数理解析研究所講究録
第 1898 巻 2014 年 174-193

174

Convolution identities for Cauchy numbers of the firstkind and of the second kind
Takao Komatsu 1
Graduate School of Science and Technology
Hirosaki University

1

Introduction

The Cauchy numbers of the first kind $c_{n}(n\geq 0)$ are defined by$c_{n}= \int_{0}^{1}x(x-1)\ldots(x-n+1)dx$

and the generating function of$c_{n}$

is given by$\frac{x}{\ln(1+x)}=\sum_{n=0}^{\infty}c_{n}\frac{x^{n}}{n!} (|x|<1)$$([4,11])$ . The first few values are$c_{0}=1,$$c_{1}= \frac{1}{2},$$c_{2}=- \frac{1}{6},$$c_{3}= \frac{1}{4},$$c_{4}=- \frac{19}{30},$$c_{5}= \frac{9}{4},$$c_{6}=- \frac{863}{84}.$

The Cauchy numbers are almost identical with the Bernoulli numbers of thesecond kind, , as introduced in [8] and later studied by various authors. In
fact, we have $c_{n}=n!b_{n}$ . Notice that the Cauchy numbers can be expressedin terms of the unsigned Stirling numbers of the first kind and:$b_{n}$$c_{n}$$\{\begin{array}{l}ni\end{array}\}$$c_{n}= \sum_{i=0}^{n}\frac{(-1)^{n-i}}{i+1}\{\begin{array}{l}ni\end{array}\}$(1)

lThis research was supported in part by the Grant-in-Aid for Scientific research (C)(No.22540005), the Japan Society for the Promotion of Science.

175([9, 11]), where the unsigned Stirling numbers of the first kindcoefficients of the rising factorial$\{\begin{array}{l}ni\end{array}\}$arise as$x(x+1) \ldots(x+n-1)=\sum_{i=0}^{n}\{\begin{array}{l}ni\end{array}\}x^{i}$(see e.g. [7]).

The Cauchy numbers of the second kind$\hat{c}_{n}(n\geq 0)$

are defined by$\hat{c}_{n}=\int_{0}^{1}(-x)(-x-1)\ldots(-x-n+1)dx$

and the generating function of$\hat{c}_{n}$is given by$\frac{x}{(1+x)\ln(1+x)}=\sum_{n=0}^{\infty}\hat{c}_{n}\frac{x^{n}}{n!} (|x|<1)$([4, 11]). The first few values are$\hat{c}_{0}=1,$$\hat{c}_{1}=-\frac{1}{2},$$\hat{c}_{2}=\frac{5}{6},$$\hat{c}_{3}=-\frac{9}{4},$$\hat{c}_{4}=\frac{251}{30},$$\hat{c}_{5}=-\frac{475}{12},$$\hat{c}_{6}=\frac{19087}{84}.$

With the classical umbral calculus notation (see e.g. $[12]$ ), defineand$(\hat{c_{1}}+\hat{c}_{m})^{n}$

for

$l,$

$m,$ $n\geq 0$$(c_{l}+\hat{c}_{m})^{n}$by$(c_{l}+c_{m})^{n}:= \sum_{j=0}^{n}(\begin{array}{l}nj\end{array})c_{l+j}c_{m+n-j},$$( \hat{c_{l}}+\hat{c}_{m})^{n}:=\sum_{j=0}^{n}(\begin{array}{l}nj\end{array})\hat{c_{l+j^{\hat{C}}m+n-j}},$respectively. The analogous concept for the Bernoulli numbersby the generating function$\frac{x}{e^{x}-1}=\sum_{n=0}^{\infty}B_{n}\frac{x^{n}}{n!} (|x|<2\pi)$$B_{n}$, defined,

has been extensively studied by many authors, including Agoh and Dilcher([1, 2, 5] and references there). Define$(B_{l}+B_{m})^{n}:= \sum_{j=0}^{n}(\begin{array}{l}nj\end{array})B_{l+j}B_{m+n-j}.$

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Then Euler’s famous formula can be written as$(B_{0}+B_{0})^{n}=-nB_{n-1}-(n-1)B_{n} (n\geq 1)$ .

(2)

In [1] an expression for $(B_{l}+B_{m})^{n}$ was found. In addition, some initial cases
were listed, including e.g.,$(B_{0}+B_{1})^{n}=- \frac{1}{2}(n+1)B_{n}-\frac{1}{2}nB_{n+1},$$(B_{0}+B_{2})^{n}=- \frac{1}{6}(n-1)B_{n}-\frac{1}{2}nB_{n+1}-\frac{1}{3}nB_{n+2},$$(B_{1}+B_{1})^{n}= \frac{1}{6}(n-1)B_{n}-B_{n+1}-\frac{1}{6}(n+3)B_{n+2}.$

On the other hand, an analogous formula to (2) is given by(3)$(c_{0}+c_{0})^{n}=-n(n-2)c_{n-1}-(n-1)c_{n} (n\geq 0)$([13]).
different direction of extended convolution identities for Cauchy
numbers is discussed in [10].
In this paper, we give an explicit formula for $(c_{l}+c_{m})^{n}$ and$(n\geq 0)$ together with some initial cases, including$A$$(\hat{c_{l}}+\hat{c}_{m})^{n}$$(c_{0}+c_{1})^{n}=- \frac{1}{2}(n+1)(n-1)c_{n}-\frac{1}{2}nc_{n+1}$(4),$(c_{0}+c_{2})^{n}= \frac{n!}{6}\sum_{k=0}^{n}\frac{(-1)^{n-k}(k-1)c_{k}}{k!}-\frac{1}{6}n(2n+1)c_{n+1}-\frac{1}{3}nc_{n+2}$,(5)$(c_{1}+c_{1})^{n}=- \frac{n!}{6}\sum_{k=0}^{n}\frac{(-1)^{n-k}(k-1)c_{k}}{k!}-\frac{1}{6}n(n+5)c_{n+1}-\frac{1}{6}(n+3)c_{n+2}$

