A Note on Fibonacci & Lucas and Bernoulli & Euler Polynomials

Claudio de Jesús Pita Ruiz Velasco Universidad Panamericana Mexico City, Mexico

                        cpita@up.edu.mx

Abstract We study certain polynomials P_{m} (x, y; t) and Q_{m} (x, y; t) of the variable t whose coefficients involve bivariate Fibonacci polynomials F_{j} (x, y) or bivariate Lucas polynomials L_{j} (x, y). By working with P_{m} (x, y; tx) and Q_{m} (x, y; tx), together with the generating functions for Bernoulli polynomials B_{i} (t) and Euler polynomials E_{i} (t), we obtain a list of eight identities connecting F_{j} (x, y) or L_{j} (x, y) with B_{i} (t) or E_{i} (t). We present also some consequences of these results.

1 Introduction

We use ℕ for the natural numbers and ℕ′ for ℕ ∪ {0}.

We recall now some definitions and basic facts of the main mathematical objects involved in this work, namely Fibonacci and Lucas numbers and polynomials (see [6] and [8]), and Bernoulli and Euler numbers and polynomials (see [4]).

We follow the standard notation F_{n} (x, y) and L_{n} (x, y) for the sequences of bivariate Fibonacci and Lucas polynomials, defined by the recurrences F_{n+2} (x, y) = xF_{n−1} (x, y) + yF_{n} (x, y), F_{0} (x, y) = 0, F_{1} (x, y) = 1, and L_{n+2} (x, y) = xL_{n−1} (x, y)+yL_{n} (x, y), L_{0} (x, y) = 2, L_{1} (x, y) = x, respectively, and extended to n ∈ ℤ as F_{−n} (x, y) = − (−y)^{−n} F_{n} (x, y) and L_{−n} (x, y) = (−y)^{−n} L_{n} (x, y). Plainly we have F_{n} (1, 1) = F_{n} and L_{n} (1, 1) = L_{n}, the Fibonacci and Lucas number sequences (\underline{A000045} and \underline{A000032} of Sloane’s Encyclopedia). Some bivariate Fibonacci polynomials are F_{2} (x, y) = x, F_{3} (x, y) = x^{2} + y, F_{4} (x, y) = x^{3} +2xy, F_{5} (x, y) = x^{4} +3x^{2}y +y^{2},…, and some bivariate Lucas polynomials are L_{2} (x, y) = x^{2} + 2y, L_{3} (x, y) = x^{3} + 3xy, L_{4} (y) = x^{4} + 4x^{2}y + 2y^{2}, L_{5} (y) = x^{5} + 5x^{3}y + 5xy^{2}, …. We will use extensively Binet’s formulas (without further comments):

F_{n} (x, y) = \frac{1}{\sqrt{x^{2} + 4y}} (α^{n} (x, y) − β^{n} (x, y)) and L_{n} (x, y) = α^{n} (x, y) + β^{n} (x, y) , (1)

where

α (x, y) = \frac{1}{2} (x + \sqrt{x^{2} + 4y}) and β (x, y) = \frac{1}{2} (x − \sqrt{x^{2} + 4y}),  (2)

together with the basic facts α (x, y) + β (x, y) = x and α (x, y) β (x, y) = −y. We will use also the following explicit formulas for bivariate Fibonacci and Lucas polynomials:

F_{n} (x, y) = \begin{matrix}⌊ & \frac{n−1}{2} & ⌋ \\ ∑ \\ k=0 \end{matrix} \binom{n − 1 − k}{k}x^{n−1−2k}y^{k} and L_{n} (x, y) = \begin{matrix}⌊ & \frac{n}{2} & ⌋ \\ ∑ \\ k=0 \end{matrix} \frac{n}{n − k} \binom{n − k}{k}x^{n−2k}y^{k}, (3)

(the first formula is valid for n ∈ ℕ′, and the second one is valid for n ∈ ℕ).

We will be working with Bernoulli and Euler polynomials, which can be defined as

B_{n} (t) = ∑_{j=0}^{n} \binom{n}{j}B_{j}t^{n−j} and E_{n} (t) = ∑_{j=0}^{n} \binom{n}{j}\frac{E_{j}}{2^{j}} (t − \frac{1}{2})^{n−j} ,  (4)

where B_{j} and E_{j} are the Bernoulli and Euler numbers, respectively, defined by the generating functions

\frac{z}{e^{z} − 1} = ∑_{j=0}^{∞} B_{j} \frac{z^{j}}{j!} and \frac{2e^{z}}{e^{2z} + 1} = ∑_{j=0}^{∞} E_{j} \frac{z^{j}}{j!} ,  (5)

The corresponding generating functions for Bernoulli and Euler polynomials are

\frac{ze^{zt}}{e^{z} − 1} = ∑_{j=0}^{∞} B_{j} (t) \frac{z^{j}}{j!} and \frac{2e^{zt}}{e^{z} + 1} = ∑_{n=0}^{∞} E_{n} (t) \frac{z^{n}}{n!} .  (6)

It is not difficult to see that for j ∈ ℕ, one has B_{2j+1} = 0 and E_{2j−1} = 0 (odd Bernoulli numbers are zero, except B_{1} = −\frac{1}{2}, and odd Euler numbers are zero). Also we have B_{0} = E_{0} = 1. Some other values are B_{2} = \frac{1}{6}, B_{4} = −\frac{1}{30}, B_{6} = \frac{1}{42}, B_{8} = −\frac{1}{30}, … and E_{2} = −1, E_{4} = 5, E_{6} = −61, E_{8} = 1385, …. The first Bernoulli polynomials are B_{0} (t) = 1, B_{1} (t) = t − \frac{1}{2}, B_{2} (t) = t^{2} − t + \frac{1}{6}, B_{3} (t) = t^{3} − \frac{3}{2}t^{2} + \frac{1}{2}t, B_{4} (t) = t^{4} − 2t^{3} + t^{2} − \frac{1}{30}, …, and the first Euler polynomials are E_{0} (t) = 1, E_{1} (t) = t− \frac{1}{2}, E_{2} (t) = t^{2} −t, E_{3} (t) = t^{3} − \frac{3}{2}t^{2} + \frac{1}{4}, E_{4} (t) = t^{4} − 2t^{3} + t, …. One can see easily that for n ∈ ℕ, one has B′_{n} (t) = nB_{n−1} (t) and E′_{n} (t) = nE_{n−1} (t).

