Convolutions with the Bernoulli and Euler Numbers

Meryem Bouzeraib and Ali Boussayoud LMAM Laboratory and Department of Mathematics

Mohamed Seddik Ben Yahia University Jijel Algeria

meryembouzeraib@gmail.com
  aboussayoud@yahoo.fr

Baghdadi Aloui

Higher Institute of Industrial Systems of Gabes

Department of Electromechanics

City of Erriadh 6072 Gabes Tunisia

Baghdadi.Aloui@fsg.rnu.tn

Abstract

We use exponential generating functions to study the relationship between Bernoulli and Euler numbers with k-Jacobsthal numbers, k-Jacobsthal-Lucas numbers, and bivariate Fibonacci, Lucas, Pell and Pell-Lucas polynomials.

1 Introduction and preliminaries

Recently, there has been some study of special numbers and polynomials, especially the Bernoulli numbers and Euler numbers, by virtue of their applications in almost all branches of mathematics, computer algorithms, engineering, and other areas. (For other numbers see [5, 6, 8, 14, 15]).

Bernoulli and Euler polynomials are famous mathematical objects and are fairly well understood. They are, respectively, defined by the exponential generating functions as follows:

∑_{n=0}^{∞} B_{n} \frac{z^{n}}{n!} = \frac{z}{exp(z) − 1}, |z| < 2π,  (1)
∑_{n⩾0} E_{n} \frac{z^{n}}{n!} = \frac{2 exp(z)}{exp(2z) + 1}, |z| < π.  (2)

The Euler and Bernoulli numbers have appeared in many important results. There has been a growing interest in deriving new relations for these two pairs of sequences. In 1975, Byrd [2] derived the following identity relating Lucas numbers to Euler numbers:

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(\frac{5}{4})^{l}L_{n−2l}E_{2l} = 2^{1−n}.  (3)

Zhang and Ma [16] proved a relation between Fibonacci polynomials and Bernoulli numbers. The following identity is a special case of their result:

∑_{l=0}^{n} \binom{n}{l}5^{\frac{n−k}{2}} F_{k}B_{n−k} = n(\frac{1 − \sqrt{5}}{2})^{n−1},

or equivalently

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}5^{l}F_{n−2l}B_{2l} = \frac{nL_{n−1}}{2} .

Patel et al. [13] obtained similar identities when Fibonacci and Lucas numbers are replaced, respectively, by the Pell and Lucas polynomials (defined below in Eqs. (6) and (7)):

∑_{l=0}^{n} \binom{n}{l}(2\sqrt{x^{2} + 1})^{l}P_{n−l}(x)B_{l} = n(x − \sqrt{x^{2} + 1})^{n−1}.

This can be stated as

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}4^{l}(x^{2} + 1)^{l}P_{n−2l}(x)B_{2l} = \frac{nQ_{n−1}(x)}{2}.

Still other relations are contained in the many articles (see [7, 8, 9]). The present paper is devoted to developing further relations between Bernoulli and Euler numbers with famous number sequences. To achieve this goal, we use elementary methods, including exponential generating functions, to study the relationship between Bernoulli and Euler numbers and between the k-Jacobsthal numbers k-Jacobsthal-Lucas numbers and bivariate Fibonacci, Lucas, Pell, and Pell-Lucas polynomials.

For every positive real number k, the k-Jacobsthal numbers (J_{n,k})_{n∈ℕ} are defined recursively by the relation [11]

J_{k,n} = kJ_{k,n−1} + 2J_{k,n−1}, for n ≥ 2,

with initial values J_{k,0} = 0 and J_{k,1} = 1.

Jhala et al. [12] defined the k-Jacobsthal-Lucas numbers as follows:

j_{k,n} = kj_{k,n−1} + 2j_{k,n−1}, for n ≥ 2, with j_{k,0} = 2, j_{k,1} = k.

The well-known Binet formulas for k-Jacobsthal (J_{n,k})_{n∈ℕ} and k-Jacobsthal-Lucas numbers (j_{n,k})_{n∈ℕ} are given as follows:

J_{n,k} = \frac{λ_{1}^{n} − λ_{2}^{n}}{λ_{1} − λ_{2}}
j_{n,k} = λ_{1}^{n} + λ_{2}^{n},

where λ_{1} = \frac{k+\sqrt{k^{2}+8}}{2} and λ_{2} = \frac{k−\sqrt{k^{2}+8}}{2} are the roots of the characteristic equation λ^{2} − kλ − 2 = 0.

If k = 1, the classical Jacobsthal and Jacobsthal-Lucas numbers are

J_{n} = J_{n−1} + 2J_{n−1}, for n ≥ 2, with J_{0} = 0, J_{1} = 1,
j_{n} = j_{n−1} + 2j_{n−1}, for n ≥ 2, with j_{0} = 2, j_{1} = 1.

The bivariate Fibonacci (F_{n,k}(x, y))_{n≥0} and bivariate Lucas (L_{n,k}(x, y))_{n≥0} polynomial sequences are, respectively, defined by the following recurrence relations [3, 4]:

F_{n}(x, y) = xF_{n−1}(x, y) + yF_{n−2}(x, y), for n ≥ 2,  (4)

with initial values F_{0}(x, y) = 0 and F_{1}(x, y) = 1, and

L_{n}(x, y) = xL_{n−1}(x, y) + yL_{n−2}(x, y), for n ≥ 2,  (5)

with initial values L_{0}(x, y) = 2 and L_{1}(x, y) = x.

The bivariate Pell and Pell-Lucas polynomials are, respectively, as follows [10]:

P_{n}(x, y) = 2xyP_{n−1}(x, y) + yP_{n−2}(x, y), n ≥ 2,  (6)

with initial values P_{0}(x, y) = 0 and P_{1}(x, y) = 1, and

Q_{n}(x, y) = 2xyQ_{n−1}(x, y) + yQ_{n−2}(x, y), n ≥ 2,  (7)

with initial values Q_{0}(x, y) = 2 and Q_{1}(x, y) = 2xy.

We present the Binet formulas of the bivariate Fibonacci, Lucas, Pell, and Pell-Lucas polynomials in Table 1.