(6)

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and
$( \hat{c}_{0}+\hat{c}_{0})^{n}=n!\sum_{k=0}^{n}(-1)^{n-k}\frac{\hat{c}_{k}}{k!}-n\hat{c}_{n}$,(7)$( \hat{c}_{0}+\hat{c}_{1})^{n}=-\frac{(n+1)!}{2}\sum_{k=0}^{n}(-1)^{n-k}\frac{\hat{c}_{k}}{k!}-\frac{1}{2}n\hat{c}_{n+1}$,(8)$( \hat{c}_{0}+\hat{c}_{2})^{n}=\frac{n!}{12}\sum_{k=0}^{n}\frac{(-1)^{n-k}}{k!}(2k(n+2k-2)+5(n-k+2)(n-k+1))\hat{c}_{k}$$- \frac{n}{3}\hat{c}_{n+2}-\frac{n}{2}\hat{c}_{n+1}$,(9)$( \hat{c}_{1}+\hat{c}_{1})^{n}=\frac{n!}{12}\sum_{k=0}^{n}\frac{(-1)^{n-k}}{k!}((n+1)(n+2)+k(8n-9k+19))\hat{c}_{k}$$- \hat{c}_{n+1}-\frac{n+3}{6}\hat{c}_{n+2}$

2

.

(10)

Fundamental results

Euler’s identity (2) is an easy consequence of the formula$b(x)^{2}=(1-x)b(x)-xb’(x)$ ,where $b(x)=x/(e^{x}-1)$ (see e.g. [6]). Similarly, the identity (3) is obtained
because $c(x)=x/\ln(1+x)$ satisfies the formula$c(x)^{2}=(1+x)c(x)-(1+x)xc’(x)$ .

From the generating function for the$c_{n}$

we obtain for$x^{i}c^{(\nu)}(x)= \sum_{n=0}^{\infty}\frac{n!}{(n-i)!}c_{n+\nu-i}\frac{x^{n}}{n!}$

$i,$

(11)
$v\geq 0,$

.

(12)

Therefore the identity (11) immediately leads to the formula (3). Differentiating both sides of (11) with respect to and dividing them by 2, we
obtain$x$$c(x)c’(x)=- \frac{1}{2}x(x+1)c"(x)-\frac{1}{2}xc’(x)+\frac{1}{2}c(x)$Then by applying (12), we have the identity (4).

.

(13)

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In general, by differentiating both sides of (11) times with respect to, we have the following. The left-hand side is due to the General Leibniz’srule. The right-hand side can be proved by induction.
$\mu$

$x$

Proposition 1. For$\mu\geq 0$$\sum_{\kappa=0}^{\mu}(\begin{array}{l}\mu\kappa\end{array})c^{(\kappa)}(x)c^{(\mu-\kappa)}(x)$.
(14)$=-(\mu-2)\mu c^{(\mu-1)}(x)-((2\mu-1)x+\mu-1)c^{(\mu)}(x)-x(1+x)c^{(\mu+1)}(x)$By applying (12) in Proposition 1, we obtain the following result.

Theorem 1. For$\mu\geq 0$

and $n\geq 0$ we have$\sum_{\kappa=0}^{\mu}(\begin{array}{l}\mu\kappa\end{array})(c_{\kappa}+c_{\mu-\kappa})^{n}=-(n+\mu-2)(n+\mu)c_{n+\mu-1}-(n+\mu-1)c_{n+\mu}.$

Examples. If we put $\mu=0,1$ in Theorem 1, we have the identities (3) and(4), respectively. If we put $\mu=2,3,4$ in Theorem 1, we obtain$(c_{0}+c_{2})^{n}+(c_{1}+c_{1})^{n}=- \frac{1}{2}n(n+2)c_{n+1}-\frac{1}{2}(n+1)c_{n+2}$

(15)

,

$(c_{0}+c_{3})^{n}+3(c_{1}+c_{2})^{n}=- \frac{1}{2}(n+1)(n+3)c_{n+2}-\frac{1}{2}(n+2)c_{n+3}$

,

(16)

$(c_{0}+c_{4})^{n}+4(c_{1}+c_{3})^{n}+3(c_{2}+c_{2})^{n}$$=- \frac{1}{2}(n+2)(n+4)c_{n+3}-\frac{1}{2}(n+3)c_{n+4}$

.

(17)

Since $\hat{c}(x)=x/(1+x)\ln(1+x)$ satisfies the formula$\hat{c}(x)^{2}=-xc\neg(x)+\frac{1}{1+x}\hat{c}(x)$

(18)

,

by $1/(1+x)= \sum_{i=0}^{\infty}(-1)^{i}x^{i}$ and the fact$x^{i} \hat{c}^{\langle\nu)}(x)=\sum_{n=0}^{\infty}\frac{n!}{(n-i)!}\hat{c}_{n+\nu-i}\frac{x^{n}}{n!} (i, \nu\geq 0)$

,

(19)

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we have,$( \hat{c}_{0}+\hat{c}_{0})^{n}=-n\hat{c}_{n}+n!\sum_{k=0}^{n}(-1)^{n-k}\frac{\hat{c}_{k}}{k!} (n\geq 0)$

which is the formula (7). Differentiating both sides of (18) bythem by 2, we obtain$x$

and dividing$\hat{c}(x)_{\tilde{\mathcal{C}}}(x)=-\frac{1}{2}x\vec{c}’(x)-\frac{x}{2(1+x)}\tilde{c}(x)-\frac{1}{2(1+x)^{2}}\hat{c}(x)$

.