Besides the trivial fact B_{n} (0) = B_{n}, we will use that

B_{n} (\frac{1}{2}) = (2^{1−n} − 1) B_{n} , E_{n} (\frac{1}{2}) = 2^{−n}E_{n}.  (7)

E_{n} (0) = −\frac{2 (2^{n+1} − 1)}{n + 1}B_{n+1}.  (8)

There are interesting papers in the literature pursuing relations among Bernoulli and Euler numbers and/or polynomials (see the 2006 works of Sun and Pan [11, 12] and references therein). On the other hand, there are certainly much research establishing relations among Fibonacci and Lucas numbers or polynomials (see references in the books of Koshy [6] and Vajda [8]). But also there has been interest in finding connections between the mathematics of Bernoulli and Euler and the mathematics of Fibonacci and Lucas, and this interest is not new: in 1975 P. F. Byrd [1] obtains some nice formulas connecting Bernoulli, Fibonacci and Lucas numbers. We also have to mention the 1957 work of Kelisky [5], the 2005 work of T. Zhang and Y. Ma [10], and the nice papers of J. Cigler [2, 3], which contains some of the results of our corollary 3 in section 3. This article responds to the interest in exploring more about these kind of connections. We work with certain kind of Appel sequences of polynomials P_{n} (x, y; t) and Q_{n} (x, y; t) in the variable t, whose coefficients are in turn bivariate Fibonacci polynomials F_{m} (x, y) or bivariate Lucas polynomials L_{m} (x, y). By working with generating functions of Bernoulli and Euler polynomials, we establish some identities involving the polynomials P_{n} (x, y; xt) and Q_{n} (x, y; xt) together with Bernoulli polynomials B_{j} (t), Euler polynomials E_{j} (t), bivariate Fibonacci F_{k} (x, y) and bivariate Lucas L_{k} (x, y) polynomials. These identities are the main results of the work (proposition 2). In section 3 we obtain some corollaries from identities of section 2.

2 The main results

We begin with a lemma with two easy identities that we will need in the proof of the main results of this work. Lemma 1. The following identities hold

(1 − e^{−xz}) ∑_{n=0}^{∞} L_{n} (x, y) \frac{z^{n}}{n!} = 2 ∑_{n=0}^{∞} L_{2n+1} (x, y) \frac{z^{2n+1}}{(2n + 1)!}.  (9)

(1 + e^{−xz}) ∑_{n=0}^{∞} L_{n} (x, y) \frac{z^{n}}{n!} = 2 ∑_{n=0}^{∞} L_{2n} (x, y) \frac{z^{2n}}{(2n)!}.  (10)

Proof. We have

e^{α(x,y)z} + e^{β(x,y)z} = e^{(x−β(x,y))z} + e^{(x−α(x,y))z} = e^{xz} (e^{−α(x,y)z} + e^{−β(x,y)z}) ,

and then

e^{−xz} ∑_{n=0}^{∞} L_{n} (x, y) \frac{z^{n}}{n!} = ∑_{n=0}^{∞} L_{n} (x, y) \frac{(−z)^{n}}{n!} .

Thus

(1 − e^{−xz}) ∑_{n=0}^{∞}L_{n} (x, y) \frac{z^{n}}{n!} = ∑_{n=0}^{∞}L_{n} (x, y) \frac{z^{n}}{n!} − ∑_{n=0}^{∞}L_{n} (x, y) \frac{(−z)^{n}}{n!}

= 2 ∑_{n=0}^{∞}L_{2n+1} (x, y) \frac{z^{2n+1}}{(2n + 1)!},

which proves (9), and

(1 + e^{−xz}) ∑_{n=0}^{∞} L_{n} (x, y) \frac{z^{n}}{n!} = ∑_{n=0}^{∞} L_{n} (x, y) \frac{z^{n}}{n!} + ∑_{n=0}^{∞} L_{n} (x, y) \frac{(−z)^{n}}{n!}

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

which proves (10).

Let us consider the polynomials

Q_{n} (x, y; t) = ∑_{k=0}^{n} \binom{n}{k} (−1)^{k} L_{k} (x, y) t^{n−k},  (11)

which can be written as

Q_{n} (x, y; t) = (t − α (x, y))^{n} + (t − β (x, y))^{n} .  (12)

Observe that

∑_{n=0}^{∞} Q_{n} (x, y; tx) \frac{z^{n}}{n!} = ∑_{n=0}^{∞} (tx − α (x, y))^{n} \frac{z^{n}}{n!} + ∑_{n=0}^{∞} (tx − β (x, y))^{n} \frac{z^{n}}{n!}
= e^{(tx−α(x,y))z} + e^{(tx−β(x,y))z}
= e^{(tx−x+β(x,y))z} + e^{(tx−x+α(x,y))z}
= e^{(t−1)xz} (e^{β(x,y)z} + e^{α(x,y)z})
= e^{(t−1)xz} ∑_{n=0}^{∞} (α^{n} (x, y) + β^{n} (x, y)) \frac{z^{n}}{n!} .

