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### Article

Kyungpook Mathematical Journal 2021; 61(3): 473-486

Published online September 30, 2021

### Lucas-Euler Relations Using Balancing and Lucas-Balancing Polynomials

Robert Frontczak, Taras Goy*

e-mail : robert.frontczak@lbbw.de

Faculty of Mathematics and Computer Science, Vasyl Stefanyk Precarpathian National University, Ivano-Frankivsk 76018, Ukraine
e-mail : taras.goy@pnu.edu.ua

Received: October 21, 2021; Revised: March 12, 2021; Accepted: March 23, 2021

We establish some new combinatorial identities involving Euler polynomials and balancing (Lucas-balancing) polynomials. The derivations use elementary techniques and are based on functional equations for the respective generating functions. From these polynomial relations, we deduce interesting identities with Fibonacci and Lucas numbers, and Euler numbers. The results must be regarded as companion results to some Fibonacci-Bernoulli identities, which we derived in our previous paper.

Keywords: Euler polynomials and numbers, Bernoulli numbers, balancing polynomials and numbers, Fibonacci numbers, generating function.

In 1975, Byrd [1] derived the following identity relating Lucas numbers to Euler numbers:

$∑ k=0 n/2n2k54kLn−2kE2k=21−n.$

In [18], Wang and Zhang obtained a more general result valid for j≥ 1 as follows

$∑ k=0 n/2n2k​54k​Fj2kLj(n−2k)E2k=21−nLjn.$

Castellanos [2] found

$∑ k=0n2n2k2−2k−1L2(n−k)jLj2kE2k=(54)nFj2n,$

which expresses even powers of Fibonacci numbers in terms of Lucas and Euler numbers.

Here, as usual, Fibonacci and Lucas numbers satisfy the recurrence relation $un=un−1+un−2$, n≥ 2, with initial conditions F0 = 0, F1 = 1 and L0 = 2, L1 = 1, respectively, whereas Euler numbers $(En)n≥0$ are given by the power series

$∑ n=0∞Enznn!=1coshz.$

Fibonacci and Lucas numbers are entries A000045 and A000032 in the On-Line Encyclopedia of Integer Sequences [17], respectively.

The Lucas-Euler pair may be regarded as the twin of the Fibonacci-Bernoulli pair. In the last years, there has been a growing interest in deriving new relations for these two pairs of sequences. For example, Zhang and Ma [21] proved a relation between Fibonacci polynomials and Bernoulli numbers $(Bn)n≥0$ defined by

$∑ n=0∞Bnznn!=zez−1.$

The following identity is a special case of their result:

$∑ k=0nnk5n−k2FkBn−k=nβn−1,$

where $β=(1−5)/2$, or, equivalently,

$∑ k=0 n/2n2k5kFn−2kB2k=nLn−12.$

See also [14, 18, 19, 20] for other results in this direction. Recently, Frontczak [5], Frontczak and Goy [7], and Frontczak and Tomovski [8] proved some generalizations of existing results. For instance, from [7] we have

$∑ k=0nnk(5Fj)n−kFjkBn−k=nFjβj(n−1),$

which holds for all j ≥ 1 and generalizes (1.4) to an arithmetic progression, and

$∑ k=0 n/2n2k(20k−5k)F2j2kL2j(n−2k)B2k=5n2F2jF2j(n−1).$

Note, since $B2n+1=0$ for n≥1, from (1.4) we get Kelisky's formula [9]

$∑ k=0 n/2n2k5kFj2kFj(n−2k)B2k=n2FjLj(n−1).$

In this paper, we present new identities linking Lucas numbers to Euler numbers (polynomials). The results stated are polynomial generalizations of (1.4) and are complements of the recent discoveries from [5, 7].

Throughout the paper, we will work with different kind of polynomials of a complex variable x: Euler polynomials $(En(x))n≥0$, Bernoulli polynomials $(Bn(x))n≥0$, balancing polynomials $(Bn*(x))n≥0$, and Lucas-balancing polynomials $(Cn(x))n≥0$.