(20)

Then by (19) and for $k\geq 1$$\frac{1}{(1+x)^{k}}=\sum_{i=0}^{\infty}(-1)^{\dot{\iota}}(k +i-i 1)x^{i}$

,

(21)

we have$\sum_{n=0}^{\infty}(\hat{c}_{0}+\hat{c}_{1})^{n}\frac{x^{n}}{n!}=-\frac{1}{2}\sum_{n=0}^{\infty}n\hat{c}_{n+1}\frac{x^{n}}{n!}-\frac{1}{2}\sum_{i=0}^{\infty}(-1)^{i}\sum_{n=0}^{\infty}\frac{n!}{(n-i-1)!}\hat{c}_{n-i}\frac{x^{n}}{n!}$$arrow\frac{1}{2}\sum_{i=0}^{\infty}(-1)^{i}(i+1)\sum_{n=0}^{\infty}\frac{n!}{(n-i)!}\hat{c}_{n-i}\frac{x^{n}}{n!}$$=- \frac{1}{2}\sum_{n=0}^{\infty}n\hat{c}_{n+1}\frac{x^{n}}{n!}-\frac{1}{2}\sum_{i=0}^{n}(-1)^{i}\sum_{n=0}^{\infty}\frac{(n+1)!}{(n-i)!}\hat{c}_{n-i}\frac{x^{n}}{n!},$yielding the formula (8).
In general, by differentiating both sides of (18) by
the following.
Proposition 2. For

$x$

at

$\mu$

times, we have$\mu\geq 0$$\sum_{\kappa=0}^{\mu}(\begin{array}{l}\mu\kappa\end{array})\hat{c}^{(\kappa)}(x)\hat{c}^{\langle\mu-\kappa)}(x)=\sum_{\nu=0}^{\mu}(-1)^{\mu-\nu}\frac{\mu!\hat{c}^{(\nu)}(x)}{\nu!(1+x)^{\mu-\nu+1}}-\mu\hat{c}^{(\mu)}(x)-x\hat{c}^{\langle\mu+1)}(x)$.

By applying (21) and (19) in Proposition 2, we obtain the following result.

Theorem 2. For$\mu\geq 0$

and $n\geq 0$ we have$\sum_{\kappa=0}^{\mu}(\begin{array}{l}\mu\kappa\end{array})(\hat{c}_{\kappa}+\hat{c}_{\mu-\kappa})^{n}=-(n+\mu)\hat{c}_{n+\mu}+\sum_{\nu=0}^{\mu}\frac{\mu!n!}{v!}\sum_{i=0}^{n}(-1)^{\mu-\nu+i}(\begin{array}{ll}\mu-v +ii \end{array}) \frac{\hat{c}_{n+\nu-i}}{(n-i)!}$

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Examples. If we put $\mu=0,1$ in Theorem 2, we have the identities (7) and(8), respectively. If we put $\mu=2,3,4$ in Theorem 2, we obtain$( \hat{c}_{0}+\hat{c}_{2})^{n}+(\hat{c}_{1}+\hat{c}_{1})^{n}=\frac{(n+2)!}{2}\sum_{k=0}^{n}(-1)^{n-k}\frac{\hat {}c_{k}}{k!}-\frac{n+2}{2}\hat{c}_{n+1}-\frac{n+1}{2}\hat{c}_{n+2},$(22)$(\hat{c}_{0}+\hat{c}_{3})^{n}+3(\hat{c}_{1}+\hat{c}_{2})^{n}$$=- \frac{(n+3)!}{2}\sum_{k=0}^{n}(-1)^{n-k}\frac{\hat {}c_{k}}{k!}+\frac{(n+2)(n+3)}{2}\hat{c}_{n+1}-\frac{n+3}{2}\hat{c}_{n+2}-\frac{n+2}{2}\hat{c}_{n+3},$(23)$(\hat{c}_{0}+\hat{c}_{4})^{n}+4(\hat{c}_{1}+\hat{c}_{3})^{n}+3(\hat{c}_{2}+\hat{c}_{2})^{n}$$= \frac{(n+4)!}{2}\sum_{k=0}^{n}(-1)^{n-k}\frac{\hat{c}_{k}}{k!}-\frac{(n+2)(n+3)(n+4)}{2}\hat{c}_{n+1}\prime+\frac{(n+3)(n+4)}{2}\hat{c}_{n+2}$$- \frac{n+4}{2}\hat{c}_{n+3}-\frac{n+3}{2}\hat{c}_{n+4}$

.

(24)

Next, notice that $c(x)$ satisfies the identity$3c(x)c”(x)=-(1+x)xc^{(3)}(x)- \frac{3}{2}xc"(x)+\frac{x}{2(1+x)}c’(x)-\frac{1}{2(1+x)}c(x)$Since by (12) we have$x^{2}c^{(3)}(x)= \sum_{n=0}^{\infty}n(n-1)c_{n+1^{\frac{x^{n}}{n!}}},$$xc^{(3)}(x)= \sum_{n=0}^{\infty}nc_{n+2^{\frac{x^{n}}{n!}}},$$xc”(x)= \sum_{n=0}^{\infty}nc_{n+1^{\frac{x^{n}}{n!}}}$

and$\frac{x}{1+x}c’(x)=(\sum_{\nu=0}^{\infty}(-x)^{\nu})(\sum_{n=0}^{\infty}nc_{n}\frac{x^{n}}{n!})$$= \sum_{n=0}^{\infty}(n!\sum_{k=0}^{n}\frac{(-1)^{n-k}kc_{k}}{k!})\frac{x^{n}}{n!}$

.

