{y}^3\left(D_0\left(x\right)+D_1\left(x\right)y\right)+D_0\left(x\right)+D_1\left(x\right)y+D_2\left(x\right)y^2\\ +D_3\left(x\right)y^3+D_4\left(x\right)y^4\\ =\left(xy+x{y}^2+x{y}^3-x^2{y}^3\right){\displaystyle \underset{n\ge 0}{\sum }{D}_n}\left(x\right)y^n+1+xy+2x^2{y}^3-3x^3{y}^3-3x^2{y}^4\\ +2x^3{y}^4-x{y}^2+2x^2{y}^2+x{y}^4.\end{array}$

推论3.4 当 $n\ge 5$时， ${\Psi }_n$中度为k的顶点个数为：

$\begin{array}{c}{d}_{n,k}={\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}j\\ n-k-j\end{array}\right)}\left(2\left(\begin{array}{c}n-2j-2\\ k-j-2\end{array}\right)+\left(\begin{array}{c}n-2j\\ k-j\end{array}\right)+\left(\begin{array}{c}n-2j-1\\ k-j-1\end{array}\right)-3\left(\begin{array}{c}n-2j-3\\ k-j-3\end{array}\right)\right)\\ +{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}j\\ n-k-j-1\end{array}\right)}\left(2\left(\begin{array}{c}n-2j-3\\ k-j-2\end{array}\right)+2\left(\begin{array}{c}n-2j-4\\ k-j-3\end{array}\right)-\left(\begin{array}{c}n-2j-2\\ k-j-1\end{array}\right)\right)\\ +{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\left(\begin{array}{c}j\\ n-j-k-3\end{array}\right)\left(\begin{array}{c}n-2j-4\\ k-j-1\end{array}\right)-3\left(\begin{array}{c}j\\ n-k-j-2\end{array}\right)\left(\begin{array}{c}n-2j-4\\ k-j-2\end{array}\right)\right)}.\end{array}$

证明 利用二项式展开，

$\frac{x^n}{{\left(1-x\right)}^{n+1}}={\displaystyle \underset{j\ge n}{\sum }\left(\begin{array}{c}j\\ n\end{array}\right)x^j}.$

我们先考虑部分展开，

$f\left(x,y\right)=\frac{1}{\left(1-xy\right)\left(1-x{y}^2\right)-x{y}^3}$

$\begin{array}{c}f\left(x,y\right)=\frac{1}{\left(1-xy\right)\left(1-x{y}^2\right)-x{y}^3}\\ =\frac{{\left(1-xy\right)}^{-1}{\left(1-x{y}^2\right)}^{-1}}{1-x{y}^3{\left(1-xy\right)}^{-1}{\left(1-x{y}^2\right)}^{-1}}\\ ={\displaystyle \underset{t\ge 0}{\sum }{x}^t{y}^{3t}}{\left(1-xy\right)}^{-t-1}{\left(1-x{y}^2\right)}^{-t-1}\end{array}$

$\begin{array}{c}={\displaystyle \underset{t\ge 0}{\sum }\frac{x{y}^t}{{\left(1-xy\right)}^{t+1}}}\frac{{\left(x{y}^2\right)}^t}{{\left(1-x{y}^2\right)}^{t+1}}{x}^{-t}\\ ={\displaystyle \underset{t\ge 0}{\sum }{\displaystyle \underset{i\ge t}{\sum }\left(\begin{array}{c}i\\ t\end{array}\right)}}{\left(xy\right)}^i{\displaystyle \underset{j\ge t}{\sum }\left(\begin{array}{c}j\\ t\end{array}\right)}{\left(x{y}^2\right)}^j{x}^{-t}\\ ={\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{k=0}{\overset{n}{\sum }}{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j\\ k-j\end{array}\right)}}}\left(\begin{array}{c}j\\ n-k-j\end{array}\right)x^k{y}^n\end{array}$

