1 Introduction

The Daehee polynomials are defined by the generating function to be

$$ \frac{\log (1+t)}{t}(1+t)^{x}=\sum _{n=0}^{\infty }D_{n}(x) \frac{t^{n}}{n!} $$
(1.1)

(see [2–7]).

For \(x=0\), \(D_{n}=D_{n}(0)\) are called the Daehee numbers.

In [1], Kim and Kim introduced the degenerate Daehee numbers which are given by the generating function:

$$ \frac{\lambda \log (1+\frac{1}{\lambda }\log (1+\lambda t) )}{ \log (1+\lambda t)}=\sum_{n=0}^{\infty }D_{n,\lambda } \frac{t^{n}}{n!}. $$
(1.2)

For \(x=0\), \(D_{n,\lambda }=D_{n,\lambda }(0)\) are called the degenerate Daehee numbers.

We observe here that \(D_{n,\lambda } \longrightarrow D_{n}\) as \(\lambda \longrightarrow 0\).

The Stirling numbers of the first kind are given by

$$ (x)_{n}= x(x-1)\cdots (x-n+1)=\sum ^{n} _{l=0}S_{1} (n,l)x^{l} \quad (x \geq 0), $$
(1.3)

and the Stirling numbers of the first kind are defined by the generating function to be

$$ \bigl( \log (x+1) \bigr) ^{n}=n!\sum _{l=n} ^{\infty }S_{1}(l,n)\frac{x ^{l}}{l!}\quad(n\geq 0) $$

(see [1, 3]).

Recently, many researchers have studied nonlinear differential equations arising from the generating functions of various special polynomials (see [1–6, 8–20]). They also investigated some identities and explicit expression of these polynomials from the solution of nonlinear differential equations. In [1], Kim and Kim have studied some results of degenerate Daehee numbers which are derived from nonlinear differential equations. From the viewpoint of the inversion formula, we study the degenerate Daehee number arising from a nonlinear differential equation. In this paper, by using the generating function and nonlinear differential equations, we deduce the explicit expression of degenerate Daehee numbers as the inversion formula of [1].

2 Some identities of degenerate Daehee numbers arising from nonlinear differential equations

Let

$$ \begin{aligned} F=F(t)=\log \biggl(1+\frac{1}{\lambda } \log (1+\lambda t) \biggr). \end{aligned} $$
(2.1)

Then, by taking the derivative with respect to t of (2.1), we get

$$\begin{aligned} F^{(1)}&=\frac{d}{dt}F(t)= \biggl(1+\frac{1}{\lambda }\log (1+ \lambda t) \biggr)^{-1}\frac{1}{1+\lambda t} \\ &=\frac{1}{1+\lambda t} e^{-\log (1+\frac{1}{\lambda }\log (1+ \lambda t) )} \\ &=\frac{1}{1+\lambda t}e^{-F}. \end{aligned}$$
(2.2)

From (2.2), we get

$$ \begin{aligned} e^{-F}=(1+\lambda t)F^{(1)}. \end{aligned} $$
(2.3)

From (2.3), we note that

$$ \begin{aligned} \bigl(-F^{(1)} \bigr)e^{-F}=\lambda F^{(1)}+(1+\lambda t)F^{(2)}. \end{aligned} $$
(2.4)

Thus, by multiplying \((1+\lambda t)\) on both sides of (2.4), we get

$$ \begin{aligned} (1+\lambda t)F^{(1)}e^{-F}=- \lambda (1+\lambda t)F^{(1)}-(1+\lambda t)^{2}F ^{(2)}. \end{aligned} $$
(2.5)

From (2.3) and (2.5), we get

$$ \begin{aligned} e^{-2F}=-\lambda (1+\lambda t)F^{(1)}-(1+\lambda t)^{2}F^{(2)}. \end{aligned} $$
(2.6)

From (2.6), we have

$$ \begin{aligned} -2F^{(1)}e^{-2F}=- \lambda^{2}F^{(1)}-3\lambda (1+\lambda t)F^{(2)}-(1+ \lambda t)^{2}F^{(3)}. \end{aligned} $$
(2.7)