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and
$\frac{1}{1+x}c(x)=\sum_{n=0}^{\infty}(n!\sum_{k=0}^{n}\frac{(-1)^{n-k_{C_{k}}}}{k!})\frac{x^{n}}{n!},$we get$3(c_{0}+c_{2})^{n}=-nc_{n+2}-n(n+ \frac{1}{2})c_{n+1}+\frac{n!}{2}\sum_{k=0}^{n}\frac{(-1)^{n-k}(k-1)c_{k}}{k!},$yielding the identity (5). In addition, by the identities (5) and (15), we hav$(c_{1}+c_{1})^{n}=- \frac{n!}{6}\sum_{k=0}^{n}\frac{(-1)^{n-k}(k-1)c_{k}}{k!}-\frac{1}{6}n(n+5)c_{n+1}-\frac{1}{6}(n+3)c_{n+2},$

which is the identity (6).
Next, notice that $c(x)$ satisfies the identity$\hat{c}(x)\tilde{c}’(x)=-\frac{x}{3}\hat{c}^{(3)}(x)-\frac{x}{2(1+x)}\hat{c}"(x)+\frac{x}{6(1+x)^{2}}\tilde{c}(x)+\frac{5}{6(1+x)^{3}}\hat{c}(x)$By (19) and (21).

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yielding the identity (9). In addition, by the identities (9) and (22), we have$( \hat{c}_{1}+\hat{c}_{1})^{n}=\frac{n!}{12}\sum_{k=0}^{n}\frac{(-1)^{n-k}}{k!}((n+1)(n+2)+k(8n-9k+19))\hat{c}_{k}-\hat{c}_{n+1}-\frac{n+3}{6}\hat{c}_{n+2}$

which is the identity (10).
Proposition 3. For $m\geq 0$ we have$\hat{c}(x)\hat{c}^{(m)}(x)$$=- \frac{x}{m+1}\hat{c}^{\langle m+1)}(x)-\sum_{k=0}^{m-1}(\begin{array}{l}mk\end{array})\frac{c_{k+1}}{k+1}\frac{x}{(1+x)^{k+1}}\hat{c}^{(m-k)}(x)+\frac{\hat{c}_{m}}{(1+x)^{m+1}}\hat{c}(x)$(25)

Lemma 1. For $n\geq 0$ we have$\hat{c}^{\langle n)}(x)=(\frac{-1}{1+x})^{n+1}\sum_{j=0}^{n}\frac{\hat{g}_{n,j}}{(\ln(1+x))^{j+1}},$

where$\hat{g}_{n,j}=j!(n\{\begin{array}{ll} nj +1\end{array}\}-\{\begin{array}{l}nj\end{array}\}x)$

Proof. Since$\hat{g}_{1,0}=1$

and $\hat{g}_{1,1}=-x$ , for $n=1$ we have$\frac{d}{dx}\frac{x}{(1+x)\ln(1+x)}=\frac{\ln(1+x)-x}{(1+x)^{2}(\ln(1+x))^{2}}$$= \frac{1}{(1+x)^{2}}(\frac{\hat{g}_{1,0}}{\ln(1+x)}+\frac{\hat{g}_{1,1}}{(\ln(1+x))^{2}})$.

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Assume that the result holds for . Then$n$$\frac{d^{n+1}}{dx^{n+1}}\frac{x}{(1+x)\ln(1+x)}$$=(n+1)( \frac{-1}{1+x})^{n+2}\sum_{j=0}^{n}\frac{\hat{g}_{n,j}}{(\ln(1+x))^{j+1}}$$+( \frac{-1}{1+x})^{n+1}\sum_{j=0}^{n}\frac{-j!\{\begin{array}{l}nj\end{array}\}(\ln(1+x))^{j+1}-\hat{g}_{n,j}(j+1)(\ln(1+x))^{j}\frac{1}{1+x}}{(\ln(1+x))^{2j+2}}$$=n( \frac{-1}{1+x})^{n+2}\sum_{j=0}^{n}\frac{\hat{g}_{n,j}}{(\ln(1+x))^{j+1}}+(\frac{-1}{1+x})^{n+2}\sum_{j=0}^{n}\frac{\hat{g}_{n,j}}{(\ln(1+x))^{j+1}}$$+( \frac{-1}{1+x})^{n+2}\sum_{j=0}^{n}\frac{j!\{\begin{array}{l}nj\end{array}\}(1+x)}{(\ln(1+x))^{j+1}}+(\frac{-1}{1+x})^{n+2}\sum_{j=0}^{n}\frac{\hat{g}_{n,j}(j+1)}{(\ln(1+x))^{j+2}}$$=( \frac{-1}{1+x})^{n+2}\sum_{j=0}^{n}\frac{n\hat{g}_{n,j}+j!\{\begin{array}{l}n+1j+1\end{array}\}}{(\ln(1+x))^{j+1}}+(\frac{-1}{1+x})^{n+2}\sum_{j=1}^{n+1}\frac{j\hat{g}_{n,j-1}}{(\ln(1+x))^{j+1}}$$=( \frac{-1}{1+x})^{n+2}\sum_{j=0}^{n+1}\frac{\hat{g}_{n+1,j}}{(\ln(1+x))^{j+1}}.$口

