Symmetric Identities on Modified Degenerate Bernoulli Polynomials

Qi-Peng Su and Hao Pan School of Applied Mathematics Nanjing University of Finance and Economics Nanjing, Jiangsu 210023 China

                   3093420060@qq.com
                   haopan79@zoho.com

Abstract

The modified degenerate Bernoulli polynomial 𝔅_{n,λ}(x) introduced by Dolgy et al. is given by

∑_{n=0}^{∞} \frac{𝔅_{n,λ}(x)}{n!}t^{n} = \frac{t(1 + λ)^{\frac{xt}{λ}}}{(1 + λ)^{\frac{t}{λ}} − 1}.

In this paper, we prove some symmetric identities on modified degenerate Bernoulli polynomials, which generalize the result of Fu, Pan, and Zhang.

1 Introduction

The Bernoulli polynomial B_{n}(x) (n ∈ ℕ) is given by

∑_{n=0}^{∞} \frac{B_{n}(x)}{n!}t^{n} = \frac{te^{xt}}{e^{t} − 1}.

In particular, B_{n} := B_{n}(0) is called the n-th Bernoulli number. A curious identity of Miki [9] says that

∑_{k=2}^{n−2} \frac{B_{k}B_{n−k}}{k(n − k)} = ∑_{k=2}^{n−2} \binom{n}{k}\frac{B_{k}B_{n−k}}{k(n − k)} + 2H_{n} · \frac{B_{n}}{n}  (1)

for n ⩾ 4, where the n-th harmonic number H_{n} := ∑_{i=1}^{n} 1/i. Different proofs of (1) were given by Shirantani and Yokoyama [12], Gessel [6], Dunne and Schubert [4]. Furthermore, for n ⩾ 4, Dunne and Schubert also proved a similar identity

(n + 2) ∑_{k=2}^{n−2} B_{k}B_{n−k} = 2 ∑_{k=2}^{n−2} \binom{n + 2}{k}B_{k}B_{n−k} + n(n + 1)B_{n},  (2)

which was conjectured by Matiyasevich [8].

In [10], Pan and Sun used the difference-differential method to establish the polynomial extensions of the Miki identity and the Matiyasevich-Dunne-Schubert identity. Moreover, Sun and Pan [14] proved the following general symmetric identity involving Bernoulli polynomials:

r [\begin{matrix}s & t \\ x & y \end{matrix}]_{n} + s [\begin{matrix}t & r \\ y & z \end{matrix}]_{n} + t [\begin{matrix}r & s \\ z & x \end{matrix}]_{n} = 0,  (3)

where r + s + t = n, x + y + z = 1 and

[\begin{matrix}s & t \\ x & y \end{matrix}]_{n} = ∑_{k=0}^{n}(−1)^{k}\binom{s}{k}\binom{t}{n − k}B_{n−k}(x)B_{k}(y).

Subsequently, Fu, Pan, and Zhang [5] gave a generalization of (3) involving sums of products of more Bernoulli polynomials as follows:

r_{m+1} ∑_{\begin{matrix}k_{1},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n \end{matrix}} ∏_{j=1}^{m} \binom{r_{j}}{k_{j}}B_{k_{j}}(x_{j})

= − ∑_{i=1}^{m} r_{i} ∑_{\begin{matrix}k_{1},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n \end{matrix}} \binom{r_{m+1}}{k_{i}}B_{k_{i}}(1 − x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{r_{j}}{k_{j}}B_{k_{j}}(x_{j} − x_{i} + 1_{j>i}),  (4)

where r_{m+1} = n − r_{1} − ⋯ − r_{m} and 1_{j>i} = 1 or 0 according to whether j > i or not.

On the other hand, in [2], Dolgy, Kim, Kwon, and Seo introduced the modified degenerate Bernoulli polynomial, which is different from Carlitz’s degenerate Bernoulli polynomial [1]. Define the modified degenerate Bernoulli polynomial 𝔅_{n,λ}(x) by

∑_{n=0}^{∞} \frac{𝔅_{n,λ}(x)}{n!}t^{n} = \frac{t(1 + λ)^{\frac{xt}{λ}}}{(1 + λ)^{\frac{t}{λ}} − 1},

where λ ∈ ℂ. Moreover, set 𝔅_{n,λ}(x) = 0 if n < 0. Since (1+λ)^{\frac{1}{λ}} tends to e as λ → 0, clearly

lim_{λ→0} 𝔅_{n,λ}(x) = B_{n}(x).