Multiplying \((1+\lambda t)\) on both sides of (2.7), we get

$$\begin{aligned} 2!(1+\lambda t)F^{(1)}e^{-2F} &=(-1)^{2} \lambda^{2}(1+\lambda t)F^{(1)}+(-1)^{2} 3\lambda (1+\lambda t)^{2}F^{(2)} \\ &\quad {}+(-1)^{2}(1+\lambda t)^{3}F^{(3)}. \end{aligned}$$
(2.8)

From (2.3) and (2.8), we get

$$\begin{aligned} 2!e^{-3F} &=(-1)^{2}\lambda^{2}(1+\lambda t)F^{(1)}+ (-1)^{2} 3\lambda (1+\lambda t)^{2}F^{(2)} +(-1)^{2}(1+\lambda t)^{3}F^{(3)}. \end{aligned}$$
(2.9)

From (2.9), we have

$$\begin{aligned} 2!\bigl(-3F^{(1)}\bigr)e^{-3F} &=(-1)^{2} \lambda^{3}F^{(1)}+ (-1)^{2} 7\lambda ^{2}(1+\lambda t)F^{(2)} \\ &\quad {}+(-1)^{2} 6\lambda (1+\lambda t)^{2}F^{(3)}+(-1)^{2}(1+ \lambda t)^{3}F ^{(4)}. \end{aligned}$$
(2.10)

Multiplying \((1+\lambda t)\) on both sides of (2.10), we get

$$\begin{aligned} 3!(1+\lambda t)F^{(1)}e^{-3F} &=(-1)^{3} \lambda^{3}(1+\lambda t)F^{(1)}+(-1)^{3} 7 \lambda^{2}(1+\lambda t)^{2}F^{(2)} \\ &\quad {}+(-1)^{3} 6\lambda (1+\lambda t)^{3}F^{(3)}+(-1)^{3}(1+ \lambda t)^{4}F ^{(4)}. \end{aligned}$$
(2.11)

From (2.3) and (2.11), we get

$$\begin{aligned} 3!e^{-4F} &=(-1)^{3}\lambda^{3}(1+\lambda t)F^{(1)}+ (-1)^{3} 7\lambda ^{2}(1+\lambda t)^{2}F^{(2)} \\ &\quad {}+(-1)^{3} 6\lambda (1+\lambda t)^{3}F^{(3)}+(-1)^{3}(1+ \lambda t)^{4}F ^{(4)}. \end{aligned}$$
(2.12)

Continuing this process, we get

$$\begin{aligned} (N-1)!e^{-NF}=(-1)^{N-1}\sum_{k=1}^{N} \lambda^{N-k}(1+\lambda t)^{k} a _{k}(N)F^{(k)}. \end{aligned}$$
(2.13)

Let us take the derivative on both sides of (2.13) with respect to t. Then we have

$$\begin{aligned} (N-1)!\bigl(-N F^{(1)}\bigr)e^{-NF} &=(-1)^{N-1}\sum _{k=1}^{N}\lambda^{N-k}a_{k}(N) \bigl\{ k\lambda (1+\lambda t)^{k-1} \\ &\quad {}\times F^{(k)} +(1+\lambda t)^{k}F^{(k+1)} \bigr\} . \end{aligned}$$
(2.14)

Multiplying \((1+\lambda t)\) on both sides of (2.14), we get

$$\begin{aligned} N!(1+\lambda t)F^{(1)}e^{-NF} &=(-1)^{N}\sum _{k=1}^{N}\lambda^{N-k}a _{k}(N) \bigl\{ k\lambda (1+\lambda t)^{k} \\ &\quad {}\times F^{(k)} +(1+\lambda t)^{k+1}F^{(k+1)} \bigr\} . \end{aligned}$$
(2.15)