Bivariate polynomials Roots (λ_{1}, λ_{1}) Binet formula
Bivariate Fibonacci polynomials λ_{1,2} = ^{\frac{x±\sqrt{x^{2}+4y}}{2}}  F_{n}(x, y) = \frac{λ_{1}^{n}−λ_{2}^{n}}{λ_{1}−λ_{2}}
Bivariate Lucas polynomials λ_{1,2} = ^{\frac{x±\sqrt{x^{2}+4y}}{2}} Q_{n}(x, y) = λ_{1}^{n} + λ_{2}^{n}
Bivariate Pell polynomials  λ_{1,2} = xy ± \sqrt{x^{2}y^{2} + y} P_{n}(x, y) = \frac{λ_{1}^{n}−λ_{2}^{n}}{λ_{1}−λ_{2}}
Bivariate Pell-Lucas polynomials λ_{1,2} = xy ± \sqrt{x^{2}y^{2} + y} Q_{n}(x, y) = λ_{1}^{n} + λ_{2}^{n}

Table 1: Binet’s formulas of the bivariate polynomials.

Putting y = 1 in the Eqs (4), (5), (6) and (7) yields, respectively, the following four polynomials:

• the Fibonacci polynomials defined by F_{0}(x) = 0, F_{1}(x) = 1, F_{n}(x) = xF_{n−1}(x) + F_{n−2}(x), for n ≥ 2,

• the Lucas polynomials defined by L_{0}(x) = 2, L_{1}(x) = x, L_{n}(x) = xL_{n−1}(x) + L_{n−2}(x), for n ≥ 2,

• the Pell polynomials defined by P_{0}(x) = 0, P_{1}(x) = 1, P_{n}(x) = 2xP_{n−1}(x) + P_{n−2}(x), for n ≥ 2,

• the Pell-Lucas polynomials defined by Q_{0}(x) = 2, Q_{1}(x) = 2x, Q_{n}(x) = 2xQ_{n−1}(x) + Q_{n−2}(x), for n ≥ 2.

Putting x = y = 1 in Eqs. (4), (5), (6) and (7), we respectively get the Fibonacci, Lucas, Pell, and Pell-Lucas numbers, which are given as follows:

• F_{n} = F_{n−1} + F_{n−2}, n ≥ 2, with initial values F_{0} = 0, F_{1} = 1,

• L_{n} = L_{n−1} + L_{n−2}, n ≥ 2, with initial values L_{0} = 2, L_{1} = 1,

• P_{n} = 2P_{n−1} + P_{n−2}, n ≥ 2, with initial values P_{0} = 0, P_{1} = 1,

• Q_{n} = 2Q_{n−1} + Q_{n−2}, n ≥ 2, with initial values Q_{0}(x) = 2, Q_{1} = 2.

2 New k-Jacobsthal-Bernoulli and k-Jacobsthal-Euler

relations

In this section, we present our first findings in three theorems, which provide some relations involving Bernoulli and Euler numbers with k-Jacobsthal and k-Jacobsthal-Lucas numbers.

Theorem 1. Let n be a positive integer. We have

∑_{l=0}^{n} \binom{n}{l}(\sqrt{k^{2} + 8})^{l}J_{k,n−l}B_{l} = n(\frac{k − \sqrt{k^{2} + 8}}{2})^{n−1},
∑_{l=0}^{n} \binom{n}{l}(\sqrt{k^{2} + 8})^{l}(2^{l} − 1)j_{k,n−l}B_{l} = −n\sqrt{k^{2} + 8}(\frac{k − \sqrt{k^{2} + 8}}{2})^{n−1}.  (8)

Proof. Using the change of variables z = \sqrt{k^{2} + 8}z in (1), we obtain

∑_{n=0}^{∞}(\sqrt{k^{2} + 8})^{n}B_{n} \frac{z^{n}}{n!} = \frac{\sqrt{k^{2} + 8}z}{exp(\sqrt{k^{2} + 8}z) − 1}.  (9)

Multiplying (9) by the exponential generating function for (J_{k,n})_{n∈ℕ}, we have

(∑_{n=0}^{∞} J_{k,n} \frac{z^{n}}{n!} )(∑_{l=0}^{∞}(\sqrt{k^{2} + 8})^{l}B_{l} \frac{z^{l}}{l!} ) = ∑_{n=0}^{∞} ∑_{l=0}^{∞}(\sqrt{k^{2} + 8})^{l}B_{l} \frac{z^{l}}{l!} \frac{z^{n}}{n!}

= ∑_{n=0}^{∞} ∑_{l=0}^{n} \binom{n}{l}(\sqrt{k^{2} + 8})^{l}B_{l}J_{k,n−l} \frac{z^{n}}{n!} .

Therefore

\frac{exp(\frac{k+\sqrt{k^{2}+8}}{2}z) − exp(\frac{k−\sqrt{k^{2}+8}}{2}z)}{\sqrt{k^{2} + 8}} × \frac{\sqrt{k^{2} + 8}z}{exp(\sqrt{k^{2} + 8}z) − 1}

= exp(\frac{k − \sqrt{k^{2} + 8}}{2}z)(exp(\sqrt{k^{2} + 8}z) − 1) × \frac{z}{exp(\sqrt{k^{2} + 8}z) − 1}

= z exp(\frac{k − \sqrt{k^{2} + 8}}{2}z)

= ∑_{n=0}^{∞} n(\frac{k − \sqrt{k^{2} + 8}}{2}z)^{n−1} \frac{z^{n}}{n!} .

By comparing the coefficients of \frac{z^{n}}{n!} , we obtain the desired result.

