目录
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[egin{array}{c}
2 n \
n
end{array}
ight]
]
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
[egin{aligned}
q^0+q^3+q^4+q^2+q^6
&=frac{1}{1+q+q^2+q^3}cdotfrac{(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
The q-Vandermonde convolution/q-范特蒙德卷积
这么定义the basic 超几何级数
[p+1 phi_{p}left(egin{array}{ccc}
a_{1}, & a_{2}, & ldots, & a_{p+1} \
& b_{1}, & ldots, & b_{p}
end{array} ; q ; z
ight)=sum_{k=0}^{infty} frac{left(a_{1}
ight)_{k} cdotsleft(a_{p+1}
ight)_{k}}{(q)_{k}left(b_{1}
ight)_{k} cdotsleft(b_{p}
ight)_{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(egin{array}{ll}
a & \
- &
end{array} ; q ; z
ight)=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}
ight)
]
推论11.2.11 The q-binomial theorem
[sum_{k=0}^{n} q^{left(egin{array}{c}
k \
2
end{array}
ight)}left[egin{array}{l}
n \
k
end{array}
ight] z^{k}=(-z ; q)_{n}
]
[sum_{k=0}^{infty}left[egin{array}{c}
n+k \
k
end{array}
ight] z^{k}=frac{1}{(z ; q)_{n+1}}
]
推论 11.2.12
[sum_{k=0}^{h} q^{(n-k)(h-k)}left[egin{array}{l}
n \
k
end{array}
ight]left[egin{array}{c}
m \
h-k
end{array}
ight]=left[egin{array}{c}
m+n \
h
end{array}
ight] ext { holds. }
]
推论11.2.13
[sum_{k=0}^{h} q^{(m+1) k}left[egin{array}{c}
n-1+k \
k
end{array}
ight]left[egin{array}{c}
m+h-k \
h-k
end{array}
ight]=left[egin{array}{c}
m+n+h \
h
end{array}
ight]
]
The q-Vandermonde convolution
。。。这个为什么是拿超几何级数来写的。。。。
剩下的空着