所以(其中符号 $\left[x^k\right]f\left(x,y\right)$表示 $f\left(x,y\right)$中 $x^k$的系数)

$\left[x^k\right]\left[y^n\right]f\left(x,y\right)={\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j\\ k-j\end{array}\right)}\left(\begin{array}{c}j\\ n-k-j\end{array}\right).$

记

$\begin{array}{l}\frac{1+xy+2x^2{y}^3-3x^3{y}^3-3x^2{y}^4+2x^3{y}^4-x{y}^2+2x^2{y}^2+x{y}^4}{\left(1-xy\right)\left(1-x{y}^2\right)-x{y}^3}\\ =f\left(x,y\right)+xyf\left(x,y\right)+2x^2{y}^{3}f\left(x,y\right)-3x^3{y}^{3}f\left(x,y\right)-3x^2{y}^{4}f\left(x,y\right)\\ +2x^3{y}^{4}f\left(x,y\right)-x{y}^{2}f\left(x,y\right)+2x^2{y}^{2}f\left(x,y\right)+x{y}^{4}f\left(x,y\right)\end{array}$

则

$\begin{array}{l}\left[x^k\right]\left[y^n\right]\frac{1+xy+2x^2{y}^3-3x^3{y}^3-3x^2{y}^4+2x^3{y}^4-x{y}^2+2x^2{y}^2+x{y}^4}{\left(1-xy\right)\left(1-x{y}^2\right)-x{y}^3}\\ =\left[x^k\right]\left[y^n\right]f\left(x,y\right)+\left[x^{k-1}\right]\left[y^{n-1}\right]f\left(x,y\right)+2\left[x^{k-2}\right]\left[y^{n-3}\right]f\left(x,y\right)\\ -3\left[x^{k-3}\right]\left[y^{n-3}\right]f\left(x,y\right)-3\left[x^{k-2}\right]\left[y^{n-4}\right]f\left(x,y\right)+2\left[x^{k-3}\right]\left[y^{n-4}\right]f\left(x,y\right)\\ -\left[x^{k-1}\right]\left[y^{n-2}\right]f\left(x,y\right)+2\left[x^{k-2}\right]\left[y^{n-2}\right]f\left(x,y\right)+\left[x^{k-1}\right]\left[y^{n-4}\right]f\left(x,y\right)\\ ={\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j\\ k-j\end{array}\right)}\left(\begin{array}{c}j\\ n-k-j\end{array}\right)+{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j-1\\ k-j-1\end{array}\right)}\left(\begin{array}{c}j\\ n-k-j\end{array}\right)\\ +2{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j-3\\ k-j-2\end{array}\right)}\left(\begin{array}{c}j\\ n-k-j-1\end{array}\right)-3{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j-3\\ k-j-3\end{array}\right)}\left(\begin{array}{c}j\\ n-k-j\end{array}\right)\\ -3{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j-4\\ k-j-2\end{array}\right)}\left(\begin{array}{c}j\\ n-k-j-2\end{array}\right)+2{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j-4\\ k-j-3\end{array}\right)}\left(\begin{array}{c}j\\ n-k-j-1\end{array}\right)\\ -{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j-2\\ k-j-1\end{array}\right)}\left(\begin{array}{c}j\\ n-k-j-1\end{array}\right)+2{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j-2\\ k-j-2\end{array}\right)}\left(\begin{array}{c}j\\ n-k-j\end{array}\right)\\ +{\displaystyle \underset{j=0}{\overset{k}{\sum }}\left(\begin{array}{c}n-2j-4\\ k-j-1\end{array}\right)}\left(\begin{array}{c}j\\ n-k-j-3\end{array}\right).\end{array}$

注释1 当 $x=1$时，求解命题3.2中 $D_n\left(x\right)$的递推关系，可得 ${\Psi }_n\left(n\ge 2\right)$的顶点数为 $f\left(n\right)+3f\left(n-1\right)$，其中 $f\left(n\right)$是第n个斐波那契数。