Similarly, we use the change of variables z = 2\sqrt{k^{2} + 8}z in (1), we obtain

∑_{n=0}^{∞}(2\sqrt{k^{2} + 8})^{n}B_{n} \frac{z^{n}}{n!} = \frac{2\sqrt{k^{2} + 8}z}{exp(2\sqrt{k^{2} + 8}z) − 1},  (10)

and multiplying (10), by the exponential generating function for (j_{n,k})_{n∈ℕ}, we get

(∑_{n=0}^{∞} j_{k,n} \frac{z^{n}}{n!} )(∑_{l=0}^{∞}(2\sqrt{k^{2} + 8})^{l}B_{l} \frac{z^{l}}{l!} ) = ∑_{n=0}^{∞} ∑_{l=0}^{∞}(2\sqrt{k^{2} + 8})^{l}B_{l}j_{k,n} \frac{z^{l}}{l!} \frac{z^{n}}{n!}

= ∑_{n=0}^{∞} ∑_{l=0}^{n} \binom{n}{l}2^{l}(\sqrt{k^{2} + 8})^{l}j_{k,n−l}B_{l} \frac{z^{n}}{n!} .

Then we obtain

(exp(\frac{(k + \sqrt{k^{2} + 8})z}{2} ) + exp(\frac{(k − \sqrt{k^{2} + 8})z}{2} ))\frac{2\sqrt{k^{2} + 8}z}{exp(2\sqrt{k^{2} + 8}z) − 1}

= exp(\frac{(k − \sqrt{k^{2} + 8})z}{2} )\frac{2\sqrt{k^{2} + 8}z}{exp(\sqrt{k^{2} + 8}z) − 1}

= (∑_{n=0}^{∞}(\frac{k − \sqrt{k^{2} + 8}}{2})^{n} \frac{z^{n}}{n!} )(∑_{n=0}^{∞} 2(\sqrt{k^{2} + 8})^{l}B_{l} \frac{z^{l}}{l!} )

= ∑_{n=0}^{∞} ∑_{l=0}^{n} \binom{n}{l}2(\sqrt{k^{2} + 8})^{l}(\frac{k − \sqrt{k^{2} + 8}}{2})^{n−l}B_{l} \frac{z^{n}}{n!} .

By comparing the coefficients of \frac{z^{n}}{n!} , we get

∑_{l=0}^{n} \binom{n}{k}2^{l}(\sqrt{k^{2} + 8})^{l}j_{k,n−l}B_{l} = ∑_{l=0}^{n} \binom{n}{l}2(\sqrt{k^{2} + 8})^{l}(\frac{k − \sqrt{k^{2} + 8}}{2})^{n−l}B_{l},

which is equivalent to

∑_{l=0}^{n} \binom{n}{k}2^{l}(\sqrt{k^{2} + 8})^{l}j_{k,n−l}B_{l} = ∑_{l=0}^{n} \binom{n}{l}2(\sqrt{k^{2} + 8})^{l}(\frac{j_{k,n−1} − \sqrt{k^{2} + 8}J_{k,n−1}}{2} )B_{l}.

Therefore

∑_{l=0}^{n} \binom{n}{l}(\sqrt{k^{2} + 8})^{l}(2^{l} − 1)j_{k,n−l}B_{l} = −\sqrt{k^{2} + 8} ∑_{l=0}^{n} \binom{n}{l}(\sqrt{k^{2} + 8})^{l}J_{k,n−l}B_{l},

= −n\sqrt{k^{2} + 8}(\frac{k − \sqrt{k^{2} + 8}}{2})^{n−1}.

Hence the desired result. Theorem 2. Let n be a positive integer. The following results hold:

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(k^{2} + 8)^{l}J_{k,n−2l}B_{2l} = \frac{nj_{k,n−1}}{2} ,

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(k^{2} + 8)^{l}(2^{2l} − 1)j_{k,n−2l}B_{2l} = \frac{(k^{2} + 8)nJ_{k,n−1}}{2} .

Proof. We have

∑_{l=0}^{n} \binom{n}{l}(\sqrt{k^{2} + 8})^{l}J_{k,n−l}B_{l} = n(\frac{k − \sqrt{k^{2} + 8}}{2})^{n−1}.

This gives

∑_{l=0}^{n} \binom{n}{l}(\sqrt{k^{2} + 8})^{n−l}J_{k,n−l}B_{l} = \frac{n(j_{k,n−1} − \sqrt{k^{2} + 8}J_{k,n−1})}{2},

with the conditions B_{0} = 1 and B_{1} = −\frac{1}{2}.

For l ≥ 0, we have B_{2l+1} = 0 and then

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(k^{2} + 8)^{l}J_{k,n−2l}B_{2l} = \frac{nj_{k,n−1}}{2} .

Similarly, from (8), we get

∑_{l=0}^{n} \binom{n}{l}(\sqrt{k^{2} + 8})^{l}(2^{l} − 1)j_{k,n−l}B_{l} = \frac{−\sqrt{k^{2} + 8}n(j_{k,n−1} − \sqrt{k^{2} + 8}J_{k,n−1})}{2},

with B_{0} = 1 and B_{1} = −\frac{1}{2}.

For l ≥ 0, we have B_{2l+1} = 0. Then we obtain

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(k^{2} + 8)^{l}(2^{2l} − 1)j_{k,n−2l}B_{2l} = \frac{(k^{2} + 8)^{2}nJ_{k,n−1}}{2} .

Which complete the proof.

• Putting k = 1 in Theorem 1 and Theorem 2 we have the following Jacobsthal-Bernoulli and Jacobsthal-Lucas-Bernoulli identities

∑_{l=0}^{n} \binom{n}{l}3^{l}J_{n−l}B_{l} = (−1)^{n−1}n, or ∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}9^{l}J_{n−2l}B_{2l} = \frac{nj_{n−1}}{2} .

∑_{l=0}^{n} \binom{n}{2l}3^{l}(2^{l} − 1)j_{n−l}B_{l} = (−1)^{n}n, or ∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}9^{l}(2^{l} − 1)j_{n−2l}B_{2l} = \frac{9nJ_{n−1}}{2} .

Theorem 3. For every positive integer n, we have

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}2(4)^{n−2l−1}(\sqrt{k^{2} + 8})^{2l+1}J_{k,n−2l}E_{2l} = (2k + \sqrt{k^{2} + 8})^{n} − (2k + \sqrt{k^{2} + 8})^{n},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(\frac{k^{2} + 8}{4})^{l}j_{k,n−2l}E_{2l} = 2(\frac{k}{2})^{n}.

