# 数学代写|组合学代写Combinatorics代考|Riordan arrays

## 数学代写|组合学代写Combinatorics代考|Riordan arrays

The Riordan arrays are extremely useful if we would like to calculate sums of the form
$$\sum_{k=0}^n d_{n, k} a_k,$$
where $a_k$ is a given sequence and $d_{n, k}$ is a triangular sequence as, for example, the binomial coefficients or the Stirling numbers. By definition, the Riordan ${ }^7$ array is the lower triangular infinite matrix
$$d_{n, k}=\left[x^n\right] d(x)(x \cdot h(x))^k \quad(n \geq 1),$$
where $d$ and $h$ are two almost arbitrary generating functions. It must hold true that $d(0) \neq 0$. Also, if we want our array to be invertible, which equivalently means that $d_{n, n} \neq 0$ for all $n$, then we must suppose that $h(0) \neq 0$. In this case, the Riordan array is proper. Above $\left[x^n\right] f(x)$ denotes the coefficient of $x^n$ in the generating function $f(x)$. Thus, for example,
$$\left[x^n\right] e^x=\frac{1}{n !}$$
If we write down $d_{n, k}$ in matrix form where $n$ is the row index and $k$ is the column index, then we see that the resulting matrix will be a lower triangular matrix as we said. This is so, because $\left[x^n\right] d(x)(x \cdot h(x))^k$ will be obviously zero if $n<k$.
In symbols, we write that
$$\mathcal{R}\left(d_{n, k}\right)=(d(x), h(x))=(d, h)$$
and we say that $d_{n, k}$ is a Riordan array or Riordan matrix. It is determined by $d$ and $h$ via (5.18).

## 数学代写|组合学代写Combinatorics代考|Some identities for the Cauchy numbers

Three more identities for the Cauchy numbers are added here. Two of them are applications of the Riordan method. This short subsection is added to show additional applications of this useful tool. The results of this section are taken from [406].
The first identity that we are going to prove is
$$\sum_{k=0}^n\left{\begin{array}{l} n \ k \end{array}\right} c_k=\frac{1}{n+1} \quad(n \geq 0)$$
A trivial transformation first:
$$\sum_{k=0}^n\left{\begin{array}{l} n \ k \end{array}\right} c_k=n ! \sum_{k=0}^n \frac{k !}{n !}\left{\begin{array}{l} n \ k \end{array}\right} \frac{c_k}{k !}$$
Then, by $(5.21)$,
$$\sum_{k=0}^n\left{\begin{array}{l} n \ k \end{array}\right} c_k=n !\left[x^n\right] d(x) f(x \cdot h(x))$$
where
$$(d(x), h(x))=\left(1, \frac{e^x-1}{x}\right), \quad \text { and } \quad f(x)=\frac{x}{\log (1+x)}$$
by (5.19) and (5.15). Therefore,
$$\sum_{k=0}^n\left{\begin{array}{l} n \ k \end{array}\right} c_k=n !\left[x^n\right] \frac{x\left(\frac{e^x-1}{x}\right)}{\log \left(1+x\left(e^x-1 x\right)\right)}=n !\left[x^n\right] \frac{e^x-1}{x} .$$
Since, as it is easy to see, $\frac{e^x-1}{x}$ is the generating function of $\frac{1}{(n+1) !}$,
$$n !\left[x^n\right] \frac{e^x-1}{x}=\frac{1}{n+1}$$
and (5.22) follows.
The dual of this identity is
$$\sum_{k=0}^n\left{\begin{array}{l} n \ k \end{array}\right}(-1)^k C_k=\frac{(-1)^n}{n+1}$$
The reader is called to prove this result.

# 组合学代写

## 数学代写|组合学代写Combinatorics代考|Riordan arrays

$$\sum_{k=0}^n d_{n, k} a_k,$$

$$d_{n, k}=\left[x^n\right] d(x)(x \cdot h(x))^k \quad(n \geq 1)$$

$$\left[x^n\right] e^x=\frac{1}{n !}$$

$$\mathcal{R}\left(d_{n, k}\right)=(d(x), h(x))=(d, h)$$

## 数学代写|组合学代写Combinatorics代考|Some identities for the Cauchy numbers

$$(d(x), h(x))=\left(1, \frac{e^x-1}{x}\right), \quad \text { and } \quad f(x)=\frac{x}{\log (1+x)}$$

$$n !\left[x^n\right] \frac{e^x-1}{x}=\frac{1}{n+1}$$

$\backslash$ |sum_{k=0}^ $\backslash \backslash$ left $\backslash$ begin ${$ array $}{}} n \backslash k \backslash$ end ${$ array $} \backslash r i g h t}(-1)^{\wedge} k C_{-} k$

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