格路边界对的Chuns-Feller定理

发布时间：2018-05-18 14:17

本文选题：Chung-Feller定理 + 格路　；参考：《华东师范大学》2017年硕士论文

【摘要】：Irving和Rattan给出了在循环平移分段线性边界控制下的格路个数的计算公式.他们的主要结论可以看作以下著名定理的一个推广:从点(0,0)到点(kn,n)且被直线x=ky控制的格路个数(当= 1时,就是Dyck路).另一方面,著名的Chung-Feller定理告诉我们,从(0,0)到(n,n)且恰有2k(k= 0,1,…,n)个步法在直线y=x上方的格路个数与k的选取无关,因此这样的格路个数为Catalan数Cn=1/n+1(?).本篇论文主要研究在格路边界对(P,a)下恰有k个瑕疵的格路个数.其中P是从点(0,0)到点(n,m)的一条格路,a表示n的一个弱m-分拆,若格路P有k个向右的步法在边界(?)a的上方,则称格路边界对(P,a)有k个瑕疵.我们通过构造一个双射证明了,对于任意一个给定的分拆a,将所有在循环平移分段线性边界(?)a控制下,满足上述条件的格路加起来恰好是(?).也就是说,我们把Ivring-Rattan公式推广到一个Chung-Feller类型的定理.我们还将此双射用于计算格路的双上升数目,从而得到更为细致的结果.
[Abstract]:Irving and Rattan give the calculation formula of the number of lattice paths under the control of the circular translation piecewise linear boundary. Their main conclusions can be regarded as a generalization of the following famous theorems: the number of lattice paths controlled by the point (0,0) to point (KN, n) and by the linear x=ky (when = 1, it is Dyck Road). On the other hand, the famous Chung-Feller theorem tells me People, from (0,0) to (n, n) and there are just 2K (k= 0,1,... N) the number of lattice paths above the line y=x is independent of the selection of K, so the number of K is Catalan Cn=1/n+1 (?). This paper mainly deals with the number of K defects in the P, a, where P is a lattice from a point (0,0) to point (n, M). The step method is above the boundary (?) a, then it is called the lattice path boundary pair (P, a) with K defects. By constructing a double fire proof, for any given partition a, we add all the lattice paths that satisfy the above conditions under the cyclic translational piecewise linear boundary (?) a control. In other words, we extend the Ivring-Rattan formula. To a theorem of Chung-Feller type, we also use this double shot to calculate the number of double rise of grid path and get more detailed results.
【学位授予单位】：华东师范大学
【学位级别】：硕士
【学位授予年份】：2017
【分类号】：O157

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