2) 入度和出度多项式

设 ${\Psi }_n$的入度多项式为 $D_n^{-}\left(x\right)={\displaystyle \underset{k=0}{\overset{n-1}{\sum }}{d}_{n,k}^{-}}{x}^k$，其中 $d_{n,k}^{-}$表示 ${{\Psi }^{\prime }}_n$中入度为k的顶点个数。部分入度多项式 $D_n^{-}\left(x\right)$如下所示：

$\begin{array}{l}{D}_0^{-}\left(x\right)=1,\\ D_1^{-}\left(x\right)=1,\\ D_2^{-}\left(x\right)=1+2x+x^2,\\ D_3^{-}\left(x\right)=1+3x+x^2,\\ D_4^{-}\left(x\right)=1+4x+3x^2+x^3,\\ D_5^{-}\left(x\right)=1+5x+6x^2+2x^3,\\ D_6^{-}\left(x\right)=1+6x+10x^2+5x^3+x^4,\\ D_7^{-}\left(x\right)=1+7x+15x^2+11x^3+3x^4,\\ D_8^{-}\left(x\right)=1+8x+21x^2+21x^3+8x^4+x^5,\\ D_9^{-}\left(x\right)=1+9x+28x^2+36x^3+19x^4+4x^5.\end{array}$

引理3.5 [12] 设L是一个有限分配格。如果K是L的一个割，则：

$d_k^{-}\left(L⊞K\right)=d_k^{-}\left(L\right)+d_{k-1}^{-}\left(K\right).$

由引理3.5我们有如下命题。

命题3.6 当 $n\ge 0$时， $d_{n,0}^{-}=1$；当 $n\ge 3$， $k\ge 1$时，

$d_{n,k}^{-}=d_{n-1,k}^{-}+d_{n-2,k-1}^{-}.$

命题3.7 当 $n\ge 3$时，

$D_n^{-}\left(x\right)=D_{n-1}^{-}\left(x\right)+x{D}_{n-2}^{-}\left(x\right).$

定理3.8 $D_n^{-}\left(x\right)$的生成函数为：

${\displaystyle \underset{n\ge 0}{\sum }{D}_n^{-}}\left(x\right)y^n=\frac{1+\left(x+x^2\right)y^2}{1-y-x{y}^2}.$

证明

$\begin{array}{c}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{D}_n^{-}}\left(x\right)y^n={\displaystyle \underset{n=3}{\overset{\infty }{\sum }}{D}_n^{-}}\left(x\right)y^n+{\displaystyle \underset{n=0}{\overset{2}{\sum }}{D}_n^{-}}\left(x\right)y^n\\ ={\displaystyle \underset{n=3}{\overset{\infty }{\sum }}\left(D_{n-1}^{-}\left(x\right)+x{D}_{n-2}^{-}\left(x\right)\right)y^n}+{\displaystyle \underset{n=0}{\overset{2}{\sum }}{D}_n^{-}}\left(x\right)y^n\\ ={\displaystyle \underset{n=3}{\overset{\infty }{\sum }}{D}_{n-1}^{-}}\left(x\right)y^n+x{\displaystyle \underset{n=3}{\overset{\infty }{\sum }}{D}_{n-2}^{-}}\left(x\right)y^n+{\displaystyle \underset{n=0}{\overset{2}{\sum }}{D}_n^{-}}\left(x\right)y^n\\ =y{\displaystyle \underset{n=3}{\overset{\infty }{\sum }}{D}_{n-1}^{-}}\left(x\right)y^{n-1}+x{y}^2{\displaystyle \underset{n=3}{\overset{\infty }{\sum }}{D}_{n-2}^{-}}\left(x\right)y^{n-2}+{\displaystyle \underset{n=0}{\overset{2}{\sum }}{D}_n^{-}}\left(x\right)y^n\\ =y{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{D}_n^{-}}\left(x\right)y^n-y\left(D_0^{-}\left(x\right)+D_1^{-}\left(x\right)y\right)+x{y}^2{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{D}_n^{-}}\left(x\right)y^n\\ -x{y}^2{D}_0^{-}\left(x\right)+D_0^{-}\left(x\right)+D_1^{-}\left(x\right)y+D_2^{-}\left(x\right)y^2\\ =\left(y+x{y}^2\right){\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{D}_n^{-}}\left(x\right)y^n+1+x{y}^2+x^2{y}^2.\end{array}$