Various results related to the modified degenerate Bernoulli polynomials have been discussed in [3, 7, 11, 13].

In this short note, we extend (4) to the modified degenerate Bernoulli polynomials.

Theorem 1. Suppose that m ≥ 2 and n ≥ 1 are integers. Let r_{1}, …, r_{m}, λ be arbitrary complex numbers and r_{0} = n − r_{1} − ⋯ − r_{m}. We have

r_{0} ∑_{\begin{matrix}k_{1},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n \end{matrix}} ∏_{j=1}^{m} \binom{r_{j}}{k_{j}}𝔅_{k_{j},λ}(x_{j})

= − ∑_{i=1}^{m} r_{i} ∑_{\begin{matrix}k_{1},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n \end{matrix}} \binom{r_{0}}{k_{i}}𝔅_{k_{i},λ}(1 − x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{r_{j}}{k_{j}}𝔅_{k_{j},λ}(x_{j} − x_{i} + 1_{j>i}).

(5)

When m = 2, we have the following extension of (3).

Corollary 2. For s, t, λ ∈ ℂ, let

𝔇_{s,t,λ}(x, y) = ∑_{k=0}^{n}(−1)^{k}\binom{s}{k}\binom{t}{n − k}𝔅_{n−k,λ}(x)𝔅_{k,λ}(y).

Then for each n ≥ 1,

r · 𝔇_{s,t,λ}(x, y) + s · 𝔇_{t,r,λ}(y, z) + t · 𝔇_{r,s,λ}(z, x) = 0,  (6)

where r, s, t, x, y, z ∈ ℂ satisfy that r + s + t = n and x + y + z = 1. In particular, we have the extensions of (1) and (2) as follows:

∑_{k=1}^{n−1} \frac{𝔅_{k,λ}}{k} · 𝔅_{n−k,λ} − ∑_{k=1}^{n−1} \binom{n}{k}\frac{𝔅_{k,λ}}{k} · 𝔅_{n−k,λ} = \frac{λ}{ln(1 + λ)} · H_{n}𝔅_{n,λ} + \frac{n}{2} · 𝔅_{n−1,λ}, (7)

and

(n + 2) ∑_{k=2}^{n−2} 𝔅_{k,λ}𝔅_{n−k,λ} = 2 ∑_{k=2}^{n−2} \binom{n + 2}{k}𝔅_{k,λ}𝔅_{n−k,λ} + \frac{λ}{ln(1 + λ)} · n(n + 1)𝔅_{n,λ}, (8)

where 𝔅_{k,λ} = 𝔅_{k,λ}(0) and n ≥ 4.

The proofs of Theorem 1 and Corollary 2 will be given in the next section.

2 Proofs of Theorem 1 and Corollary 2

First, we need the following lemma.

Lemma 3. Suppose that m ≥ 2 and s_{1}, …, s_{m} be non-negative integers. Then

∑_{i=1}^{m} s_{i}𝔅_{s_{i}−1,λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} 𝔅_{s_{j},λ}(x_{j})

= ∑_{i=1}^{m} s_{i} ∑_{\begin{matrix}k_{1},k_{2},...,k_{m}≥0 \\ k_{1}+···+k_{m}=s_{1}+···+s_{m} \end{matrix}} 𝔅_{k_{i}−1,λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{s_{j}}{k_{j}}𝔅_{k_{j},λ}(x_{j} − x_{i} + 1_{j>i}).  (9)

Proof. Clearly

(t_{1} + ⋯ + t_{m}) ∏_{j=1}^{m} \frac{t_{j}(λ + 1)^{\frac{x_{j} t_{j}}{λ}}}{(λ + 1)^{\frac{t_{j}}{λ}} − 1}

= \frac{(t_{1} + ⋯ + t_{m})}{(λ + 1)^{\frac{t_{1}+···+tm}{λ}} − 1}( ∑_{i=1}^{m}((λ + 1)^{\frac{t_{i}}{λ}} − 1)(λ + 1)^{∑^{i<j≤m} \frac{t_{j}}{λ}} ) ∏_{j=1}^{m} \frac{t_{j}(λ + 1)^{\frac{x_{j} t_{j}}{λ}}}{(λ + 1)^{\frac{t_{j}}{λ}} − 1}