That is, we have

∑_{n=0}^{∞} Q_{n} (x, y; tx) \frac{z^{n}}{n!} = e^{(t−1)xz} ∑_{n=0}^{∞} L_{n} (x, y) \frac{z^{n}}{n!} .  (13)

We can use (9) to write (13) as

∑_{n=0}^{∞} Q_{n} (x, y; tx) \frac{z^{n}}{n!} = 2\frac{e^{txz}}{e^{xz} − 1} ∑_{n=0}^{∞} L_{2n+1} (x, y) \frac{z^{2n+1}}{(2n + 1)!}.  (14)

By using the generating function for Bernoulli polynomials (6) we can write (14) as

∑_{n=0}^{∞} Q_{n} (x, y; tx) \frac{z^{n}}{n!} = 2 ∑_{j=0}^{∞} ∑_{n=0}^{∞} \binom{2n + j}{j}B_{j} (t) x^{j−1} \frac{L_{2n+1} (x, y)}{2n + 1} \frac{z^{2n+j}}{(2n + j)!}.  (15)

By equating the coefficients of similar powers of z in (15) we get

Q_{2m} (x, y; tx) = 2 ∑_{j=0}^{m} \binom{2m}{2j}B_{2j} (t) \frac{x^{2j−1}}{2m + 1 − 2j} L_{2m+1−2j} (x, y) ,  (16)

and

Q_{2m+1} (x, y; tx) = 2 ∑_{j=0}^{m} \binom{2m + 1}{2j + 1}B_{2j+1} (t) \frac{x^{2j}}{2m + 1 − 2j} L_{2m+1−2j} (x, y) .  (17)

Similarly, observe that

∑_{n=0}^{∞} Q_{n} (x, y; tx) \frac{(−z)^{n}}{n!} = ∑_{n=0}^{∞} (−tx + α (x, y))^{n} \frac{z^{n}}{n!} + ∑_{n=0}^{∞} (−tx + β (x, y))^{n} \frac{z^{n}}{n!}
= e^{(−tx+α(x,y))z} + e^{(−tx+β(x,y))z}
= e^{−txz} (e^{α(x,y)z} + e^{β(x,y)z})
= e^{−txz} ∑_{n=0}^{∞} (α^{n} (x, y) + β^{n} (x, y)) \frac{z^{n}}{n!} .

That is, we have

∑_{n=0}^{∞} Q_{n} (x, y; tx) \frac{(−z)^{n}}{n!} = e^{−txz} ∑_{n=0}^{∞} L_{n} (x, y) \frac{z^{n}}{n!} .  (18)

We can use (10) to write (18) as

∑_{n=0}^{∞} Q_{n} (x, y; tx) \frac{(−z)^{n}}{n!} = \frac{2e^{−txz}}{1 + e^{−xz}} ∑_{n=0}^{∞} L_{2n} (x, y) \frac{z^{2n}}{(2n)!}.  (19)

By using the generating functions of Euler polynomials (6), we can write (19) as

∑_{n=0}^{∞} Q_{n} (x, y; tx) \frac{(−z)^{n}}{n!} = ∑_{j=0}^{∞} ∑_{n=0}^{∞} \binom{2n + j}{j}E_{j} (t) (−x)^{j} L_{2n} (x, y) \frac{z^{2n+j}}{(2n + j)!}. (20)

By equating the coefficients of similar powers of z in (20) we obtain

Q_{2m} (x, y; tx) = ∑_{j=0}^{m} \binom{2m}{2j}E_{2j} (t) x^{2j}L_{2m−2j} (x, y) ,  (21)

and

Q_{2m+1} (x, y; tx) = ∑_{j=0}^{m} \binom{2m + 1}{2j + 1}E_{2j+1} (t) x^{2j+1}L_{2m−2j} (x, y) .  (22)

Thus, (16) and (21) give us

Q_{2m} (x, y; tx) = 2 ∑_{j=0}^{m} \binom{2m}{2j}B_{2j} (t) \frac{x^{2j−1}}{2m + 1 − 2j} L_{2m+1−2j} (x, y)  (23)

= ∑_{j=0}^{m} \binom{2m}{2j}E_{2j} (t) x^{2j}L_{2m−2j} (x, y) ,

and (17) and (22) give us

Q_{2m+1} (x, y; tx) = 2 ∑_{j=0}^{m} \binom{2m + 1}{2j + 1}B_{2j+1} (t) \frac{x^{2j}}{2m + 1 − 2j} L_{2m+1−2j} (x, y)  (24)

= ∑_{j=0}^{m} \binom{2m + 1}{2j + 1}E_{2j+1} (t) x^{2j+1}L_{2m−2j} (x, y) .

If we consider now the polynomials

P_{n} (x, y; t) = ∑_{k=0}^{n} \binom{n}{k} (−1)^{k+1} F_{k} (x, y) t^{n−k}  (25)
= −\frac{1}{\sqrt{x^{2} + 4y}} ((t − α (x, y))^{n} − (t − β (x, y))^{n}) ,

it is possible to establish similar results to (23) and (24) (involving Fibonacci polynomials F_{k} (x, y) instead of Lucas polynomials L_{k} (x, y)). We describe the main steps of the procedure to obtain these new results and leave the reader to complete the details of the proofs. Firstly one proves that

(1 − e^{−xz}) ∑_{n=0}^{∞} F_{n} (x, y) \frac{z^{n}}{n!} = 2 ∑_{n=0}^{∞} F_{2n} (x, y) \frac{z^{2n}}{(2n)!}.  (26)

(1 + e^{−xz}) ∑_{n=0}^{∞} F_{n} (x, y) \frac{z^{n}}{n!} = 2 ∑_{n=0}^{∞} F_{2n+1} (x, y) \frac{z^{2n+1}}{(2n + 1)!}.  (27)