Lemma 2. For integers

$m$

and

$j$

with $m\geq j\geq 0$ we have$\hat{g}_{m+1,j}-(m+1)\hat{g}_{m,j-1}=\sum_{k=0}^{m-j}(\begin{array}{ll}m +1k +1\end{array})(-1)^{k}c_{k+1} \hat{g}_{m-k,j}.$Proof. It is enough to prove$\{\begin{array}{ll}m +1j +1\end{array}\}-\frac{m}{j}\{\begin{array}{l}mj\end{array}\}=\sum_{k=0}^{m-j}(\begin{array}{ll}m +1k +1\end{array})(-1)^{k}c_{k+1} \frac{m-k}{m+1}\{\begin{array}{l}m-kj+1\end{array}\}$

(26)

and
$\{\begin{array}{ll}m +1 j\end{array}\}-\frac{m+1}{j}\{\begin{array}{ll} mj -1\end{array}\}= \sum_{k=0}^{m-j}(\begin{array}{ll}m +1k +1\end{array})(-1)^{k}c_{k+1} \{\begin{array}{l}m-kj\end{array}\}$

(27)

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The identity (26) holds because by (1) formulae in Table 241 in [7]$\sum_{k=0}^{m-j+1}(\begin{array}{ll}m +1 k\end{array})$$(-1)^{k}c_{k} \frac{m-k+1}{m+1}\{\begin{array}{ll}m-k +1+1j \end{array}\}$$= \sum_{k=0}^{m-j+1}(\begin{array}{ll}m +1 k\end{array})(-1)^{k} \sum_{i=0}^{k}\frac{(-1)^{k-i}}{i+1}\{\begin{array}{l}ki\end{array}\}\frac{m-k+1}{m+1}\{\begin{array}{ll}m-k +1+1j \end{array}\}$$= \sum_{i=0}^{m-j+1}\frac{(-1)^{i}}{i+1}\sum_{k=i}^{m-j+1}(\begin{array}{ll}m +1 k\end{array}) \{\begin{array}{l}ki\end{array}\}\frac{m-k+1}{m+1}\{\begin{array}{ll}m-k +1+1j \end{array}\}$$= \sum_{i=0}^{m-j+1}\frac{(-1)^{i}}{i+1}(\begin{array}{ll}i+ jj \end{array}) \{\begin{array}{ll}m +1i+ +1j\end{array}\}$$= \frac{m}{j}\{\begin{array}{l}mj\end{array}\}$Similarly, the identity (27) holds because$\sum_{k=0}^{m-j+1}(\begin{array}{ll}m +1 k\end{array})(-1)^{k}c_{k} \{\begin{array}{ll}m-k +1j \end{array}\}$$= \sum_{k=0}^{m-j+1}(\begin{array}{ll}m +1 k\end{array})(-1)^{k} \sum_{i=0}^{k}\frac{(-1)^{k-i}}{i+1}\{\begin{array}{l}ki\end{array}\}\{\begin{array}{ll}m-k +1j \end{array}\}$$= \sum_{i=0}^{m-j+1}\frac{(-1)^{i}}{i+1}\sum_{k=i}^{m-j+1}(\begin{array}{ll}m +1 k\end{array}) \{\begin{array}{l}ki\end{array}\}\{\begin{array}{ll}m-k +1j \end{array}\}$$= \sum_{i=0}^{m-j+1}\frac{(-1)^{i}}{i+1}(\begin{array}{ll}i+ jj \end{array}) \{\begin{array}{ll}+1m i+ j\end{array}\}$$= \frac{m+1}{j}\{\begin{array}{ll} mj -1\end{array}\}$口We also need the following relation [11, Theorem 2.4 (2.1)] in order toprove Proposition 3.
$($

Lemma 3. For $m\geq 0$$(-1)^{m} \frac{\hat{c}_{m}}{m!}=\sum_{k=0}^{m}(-1)^{k}\frac{c_{k}}{k!}$

$)$

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Proof of Proposition 3. The identity (25) holds for $m=0$ and $m=1$ becauseof (18) and (20), respectively. Let $m\geq 2$ . By Lemma 1 with $\hat{g}_{n,n}=-n!x$$(n\geq 0),\hat{g}_{n,0}=n!(n\geq 1)$ and $\hat{g}_{0,0}=-x$ together with Lemmata 2 and 3$\hat{c}(x)\hat{c}^{\{m)}(x)+\frac{x}{m+1}\hat{c}^{\langle m+1)}(x)$$= \frac{x}{m+1}(\frac{-1}{1+x})^{m+2}(\sum_{j=0}^{m-1}\frac{\hat{g}_{m+1,j+1}-(m+1)\hat{g}_{m,j})}{(\ln(1+x))^{j+2}}+\frac{(m+1)!}{\ln(1+x)})$$= \frac{x}{m+1}(\frac{-1}{1+x})^{m+2}(\sum_{j=0}^{m-1}\frac{1}{(\ln(1+x))^{j+2}}\sum_{k=0}^{m-j-1}(\begin{array}{ll}m +1k +1\end{array})(-1)^{k}c_{k+1} \hat{g}_{m-k,j+1}$$+ \frac{(m+1)!}{\ln(1+x)}\sum_{k=0}^{m-1}(-1)^{k}\frac{c_{k+1}}{(k+1)!}+\frac{(-1)^{m}(m+1)\hat{c}_{m}}{\ln(1+x)})$$= \frac{x}{m+1}(\frac{-1}{1+x})^{m+2}(\sum_{j=0}^{m}\frac{m+1}{(\ln(1+x))^{j+1}}\sum_{k=0}^{m-j}\frac{c_{k+1}}{k+1}(\begin{array}{l}mk\end{array})(-1)^{k}\hat{g}_{m-k,j}$$+ \frac{(-1)^{m}}{\ln(1+x)}(c_{m+1^{X}}+(m+1)\hat{c}_{m}))$$=( \frac{-1}{1+x})^{m+2}\sum_{k=0}^{m}(\begin{array}{l}mk\end{array})\frac{c_{k+1}}{k+1}(-1)^{k}\sum_{j=0}^{m-k}\frac{x\hat{g}_{m-k,j}}{(\ln(1+x))^{j+1}}$$+( \frac{-1}{1+x})^{m+2}\frac{x}{\ln(1+x)}(-1)^{m}(\frac{c_{m+1}}{m+1}x+\hat{c}_{m})$$=- \sum_{k=0}^{m-1}(\begin{array}{l}mk\end{array})\frac{c_{k+1}}{k+1}\frac{x}{(1+x)^{k+1}}\hat{c}^{\langle m-k)}(x)+\frac{\hat{c}_{m}}{(1+x)^{m+1}}\hat{c}(x)$.
口By Proposition 3, we obtain the following theorem.