Proof. Using the change of variables z = \sqrt{k^{2} + 8}z in (2) and z = 4z in the exponential generating function for (J_{k,n})_{n∈ℕ}, we obtain

∑_{n=0}^{∞}(\sqrt{k^{2} + 8})^{l}E_{l} \frac{z^{l}}{l!} = \frac{2 exp(\sqrt{k^{2} + 8}z)}{exp(2\sqrt{k^{2} + 8}z) + 1},  (11)

∑_{n=0}^{∞} 4^{n}J_{k,n} \frac{z^{n}}{n!} = \frac{exp(k − \sqrt{k^{2} + 8})z)}{\sqrt{k^{2} + 8}}(exp(2\sqrt{k^{2} + 8}z) − 1)(exp(2\sqrt{k^{2} + 8}z) + 1). (12)

Multiplying (12) by (11), this gives

(∑_{n=0}^{∞} 4^{n}J_{k,n} \frac{z^{n}}{n!} )(∑_{l=0}^{∞}(\sqrt{k^{2} + 8})^{l}E_{l} \frac{z^{l}}{l!} ) = ∑_{n=0}^{∞} ∑_{l=0}^{n} \binom{n}{l}4^{n−l}(\sqrt{k^{2} + 8})^{l}J_{k,n−l}E_{l} \frac{z^{n}}{n!} .

Then we get

\frac{exp(k − \sqrt{k^{2} + 8})z)}{\sqrt{k^{2} + 8}}(exp(2\sqrt{k^{2} + 8}z) − 1)(exp(2\sqrt{k^{2} + 8}z) + 1) × \frac{2 exp(\sqrt{k^{2} + 8}z)}{(exp 2\sqrt{k^{2} + 8}z) + 1}

= \frac{2}{\sqrt{k^{2} + 8}}(exp((2k + \sqrt{k^{2} + 8})z) − exp((2k − \sqrt{k^{2} + 8})z))

= \frac{2}{\sqrt{k^{2} + 8}} ∑_{n=0}^{∞}(2k + \sqrt{k^{2} + 8})^{n} − (2k − \sqrt{k^{2} + 8})^{n} \frac{z^{n}}{n!} .

By comparing the coefficients of \frac{z^{n}}{n!} , we obtain

∑_{l=0}^{n} \binom{n}{l}2(4)^{n−l−1}(\sqrt{k^{2} + 8})^{l+1}J_{k,n−l}E_{l} = (2k + \sqrt{k^{2} + 8})^{n} − (2k − \sqrt{k^{2} + 8})^{n}.

For l ≥ 0, we have E_{2l+1} = 0, which gives the desired result.

Similarly, we use the change of variables z = \frac{\sqrt{k^{2}+8}}{2}z in (2). We have

∑_{n=0}^{∞}(\frac{\sqrt{k^{2} + 8}}{2})^{n}E_{n} \frac{z^{n}}{n!} = \frac{2 exp(\frac{\sqrt{k^{2}+8}}{2}z)}{exp(\sqrt{k^{2} + 8}z) + 1},  (13)

and multiplying (13) by the exponential generating function for (j_{k,n})_{n∈ℕ}, we get

(∑_{n=0}^{∞} j_{k,n} \frac{z^{n}}{n!} )(∑_{l=0}^{∞}(\frac{\sqrt{k^{2} + 8}}{2})^{l}E_{l} \frac{z^{l}}{l!} ) = ∑_{l=0}^{∞} ∑_{l=0}^{n} \binom{n}{l}(\frac{\sqrt{k^{2} + 8}}{2})^{l}j_{k,n−l}E_{l} \frac{z^{n}}{n!} ,

and

\frac{2 exp(\frac{\sqrt{k^{2}+8}}{2}z)}{exp(\sqrt{k^{2} + 8}z) + 1}(exp(\frac{k + \sqrt{k^{2} + 8}}{2}z) + exp(\frac{k − \sqrt{k^{2} + 8}}{2}z)) = 2 ∑_{k=0}^{∞}(\frac{k}{2})^{n} \frac{z^{n}}{n!} .

By comparing the coefficients of \frac{z^{n}}{n!} , we get

∑_{l=0}^{n} \binom{n}{l}(\frac{\sqrt{k^{2} + 8}}{2})^{l}j_{k,n−l}E_{l} = 2(\frac{k}{2})^{n}.

For k ≥ 0, we obtain E_{2k+1} = 0; hence the desired result.

• By taking k = 1 in the Theorem 3, we can state the following identities involving Euler numbers with Jacobsthal and Jacobsthal-Lucas numbers

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}2(4)^{n−2l−1}3^{2l+1}J_{n−2l}E_{2l} = 5^{n} − (−1)^{n},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(\frac{9}{4})^{l}j_{n−2l}E_{2l} = 2^{1−n}.

3 Bivariate polynomial identities

The aim of this section is to establish some new identities involving two variables x and y. We start with three theorems involving Bernoulli and Euler numbers with bivariate Fibonacci and Lucas polynomials.

Theorem 4. For every positive integer n, we have

∑_{l=0}^{n} \binom{n}{l}(\sqrt{x^{2} + 4y})^{l}F_{n−l}(x, y)B_{l} = n(\frac{x − \sqrt{x^{2} + 4y}}{2} )^{n−1},  (14)

∑_{l=0}^{n} \binom{n}{l}(\sqrt{x^{2} + 4y})^{l}(2^{l} − 1)L_{n−l}(x, y)B_{l} = −2\sqrt{x^{2}y^{2} + y}n(\frac{x − \sqrt{x^{2} + 4y}}{2} )^{n−1}.