= ∑_{i=1}^{m} \frac{(t_{1} + ⋯ + t_{m})}{(λ + 1)^{\frac{t_{1}+···+tm}{λ}} − 1} · (λ + 1)^{∑^{i<j≤m} \frac{t_{j}}{λ}} · t_{i}(λ + 1)^{\frac{x_{i}t_{i}}{λ}} ∏_{\begin{matrix}j=1 \\ j≠i \end{matrix}}^{m} \frac{t_{j}(λ + 1)^{\frac{x_{j} t_{j}}{λ}}}{(λ + 1)^{\frac{t_{j}}{λ}} − 1}

= ∑_{i=1}^{m} \frac{t_{i}(t_{1} + ⋯ + t_{m})(λ + 1)^{\frac{x_{i}(t_{1}+···+tm)}{λ}}}{(λ + 1)^{\frac{t_{1}+···+tm}{λ}} − 1} ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \frac{t_{j}(λ + 1)^{\frac{(x_{j} −x_{i}+1_{j>i})t_{j}}{λ}}}{(λ + 1)^{\frac{t_{j}}{λ}} − 1} .  (10)

Let [t_{1}^{s_{1}} ⋯ t_{m}^{s_{m}}]f(t_{1}, …, t_{m}) denote the coefficient of t_{1}^{s_{1}} ⋯ t_{m}^{s_{m}} in the formal power series f(t_{1}, …, t_{m}). Then

[t_{1}^{s_{1}} ⋯ t_{m}^{s_{m}}](t_{1} + ⋯ + t_{m}) ∏_{j=1}^{m} \frac{t_{j}(λ + 1)^{\frac{x_{j} t_{j}}{λ}}}{(λ + 1)^{\frac{t_{j}}{λ}} − 1}

= [t_{1}^{s_{1}} ⋯ t_{m}^{s_{m}}](t_{1} + ⋯ + t_{m}) ∑_{k_{j}=0}^{∞} \frac{𝔅_{s_{j},λ}(x)}{s_{j}!}t_{j}^{s_{j}}

= ∑_{i=1}^{m} \frac{𝔅_{s_{i}−1,λ}(x_{i}, λ)}{(s_{i} − 1)!} ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \frac{𝔅_{s_{j},λ}(x_{j})}{s_{j}!}.  (11)

On the other hand, we have

[t_{1}^{s_{1}} ⋯ t_{m}^{s_{m}}] ∑_{i=1}^{m} \frac{t_{i}(t_{1} + ⋯ + t_{m})(λ + 1)^{\frac{x_{i}(t_{1}+···+tm)}{λ}}}{(λ + 1)^{\frac{t_{1}+···+tm}{λ}} − 1} ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \frac{t_{j}(λ + 1)^{\frac{t_{j} (x_{j} −x_{i}+1_{j>i})}{λ}}}{(λ + 1)^{\frac{t_{j}}{λ}} − 1}

(12)

= [t_{1}^{s_{1}} ⋯ t_{m}^{s_{m}}] ∑_{i=1}^{m} t_{i} ∑_{k_{1},...,k_{m}≥0}^{∞} \frac{𝔅_{k_{1}+···+k_{m},λ}(x_{i})}{k_{1}!k_{2}! ⋯ k_{m}!}t_{1}^{k_{1}} ⋯ t_{m}^{k_{m}} ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} ∑_{s_{j}=0}^{∞} \frac{𝔅_{s_{j},λ}(x_{j} − x_{i} + 1_{j>i})}{s_{j}!}t_{j}^{s_{j}}

= [t_{1}^{s_{1}} ⋯ t_{m}^{s_{m}}] ∑_{i=1}^{m} t_{i} ∑_{k_{1},k_{2},...,k_{m}≥0}^{∞} \frac{𝔅_{k_{1}+···+k_{m},λ}(x_{i})}{k_{1}!k_{2}! ⋯ k_{m}!}t_{1}^{k_{1}}t_{2}^{k_{2}} ⋯ t_{m}^{k_{m}}·

· ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} ∑_{s_{j}−k_{j}=0}^{∞} \frac{𝔅_{s_{j}−k_{j},λ}(x_{j} − x_{i} + 1_{j>i})}{(s_{j} − k_{j})!}t_{j}^{s_{j}−k_{j}}

= ∑_{i=1}^{m} ∑_{k_{1},...,k_{i−1},k_{i+1},...,k_{m}≥0} \frac{𝔅_{k_{1}+···+k_{i−1}+s_{i}−1+k_{i+1}+···+k_{m},λ}(x_{i})}{k_{1}! ⋯ k_{i−1}!(s_{i} − 1)!k_{i+1}! ⋯ k_{m}!} ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \frac{𝔅_{s_{j}−k_{j},λ}(x_{j} − x_{i} + 1_{j>i})}{(s_{j} − k_{j})!}. (13)

Combining (10), (11), and (12), we obtain that

∑_{i=1}^{m} s_{i}𝔅_{s_{i}−1,λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} 𝔅_{s_{j},λ}(x_{j}) = s_{1}!s_{2}! ⋯ s_{m}! ∑_{i=1}^{m} \frac{𝔅_{i−1,λ}(x_{i})}{(s_{i} − 1)!} ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \frac{𝔅_{s_{j},λ}(x_{j})}{s_{j}!}

= s_{1}! ⋯ s_{m}! ∑_{i=1}^{m} ∑_{k_{1},...,k_{i−1},k_{i+1},...,k_{m}≥0} \frac{𝔅_{k_{1}+···+k_{i−1}+s_{i}−1+k_{i+1}+···+k_{m}}(x_{i}, λ_{1}λ_{2} ⋯ λ_{m})}{k_{1}!k_{2}! ⋯ k_{i−1}!(s_{i} − 1)!k_{i+1}! ⋯ k_{m}!}·

· ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \frac{𝔅_{s_{j}−k_{j}}(x_{j} − x_{i} + 1_{j>i}, λ_{j})}{(s_{j} − k_{j})!}

= ∑_{i=1}^{m} s_{i} ∑_{\begin{matrix}k_{1},k_{2},...,k_{m}≥0 \\ k_{1}+···+k_{m}=s_{1}+···+s_{m} \end{matrix}} 𝔅_{k_{i}−1,λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{s_{j}}{k_{j}}𝔅_{k_{j},λ}(x_{j} − x_{i} + 1_{j>i}).

Thus we get the desired result.

Now we are ready to complete the proof of Theorem 1. Proof of Theorem 1. Clearly

(n − r_{1} − ⋯ − r_{m}) ∑_{\begin{matrix}k_{1},k_{2},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n \end{matrix}} ∏_{j=1}^{m} \binom{r_{j}}{k_{j}}𝔅_{k_{j},λ}(x_{j})

= ∑_{\begin{matrix}k_{1},k_{2},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n \end{matrix}} ∑_{i=1}^{m}(k_{i} − r_{i})\binom{r_{i}}{k_{i}}𝔅_{k_{i},λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{r_{j}}{k_{j}}𝔅_{k_{j},λ}(x_{j})

= − ∑_{\begin{matrix}k_{1},k_{2},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n \end{matrix}} ∑_{i=1}^{m}(k_{i} + 1)\binom{r_{i}}{k_{i} + 1}𝔅_{k_{i},λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{r_{j}}{k_{j}}𝔅_{k_{j},λ}(x_{j})

= − ∑_{\begin{matrix}k_{1},k_{2},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n+1 \end{matrix}} ∑_{i=1}^{m} k_{i}\binom{r_{i}}{k_{i}}𝔅_{k_{i}−1,λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{r_{j}}{k_{j}}𝔅_{k_{j},λ}(x_{j}).

By Lemma 2.1, we have

∑_{\begin{matrix}k_{1},k_{2},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n+1 \end{matrix}} ∑_{i=1}^{m} k_{i}\binom{r_{i}}{k_{i}}𝔅_{k_{i}−1,λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{r_{j}}{k_{j}}𝔅_{k_{j},λ}(x_{j})

= ∑_{\begin{matrix}k_{1},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n+1 \end{matrix}} ∏_{j=1}^{m} \binom{r_{j}}{k_{j}} · ( ∑_{i=1}^{m} k_{i}𝔅_{k_{i}−1,λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} 𝔅_{k_{j},λ}(x_{j}))