(Similar to (9) and (10).) Secondly one proves that

∑_{n=0}^{∞} P_{n} (x, y; tx) \frac{z^{n}}{n!} = e^{(t−1)xz} ∑_{n=0}^{∞} F_{n} (x, y) \frac{z^{n}}{n!} ,  (28)

then uses (26) to write this expression as

∑_{n=0}^{∞} P_{n} (x, y; tx) \frac{z^{n}}{n!} = 2\frac{e^{txz}}{e^{xz} − 1} ∑_{n=0}^{∞} F_{2n} (x, y) \frac{z^{2n}}{(2n)!},  (29)

and finally uses the generating function for Bernoulli polynomials (6) to write (29) as

∑_{n=0}^{∞} P_{n} (x, y; tx) \frac{z^{n}}{n!} = ∑_{n=0}^{∞} ∑_{j=0}^{∞} \binom{2n + 1 + j}{j}B_{j} (t) \frac{F_{2n+2} (x, y)}{n + 1}x^{j−1} \frac{z^{2n+j+1}}{(2n + 1 + j)!}. (30)

By equating the coefficients of similar powers of z in (30) one obtains that

P_{2m+1} (x, y; tx) = ∑_{j=0}^{m} \binom{2m + 1}{2j}B_{2j} (t) \frac{x^{2j−1}}{m + 1 − j} F_{2m+2−2j} (x, y) ,  (31)

and

P_{2m} (x, y; tx) = ∑_{j=0}^{m−1} \binom{2m}{2j + 1}B_{2j+1} (t) \frac{x^{2j}}{m − j} F_{2m−2j} (x, y) .  (32)

On the other hand, one proves first that

∑_{n=0}^{∞} P_{n} (x, y; tx) \frac{(−z)^{n}}{n!} = −e^{−txz} ∑_{n=0}^{∞} F_{n} (x, y) \frac{z^{n}}{n!} ,  (33)

then uses (27) to write (33) as

∑_{n=0}^{∞} P_{n} (x, y; tx) \frac{(−z)^{n}}{n!} = −\frac{2e^{−txz}}{1 + e^{−xz}} ∑_{n=0}^{∞} F_{2n+1} (x, y) \frac{z^{2n+1}}{(2n + 1)!},  (34)

and finally uses the generating function of Euler polynomials (6) to write (34) as

∑_{n=0}^{∞} P_{n} (x, y; tx) \frac{(−z)^{n}}{n!} = − ∑_{j=0}^{∞} ∑_{n=0}^{∞} \binom{2n + 1 + j}{j}E_{j} (t) (−x)^{j} F_{2n+1} (x, y) \frac{z^{2n+j+1}}{(2n + 1 + j)!}.

(35)

By equating the coefficients of similar powers of z in (35) one gets that

P_{2m} (x, y; tx) = ∑_{j=0}^{m−1} \binom{2m}{2j + 1}E_{2j+1} (t) x^{2j+1}F_{2m−1−2j} (x, y) ,  (36)

and

P_{2m+1} (x, y; tx) = ∑_{j=0}^{m} \binom{2m + 1}{2j}E_{2j} (t) x^{2j}F_{2m+1−2j} (x, y) .  (37)

Thus, we have identities (32) and (36) similar to (23), namely

P_{2m} (x, y; tx) = ∑_{j=0}^{m−1} \binom{2m}{2j + 1}B_{2j+1} (t) \frac{x^{2j}}{m − j} F_{2m−2j} (x, y)  (38)

= ∑_{j=0}^{m−1} \binom{2m}{2j + 1}E_{2j+1} (t) x^{2j+1}F_{2m−1−2j} (x, y) ,

and identities (31) and (37) similar to (24), namely

P_{2m+1} (x, y; tx) = ∑_{j=0}^{m} \binom{2m + 1}{2j}B_{2j} (t) \frac{x^{2j−1}}{m + 1 − j} F_{2m+2−2j} (x, y)  (39)

= ∑_{j=0}^{m} \binom{2m + 1}{2j}E_{2j} (t) x^{2j}F_{2m+1−2j} (x, y) .

In the following proposition we collect the main identities (23), (24), (38) and (39) obtained in this section, which are the main results of this work.

Proposition 2. The following identities hold

∑_{k=0}^{2m}\binom{2m}{k}(−1)^{k+1}F_{k} (x, y)(tx)^{2m−k} = ∑_{j=0}^{m−1} \binom{2m}{2j + 1}B_{2j+1} (t) \frac{x^{2j}}{m − j} F_{2m−2j} (x, y) (40)

= ∑_{j=0}^{m−1} \binom{2m}{2j + 1}E_{2j+1} (t) x^{2j+1}F_{2m−1−2j} (x, y) .

∑_{k=0}^{2m+1}\binom{2m+1}{k}(−1)^{k+1}F_{k}(x, y)(tx)^{2m+1−k} = ∑_{j=0}^{m}\binom{2m+1}{2j}\frac{B_{2j}(t) x^{2j−1}}{m+1−j}F_{2m+2−2j}(x, y) (41)

= ∑_{j=0}^{m} \binom{2m + 1}{2j}E_{2j} (t) x^{2j}F_{2m+1−2j} (x, y) .

∑_{k=0}^{2m}\binom{2m}{k}(−1)^{k}L_{k}(x, y)(tx)^{2m−k} = 2 ∑_{j=0}^{m} \binom{2m}{2j}\frac{B_{2j} (t) x^{2j−1}}{2m + 1 − 2j} L_{2m+1−2j} (x, y) (42)

= ∑_{j=0}^{m} \binom{2m}{2j}E_{2j} (t) x^{2j}L_{2m−2j} (x, y) .

∑_{k=0}^{2m+1}\binom{2m+1}{k}(−1)^{k}L_{k}(x, y)(tx)^{2m+1−k} = 2∑_{j=0}^{m}\binom{2m+1}{2j+1}\frac{B_{2j+1}(t) x^{2j}}{2m+1−2j} L_{2m+1−2j}(x, y) (43)

= ∑_{j=0}^{m} \binom{2m + 1}{2j + 1}E_{2j+1} (t) x^{2j+1}L_{2m−2j} (x, y) .