Proof. Using the change of variables z = \sqrt{x^{2} + 4y}z in (1), we obtain

∑_{n=0}^{∞}(\sqrt{x^{2} + 4y})^{n}B_{n} \frac{z^{n}}{n!} = \frac{\sqrt{x^{2} + 4y}z}{exp(\sqrt{x^{2} + 4y}z) − 1}.  (15)

Multiplying (15) by the exponential generating function for (F_{n}(x, y))_{n∈ℕ}, we get

(∑_{n=0}^{∞} F_{n}(x, y)\frac{z^{n}}{n!} )(∑_{l=0}^{∞}(\sqrt{x^{2} + 4y})^{l}B_{l} \frac{z^{l}}{l!} ) = ∑_{n=0}^{∞} ∑_{l=0}^{∞}(\sqrt{x^{2} + 4y})^{l}B_{l}F_{n}(x, y)\frac{z^{l}}{l!} \frac{z^{n}}{n!}

= ∑_{n=0}^{∞} ∑_{l=0}^{n} \binom{n}{l}(\sqrt{x^{2} + 4y})^{l}B_{l}F_{n}(x, y)\frac{z^{n}}{n!} .

Then

\frac{exp(^{\frac{x−\sqrt{x^{2}+4y}}{2}}z) − exp(^{\frac{x−\sqrt{x^{2}+4y}}{2}}z)}{\sqrt{x^{2} + 4y}} × \frac{\sqrt{x^{2} + 4y}z}{exp(\sqrt{x^{2} + 4y}z) − 1}

= z exp(\frac{x − \sqrt{x^{2} + 4y}}{2} z) = ∑_{n=0}^{∞} n(\frac{x − \sqrt{x^{2} + 4y}}{2} )^{n−1} \frac{z^{n}}{n!} .

By comparing the coefficients of \frac{z^{n}}{n!} , we obtain the desired result.

Similarly, by using the change of variables z = 2\sqrt{x^{2} + 4y}z in (1), we obtain

∑_{n=0}^{∞}(2\sqrt{x^{2} + 4y})^{n}B_{n} \frac{z^{n}}{n!} = \frac{2\sqrt{x^{2} + 4y}z}{exp(2\sqrt{x^{2} + 4y}z) − 1},  (16)

and multiplying (16) by the exponential generating function for (L_{n}(x, y))_{n∈ℕ}, we get

(∑_{n=0}^{∞} L_{n}(x, y)\frac{z^{n}}{n!} )(∑_{l=0}^{∞}(2\sqrt{x^{2} + 4y})^{l}B_{l} \frac{z^{l}}{l!} ) = ∑_{n=0}^{∞} ∑_{l=0}^{∞}(2\sqrt{x^{2} + 4y})^{l}B_{l}L_{n}(x, y)\frac{z^{l}}{l!} \frac{z^{n}}{n!}

= ∑_{n=0}^{∞} ∑_{l=0}^{n} \binom{n}{l}2^{l}(\sqrt{x^{2} + 4y})^{l}L_{n−l}(x, y)B_{l} \frac{z^{n}}{n!} .

Therefore

(exp(\frac{x + \sqrt{x^{2} + 4y}}{2} z) + exp(\frac{x − \sqrt{x^{2} + 4y}}{2} z))\frac{2\sqrt{x^{2} + 4y}z}{exp(2\sqrt{x^{2} + 4y}z) − 1}

= exp(\frac{x − \sqrt{x^{2} + 4y}}{2} z)\frac{2\sqrt{x^{2} + 4y}z}{exp(\sqrt{x^{2} + 4y}z) − 1}

= ∑_{n=0}^{∞} ∑_{l=0}^{n} \binom{n}{l}2(\sqrt{x^{2} + 4y})^{l}(\frac{x − \sqrt{x^{2} + 4y}}{2} )^{n−l}B_{l} \frac{z^{n}}{n!} .

By comparing the coefficients of \frac{z^{n}}{n!} , we obtain

∑_{l=0}^{n} \binom{n}{k}2^{l}(\sqrt{x^{2} + 4y})^{l}L_{n−l}(x, y)B_{l} = ∑_{l=0}^{n} \binom{n}{l}2(\sqrt{x^{2} + 4y})^{l}(\frac{x − \sqrt{x^{2} + 4y}}{2} )^{n−l}B_{l}.

We know that

(\frac{x − \sqrt{x^{2} + 4y}}{2} )^{n−l} = \frac{L_{n−l}(x, y) − \sqrt{x^{2} + 4y}F_{n−l}(x, y)}{2},

so we get

∑_{l=0}^{n} \binom{n}{l}(\sqrt{x^{2} + 4y})^{l}(2^{l} − 1)L_{n−l}(x, y)B_{l} = −\sqrt{x^{2} + 4y}n(\frac{x − \sqrt{x^{2} + 4y}}{2} )^{n−1}.

Hence the result.

Theorem 5. For every positive integer n, we have

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(x^{2} + 4y)^{l}F_{n−2l}(x, y)B_{2l} = \frac{nL_{n−1}(x, y)}{2},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(x^{2} + 4y)^{l}(2^{2l} − 1)L_{n−2l}(x, y)B_{2l} = \frac{n(x^{2} + 4y)F_{n−1}(x, y)}{2}.  (17)

Proof. From (14) and on account of the identity

(\frac{x − \sqrt{x^{2} + 4y}}{2} )^{n−l} = \frac{L_{n−l}(x, y) − \sqrt{x^{2} + 4y}F_{n−l}(x, y)}{2},

we get

∑_{l=0}^{n} \binom{n}{l}(\sqrt{x^{2} + 4y})^{n−l}F_{n−l}(x, y)B_{l} = \frac{n(L_{n−1}(x, y) − \sqrt{x^{2} + 4y}F_{n−1}(x, y))}{2},

with B_{0} = 1 and B_{1} = −\frac{1}{2}.

For l ≥ 0, we have B_{2l+1} = 0, and then

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(x^{2} + 4y)^{l}F_{n−2l}(x, y)B_{2l} = \frac{nL_{n−1}(x, y)}{2}.

Similarly, from (17), we get

∑_{l=0}^{n} \binom{n}{l}(\sqrt{x^{2} + 4y})^{l}(2^{l}−1)L_{n−l}(x, y)B_{l} = \frac{−\sqrt{x^{2} + 4y}n(L_{n−1}(x, y) − \sqrt{x^{2} + 4y}F_{n−1}(x, y))}{2},

with B_{0} = 1, B_{1} = −\frac{1}{2}.