Then, by (2.3) and (2.15), we get

$$\begin{aligned} N!e^{-(N+1)F}&=(-1)^{N}\sum_{k=1}^{N} \lambda^{N-K}a_{k}(N) \bigl\{ k \lambda (1+\lambda t)^{k} \\ &\quad {}\times F^{(k)} +(1+\lambda t)^{k+1}F^{(k+1)} \bigr\} \\ &=(-1)^{N}\sum_{k=1}^{N} \lambda^{N-k+1}(1+\lambda t)^{k}ka_{k}(N)F^{(k)} \\ &\quad {}+(-1)^{N}\sum_{k=1}^{N} \lambda^{N-k}(1+\lambda t)^{k+1}a_{k}(N)F ^{(k+1)} \\ &=(-1)^{N}\sum_{k=1}^{N} \lambda^{N-k+1}(1+\lambda t)^{k}ka_{k}(N)F ^{(k)} \\ &\quad {}+(-1)^{N}\sum_{k=2}^{N+1} \lambda^{N-k+1}(1+\lambda t)^{k}a_{k-1}(N)F ^{(k)} \\ &=(-1)^{N}\lambda^{N}(1+\lambda t)a_{1}{(N)}F^{(1)}+(-1)^{N}(1+1 \lambda t)^{N+1} \\ &\quad {}\times a_{N}{(N)} F^{(N+1)}+(-1)^{N}\sum _{k=2}^{N}\lambda^{N-k+1}(1+ \lambda t)^{k} \\ &\quad {}\times \bigl(ka_{k}(N)+a_{k-1}(N) \bigr)F^{(k)}. \end{aligned}$$
(2.16)

By replacing N by \(N+1\) in (2.13), we get

$$\begin{aligned} N!e^{-(N+1)F} &=(-1)^{N}\sum_{k=1}^{N+1} \lambda^{N-k+1}(1+\lambda t)^{k} a_{k}(N+1)F^{(k)} \\ &=(-1)^{N}\lambda^{N}(1+\lambda t)a_{1}{(N+1)}F^{(1)}+(-1)^{N}(1+ \lambda t)^{N+1} \\ &\quad {}\times a_{N+1}{(N+1)}F^{(N+1)}+(-1)^{N}\sum _{k=2}^{N}\lambda^{N-k+1}(1+ \lambda t)^{k} \\ &\quad {}\times a_{k}{(N+1)}F^{k}. \end{aligned}$$
(2.17)

Comparing the coefficients on both sides of (2.16) and (2.17), we have

$$\begin{aligned} a_{1}(N+1)=a_{1}(N),\qquad a_{N+1}(N+1)=a_{N}(N), \end{aligned}$$
(2.18)

and

$$ a_{k}(N+1)=ka_{k}(N)+a_{k-1}(N), \quad \mbox{for } 2\leq k\leq N. $$
(2.19)

From (2.3) and (2.13), we have

$$ e^{-F}=(1+\lambda t)F^{(1)}=(1+\lambda t)a_{1}(1)F^{(1)}. $$
(2.20)

By (2.20), we get

$$ a_{1}(1)=1. $$
(2.21)

Thus, by (2.18) and (2.21), we have

$$ a_{1}(N+1)=a_{1}(N)=a_{1}(N-1)= \cdots =a_{1}(1)=1 $$
(2.22)

and

$$ a_{N+1}(N+1)=a_{N}(N)=a_{N-1}(N-1)= \cdots =a_{1}(1)=1. $$
(2.23)

For \(k=2\) in (2.19), we have

$$\begin{aligned} a_{2}(N+1) &=2a_{2}(N)+a_{1}(N) \\ &=2 \bigl(2a_{2}(N-1)+a_{1}(N-1) \bigr)+a_{1}(N) \\ &=2^{2}a_{2}(N-1)+2a_{1}(N-1)+a_{1}(N) \\ &=\cdots \\ &=2^{N-1}a_{2}(2)+2^{N-2}a_{1}(2)+\cdots +a_{1}(N). \end{aligned}$$
(2.24)