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Theorem 3. For $m\geq 0$ and $n\geq 0$ we have$(\hat{c}_{0}+\hat{c}_{m})^{n}$$n$$=-\overline{m+1}^{\hat{\mathcal{C}}_{n+m}}$$- \frac{1}{m+1}\sum_{k=0}^{m-1}(\begin{array}{ll}m +1k +1\end{array})c_{k+1} \sum_{i=0}^{n-1}(-1)^{i}(k +ii) \frac{n!}{(n-i-1)!}\hat{c}_{n+m-k-i-1}$$+ \hat{c}_{m}\sum_{i=0}^{n}(-1)^{\dot{\iota}}(\begin{array}{ll}m +i i\end{array}) \frac{n!}{(n-i)!}\hat{c}_{n-i}$.

Examples. When $m=0,$ $m=1$ and $m=2$ , we have the formulae (7), (8)and (9), respectively. If $m=3$ , we have$( \hat{c}_{0}+\hat{c}_{3})^{n}=-\frac{(n+1)!}{8}\sum_{k=0}^{n}\frac{(-1)^{n-k}(2(2n^{2}-6n+9)-(n-k)(n+9k-27))\hat{c}_{k}}{k!}$$+ \frac{n(2n-1)\hat{c}_{n+1}}{4}-\frac{n\hat{c}_{n+2}}{2}-\frac{n\hat{c}_{n+3}}{4}.$Proof of Theorem 3. By (19)$x \hat{c}^{\langle m+1)}(x)=\sum_{n=0}^{\infty}n\hat{c}_{n+m^{\frac{x^{n}}{n!}}}$By (21) we get$\sum_{k=0}^{m-1}(\begin{array}{ll}m +1k +1\end{array}) \frac{c_{k+1^{X}}}{(1+x)^{k+1}}\hat{c}^{\langle m-k)}(x)$$= \sum_{k=0}^{m-1}(\begin{array}{l}m+1k+1\end{array})c_{k+1}\sum_{i=0}^{\infty}(-1)^{i}(k +ii) \dot{\iota}+1\triangleleft m-k)$

砺 $+$ m-k-i-l$= \sum_{k=0}^{m-1}(\begin{array}{ll}m +1k +1\end{array})c_{k+1} \sum_{i=0}^{n-1}(-1)^{i}(k +ii) \sum_{n=0}^{\infty}\frac{n!}{(n-i-1)!}$

Finally, for $m\geq 0$$\frac{\hat{c}_{m}}{(1+x)^{m+1}}\hat{c}(x)=\hat{c}_{m}\sum_{i=0}^{n}(-1)^{i}(\begin{array}{ll}m +i i\end{array}) \sum_{n=0}^{\infty}\frac{n!}{(n-i)!}\hat{c}_{n-i}\frac{x^{n}}{n!}.$$\frac{x^{n}}{n!}$187
口Similarly to Theorem 3 for Cauchy numbers of the second kind, we havethe following for Cauchy numbers of the first kind.

Theorem 4. For $m\geq 2$ and $n\geq 0$ we have$(c_{0}+c_{m})^{n}=$$- \frac{n}{m+1}(\frac{2n+m-1}{2}c_{n+m-1}+c_{n+m})$$- \sum_{k=1}^{m-1}\frac{c_{k+1}}{k+1}(\begin{array}{l}mk\end{array})\sum_{i=0}^{n-1}(-1)^{i}(k +i-1i) \frac{n!}{(n-i-1)!}c_{n+m-k-i-1}$$+c_{m} \sum_{i=0}^{n}(-1)^{i}(\begin{array}{ll}m +i-2 i\end{array}) \frac{n!}{(n-i)!}c_{n-i}.$

The expression of $(c_{0}+c_{m})^{n}$ is based upon the following relation.
Proposition 4. For $m\geq 0$ we have$c(x)c^{(m)}(x)$$=- \frac{x(1+x)}{m+1}c^{(m+1)}(x)-\sum_{k=0}^{m-1}\frac{c_{k+1}}{k+1}(\begin{array}{l}mk\end{array})\frac{x}{(1+x)^{k}}c^{(m-k)}(x)$$+ \frac{c_{m}}{(1+x)^{m-1}}c(x)$We need the following Lemma.