For l ≥ 0, we have B_{2l+1} = 0. Hence we obtain

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(x^{2} + 4y)^{l}(2^{2l} − 1)L_{n−2l}(x, y)B_{2l} = \frac{(x^{2} + 4y)^{2}nF_{n−1}(x, y)}{2}.

Hence the desired result is proved.

• By putting y = 1 in Theorem 4 and Theorem 5, we obtain the following identities

∑_{l=0}^{n} \binom{n}{l}(\sqrt{x^{2} + 4})^{l}F_{n−l}(x)B_{l} = n(\frac{x − \sqrt{x^{2} + 4}}{2})^{n−1},  (18)

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(x^{2} + 4)^{l}F_{n−2l}(x)B_{2l} = \frac{nL_{n−1}(x)}{2},  (19)

∑_{l=0}^{n} \binom{n}{l}(\sqrt{x^{2} + 4})^{l}(2^{l} − 1)L_{n−l}(x)B_{l} = −2(\sqrt{x^{2} + 4})n(\frac{x − \sqrt{x^{2} + 4}}{2})^{n−1}, (20)

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(x^{2} + 4)^{l}(2^{2l} − 1)L_{n−2l}(x)B_{2l} = \frac{n(x^{2} + 4)F_{n−1}(x)}{2}.  (21)

• By setting x = y = 1 in Theorem 4 and Theorem 5 or x = 1 in the relationships (18), (19), (20) and (21) we obtain the following identities involving Bernoulli numbers with Fibonacci and Lucas numbers (see [1, 2]):

∑_{l=0}^{n} \binom{n}{l}(\sqrt{5})^{l}F_{n−l}B_{l} = n(\frac{1 − \sqrt{5}}{2})^{n−1},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}5^{l}F_{n−2l}B_{2l} = \frac{nL_{n−1}}{2} ,

∑_{l=0}^{n} \binom{n}{l}(\sqrt{5})^{l}(2^{l} − 1)L_{n−l}B_{l} = −2\sqrt{5}n(\frac{1 − \sqrt{5}}{2})^{n−1},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}5^{l}(2^{2l} − 1)L_{n−2l}B_{2l} = \frac{n5F_{n−1}}{2} .

Theorem 6. For every positive integer n, we have

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}2(4)^{n−2k−1}(\sqrt{x^{2} + 4y})^{2l+1}F_{n−2l}(x, y)E_{2l} = (2x + \sqrt{x^{2} + 4y})^{n} − (2x − \sqrt{x^{2} + 4y})^{n},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(\frac{x^{2} + 4y}{4} )^{2l}L_{n−2l}(x, y)E_{2l} = 2(\frac{x}{2})^{n}.

Proof. Using the change of variables z = \sqrt{x^{2} + 4y}z in (2), we obtain

∑_{l=0}^{∞}(\sqrt{x^{2} + 4y})^{l}E_{l} \frac{z^{l}}{l!} = \frac{2 exp(\sqrt{x^{2} + 4y}z)}{exp(2\sqrt{x^{2} + 4y}z) + 1}.  (22)

By multiplying (22) by the exponential generating function for the sequence (4^{n}F_{n}(x, y))_{n∈ℕ}, we obtain

(∑_{k=0}^{∞} 4^{n}F_{n}(x, y)\frac{z^{n}}{n!} )(∑_{l=0}^{∞}(\sqrt{x^{2} + 4y})^{l}E_{l} \frac{z^{l}}{l!} ) = ∑_{n=0}^{∞} ∑_{l=0}^{n} \binom{n}{l}4^{n−l}(\sqrt{x^{2} + 4y})^{l}F_{n−l}(x, y)E_{l} \frac{z^{n}}{n!} .

Then we get

\frac{exp(2(x + \sqrt{x^{2} + 4y})z) − exp(2(x − \sqrt{x^{2} + 4y})z)}{\sqrt{x^{2} + 4y}} × \frac{2 exp(\sqrt{x^{2} + 4y}z)}{exp(2\sqrt{x^{2} + 4y}z) + 1}

= \frac{2 exp(2(x − \sqrt{x^{2} + 4y})z) exp(\sqrt{x^{2} + 4y}z)}{\sqrt{x^{2} + 4y}(exp(2\sqrt{x^{2} + 4y}z) + 1)}(exp(4\sqrt{x^{2} + 4y}z) − 1)

= \frac{2}{\sqrt{x^{2} + 4y}} (exp((2x + \sqrt{x^{2} + 4y})z) + exp((2x − \sqrt{x^{2} + 4y})z))

= \frac{2}{\sqrt{x^{2} + 4y}} ∑_{n=0}^{∞}((2x + \sqrt{x^{2} + 4y})^{n} − (2x − \sqrt{x^{2} + 4y})^{n})\frac{z^{n}}{n!} .

By comparing the coefficients of \frac{z^{n}}{n!} , we get

∑_{l=0}^{n} \binom{n}{l}2(4)^{n−l−1}(\sqrt{x^{2} + 4y})^{l+1}F_{n−l}(x, y)E_{l} = (2x + \sqrt{x^{2} + 4y})^{n} − (2x − \sqrt{x^{2} + 4y})^{n}.

For l ≥ 0, we have E_{2l+1} = 0, which gives the result.

Similarly, using the change of variables z = ^{\frac{\sqrt{x^{2}+4y}}{2}}z in (2), we obtain

∑_{n=0}^{∞}(\frac{\sqrt{x^{2} + 4y}}{2} )^{n}E_{n} \frac{z^{n}}{n!} = \frac{2 exp(^{\frac{\sqrt{x^{2}+4y}}{2}}z)}{exp(\sqrt{x^{2} + 4y}z) + 1},  (23)

and multiplying (23) by the exponential generating function for (L_{n}(x, y))_{n∈ℕ}, we get

(∑_{n=0}^{∞} L_{n}(x, y)\frac{z^{n}}{n!} )(∑_{l=0}^{∞}(\frac{\sqrt{x^{2} + 4y}}{2} )^{l}E_{l} \frac{z^{l}}{l!} ) = ∑_{n=0}^{∞} ∑_{l=0}^{n} \binom{n}{l}(\frac{\sqrt{x^{2} + 4y}}{2} )^{l}L_{n−l}(x, y)E_{l} \frac{z^{n}}{n!} ,

and

(exp(\frac{x + \sqrt{x^{2} + 4y}}{2} z) + exp(\frac{x − \sqrt{x^{2} + 4y}}{2} z)) × \frac{2 exp(^{\frac{\sqrt{x^{2}+4y}}{2}}z)}{exp(\sqrt{x^{2} + 4y}z) + 1} = 2 exp(\frac{x}{2}z)

= 2 ∑_{n=0}^{∞}(\frac{x}{2})^{n} \frac{z^{n}}{n!} .