推论3.9 当 $n\ge 0$时，

$D_n^{-}\left(x\right)=\left({\displaystyle \underset{k=0}{\overset{\frac{n}{2}}{\sum }}\left(\begin{array}{c}n-k\\ k\end{array}\right)}+{\displaystyle \underset{k=1}{\overset{\frac{n}{2}}{\sum }}\left(\begin{array}{c}n-k-1\\ k-1\end{array}\right)}+{\displaystyle \underset{k=2}{\overset{\frac{n}{2}+1}{\sum }}\left(\begin{array}{c}n-k\\ k-2\end{array}\right)}\right)x^k;$

利用二项式展开有如下证明。

证明

$\begin{array}{c}{\displaystyle \underset{n\ge 0}{\sum }{D}_n^{-}}\left(x\right)y^n=\frac{1+x{y}^2+x^2{y}^2}{1-y-x{y}^2}\\ =\left(1+x{y}^2+x^2{y}^2\right){\displaystyle \underset{j\ge 0}{\sum }{\left(y+x{y}^2\right)}^j}\\ ={\displaystyle \underset{j\ge 0}{\sum }{\left(y+x{y}^2\right)}^j}+x{y}^2{\displaystyle \underset{j\ge 0}{\sum }{\left(y+x{y}^2\right)}^j}+x^2{y}^2{\displaystyle \underset{j\ge 0}{\sum }{\left(y+x{y}^2\right)}^j}\\ ={\displaystyle \underset{j\ge 0}{\sum }{y}^j}{\left(1+xy\right)}^j+{\displaystyle \underset{j\ge 0}{\sum }x}{y}^{j+2}{\left(1+xy\right)}^j+{\displaystyle \underset{j\ge 0}{\sum }{x}^2}{y}^{j+2}{\left(1+xy\right)}^j\\ ={\displaystyle \underset{j\ge 0}{\sum }{\displaystyle \underset{n-j=0}{\overset{j}{\sum }}\left(\begin{array}{c}j\\ n-j\end{array}\right)x^{n-j}{y}^n}}+{\displaystyle \underset{j\ge 0}{\sum }{\displaystyle \underset{n-j-2=0}{\overset{j}{\sum }}\left(\begin{array}{c}j\\ n-j-2\end{array}\right)x^{n-j-1}{y}^n}}+{\displaystyle \underset{j\ge 0}{\sum }{\displaystyle \underset{n-j-2=0}{\overset{j}{\sum }}\left(\begin{array}{c}j\\ n-j-2\end{array}\right)x^{n-j}{y}^n}}\\ ={\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{j=\frac{n}{2}}{\overset{n}{\sum }}\left(\begin{array}{c}j\\ n-j\end{array}\right)x^{n-j}{y}^n}}+{\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{j=\frac{n}{2}-1}{\overset{n-2}{\sum }}\left(\begin{array}{c}j\\ n-j-2\end{array}\right)x^{n-j-1}{y}^n}}+{\displaystyle \underset{n\ge 0}{\sum }{\displaystyle \underset{j=\frac{n}{2}-1}{\overset{n-2}{\sum }}\left(\begin{array}{c}j\\ n-j-2\end{array}\right)x^{n-j}{y}^n}}\\ ={\displaystyle \underset{n\ge 0}{\sum }\left({\displaystyle \underset{k=0}{\overset{\frac{n}{2}}{\sum }}\left(\begin{array}{c}n-k\\ k\end{array}\right)}+{\displaystyle \underset{k=1}{\overset{\frac{n}{2}}{\sum }}\left(\begin{array}{c}n-k-1\\ k-1\end{array}\right)}+{\displaystyle \underset{k=2}{\overset{\frac{n}{2}+1}{\sum }}\left(\begin{array}{c}n-k\\ k-2\end{array}\right)}\right)x^k{y}^n}.\end{array}$