We want to mention that (40) and (41) can be obtained as consequences of (43) and (42), respectively. Indeed, if we use the well-known fact that \frac{∂}{∂y}L_{s} (x, y) = sF_{s−1} (x, y) (see [9]) and take the derivative with respect to y in (43), we get

∑_{k=1}^{2m+1} \binom{2m + 1}{k} (−1)^{k} kF_{k−1} (x, y) (tx)^{2m+1−k}

= 2 ∑_{j=0}^{m−1} \binom{2m + 1}{2j + 1}B_{2j+1} (t) x^{2j}F_{2m−2j} (x, y)

= ∑_{j=0}^{m−1} \binom{2m + 1}{2j + 1}E_{2j+1} (t) x^{2j+1} (2m − 2j) F_{2m−1−2j} (x, y) ,

which is (40) (after some easy simplifications). Similarly, the derivative with respect to y of (42) is

∑_{k=1}^{2m} \binom{2m}{k} (−1)^{k} kF_{k−1} (x, y) (tx)^{2m−k} = 2 ∑_{j=0}^{m−1} \binom{2m}{2j}B_{2j} (t) x^{2j−1}F_{2m−2j} (x, y)

= ∑_{j=0}^{m−1} \binom{2m}{2j}E_{2j} (t) x^{2j} (2m − 2j) F_{2m−1−2j} (x, y) ,

which, after some simplifications, becomes (41).

3 Some Corollaries

In this section we obtain some other identities that are contained in our main results (40) to (43). Corollary 3. The following identities hold

F_{2m} (x, y) = ∑_{j=0}^{m−1} \binom{2m}{2j + 1}\frac{(4^{j+1} − 1) B_{2j+2}}{j + 1} x^{2j+1}F_{2m−1−2j} (x, y) .  (44)

F_{2m+1} (x, y) = ∑_{j=0}^{m} \binom{2m + 1}{2j}\frac{B_{2j}}{m + 1 − j} x^{2j−1}F_{2m+2−2j} (x, y) .  (45)

L_{2m} (x, y) = 2 ∑_{j=0}^{m} \binom{2m}{2j}\frac{B_{2j}}{2m + 1 − 2j} x^{2j−1}L_{2m+1−2j} (x, y) .  (46)

L_{2m+1} (x, y) = ∑_{j=0}^{m} \binom{2m + 1}{2j + 1}\frac{(4^{j+1} − 1) B_{2j+2}}{j + 1} x^{2j+1}L_{2m−2j} (x, y) .  (47)

Proof. If we set t = 0 in (40) we obtain (by using (8))

− F_{2m} (x, y) = ∑_{j=0}^{m−1} \binom{2m}{2j + 1}B_{2j+1} \frac{x^{2j}}{m − j} F_{2m−2j} (x, y)

= − ∑_{j=0}^{m−1} \binom{2m}{2j + 1}\frac{2 (2^{2j+2} − 1)}{2j + 2}x^{2j+1}F_{2m−1−2j} (x, y) .

The first equality is trivial. The second one is (44). Similarly, by setting t = 0 in (41),

(42) and (43), and using (8), we obtain (45), (46) and (47), respectively.

Formulas (44) to (47) express even (odd) indexed bivariate Fibonacci and Lucas polynomials as linear combinations of odd (even, respectively) indexed polynomials of the same kind. Some versions of these results appear in [2] (see also [3]). Corollary 4. The following identities hold

∑_{j=0}^{m} \binom{2m + 1}{2j} (2^{1−2j} − 1) \frac{B_{2j}}{m + 1 − j} x^{2j−1}F_{2m+2−2j} (x, y)

= ∑_{j=0}^{m} \binom{2m}{2j} (2^{1−2j} − 1) \frac{B_{2j}}{2m + 1 − 2j} x^{2j−1}L_{2m+1−2j} (x, y)

= ∑_{j=0}^{m} \binom{2m + 1}{2j}2^{−2j}E_{2j}x^{2j}F_{2m+1−2j} (x, y)

= \frac{1}{2} ∑_{j=0}^{m} \binom{2m}{2j}2^{−2j}E_{2j}x^{2j}L_{2m−2j} (x, y)

= \frac{1}{2^{2m}} (x^{2} + 4y)^{m} .  (48)

Proof. Observe that (from (25))

P_{n} (x, y; \frac{x}{2}) = −\frac{1}{\sqrt{x^{2} + 4y}} ((\frac{x}{2} − α (x, y))^{n} − (\frac{x}{2} − β (x, y))^{n})
= \frac{1 − (−1)^{n}}{2^{n}} (x^{2} + 4y)^{\frac{n−1}{2}} ,

and (from (12))

Q_{n} (x, y; \frac{x}{2}) = (\frac{x}{2} − α (x, y))^{n} + (\frac{x}{2} − β (x, y))^{n}
= \frac{1 + (−1)^{n}}{2^{n}} (x^{2} + 4y)^{\frac{n}{2}} .

Then, by setting t = \frac{1}{2} in (41) and (42) (and using (7) and (8)) we obtain that

P_{2m+1} (x, y; \frac{x}{2}) = \frac{1}{2^{2m}} (x^{2} + 4y)^{m}
= ∑_{j=0}^{m} \binom{2m + 1}{2j} (2^{1−2j} − 1) \frac{B_{2j}}{m + 1 − j} x^{2j−1}F_{2m+2−2j} (x, y)

= ∑_{j=0}^{m} \binom{2m + 1}{2j}2^{−2j}E_{2j}x^{2j}F_{2m+1−2j} (x, y) ,

and

Q_{2m} (x, y; \frac{x}{2}) = \frac{2}{2^{2m}} (x^{2} + 4y)^{m}
= 2 ∑_{j=0}^{m} \binom{2m}{2j} (2^{1−2j} − 1) \frac{B_{2j}}{2m + 1 − 2j} x^{2j−1}L_{2m+1−2j} (x, y)

= ∑_{j=0}^{m} \binom{2m}{2j}2^{−2j}E_{2j}x^{2j}L_{2m−2j} (x, y) ,

respectively. These are identities (48). (Note that identities (40) and (43) produce only trivial cases with t = \frac{1}{2}.)