Comparing the coefficients of \frac{z^{n}}{n!} , we obtain

∑_{l=0}^{n} \binom{n}{l}(\frac{\sqrt{x^{2} + 4y}}{2} )^{l}L_{n−l}(x, y)E_{l} = 2(\frac{x}{2})^{n},

For k ≥ 0, we have E_{2k+1} = 0, and the result follows.

• Putting y = 1 in Theorem 6, we obtain the following identities linking Euler numbers to Fibonacci and Lucas polynomials

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}2(4)^{n−2k−1}(\sqrt{x^{2} + 4})^{2l+1}F_{n−2l}(x)E_{2l} = (2x + \sqrt{x^{2} + 4})^{n} − (2x − \sqrt{x^{2} + 4})^{n}, (24)

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(\frac{x^{2} + 4}{4})^{l}L_{n−2l}(x)E_{2l} = 2(\frac{x}{2})^{n}.  (25)

• Putting y = x = 1 in Theorem 6 or x = 1 in the relationships (24) and (25), we obtain the following identities linking Euler numbers to Fibonacci and Lucas numbers

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}2(4)^{n−2l−1}(\sqrt{5})^{2l+1}F_{n−2l}E_{2l} = (2 + \sqrt{5})^{n} − (2 − \sqrt{5})^{n},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(\frac{5}{4})^{l}L_{n−2l}E_{2l} = 2^{1−n}.  (26)

Note that Eq. (26) is Byrd’s result (3).

Now we present the analogue results for the relation between Bernoulli numbers with bivariate Pell and bivariate Pell-Lucas polynomials.

Theorem 7. For every positive integer n, we have

∑_{l=0}^{n} \binom{n}{l}(2\sqrt{x^{2}y^{2} + y})^{l}P_{n−l}(x, y)B_{l} = n(xy − \sqrt{x^{2}y^{2} + y})^{n−1},  (27)

∑_{l=0}^{n} \binom{n}{l}(2\sqrt{x^{2}y^{2} + y})^{l}(2^{l} − 1)Q_{n−l}(x, y)B_{l} = −2\sqrt{x^{2}y^{2} + y}n(xy − \sqrt{x^{2}y^{2} + y})^{n−1}. (28)

Proof. The proof is similar to the proof of Theorem 4, by using the change of variables z = 2\sqrt{x^{2}y^{2} + y}z in (1) to prove Eq. (27) and z = 4\sqrt{x^{2}y^{2} + y}z in (1) to prove the relationship (28).

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}4^{l}(x^{2}y^{2} + y)^{l}P_{n−2l}(x, y)B_{2l} = \frac{nQ_{n−1}(x, y)}{2},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}4^{l}(x^{2}y^{2} + y)^{l}(2^{2l} − 1)Q_{n−2l}(x, y)B_{2l} = 2n(x^{2}y^{2} + y)P_{n−1}(x, y).

Proof. We have

∑_{l=0}^{n} \binom{n}{l}(2\sqrt{x^{2}y^{2} + y})^{l}P_{n−l}(x, y)B_{l} = n(xy − \sqrt{x^{2}y^{2} + y})^{n−1},

which gives

∑_{l=0}^{n} \binom{n}{l}(2\sqrt{x^{2}y^{2} + y})^{n−l}P_{n−l}(x, y)B_{l} = \frac{n(Q_{n−1}(x, y) − 2\sqrt{x^{2}y^{2} + y}P_{n−1}(x, y))}{2},

with B_{0} = 1 and B_{1} = −\frac{1}{2}.

For l ≥ 0, we have B_{2l+1} = 0. Then we get

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}4^{l}(x^{2}y^{2} + y)^{l}P_{n−2l}(x, y)B_{2l} = \frac{nQ_{n−1}(x, y)}{2}.

Similarly, from (28), we get

∑_{l=0}^{n} \binom{n}{l}4^{l}(x^{2}y^{2}+y)^{l}(2^{l}−1)Q_{n−l}(x, y)B_{l} = \frac{−2\sqrt{x^{2}y^{2} + y}n(Q_{n−1}(x, y) − 2\sqrt{x^{2}y^{2} + y}P_{n−1}(x, y))}{2},

with B_{0} = 1 and B_{1} = −\frac{1}{2}.

For l ≥ 0, we have B_{2l+1} = 0. Then we obtain

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}4^{l}(x^{2}y^{2} + y)^{l}(2^{2l} − 1)Q_{n−2l}(x, y)B_{2l} = \frac{(x^{2}y^{2} + y)nF_{n−1}(x, y)}{2},

and the result follows.

• By putting y = 1 in Theorem 7 and Theorem 8, we start with the following results involving Bernoulli numbers and Pell, Pell-Lucas polynomials:

∑_{l=0}^{n} \binom{n}{l}(2\sqrt{x^{2} + 1})^{l}P_{n−l}(x)B_{l} = n(x − \sqrt{x^{2} + 1})^{n−1},  (29)

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}4^{l}(x^{2} + 1)^{l}P_{n−2l}(x)B_{2l} = \frac{nQ_{n−1}(x)}{2},  (30)

∑_{l=0}^{n} \binom{n}{l}(2\sqrt{x^{2} + 1})^{l}(2^{l} − 1)Q_{n−l}(x)B_{l} = −2n\sqrt{x^{2} + 1}(x − \sqrt{x^{2} + 1})^{n−1}, (31)

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}4^{l}(x^{2} + 1)^{l}(2^{2l} − 1)Q_{n−2l}(x)B_{2l} = 2n(x^{2} + 1)P_{n−1}(x).  (32)

• By setting x = y = 1 in Theorem 7 and Theorem 8 or x = 1 in the relationships (29), (30), (31) and (32) we deduce the following Bernoulli-Pell and the Bernoulli-Pell-Lucas identities:

∑_{l=0}^{n} \binom{n}{l}(2\sqrt{2})^{l}P_{n−l}B_{l} = n(1 − \sqrt{2})^{n−1},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}8^{l}P_{n−2l}B_{2l} = \frac{nQ_{n−1}}{2} ,

∑_{l=0}^{n} \binom{n}{l}(2\sqrt{2})^{l}(2^{l} − 1)Q_{n−l}B_{l} = −2n\sqrt{2}(1 − \sqrt{2})^{n−1},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}8^{l}(2^{2l} − 1)Q_{n−2l}B_{2l} = 2n(1 + 1)P_{n−1}.