Lemma 4. For $n\geq 0$ we have$c^{(n)}(x)=( \frac{-1}{1+x})^{n}\sum_{j=0}^{n}\frac{g_{n,j}}{(\ln(1+x))^{j+1}},$where$g_{n,j}=j!(x\{\begin{array}{l}nj\end{array}\}-(x+1)n\{\begin{array}{ll}n -1 j\end{array}\})$

Proof. It is proven inductively that$\frac{d^{n}}{dx^{n}}\frac{1}{\ln(1+x)}=\frac{(-1)^{n}}{(1+x)^{n}}\sum_{j=0}^{n}\{\begin{array}{l}nj\end{array}\}\frac{j!}{(\ln(1+x))^{j+1}} (n\geq 0)$.

188(Cf. [3, Lemma 4.1]). Using Leibniz’s rule, we have$\frac{d^{n}}{dx^{n}}\frac{x}{\ln(1+x)}=x\frac{d^{n}}{dx^{n}}\frac{1}{\ln(1+x)}+n\frac{d^{n-1}1}{dx^{n-1}\ln(1+x)}$$=x \frac{(-1)^{n}}{(1+x)^{n}}\sum_{j=0}^{n}\{\begin{array}{l}nj\end{array}\}\frac{j!}{(\ln(1+x))^{j+1}}$$+n \frac{(-1)^{n-1}}{(1+x)^{n-1}}\sum_{j=0}^{n-1}\{\begin{array}{ll}n -1 j\end{array}\} \frac{j!}{(\ln(1+x))^{j+1}}$$=( \frac{-1}{1+x})^{n}\sum_{j=0}^{n}\frac{j!}{(\ln(1+x))^{j+1}}(x\{\begin{array}{l}nj\end{array}\}-(x+1)n\{\begin{array}{ll}n -1 j\end{array}\})$

口

3

Main result

We can show the following proposition in order to obtain the main result.
Proposition 5. Let

$l,$

$m$

be nonnegative integers with $m\geq l\geq 1$ . Then$c^{(l)}(x)c^{(m)}(x)=- \frac{l!m!}{(l+m+1)!}x(1+x)c^{(\iota+m+1)}(x)-\frac{l!m!}{(l+m)!}(2x+1)c^{(\iota+m)}(x)$$- \frac{l!m!}{(l+m-1)!}c^{(l+m-1)}(x)+\sum_{k=0}^{m-l-1}\frac{a_{l,m.m-k}x+b_{l,m.m-k}}{(1+x)^{l+k}}c^{(m-k)}$(置)$+ \sum_{k=0}^{l-2}a_{l,m’ l-k}x+b_{l,m,l-k}c^{(l-k)}(x)(1+x)^{m+k}+\frac{a_{l,m,1^{X}}}{(1+x)^{l+m-1}}c’(x)+\frac{b_{l,m,0}}{(1+x)^{l+m-1}}c(x)$where for $l+1\leq r=m-k\leq m$$a_{l,m,r}=(-1)^{\iota+1} (\begin{array}{l}mk\end{array})\sum_{i=0}^{l}i!(\begin{array}{l}li\end{array})(l+ ki -2) \frac{c_{l+k-i+1}}{l+k-i+1},$$b_{l,m,r}=(-1)^{\iota}l (\begin{array}{l}mk\end{array})\sum_{i=0}^{\iota-1}i!(\begin{array}{ll}l- 1i \end{array}) (l+ ki -2)\frac{c_{l+k-i}}{l+k-i},$,

189
and for $1\leq r=l-k\leq l$$a_{l,m,r}= \sum_{j=0}^{r-2}\frac{(-1)^{l+j+1}}{r}(\begin{array}{l}lj\end{array})(\begin{array}{ll}m r-j -1\end{array}) (\begin{array}{ll}r -1 j\end{array})$$\cross\sum_{i=0}^{l-j}i!(\begin{array}{ll}l- ji \end{array}) (l+ m-ri -2)c_{l+m-i-r+1}$$+ \frac{(-1)^{l+r}}{r}(\begin{array}{ll} lr -1\end{array}) \sum_{i=0}^{\iota-r+1}i!(l- ri +1) (l+ m_{i}-r -1)c_{l+m-i-r+1}$

except$a_{l,m,3}=- \frac{m(m-1)}{3!}\sum_{i=0}^{4}i|(\begin{array}{l}4i\end{array})(\begin{array}{ll}m -l i\end{array})c_{m-i+2}$$+ \frac{4(m-3)}{3!}\sum_{i=0}^{3}i!(\begin{array}{l}3i\end{array})(\begin{array}{l}mi\end{array})c_{m-i+2} (l=4)$

and$a_{l,m,2}=(-1)^{l+1} \frac{m-l}{2}\sum_{i=0}^{\iota}i!(\begin{array}{l}li\end{array})(l+ m_{i} -3)c_{l+m-i-1}$$(l=2,3)$ ,and for $2\leq r=l-k\leq l$$b_{l,m,r}= \sum_{j=1}^{r-1}(-1)^{l+j+1}(\begin{array}{l}lj\end{array})(\begin{array}{l}mr-j\end{array})(\begin{array}{l}rj\end{array})\sum_{i=0}^{l-j}i!(\begin{array}{ll}l- ji \end{array}) (l+ m-ri -2)c_{l+m-i-r}$

except$b_{l,m,0}=(-1)^{l} \sum_{i=0}^{l}i!(\begin{array}{l}li\end{array})(l+ m_{i} -2)c_{l+m-i}=-a_{l,m,1}.$

With this proposition we now obtain our main result.