且易得

$d_{n,k}^{-}=\left(\begin{array}{c}n-k\\ k\end{array}\right)+\left(\begin{array}{c}n-k-1\\ k-1\end{array}\right)+\left(\begin{array}{c}n-k\\ k-2\end{array}\right).$

注释2 因为在有限分配格 $ℱ\left(P\right)$中， $d_{n,k}^{+}$和 $d_{n,k}^{-}$都等于P中只有k个元素的极大反链的个数 [12] ，也就是说， $d_{n,k}^{+}=d_{n,k}^{-}$，所以出度多项式与入度多项式是相同的。

3) 出入度多项式

称如下计数多项式

$P\left({\Psi }_n\right):=P\left({\Psi }_n;x,y\right)={\displaystyle \underset{i,j}{\sum }{p}_{ij}}{x}^i{y}^j$

为 ${\Psi }_n$的出入度多项式，其中 $p_{ij}$表示 ${{\Psi }^{\prime }}_n$中出度为i、入度为j的顶点个数。部分出入度多项式 $P\left({\Psi }_n\right)$如下所示：

$\begin{array}{l}P\left({\Psi }_0\right)=1,\\ P\left({\Psi }_1\right)=x+y,\\ P\left({\Psi }_2\right)=x^2+2xy+y^2,\\ P\left({\Psi }_3\right)=x^2+y+2xy+x{y}^2,\\ P\left({\Psi }_4\right)=x^3+y+3x^{2}y+3x{y}^2+x{y}^3,\\ P\left({\Psi }_5\right)=x^3+xy+3x^{2}y+x^{3}y+y^2+3x{y}^2+2x^2{y}^2+x{y}^3+x^2{y}^3,\\ P\left({\Psi }_6\right)=x^4+xy+x^{2}y+4x^{3}y+y^2+3x{y}^2+5x^2{y}^2+x^3{y}^2+2x{y}^3+3x^2{y}^3+x^2{y}^4.\end{array}$(1)

类似于 图6 的方式，在 图4 ${\Psi }_n$的递归图中虚补一个 ${\Psi }_{n-4}$，之后用容斥原理就可得到如下结论。

命题3.10

当 $n\ge 6$时，

$P\left({\Psi }_n\right)=\{\begin{array}{l}x\left(y+1\right)P\left({\Psi }_{n-2}\right)+yP\left({\Psi }_{n-3}\right)+xy\left(1-x\right)P\left({\Psi }_{n-4}\right),若n为偶数,\\ y\left(x+1\right)P\left({\Psi }_{n-2}\right)+xP\left({\Psi }_{n-3}\right)+xy\left(1-y\right)P\left({\Psi }_{n-4}\right),若n为奇数.\end{array}$

令 $A_m\left(x,y\right)=P\left({\Psi }_{2m};x,y\right)$， $B_m\left(x,y\right)=P\left({\Psi }_{2m+1};x,y\right)$，则

$\{\begin{array}{l}{A}_m\left(x,y\right)=x\left(y+1\right)A_{m-1}\left(x,y\right)+y{B}_{m-2}\left(x,y\right)+xy\left(1-x\right)A_{m-2}\left(x,y\right),\left(m\ge 3\right),\\ B_m\left(x,y\right)=y\left(x+1\right)B_{m-1}\left(x,y\right)+x{A}_{m-1}\left(x,y\right)+xy\left(1-y\right)B_{m-2}\left(x,y\right),\left(m\ge 3\right).\end{array}$