In the rest of this section we write ξ (x, y) to denote any of α (x, y) or β (x, y). When we write an expression involving ξ (x, y) together with the plus-minus sign ±, we understand that the plus sign corresponds to the case ξ = α, and the minus sign corresponds to the case ξ = β.

Corollary 5. The following identities hold

∑_{j=0}^{m} \binom{2m + 1}{2j}B_{2j} (\frac{ξ (x, y)}{x}) \frac{x^{2j−1}}{m + 1 − j} F_{2m+2−2j} (x, y)

= 2 ∑_{j=0}^{m} \binom{2m}{2j}B_{2j} (\frac{ξ (x, y)}{x}) \frac{x^{2j−1}}{2m + 1 − 2j} L_{2m+1−2j} (x, y)
= ∑_{j=0}^{m} \binom{2m + 1}{2j}E_{2j} (\frac{ξ (x, y)}{x}) x^{2j}F_{2m+1−2j} (x, y)
= ∑_{j=0}^{m} \binom{2m}{2j}E_{2j} (\frac{ξ (x, y)}{x}) x^{2j}L_{2m−2j} (x, y)
= (x^{2} + 4y)^{m} .  (49)

(x^{2} + 4y) ∑_{j=0}^{m−1} \binom{2m}{2j + 1}B_{2j+1} (\frac{ξ (x, y)}{x}) \frac{x^{2j}}{m − j} F_{2m−2j} (x, y)

= 2 ∑_{j=0}^{m} \binom{2m + 1}{2j + 1}B_{2j+1} (\frac{ξ (x, y)}{x}) \frac{x^{2j}}{2m + 1 − 2j} L_{2m+1−2j} (x, y)
= (x^{2} + 4y) ∑_{j=0}^{m−1} \binom{2m}{2j + 1}E_{2j+1} (\frac{ξ (x, y)}{x}) x^{2j+1}F_{2m−1−2j} (x, y)
= ∑_{j=0}^{m} \binom{2m + 1}{2j + 1}E_{2j+1} (\frac{ξ (x, y)}{x}) x^{2j+1}L_{2m−2j} (x, y)
= ± (x^{2} + 4y)^{\frac{2m+1}{2}} .  (50)

Proof. If we set t = α (x, y) in (25) and (12) we get

P_{n} (x, y; α (x, y)) = (x^{2} + 4y)^{\frac{n−1}{2}} ,

and

Q_{n} (x, y; α (x, y)) = (x^{2} + 4y)^{\frac{n}{2}} .

Similarly, with t = β (x, y) we get

P_{n} (x, y; β (x, y)) = (−1)^{n+1} (x^{2} + 4y)^{\frac{n−1}{2}} ,

and

Q_{n} (x, y; β (x, y)) = (−1)^{n} (x^{2} + 4y)^{\frac{n}{2}} .

Thus, we have P_{2m+1} (α (x, y)) = P_{2m+1} (β (x, y)) = Q_{2m} (α (x, y)) = Q_{2m} (β (x, y)) = (x^{2} + 4y)^{m} , and then identity (49) is obtained by setting t = \frac{α(x,y)}{x} and t = \frac{β(x,y)}{x} in (41) and (42).

In the same way we obtain (50) by setting t = \frac{α(x,y)}{x} and t = \frac{β(x,y)}{x} in identities (40) and (43).

Corollary 6. (a) For r = 0, 1, …, m, the following identities hold

∑_{j=0}^{m−1−r} \binom{2m}{2j + 1}\binom{2m − 2j − 1 − r}{r}\frac{1}{m − j} B_{2j+1} (t)  (51)

= ∑_{j=0}^{m−1−r} \binom{2m}{2j + 1}\binom{2m − 2j − 2 − r}{r}E_{2j+1} (t)

= ∑_{j=0}^{2m−1−r} \binom{2m}{j}\binom{2m − j − 1 − r}{r} (−1)^{j+1} t^{j}.

∑_{j=0}^{m−r} \binom{2m + 1}{2j}\binom{2m − 2j + 1 − r}{r}\frac{1}{m + 1 − j} B_{2j} (t)  (52)
= ∑_{j=0}^{m−r} \binom{2m + 1}{2j}\binom{2m − 2j − r}{r}E_{2j} (t)

= ∑_{j=0}^{2m−r} \binom{2m + 1}{j}\binom{2m − j − r}{r} (−1)^{j} t^{j}.

(b) For r = 1, 2, …, m, the following identities hold

2 ∑_{j=0}^{m−r} \binom{2m}{2j}\binom{2m + 1 − 2j − r}{r}\frac{1}{2m + 1 − 2j − r}B_{2j} (t)  (53)
= ∑_{j=0}^{m−r} \binom{2m}{2j}\binom{2m − 2j − r}{r}\frac{2m − 2j}{2m − 2j − r}E_{2j} (t)

= ∑_{j=2r}^{2m} \binom{2m}{j}\binom{j − r}{r}\frac{(−1)^{j} j}{j − r} t^{2m−j}.

