前面写了这篇q-analog的博客，
这里趁热打铁 2020-08-27 18:19
下面的内容来自Handbook of Enumerative Combinatorics by Miklos Bona
的第11章 Catalan Paths and q,t-enumeration
写不完写不完看不懂看不懂 2020-09-05 20:46
卡特兰The1st q-analogue of \(C_n\)
\[\sum_{\pi \in L_{n, n}^{+}} q^{\operatorname{maj}(\sigma(\pi))}=\frac{1}{[n+1]}\left[\begin{array}{c} 2 n \\ n \end{array}\right] \]这个上篇博客写过
拿下面的图举例子，\(n=3\)，000111，001011，001101，010011，010101
maj分别是0，3，4，2，6
\[\begin{aligned} q^0+q^3+q^4+q^2+q^6 &=\frac{1}{1+q+q^2+q^3}\cdot\frac{(1-q^6)(1-q^5)(1-q^4)}{(1-q^3)(1-q^2)(1-q^1)} \\ &=\frac{1}{1+q+q^2+q^3}\cdot(1+q^2)(1+q^3)(1+q+q^2+q^3+q^4) \\ &=\frac{1}{1+q+q^2+q^3}\cdot(1+q+2q^2+3q^3+3q^4+3q^5+3q^6+2q^7+q^8+q^9) \\ &=1+q^2+q^3+q^4+q^6 \end{aligned} \]The 2nd q-analogue of \(C_n\) /定义\(C_n(q)\)
我理解没错的话，Carlitz-Riordan area 是说路径和\(y=x\)对角线相交区域中完整的正方形个数
这么定义\(C_n(q)=\sum_{\pi \in L_{n, n}^{+}} q^{\operatorname{area}(\pi)}\)的话，
有递归方程
\[C_{n}(q)=\sum_{k=1}^{n} q^{k-1} C_{k-1}(q) C_{n-k}(q), \quad n \geq 1 \]这么定义的\(C_n(q)\)还能和co-inversion联系起来
这么定义the basic 超几何级数
\[p+1 \phi_{p}\left(\begin{array}{ccc} a_{1}, & a_{2}, & \ldots, & a_{p+1} \\ & b_{1}, & \ldots, & b_{p} \end{array} ; q ; z\right)=\sum_{k=0}^{\infty} \frac{\left(a_{1}\right)_{k} \cdots\left(a_{p+1}\right)_{k}}{(q)_{k}\left(b_{1}\right)_{k} \cdots\left(b_{p}\right)_{k}} z^{k} \]其中\((a)_k\)表示升阶乘，即\((a)_k=a(a+1)...(a+k-1)\)
Cauchy's q-binomial theorem
\[{ }_{1} \phi_{0}\left(\begin{array}{ll} a & \\ - & \end{array} ; q ; z\right)=\sum_{k=0}^{\infty} \frac{(a)_{k}}{(q)_{k}} z^{k}=\frac{(a z)_{\infty}}{(z)_{\infty}}, \quad|z|<1,|q|<1 \]where,
\[(a ; q)_{\infty}=(a)_{\infty}=\prod_{i=0}^{\infty}\left(1-a q^{i}\right) \]推论11.2.11 The q-binomial theorem
\[\sum_{k=0}^{n} q^{\left(\begin{array}{c} k \\ 2 \end{array}\right)}\left[\begin{array}{l} n \\ k \end{array}\right] z^{k}=(-z ; q)_{n} \] \[\sum_{k=0}^{\infty}\left[\begin{array}{c} n+k \\ k \end{array}\right] z^{k}=\frac{1}{(z ; q)_{n+1}} \]推论 11.2.12
\[\sum_{k=0}^{h} q^{(n-k)(h-k)}\left[\begin{array}{l} n \\ k \end{array}\right]\left[\begin{array}{c} m \\ h-k \end{array}\right]=\left[\begin{array}{c} m+n \\ h \end{array}\right] \text { holds. } \]推论11.2.13
\[\sum_{k=0}^{h} q^{(m+1) k}\left[\begin{array}{c} n-1+k \\ k \end{array}\right]\left[\begin{array}{c} m+h-k \\ h-k \end{array}\right]=\left[\begin{array}{c} m+n+h \\ h \end{array}\right] \]The q-Vandermonde convolution
。。。这个为什么是拿超几何级数来写的。。。。
剩下的空着