命题3.11 当 $m\ge 5$时，

$\{\begin{array}{l}{A}_m=\left(2xy+x+y\right)A_{m-1}-xy\left(xy+2x+2y-1\right)A_{m-2}+xy\left(x^{2}y+x{y}^2-x-y+1\right)A_{m-3}\\ -x^2{y}^2\left(1-x\right)\left(1-y\right)A_{m-4}\\ B_m=\left(2xy+x+y\right)B_{m-1}-xy\left(xy+2x+2y-1\right)B_{m-2}+xy\left(x^{2}y+x{y}^2-x-y+1\right)B_{m-3}\\ -x^2{y}^2(1-x)(1-y)B_{m-4}.\end{array}$

证明 由命题3.10，当 $m\ge 3$时，

$\begin{array}{c}x{A}_{m-1}=B_m-\left(x+1\right)y{B}_{m-1}-xy\left(1-y\right)B_{m-2}\\ =\frac{A_{m+2}-x\left(y+1\right)A_{m+1}-x\left(1-x\right)y{A}_m}{y}-\frac{\left(x+1\right)y\left(A_{m+1}-x\left(y+1\right)A_m-x\left(1-x\right)y{A}_{m-1}\right)}{y}\\ -\frac{xy\left(1-y\right)\left(A_m-x\left(y+1\right)A_{m-1}-x\left(1-x\right)y{A}_{m-2}\right)}{y}\\ =\frac{A_{m+2}-\left(2xy+x+y\right)A_{m+1}+xy\left(xy+2x+2y-1\right)A_m}{y}\\ +\frac{xy(x+y-x^{2}y-x{y}^2)A_{m-1}+x^2{y}^2\left(1-x\right)\left(1-y\right)A_{m-2}}{y}\end{array}$

因此

$\begin{array}{c}{A}_{m+2}=\left(2xy+x+y\right)A_{m+1}-xy\left(xy+2x+2y-1\right)A_m+xy\left(x^{2}y+x{y}^2-x-y+1\right)A_{m-1}\\ -x^2{y}^2\left(1-x\right)\left(1-y\right)A_{m-2}.\end{array}$

所以，当 $m\ge 5$时，

$\begin{array}{c}{A}_m=\left(2xy+x+y\right)A_{m-1}-xy\left(xy+2x+2y-1\right)A_{m-2}+xy\left(x^{2}y+x{y}^2-x-y+1\right)A_{m-3}\\ -x^2{y}^2\left(1-x\right)\left(1-y\right)A_{m-4}.\end{array}$

类似地，可以得到 $B_m$的递推关系。

由命题3.11，可以得到 $P\left({\Psi }_n\right)$的递推公式。

定理3.12 当 $n\ge 10$时，

$\begin{array}{c}P\left({\Psi }_n\right)=\left(2xy+x+y\right)P\left({\Psi }_{n-2}\right)-xy\left(xy+2x+2y-1\right)P\left({\Psi }_{n-4}\right)\\ +xy\left(x^{2}y+x{y}^2-x-y+1\right)P\left({\Psi }_{n-6}\right)-x^2{y}^2(1-x)(1-y)P\left({\Psi }_{n-8}\right).\end{array}$

根据定理3.12可以得到 $P\left({\Psi }_n\right)$的生成函数。

定理3.13 $P\left({\Psi }_n\right)$的生成函数为：

${\displaystyle \underset{n\ge 0}{\sum }P}\left({\Psi }_n\right)t^n=\frac{N\left(x,y,t\right)}{D\left(x,y,t\right)},$

其中，

$\begin{array}{l}N\left(x,y,t\right)=1+\left(x+y\right)t-\left(x^2-x+xy-y+y^2\right)t^2-\left(x^{2}y-y+y^2\right)t^3\\ -\left(xy\left(1-x+x^2-y+2xy\right)+y\left(y^2-1\right)\right)t^4+xy\left(-x+x^2-y+xy+y^2\right)t^5\\ +xy\left(x-1\right)\left(y-1\right)\left(x+2y-1\right)t^6-xy\left(x-1\right)\left(y-1\right)\left(x+y-1\right)t^7\\ D\left(x,y,t\right)=1-\left(x+y+xy\right)t^2-xy\left(1-x-y\right)t^4-xy\left(x-1\right)\left(y-1\right)t^6\end{array}$