= ∑_{\begin{matrix}k_{1},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n+1 \end{matrix}} ∏_{j=1}^{m} \binom{r_{j}}{k_{j}} · ( ∑_{i=1}^{m} k_{i} ∑_{\begin{matrix}l_{1},...,l_{m}≥0 \\ l_{1}+···+l_{m}=n \end{matrix}} 𝔅_{l_{i},λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{k_{j}}{l_{j}}𝔅_{l_{j},λ}(x_{j} − x_{i} + 1_{j>i}))

= ∑_{\begin{matrix}k_{1},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n+1 \end{matrix}} ∑_{i=1}^{m} r_{i}\binom{r_{i} − 1}{k_{i} − 1} ∑_{\begin{matrix}l_{1},...,l_{m}≥0 \\ l_{1}+···+l_{m}=n \end{matrix}} 𝔅_{l_{i},λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{r_{j}}{l_{j}}\binom{r_{j} − l_{j}}{k_{j} − l_{j}}𝔅_{l_{j},λ}(x_{j} − x_{i} + 1_{j>i})

= ∑_{i=1}^{m} r_{i} ∑_{\begin{matrix}l_{1},...,l_{m}≥0 \\ l_{1}+···+l_{m}=n \end{matrix}} 𝔅_{l_{i},λ}(x_{i}) ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{r_{j}}{l_{j}}𝔅_{l_{j},λ}(x_{j} − x_{i} + 1_{j>i})·

· ( ∑_{\begin{matrix}k_{1},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n+1 \end{matrix}} \binom{r_{i} − 1}{k_{i} − 1} ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{r_{j} − l_{j}}{k_{j} − l_{j}}).

According to the Chu-Vandermonde identity,

∑_{\begin{matrix}k_{1},...,k_{m}≥0 \\ k_{1}+···+k_{m}=n+1 \end{matrix}} \binom{r_{i} − 1}{k_{i} − 1} ∏_{\begin{matrix}1≤j≤m \\ j≠i \end{matrix}} \binom{r_{j} − l_{j}}{k_{j} − l_{j}}

= \binom{r_{1} + ⋯ + r_{m} − 1 − l_{1} − ⋯ − l_{i−1} − l_{i+1} − ⋯ − l_{m}}{k_{1} + ⋯ + k_{m} − 1 − l_{1} − ⋯ − l_{i−1} − l_{i+1} − ⋯ − l_{m}}

= \binom{r_{1} + ⋯ + r_{m} − 1 − n + l_{i}}{l_{i}} = (−1)^{l_{i}} \binom{n − r_{1} − ⋯ − r_{m}}{l_{i}}.

Finally, since

\frac{t(1 + λ)^{\frac{(1−x)t}{λ}}}{(1 + λ)^{\frac{t}{λ}} − 1} = \frac{(−t)(1 + λ)^{\frac{x(−t)}{λ}}}{(1 + λ)^{\frac{−t}{λ}} − 1} ,

we have

𝔅_{l,λ}(1 − x) = (−1)^{l}𝔅_{l,λ}(x)  (14)

for each l ≥ 0. All are done.

Proof of Corollary 2. Using (14) and applying to Theorem 1 with m = 2, r_{1} = s, r_{2} = t, x_{1} = 1 − y and x_{2} = x, we have

r · 𝔇_{s,t,λ}(x, y) = r ∑_{k=0}^{n} \binom{s}{k}\binom{t}{n − k}𝔅_{k,λ}(1 − y)𝔅_{n−k,λ}(x)

= −s ∑_{k=0}^{n} \binom{r}{k}\binom{t}{n − k}𝔅_{k,λ}(y)𝔅_{n−k,λ}(y + x)

− t ∑_{k=0}^{n} \binom{r}{k}\binom{s}{n − k}𝔅_{k,λ}(1 − x)𝔅_{n−k,λ}(1 − x − y)

= −s · 𝔇_{t,r,λ}(y, 1 − x − y) − t · 𝔇_{r,s,λ}(1 − x − y, x).

Thus, Eq. (6) is concluded.

However, the proof of (7) is a little more complicated. Note that \binom{−1}{k} = (−1)^{k} and

\frac{d}{ds}\binom{s}{k}|_{s=0} = \frac{(−1)^{k}}{k} .