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Theorem 5. Let$l,$$m,$ $n$

be integers with $m\geq l\geq 1$ and $n\geq 0$ . Then$(c_{l}+c_{m})^{n}$$=- \frac{l!m!}{(l+m+1)!}((n+l+m+1)c_{n+l+m}+(n+l+m+1)(n+l+m)c_{n+l+m-1})$$+ \sum_{k=0}^{m-l-1}\sum_{i=0}^{n}\frac{(-1)^{i-1}n!}{(n-i)!}$$\cross((l+ k +i-2i-1)a_{l,m,m-k}-(l+ k +i-1i)b_{l,m,m-k})c_{n+m-k-i}$$+ \sum_{k=0}^{l-2}\sum_{i=0}^{n}\frac{(-1)^{i-1}n!}{(n-i)!}$$\cross((\begin{array}{ll}m+k +i-2i-1 \end{array})a_{l,m,l-k}-(m+k +i-1i)b_{l,m,l-k})c_{n+l-k-i}$$+ \sum_{i=0}^{n}(-1)^{i}(\begin{array}{lll}l+ m +i-2 i\end{array}) \frac{n!}{(n-i)!}(n-i-1)a_{l,m,1}c_{n-i}.$where$a_{l,m,r}$and$b_{l,m,r}$

are defined in Proposition 5.

Similarly, for general integers

Proposition 6. Let

$l,$

$m$

$l$

and $m$ , we obtain the following result.

be fixed nonnegative integers with $m\geq l\geq 1$ . Then$\hat{c}^{(l)}(x)\hat{c}^{(m)}(x)=-\frac{l!m!}{(l+m+1)!}x\hat{c}^{\langle l+m+1)}(x)-\frac{l!m!}{(l+m)!}\hat{c}^{\langle\iota+m)}(x)$$+ \sum_{k=0}^{m-l-1}\frac{a_{l,m.m-k}x+b_{l,m.m-k}}{(1+x)^{l+k+1}}\hat{c}^{\langle m-k)}(x)+\sum_{k=0}^{\iota}\frac{a_{l,m,l-k}x+b_{l,m,l-k}}{(1+x)^{m+k+1}}\hat{c}^{(l-k)}(x)$where for $l+1\leq r=m-k\leq m$$a_{l,m,r}=(-1)^{\iota+1} (\begin{array}{l}mk\end{array})\sum_{i=0}^{l}i!(\begin{array}{l}li\end{array})(l+ ki -1) \frac{c_{l+k-i+1}}{l+k-i+1},$$b_{l,m,r}=(-1)^{l}l (\begin{array}{l}mk\end{array})\sum_{i=0}^{l-1}i!(\begin{array}{l}l-1i\end{array})(l+ ki -1) \frac{c_{l+k-i}}{l+k-i},$,

191
and for $1\leq r=l-k\leq l$$a_{l,m,r}= \sum_{j=0}^{r-2}\frac{(-1)^{l+j+1}}{r}(\begin{array}{l}lj\end{array})(\begin{array}{ll}m r-j -1\end{array}) (\begin{array}{ll}r -1 j\end{array})$$\cross\sum_{i=0}^{l-j}i](\begin{array}{ll}l- ji \end{array}) (l+ m-ri -1)c_{l+m-i-r+1}$$+ \frac{(-1)^{l+r}}{r}(\begin{array}{ll} lr -1\end{array}) \sum_{i=0}^{l-r+1}i!(l- r_{i} +1) (l+ m_{i}-r -1)c_{l+m-i-r+1}$

with $a_{l,m,0}=0$ and for $0\leq r=l-k\leq l$$b_{l,m,r}= \sum_{j=1}^{r}(-1)^{l+j+1}(\begin{array}{l}lj\end{array})(\begin{array}{l}mr-j\end{array})(\begin{array}{l}rj\end{array})\sum_{i=0}^{l-j}i!(\begin{array}{ll}l- ji \end{array}) (l+ m-ri -1)c_{l+m-i-r}$$+(-1)^{l+r} (\begin{array}{l}lr\end{array})\sum_{i=0}^{l-r}i!(\begin{array}{ll}l- ri \end{array}) (l+ m_{i}-r)\hat{c_{l+m-i-r}}.$By using Proposition 6, we obtain explicit expressions for

Theorem 6. Let$l,$$m,$ $n$$(\hat{c_{l}}+\hat{c}_{m})^{n}.$be integers with $m\geq l\geq 1$ and $n\geq 0$ . Then$(\hat{c_{l}}+\hat{c}_{m})^{n}$$=- \frac{l!m!}{(l+m+1)!}(n+l+m+1)\hat{c}_{n+l+m}$$+ \sum_{k=0}^{m-l-1}\sum_{i=0}^{n}\frac{(-1)^{i-1}n!}{(n-i)!}$$\cross((l+ k +i-1i-1)a_{l,m,m-k}-(l+ ki +i)b_{l,m,m-k})\hat{c}_{n+m-k-i}$$\iota$$n$$+ \sum_{k=0}\sum_{i=0}\frac{(-1)^{i-1}n!}{(n-i)!}$$\cross((\begin{array}{ll}m+k +i-1i-1 \end{array})a_{l,m,l-k}-(\begin{array}{ll}m+k +ii \end{array})b_{l,m,l-k})\hat{c}_{n+l-k-i}.$where$a_{l,m,r}$and$b_{l,m,r}$

are defined in Proposition 6, and$(\begin{array}{l}n-1\end{array})=0(n\geq 0)$

.

192

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$F$

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Graduate School of Science and Technology
Hirosaki University
Hirosaki 036-8561

JAPAN-mail address: komatsu@cc.hirosaki-u.ac.jp$E$

弘前大学大学院理工学研究科 小松尚夫