2 ∑_{j=0}^{m−r} \binom{2m + 1}{2j + 1}\binom{2m + 1 − 2j − r}{r}\frac{1}{2m + 1 − 2j − r}B_{2j+1} (t)  (54)
= ∑_{j=0}^{m−r} \binom{2m + 1}{2j + 1}\binom{2m − 2j − r}{r}\frac{2m − 2j}{2m − 2j − r}E_{2j+1} (t)

= ∑_{j=2r}^{2m+1} \binom{2m + 1}{j}\binom{j − r}{r}\frac{(−1)^{j} j}{j − r} t^{2m+1−j}.

Proof. (a) First substitute the explicit formula (3) for F_{n} (x, y) in (40) and (41), then equate the coefficients of similar terms x^{2m+1−2r}y^{r}, r = 0, 1, …, m to obtain (51) and (52), respectively.

(b) First substitute the explicit formula (3) for L_{n} (x, y) in (42) and (43), then equate the coefficients of similar terms x^{2m+1−2r}y^{r}, r = 1, 2, …, m to obtain (53) and (54), respectively.

Fibonacci and Lucas polynomials are hidden in identities (51) to (54). To let them appear we just have to write the right-hand side polynomials of these identities in a special form. The following lemma tells us how to do this.

Lemma 7. (a) For r = 0, 1, …, m we have

∑_{j=0}^{2m−1−r} \binom{2m}{j}\binom{2m − j − 1 − r}{r} (−1)^{j+1} t^{j}  (55)

= (−1)^{r+1} ∑_{j=0}^{r} \binom{2m}{j}\binom{2r − j}{r} ((t − 1)^{2m−j} + (−1)^{j+1} t^{2m−j}) ,

and

∑_{j=0}^{2m−r} \binom{2m + 1}{j}\binom{2m − j − r}{r} (−1)^{j} t^{j}  (56)

= (−1)^{r+1} ∑_{j=0}^{r} \binom{2m + 1}{j}\binom{2r − j}{r} ((t − 1)^{2m+1−j} + (−1)^{j+1} t^{2m+1−j}) .

(b) For r = 1, 2, …, m we have

∑_{j=2r}^{2m} \binom{2m}{j}\binom{j − r}{r}\frac{(−1)^{j} j}{j − r} t^{2m−j}  (57)

= (−1)^{r} ∑_{j=1}^{r} \binom{2m}{j}\binom{2r − j − 1}{r − 1}\frac{j}{r} ((t − 1)^{2m−j} + (−1)^{j} t^{2m−j}) ,

and

∑_{j=2r}^{2m+1} \binom{2m + 1}{j}\binom{j − r}{r}\frac{(−1)^{j} j}{j − r} t^{2m+1−j}  (58)

= (−1)^{r} ∑_{j=1}^{r} \binom{2m + 1}{j}\binom{2r − j − 1}{r − 1}\frac{j}{r} ((t − 1)^{2m+1−j} + (−1)^{j} t^{2m+1−j}) .

Proof. We present only the proof of (55). The corresponding proofs of (56), (57) and (58) are similar and left to the reader.

We have

∑_{j=0}^{r} \binom{2m}{j}\binom{2r − j}{r} ((t − 1)^{2m−j} + (−1)^{j+1} t^{2m−j})
= ∑_{j=0}^{r}\binom{2m}{j}\binom{2r − j}{r} ∑_{k=0}^{2m−j}\binom{2m − j}{k}(−1)^{k} t^{2m−j−k} + ∑_{j=0}^{r}\binom{2m}{j}\binom{2r − j}{r}(−1)^{j+1} t^{2m−j}
= ∑_{j=0}^{2m} ∑_{i=0}^{min(r,j)}\binom{2m}{i}\binom{2r − i}{r}\binom{2m − i}{j − i}(−1)^{j−i} t^{2m−j} + ∑_{j=0}^{r}\binom{2m}{j}\binom{2r − j}{r}(−1)^{j+1} t^{2m−j}
= ∑_{j=0}^{r} \binom{2m}{j} (∑_{i=0}^{j} \binom{2r − i}{r}\binom{j}{i} (−1)^{i} − \binom{2r − j}{r}) (−1)^{j} t^{2m−j}
+ ∑_{j=r+1}^{2m} \binom{2m}{j} (∑_{i=0}^{r} \binom{2r − i}{r}\binom{j}{i} (−1)^{i}) (−1)^{j} t^{2m−j}.  (59)

For j = 0, 1, …, r we have

∑_{i=0}^{j} \binom{2r − i}{r}\binom{j}{i} (−1)^{i} = \binom{2r − j}{r},

and for j = r + 1, …, 2m we have

∑_{i=0}^{r} \binom{2r − i}{r}\binom{j}{i} (−1)^{i} = \binom{j − 1 − r}{r} (−1)^{r} .

(See [7], identity (3.50), p. 65.) Thus (59) can be written as

∑_{j=0}^{r}\binom{2m}{j}\binom{2r − j}{r}((t − 1)^{2m−j} + (−1)^{j+1} t^{2m−j}) = ∑_{j=r+1}^{2m}\binom{2m}{j}\binom{j − 1 − r}{r}(−1)^{r+j} t^{2m−j},

and finally we can write the right-hand side of (55) as

(−1)^{r+1} ∑_{j=0}^{r} \binom{2m}{j}\binom{2r − j}{r} ((t − 1)^{2m−j} + (−1)^{j+1} t^{2m−j})
= (−1)^{r+1} ∑_{j=r+1}^{2m} \binom{2m}{j}\binom{j − 1 − r}{r} (−1)^{r} (−1)^{j} t^{2m−j}
= ∑_{j=r+1}^{2m} \binom{2m}{j}\binom{j − 1 − r}{r} (−1)^{j+1} t^{2m−j}
= ∑_{j=0}^{2m−1−r} \binom{2m}{j}\binom{2m − j − 1 − r}{r} (−1)^{j+1} t^{j},

as wanted.

