The Riordan Group and Applications 第1章 笔记

2022年的一本书，只有376页。证明直接去书里面找。

目录

• 1 介绍

• 1.1 啥是RiordanArray
• 1.2 起源和研究动机
• 1.3 基础的应用
• 章后练习
• 参考

1 介绍

1.1 啥是RiordanArray

1.2 起源和研究动机

1.3 基础的应用

常见的如下：

(*用于快速验证的Mathematica代码*)

Clear["Global`*"]

RiordanArray[p_, q_, numOfRows_Integer] := 
  Table[SeriesCoefficient[p*q^k, {z, 0, n}], {n, 0, numOfRows}, {k, 0,
      n}] // Grid;
GenerateMatricesVLT[T_, L1_, numOfRows_Integer] := 
 Module[{MatrixV, MatrixL, matrixT}, 
  MatrixV = RiordanArray[T, L1, numOfRows]; 
  MatrixL = RiordanArray[1, L1, numOfRows]; 
  MatrixT = RiordanArray[T, z, numOfRows];
  {MatrixV, MatrixL, MatrixT}]


CatalanGF = Sum[CatalanNumber[n]*z^n, {n, 0, Infinity}];
GenerateMatricesVLT[CatalanGF, z*CatalanGF^2, 8]


TenaryNumberGF = Sum[Binomial[3 n, n]/(2*n + 1)*z^n, {n, 0, Infinity}];
GenerateMatricesVLT[TenaryNumberGF^2, 2 (TenaryNumberGF - 1), 8]

Clear["Global`*"];
ASequenceOfTheUURTree[z_] := 1 + z^2;
DerivativeASequenceOfTheUURTree[z_] = D[ASequenceOfTheUURTree[z], z];
T = GF /. Solve[GF == ASequenceOfTheUURTree[z*GF], GF][[1]]
L1 = z*DerivativeASequenceOfTheUURTree[z*T]
RiordanArray[p_, q_, numOfRows_Integer] := 
  Table[SeriesCoefficient[p*q^k, {z, 0, n}], {n, 0, numOfRows}, {k, 0,
      n}] // Grid;
GenerateMatricesVLT[T_, L1_, numOfRows_Integer] := 
 Module[{MatrixV, MatrixL, matrixT}, 
  MatrixV = RiordanArray[T, L1, numOfRows];
  MatrixL = RiordanArray[1, L1, numOfRows];
  MatrixT = RiordanArray[T, z, numOfRows];
  {MatrixV, MatrixL, MatrixT}]
GenerateMatricesVLT[T, L1, 8]