Theorem 9. For every positive integer n, we have

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}2^{n−2l}(\sqrt{x^{2}y^{2} + y})^{2l+1}P_{n−2l}(x, y)E_{2l} = (2xy + \sqrt{x^{2}y^{2} + y})^{n} − (2xy − \sqrt{x^{2}y^{2} + y})^{n}, (33)

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(x^{2}y^{2} + y)^{l}Q_{n−2l}(x, y)E_{2l} = 2(xy)^{n}.  (34)

Proof. The proof is similar to the proof of Theorem 6, by using the change of variables z = 2\sqrt{x^{2}y^{2} + y}z in (2) to prove (33) and z = 4\sqrt{x^{2}y^{2} + y}z in (2) to prove (34).

• By putting y = 1 in Theorem 9, we obtain some relations between the Euler numbers and the Pell and Pell-Lucas polynomials:

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}2^{n−2l}(\sqrt{x^{2} + 1})^{2l+1}P_{n−2l}(x)E_{2l} = (2x + \sqrt{x^{2} + 1})^{n} − (2x − \sqrt{x^{2} + 1})^{n}, (35)

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(x^{2} + 1)^{l}Q_{n−2l}(x)E_{2l} = 2x^{n}.  (36)

• By putting x = y = 1 in Theorem 9 or x = 1 in the relationships (35) and (36), we can state the following Pell-Euler and Pell-Lucas-Euler identities:

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}(\sqrt{2})2^{n−l}P_{n−2l}E_{2l} = (2 + \sqrt{2})^{n} − (2 − \sqrt{2})^{n},

∑_{l=0}^{⌊n/2⌋} \binom{n}{2l}2^{l}P_{n−2l}E_{2l} = 2.

4 Acknowledgments

The authors would like to thank the anonymous referees for their valuable comments and suggestions for improving the original version of this manuscript.

References

[1] P. F. Byrd, New relations between Fibonacci and Bernoulli numbers, Fibonacci Quart. 13 (1975), 59–69.

[2] P. F. Byrd, Relations between Euler and Lucas numbers, Fibonacci Quart. 13 (1975), 114–114.

[3] M. Catalani, Generalized bivariate Fibonacci polynomials, arxiv preprint arXiv:math/0211366 [math.CO], 2004. Available at https://arxiv.org/abs/

math/0211366.

[4] M. Catalani, Some formulae for bivariate Fibonacci and Lucas polynomials, arxiv preprint arXiv:math/0406323 [math.CO], 2004. Available at https://arxiv.org/abs/

math/0406323.

[5] M. Chelgham and A. Boussayoud, On the k-Mersenne-Lucas numbers, Notes Numb. Theo. Disc. Math. 27 (2021), 7–13.

[6] K. Dilcher, Bernoulli and Euler polynomials, in F. W. J. Olver et al., eds., NIST Handbook of Mathematical Functions Cambridge University Press, 2010.

[7] R. Frontczak, Relating Fibonacci numbers to Bernoulli numbers via balancing polynomials, J. Integer Sequences 22 (2019), Article 19.5.3.

[8] S. Falcon, On the k-Lucas numbers of arithmetic indexes, Appl. Math. 3 (2012), 1202– 1206.

[9] S. Falcon and A. Plaza, The k-Fibonacci sequence and the Pascal 2-triangle, Chaos Solitons Fractals 33 (2007), 38–49.

[10] S. Halici and Z. Akyuz, On Some formulae for bivariate Pell polynomials, Far East J. Appl. Math. 41 (2010), 101–110.

[11] D. Jhala, G. P. S. Rathore, and K. Sisodiya, Some properties of k-Jacobsthal numbers with arithmetic indexes, Turkish J. Analysis Numb. Theo. 2 (2014), 119–124.

[12] D. Jhala, G. P. S. Rathore, and B. Singh, Some identities involving common factors of k-Fibonacci and k-Lucas numbers, American J. Math. Anal. 2 (2014), 33–35.

[13] B. K. Patel, N. Irmak, and P. K. Ray, Incomplete balancing and Lucas-balancing numbers, Math. Rep. 20 (2018), 59–72.

[14] K. Uslu and V. Deniz, Some identities of k-Mersenne numbers, Adv. Appl. Disc. Math. 4 (2017), 413–423.

[15] H. Zerroug, A. Boussayoud, B. Aloui, and M. Kerada, Gaussian Padovan, Gaussian Pell-Padovan numbers and new generating functions with some numbers and polynomials, J. Inf. Optim. Sci. 43 (2022), 745–760.

[16] T. Zhang and Y. Ma, On generalized Fibonacci polynomials and Bernoulli numbers, J. Integer Sequences 8 (2005), Article 05.5.3.

2020 Mathematics Subject Classification: Primary 05E05; Secondary 11B39. Keywords: Bernoulli number, Bernoulli polynomial, Euler number, Euler polynomial, convolution, generating function.

(Concerned with sequences \underline{A000032}, \underline{A000045}, \underline{A000364}, \underline{A027641}, and \underline{A027642}.)

Received December 12 2022; revised versions December 13 2022; February 17 2023; February 20 2023. Published in Journal of Integer Sequences, February 24 2023.

Return to Journal of Integer Sequences home page.