Taking the derivative with respect to s on both sides of (6) and letting s = 0 and t = −1, we obtain that

0 = −(n + 1) ∑_{k=1}^{n}(−1)^{n−k} \frac{𝔅_{k,λ}(y)}{k} · 𝔅_{n−k,λ}(x) − (−1)^{n}𝔅_{n,λ}(x)𝔅_{0,λ}(y)

+ ∑_{k=0}^{n} \binom{n + 1}{n − k}𝔅_{n−k,λ}(y)𝔅_{k,λ}(z) + (−1)^{n} ∑_{k=0}^{n−1} \binom{n + 1}{k}\frac{𝔅_{n−k,λ}(z)}{n − k} · 𝔅_{k,λ}(x)

+ (−1)^{n}(1 + \frac{n + 1}{n} + ⋯ + \frac{n + 1}{2})𝔅_{0,λ}(z)𝔅_{n,λ}(x).  (15)

Note that 𝔅_{n,λ}(1) = (−1)^{n}𝔅_{n,λ} by (14). Substituting x = 1 and y = z = 0 in (15), we get

0 = −(n + 1) ∑_{k=1}^{n} \frac{𝔅_{k,λ}}{k} · 𝔅_{n−k,λ} − 𝔅_{n,λ}𝔅_{0,λ}

+ ∑_{k=0}^{n} \binom{n + 1}{n − k}𝔅_{n−k,λ}𝔅_{k,λ} + ∑_{k=0}^{n−1}(−1)^{n−k}\binom{n + 1}{k}\frac{𝔅_{n−k,λ}}{n − k} · 𝔅_{k,λ}

+ (1 + \frac{n + 1}{n} + ⋯ + \frac{n + 1}{2})𝔅_{0,λ}(z)𝔅_{n,λ}.  (16)

Since

t = \frac{t(1 + λ)^{\frac{t}{λ}} − t}{(1 + λ)^{\frac{t}{λ}} − 1} = ∑_{k=0}^{∞} \frac{𝔅_{k,λ}(1) − 𝔅_{k,λ}}{k!} t^{k},

we have 𝔅_{k,λ}(1) = 𝔅_{k,λ} for each k ≥ 2. Thus by (14), 𝔅_{k,λ} = 0 if k ≥ 3 is odd, i.e., (−1)^{k}𝔅_{k,λ} = 0 for each k ≥ 2. It follows from (16) that

(n + 1) ∑_{k=1}^{n−1} \frac{𝔅_{k,λ}}{k} · 𝔅_{n−k,λ} + n(n + 1)𝔅_{1,λ}𝔅_{n−1,λ}

= (n + 1) ∑_{k=1}^{n−1} (\binom{n}{k + 1} + \frac{1}{n + 1}\binom{n + 1}{k})\frac{𝔅_{n−k,λ}}{n − k} · 𝔅_{k,λ} + (n + 1)H_{n}𝔅_{n,λ}𝔅_{0,λ}

= (n + 1) ∑_{k=1}^{n−1} \binom{n}{k}\frac{𝔅_{k,λ}}{k} · 𝔅_{n−k,λ} + (n + 1)H_{n}𝔅_{n,λ}𝔅_{0,λ}.  (17)

Clearly,

𝔅_{0,λ} = lim_{t→0} \frac{t}{(1 + λ)^{\frac{t}{λ}} − 1} = \frac{λ}{ln(1 + λ)}, 𝔅_{1,λ} = lim_{t→0} \frac{d}{dt}(\frac{t}{(1 + λ)^{\frac{t}{λ}} − 1}) = −\frac{1}{2}.

Thus (7) is concluded.

Finally, substituting r = n + 2, s = t = −1, x = 1 and y = z = 0 in (6), we can easily get (8).

3 Acknowledgment

The authors thank the anonymous referee for very helpful suggestions. The authors are supported by the National Natural Science Foundation of China (Grant No. 12071208).

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2020 Mathematics Subject Classification: Primary 11B68; Secondary 05A10, 11B65. Keywords: modified degenerate Bernoulli polynomial, symmetric identity, Miki’s identity, Matiyasevich’s identity.

(Concerned with sequences \underline{A027641} and \underline{A027642}.)

Received August 27 2023; revised version received April 11 2024; April 24 2024. Published in Journal of Integer Sequences, May 12 2024.

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