Then by (2.22), (2.23) and (2.24), we get

$$\begin{aligned} a_{2}(N+1) &=2^{N-1}a_{2}(2)+2^{N-2}a_{1}(2)+ \cdots +a_{1}(N) \\ &=2^{N-1}a_{1}(1)+2^{N-2}a_{1}(2)+\cdots +a_{1}(N) \\ &=2^{N-1}+2^{N-2}+\cdots +1 \\ &=\sum_{i_{1}=0}^{N-1}2^{i_{1}}. \end{aligned}$$
(2.25)

For \(k=3\) in (2.19), we have

$$\begin{aligned} a_{3}(N+1) &=3a_{3}(N)+a_{2}(N) \\ &=3 \bigl(3a_{3}(N-1)+a_{2}(N-1) \bigr)+a_{2}(N) \\ &=3^{2}a_{3}(N-1)+3a_{2}(N-1)+a_{2}(N) \\ &=\cdots \\ &=3^{N-2}a_{3}(3)+3^{N-3}a_{2}(3)+\cdots +a_{2}(N) \\ &=3^{N-2}a_{2}(2)+3^{N-3}a_{2}(3)+\cdots +a_{2}(N) \\ &=\sum_{i_{2}=0}^{N-2}3^{i_{2}}a_{2}(N-i_{2}). \end{aligned}$$
(2.26)

Then by (2.25) and (2.26), we get

$$\begin{aligned} a_{3}(N+1) &=\sum_{i_{2}=0}^{N-2}3^{i_{2}}a_{2}(N-i_{2}) \\ &=\sum_{i_{2}=0}^{N-2}3^{i_{2}}\sum _{i_{1}=0}^{N-2-i_{2}}2^{i_{1}} \\ &=\sum_{i_{2}=0}^{N-2}\sum _{i_{1}=0}^{N-2-i_{2}}3^{i_{2}}2^{i_{1}}. \end{aligned}$$
(2.27)

For \(k=4\) in (2.19), we have

$$\begin{aligned} a_{4}(N+1)&=4a_{4}(N)+a_{3}(N) \\ &=4 \bigl(4a_{4}(N-1)+a_{3}(N-1) \bigr)+a_{3}(N) \\ &=4^{2}a_{4}(N-1)+4a_{3}(N-1)+a_{3}(N) \\ &=\cdots \\ &=4^{N-3}a_{4}(4)+4^{N-4}a_{3}(4)+\cdots +a_{3}(N) \\ &=4^{N-3}a_{3}(3)+4^{N-4}a_{3}(4)+\cdots +a_{3}(N) \\ &=\sum_{i_{3}=0}^{N-3}4^{i_{3}}a_{3}(N-i_{3}). \end{aligned}$$
(2.28)

By (2.27) and (2.28), we have

$$\begin{aligned} a_{4}(N+1) &=\sum_{i_{3}=0}^{N-3}4^{i_{3}}a_{3}(N-i_{3}) \\ &=\sum_{i_{3}=0}^{N-3}4^{i_{3}}\sum _{i_{2}=0}^{N-3-i_{3}}\sum_{i_{1}=0} ^{N-3-i_{3}-i_{2}}3^{i_{2}}2^{i_{1}} \\ &=\sum_{i_{3}=0}^{N-3}\sum _{i_{2}=0}^{N-3-i_{3}}\sum_{i_{1}=0}^{N-3-i _{3}-i_{2}}4^{i_{3}}3^{i_{2}}2^{i_{1}}. \end{aligned}$$
(2.29)

Continuing this process, for \(2\leq k \leq N\), we have

$$ a_{k}(N+1)=\sum_{i_{k-1}=0}^{N-k+1}\sum _{i_{k-2}=0}^{N-k+1-i_{k-1}} \cdots \sum _{i_{1}=0}^{N-k+1-i_{k-1}-\cdots -i_{2}}k^{i_{k-1}}\cdots 2^{i_{1}}. $$
(2.30)

Therefore, we obtain the following differential equations.