证明 由命题3.10和1)，我们有：

$\begin{array}{c}{\displaystyle \underset{n\ge 0}{\sum }P}\left({\Psi }_n\right)t^n={\displaystyle \underset{n\ge 10}{\sum }P}\left({\Psi }_n\right)t^n+{\displaystyle \underset{n=0}{\overset{9}{\sum }}P}\left({\Psi }_n\right)t^n\\ ={\displaystyle \underset{n\ge 10}{\sum }(\left(2xy+x+y\right)P\left({\Psi }_{n-2}\right)-xy\left(xy+2x+2y-1\right)P\left({\Psi }_{n-4}\right)}\\ +xy\left(x^{2}y+x{y}^2-x-y+1\right)P\left({\Psi }_{n-6}\right)-x^2{y}^2\left(1-x\right)\left(1-y\right)P\left({\Psi }_{n-8}\right))t^n+{\displaystyle \underset{n=0}{\overset{9}{\sum }}P}\left({\Gamma }_n\right)t^n\\ =(\left(2xy+x+y\right)t^2-xy\left(xy+2x+2y-1\right)t^4+xy\left(x{y}^2+x^{2}y-x-y+1\right)t^6\\ -x^2{y}^2\left(1-x\right)\left(1-y\right)t^8){\displaystyle \underset{n\ge 0}{\sum }P}\left({\Psi }_n\right)t^n-\left(2xy+x+y\right)t^2{\displaystyle \underset{n=0}{\overset{7}{\sum }}P}\left({\Psi }_n\right)t^n\\ +xy\left(xy+2x+2y-1\right)t^4{\displaystyle \underset{n=0}{\overset{5}{\sum }}P}\left({\Psi }_n\right)t^n-xy\left(x^{2}y+x{y}^2-x-y+1\right)t^6{\displaystyle \underset{n=0}{\overset{3}{\sum }}P}\left({\Psi }_n\right)t^n\\ +x^2{y}^2\left(1-x\right)\left(1-y\right)t^8{\displaystyle \underset{n=0}{\overset{1}{\sum }}P}\left({\Psi }_n\right)t^n+{\displaystyle \underset{n=0}{\overset{9}{\sum }}P}\left({\Psi }_n\right)t^n\\ =(\left(2xy+x+y\right)t^2-xy\left(xy+2x+2y-1\right)t^4+xy\left(x{y}^2+x^{2}y-x-y+1\right)t^6\\ -x^2{y}^2\left(1-x\right)\left(1-y\right)t^8){\displaystyle \underset{n\ge 0}{\sum }P}\left({\Psi }_n\right)t^n+1+\left(x+y\right)t+\left(x^2-x-y+y^2\right)t^2\\ +\left(y-2x^{2}y-y^2-x{y}^2\right)t^3+\left(y-y^3+xy\left(-1+2x-2x^2+2y-3xy-y^2\right)\right)t^4\\ +xy\left(-x+x^2-2y+xy+x^{2}y+2y^2\right)t^5\\ +xy\left(-1+2x-x^2+2y-3xy+x^{3}y-2y^2+x{y}^2+2x^2{y}^2+y^3\right)t^6\\ +xy\left(1-2x+x^2-2y+3xy-x^{3}y+y^2-x^2{y}^2-x{y}^3\right)t^7\\ +x^2{y}^2\left(1-2x+x^2-3y+4xy-x^{2}y+2y^2-2x{y}^2\right)t^8\\ +x^2{y}^2\left(-1+2x-x^2+2y-3xy+x^{2}y-y^2+x{y}^2\right)t^9\end{array}$