Corollary 8. (a) For r = 0, 1, …, m we have that

∑_{j=0}^{m−1−r} \binom{2m}{2j + 1}\binom{2m − 2j − 1 − r}{r}\frac{1}{m − j} B_{2j+1} (\frac{ξ (x, y)}{x})  (60)

= ∑_{j=0}^{m−1−r} \binom{2m}{2j + 1}\binom{2m − 2j − 2 − r}{r}E_{2j+1} (\frac{ξ (x, y)}{x})

= ± (−1)^{r+1} \sqrt{x^{2} + 4y} ∑_{j=0}^{r} \binom{2m}{j}\binom{2r − j}{r}\frac{(−1)^{j+1}}{x^{2m−j}} F_{2m−j} (x, y) ,

and

∑_{j=0}^{m−r} \binom{2m + 1}{2j}\binom{2m − 2j + 1 − r}{r}\frac{1}{m + 1 − j} B_{2j} (\frac{ξ (x, y)}{x})  (61)

= ∑_{j=0}^{m−r} \binom{2m + 1}{2j}\binom{2m − 2j − r}{r}E_{2j} (\frac{ξ (x, y)}{x})

= (−1)^{r+1} ∑_{j=0}^{r} \binom{2m + 1}{j}\binom{2r − j}{r}\frac{(−1)^{j+1}}{x^{2m+1−j}} L_{2m+1−j} (x, y) .

(b) For r = 1, 2, …, m we have that

2 ∑_{j=0}^{m−r} \binom{2m}{2j}\binom{2m + 1 − 2j − r}{r}\frac{1}{2m + 1 − 2j − r}B_{2j} (\frac{ξ (x, y)}{x})  (62)

= ∑_{j=0}^{m−r} \binom{2m}{2j}\binom{2m − 2j − r}{r}\frac{2m − 2j}{2m − 2j − r}E_{2j} (\frac{ξ (x, y)}{x})

= \frac{(−1)^{r}}{r} ∑_{j=1}^{r} \binom{2m}{j}\binom{2r − j − 1}{r − 1}\frac{j (−1)^{j}}{x^{2m−j}} L_{2m−j} (x, y) ,

and

2 ∑_{j=0}^{m−r} \binom{2m + 1}{2j + 1}\binom{2m + 1 − 2j − r}{r}\frac{1}{2m + 1 − 2j − r}B_{2j+1} (\frac{ξ (x, y)}{x}) (63)

= ∑_{j=0}^{m−r} \binom{2m + 1}{2j + 1}\binom{2m − 2j − r}{r}\frac{2m − 2j}{2m − 2j − r}E_{2j+1} (\frac{ξ (x, y)}{x})

= ±\frac{(−1)^{r}}{r} \sqrt{x^{2} + 4y} ∑_{j=1}^{r} \binom{2m + 1}{j}\binom{2r − j − 1}{r − 1}\frac{j (−1)^{j}}{x^{2m+1−j}} F_{2m+1−j} (x, y) .

Proof. By using (55), (56), (57) and (58), rewrite identities (51), (52), (53) and (54), respectively. Set t = \frac{α(x,y)}{x} and t = \frac{β(x,y)}{x} in the resulting identities, to obtain (60), (61), (62) and (63), respectively.

4 Acknowledgements

The first version of this work contained some versions of identities (40) to (43), which were demonstrated by means of Fourier expansions of certain periodic extensions of certain restrictions of the polynomials P_{n} (x, 1; t) and Q_{n} (x, 1; t). In this version we present a larger list of identities, and the main ideas of the corresponding proofs are different (now the proofs follow to Cigler [2]). I thank the referee for call my attention to this way of proving those identities. I also thank him/her for the additional comments in his/her report, that certainly helped me to present a more readable version of the article.

References

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

[2] Johann Cigler, Fibonacci polynomials, generalized Stirling numbers, and Bernoulli,

Genocchi and tangent numbers, http://arxiv.org/abs/1103.2610.

[3] Johann Cigler, q-Fibonacci polynomials and q-Genocchi numbers,

http://arxiv.org/abs/0908.1219.

[4] Karl Dilcher, Bernoulli and Euler Polynomials, Digital Library of Mathematical Func-

tions, Ch. 24, http://dlmf.nist.gov/24.

[5] Richard P. Kelisky, Congruences involving combinations of the Bernoulli and Fibonacci numbers, Proc. Nat. Acad. Sci. USA, 43 (1957), 1066–1069.

[6] Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley & Sons, Inc., 2001.

[7] Renzo Sprugnoli, Riordan Array Proofs of Identities in Gould’s Book, 2006. Available

at http://www.dsi.unifi.it/^{∼}resp/GouldBK.pdf.

[8] S. Vajda, Fibonacci and Lucas Numbers, and the Golden Section, Dover, 1989.

[9] Hongquan Yu and Chuanguang Liang, Identities involving partial derivatives of bivariate Fibonacci and Lucas polynomials, Fib. Quart. 35 (1997), 19–23.

[10] Tianping Zhang and Yuankui Ma, On generalized Fibonacci polynomials and Bernoulli numbers, J. Integer Seq., 8 (2005), Article 05.5.3.

[11] Zhi-Wei Sun and Hao Pan, Identities concerning Bernoulli and Euler polynomials, Acta Arith. 125 (2006), 21–39.

[12] Zhi-Wei Sun and Hao Pan, New identities concerning Bernoulli and Euler polynomials, J. Comb. Theory A. 113 (2006), 156–175.

2010 Mathematics Subject Classification: Primary 11B39; Secondary 11B68. Keywords: Fibonacci polynomial, Lucas polynomial, Bernoulli polynomial, Euler polynomial.

(Concerned with sequence \underline{A0xxxxx}.)

Received September 9 2011; revised versions received December 8 2011; January 13 2012. Published in Journal of Integer Sequences, January 14 2012.

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