Theorem 2.1

Let \(N \in \mathbb{N}\). Then the differential equations

$$ (N-1)!e^{-NF}=(-1)^{N-1}\sum_{k=1}^{N} \lambda^{N-k}(1+\lambda t)^{k} a _{k}(N)F^{(k)} $$

have a solution \(F=F(t)=\log (1+\frac{1}{\lambda }\log (1+\lambda t) )\), where

$$ a_{N}(N)=1,\qquad a_{1}(N)=1 $$

and

$$ a_{k}(N)=\sum_{i_{k-1}=0}^{N-k}\sum _{i_{k-2}=0}^{N-k-i_{k-1}}\cdots \sum _{i_{1}=0}^{N-k-i_{k-1}-\cdots -i_{2}}k^{i_{k-1}}\cdots 2^{i_{1}}. $$

From (2.1), we easily get

$$\begin{aligned} F &=\log \biggl(1+\frac{1}{\lambda }\log (1+\lambda t) \biggr) \\ &=\frac{\lambda \log (1+\frac{1}{\lambda }\log (1+\lambda t) )}{\log (1+\lambda t)}\cdot \frac{\log (1+\lambda t)}{\lambda } \\ &= \Biggl(\sum_{l_{1}=0}^{\infty }D_{l_{1},\lambda } \frac{t^{l_{1}}}{l _{1}!} \Biggr) \Biggl(\frac{1}{\lambda }\sum _{l_{2}=1}^{\infty }\frac{(-1)^{l _{2}-1}\lambda^{l_{2}}}{l_{2}}t^{l_{2}} \Biggr) \\ &= \Biggl(\sum_{l_{1}=0}^{\infty }D_{l_{1},\lambda } \frac{t^{l_{1}}}{l _{1}!} \Biggr) \Biggl(\sum_{l_{2}=1}^{\infty } \frac{(-\lambda )^{l_{2}-1}}{l _{2}}t^{l_{2}} \Biggr) \\ &=\sum_{l_{3}=1}^{\infty } \Biggl(\sum _{l_{1}=0}^{l_{3}-1}\frac{D_{l _{1},\lambda }}{l_{1}!}\cdot \frac{(-\lambda )^{l_{3}-l_{1}-1}}{(l _{3}-l_{1})} \Biggr)t^{l_{3}}. \end{aligned}$$
(2.31)

From (2.31), we get

$$\begin{aligned} F^{(k)} &= \biggl(\frac{d}{dt} \biggr)^{k} \Biggl\{ \sum _{l_{3}=1}^{\infty } \Biggl(\sum _{l_{1}=0}^{l_{3}-1}\frac{D_{l_{1},\lambda }}{l_{1}!}\cdot \frac{(-\lambda )^{l_{3}-l_{1}-1}}{(l_{3}-l_{1})} \Biggr)t^{l_{3}} \Biggr\} \\ &=\sum_{l_{3}=k}^{\infty } \Biggl(\sum _{l_{1}=0}^{l_{3}-1}\frac{D_{l _{1},\lambda }}{l_{1}!}\cdot \frac{(-\lambda )^{l_{3}-l_{1}-1}}{(l _{3}-l_{1})} \Biggr) (l_{3})_{k}t^{l_{3}-k} \\ &=\sum_{l_{3}=0}^{\infty } \Biggl(\sum _{l_{1}=0}^{l_{3}+k-1}\frac{D _{l_{1},\lambda }}{l_{1}!}\cdot \frac{(-\lambda )^{l_{3}+k-l_{1}-1}}{(l _{3}+k-l_{1})} \Biggr) (l_{3}+k)_{k}t^{l_{3}}. \end{aligned}$$
(2.32)

From (2.32), we get

$$\begin{aligned} (1+\lambda t)^{k} F^{(k)} &= \Biggl(\sum _{l=0}^{\infty }{k \choose l} \lambda^{l}t^{l} \Biggr) \Biggl\{ \sum _{l_{3}=0}^{\infty } \Biggl(\sum_{l _{1}=0}^{l_{3}+k-1} \frac{D_{l_{1},\lambda }}{l_{1}!}\cdot \frac{(- \lambda )^{l_{3}+k-l_{1}-1}}{(l_{3}+k-l_{1})} \Biggr) (l_{3}+k)_{k}t^{l_{3}} \Biggr\} \\ &= \Biggl(\sum_{l=0}^{\infty }(k)_{l} \lambda^{l}\frac{t^{l}}{l!} \Biggr) \Biggl\{ \sum _{l_{3}=0}^{\infty } \Biggl(\sum_{l_{1}=0}^{l_{3}+k-1} \frac{D _{l_{1},\lambda }}{l_{1}!}\cdot \frac{(-\lambda )^{l_{3}+k-l_{1}-1}}{(l _{3}+k-l_{1})} \Biggr) (l_{3}+k)!\frac{t^{l_{3}}}{l_{3}!} \Biggr\} \\ &=\sum_{n=0}^{\infty } \Biggl(\sum _{l_{3}=0}^{n}\sum_{l_{1}=0}^{l_{3}+k-1} \frac{(l _{3}+k)!{n \choose l_{3}}(k)_{n-l_{3}}}{l_{1}!(l_{3}+k-l_{1})}D_{l _{1},\lambda } \\ &\quad {}\times (-1)^{l_{3}+k-l_{1}-1}\lambda^{n+k-l_{1}-1} \Biggr) \frac{t^{n}}{n!} \end{aligned}$$
(2.33)

and also

$$\begin{aligned} e^{-NF} &=\sum_{m_{1}=0}^{\infty } \frac{(-N)^{m_{1}}}{m_{1}!}F^{m_{1}} \\ &=\sum_{m_{1}=0}^{\infty }\frac{(-N)^{m_{1}}}{m_{1}!} \biggl( \log \biggl(1+\frac{1}{\lambda }\log (1+\lambda t) \biggr) \biggr)^{m_{1}} \\ &=\sum_{m_{1}=0}^{\infty }(-N)^{m_{1}}\sum _{m_{2}=m_{1}}^{\infty }S _{1}(m_{2},m_{1}) \frac{1}{m_{2}!} \biggl(\frac{1}{\lambda }\log (1+ \lambda t) \biggr)^{m_{2}} \\ &=\sum_{m_{2}=0}^{\infty } \Biggl(\sum _{m_{1}=0}^{m_{2}}(-N)^{m_{1}}S _{1}(m_{2},m_{1}) \lambda^{-m_{2}} \Biggr)\frac{1}{m_{2}!} \bigl(\log (1+ \lambda t) \bigr)^{m_{2}} \\ &=\sum_{m_{2}=0}^{\infty } \Biggl(\sum _{m_{1}=0}^{m_{2}}(-N)^{m_{1}}S _{1}(m_{2},m_{1}) \lambda^{-m_{2}} \Biggr) \Biggl(\sum_{n=m_{2}}^{\infty }S _{1}(n,m_{2})\frac{\lambda^{n}t^{n}}{n!} \Biggr) \\ &=\sum_{n=0}^{\infty } \Biggl(\sum _{m_{2}=0}^{n}\sum_{m_{1}=0}^{m_{2}}(-N)^{m _{1}}S_{1}(m_{2},m_{1})S_{1}(n,m_{2}) \lambda^{n-m_{2}} \Biggr)\frac{t ^{n}}{n!}. \end{aligned}$$
(2.34)

Here \(S_{1}(n,k)\) is the Stirling number of the first kind.

Thus, by (2.13) and (2.33), we get

$$\begin{aligned} &(N-1)!e^{-NF} \\ &\quad =(-1)^{N-1}\sum_{k=1}^{N}\lambda^{N-k}a_{k}(N) \\ &\quad \quad {}\times\sum_{n=0} ^{\infty } \Biggl(\sum_{l_{3}=0}^{n}\sum _{l_{1}=0}^{l_{3}+k-1} \frac{(l_{3}+k)!{n \choose l_{3}}(k)_{n-l_{3}}}{l_{1}!(l_{3}+k-l _{1})}D_{l_{1},\lambda }(-1)^{l_{3}+k-l_{1}-1} \lambda^{n+k-l_{1}-1} \Biggr)\frac{t^{n}}{n!} \\ &\quad =\sum_{n=0}^{\infty } \Biggl((-1)^{N-1} \sum_{k=1}^{N}\lambda^{N-k}a _{k}(N)\sum_{l_{3}=0}^{n}\sum_{l_{1}=0}^{l_{3}+k-1} \frac{(l_{3}+k)!{n \choose l_{3}}(k)_{n-l_{3}}}{l_{1}!(l_{3}+k-l_{1})} \end{aligned}$$
(2.35)
$$\begin{aligned} &\quad \quad {}\times D_{l_{1},\lambda }(-1)^{l_{3}+k-l_{1}-1}\lambda^{n+k-l_{1}-1} \Biggr)\frac{t^{n}}{n!} \\ &\quad =\sum_{n=0}^{\infty } \Biggl(\sum _{k=1}^{N}\sum_{l_{3}=0}^{n} \sum_{l_{1}=0}^{l_{3}+k-1}(-1)^{N+l_{3}+k-l_{1}} \lambda^{N+n-l_{1}-1}a _{k}(N) \\ &\quad \quad {}\times \frac{(l_{3}+k)!{n \choose l_{3}}(k)_{n-l_{3}}}{l_{1}!(l_{3}+k-l _{1})}D_{l_{1},\lambda } \Biggr)\frac{t^{n}}{n!}. \end{aligned}$$
(2.36)

By (2.34) and (2.36), we get

$$\begin{aligned} &(N-1)! \sum _{n=0}^{\infty } \Biggl(\sum_{m_{2}=0}^{n} \sum_{m_{1}=0}^{m _{2}}(-N)^{m_{1}}S_{1}(m_{2},m_{1})S_{1}(n,m_{2}) \lambda^{n-m_{2}} \Biggr)\frac{t^{n}}{n!} \\ &\quad =\sum_{n=0}^{\infty } \Biggl(\sum _{k=1}^{N}\sum_{l_{3}=0}^{n} \sum_{l_{1}=0}^{l_{3}+k-1}(-1)^{N+l_{3}+k-l_{1}} \lambda^{N+n-l_{1}-1}a _{k}(N) \frac{(l_{3}+k)!{n \choose l_{3}}(k)_{n-l_{3}}}{l_{1}!(l_{3}+k-l _{1})}D_{l_{1},\lambda } \Biggr)\frac{t^{n}}{n!}. \end{aligned}$$
(2.37)

By equation (2.37), we finally get the following theorem.

Theorem 2.2

For \(N=1,2,3,\ldots \) , and \(n=0,1,2,\ldots \) , we have

$$\begin{aligned} &(N-1)! \sum_{m_{2}=0}^{n}\sum _{m_{1}=0}^{m_{2}}(-N)^{m_{1}}S_{1}(m _{2},m_{1})S_{1}(n,m_{2}) \lambda^{n-m_{2}} \\ &\quad =\sum_{k=1}^{N}\sum _{l_{3}=0}^{n}\sum_{l_{1}=0}^{l_{3}+k-1}(-1)^{N+l _{3}+k-l_{1}} \lambda^{N+n-l_{1}-1}a_{k}(N) \\ &\quad \quad {}\times \frac{(l_{3}+k)!{n \choose l_{3}}(k)_{n-l_{3}}}{l_{1}!(l_{3}+k-l _{1})}D_{l_{1},\lambda }. \end{aligned}$$

3 Conclusion

Kim and Kim have studied some identities of degenerate Daehee numbers which are derived from the generating function using nonlinear differential equation (see [1]). In this paper, from the viewpoint of the inversion formula to [1], we study the degenerate Daehee number arising from nonlinear differential equation. Therefore we obtain the inversion formula of degenerate Daehee numbers which are related to the some identities of those numbers. In Theorem 2.1, we get the solution of nonlinear differential equation arising from the generating function of the degenerate Daehee number. In Theorem 2.2, we have an explicit expression of the degenerate Daehee number from the result of Theorem 2.1 using the generating function and nonlinear differential equations.