Euler and Bernoulli polynomials are famous mathematical objects and are fairly well understood. They are defined by [3, Chapter 24]

$H(x,z)=∑ n=0∞Bn(x)znn!=zexzez−1 (|z|<2π)$

and

$I(x,z)=∑ n=0∞En(x)znn!=2exzez+1 (|z|<π).$

The numbers $Bn(0)=Bn$ are the famous Bernoulli numbers. Bernoulli numbers are rational numbers starting with B0=1, B1=-1/2, B2=1/6, B4=-1/30, B6=1/42, and so on. Also, as already mentioned, B2n+1=0 for n≥ 1. Euler numbers En are obtained from I(1/2,2z) that is

$En=2nEn(1/2).$

In contrast to Bernoulli numbers, Euler numbers are integers where E0=1, E2=-1, E4=5 and E2n+1=0 for n≥ 0. Explicit formulas for the polynomials are

$Bn(x)=∑ k=0 nnkBkxn−k and En(x)=∑ k=0 nnkEk2k(x−12)n−k.$

Euler polynomials can be expressed in terms of Bernoulli polynomials via

$En(x)=2n+1(Bn+1(x)−2n+1Bn+1(x2)).$

Particularly,

$En(0)=2(1−2n+1)n+1Bn+1.$

Balancing polynomials are of younger age and are introduced in the next section.

Balancing polynomials $Bn*(x)$ and Lucas-balancing polynomials Cn(x) are generalizations of balancing and Lucas-balancing numbers [4]. These polynomials satisfy the recurrence $wn(x)=6xwn−1(x)−wn−2(x)$, n≥ 2, but with the respectively initial conditions $B0*(x)=0$, $B1*(x)=1$ and C0(x)=1, C1(x)=3x. The Binet formulas for these polynomials are

$Bn*(x)=λn(x)−λ−n(x)29x2−1 and Cn(x)=λn(x)+λ−n(x)2,$

where $λ(x)=3x+9x2−1$. Also, the following explicit formulas hold [15, 16]

$Bn*(x)=∑k=0(n−1)/2(−1)kn−1−kk(6x)n−1−2k, n≥0,Cn(x)=n2∑k=0n/2(−1)kn−kn−kk(6x)n−2k, n≥1.$

Consult the papers [4, 6, 10, 11, 12, 13, 16] for more information about these polynomials. The numbers $Bn*(1)=Bn*$ and Cn(1)=Cn are called balancing and Lucas-balancing numbers, respectively. These numbers are indexed in [17] under entries A001109 and A001541.

Balancing and Lucas-balancing polynomials possess interesting properties. They are related to Chebyshev polynomials by simple scaling [4][Lemma 2.1]. The exponential generating functions for balancing and Lucas-balancing polynomials are derived in [4, 6]. Here, however, we will only need the results from [6]: Let $b1(x,z)$ and $b2(x,z)$ be the exponential generating functions of odd and even indexed balancing polynomials, respectively. Then

$b1(x,z)=∑n=0∞B 2n+1*(x)znn! =e (18x2−1)z 9x2−1(3xsinh(6x9x2−1z)+9x2−1cosh(6x9x2−1z))$

and

$b2(x,z)=∑ n=0∞B2n*(x)znn!=e(18x2−1)z9x2−1sinh(6x9x2−1z).$

Similarly, the exponential generating functions for Lucas-balancing polynomials are found to be

$c1(x,z)=∑n=0∞C 2n+1(x)znn! =e(18x2−1)z(3xcosh(6x9x2−1z)+9x2−1sinh(6x9x2−1z))$

and

$c2(x,z)=∑ n=0∞C2n(x)znn!=e(18x2−1)zcosh(6x9x2−1z).$

Connections between Bernoulli polynomials Bn(x) and balancing polynomials Bn*(x) have been established in the recent papers [5, 7]. They are interesting, as they instantly give relations between Bernoulli numbers and Fibonacci and Lucas numbers. The links are the following evaluations [4]

$Bn*(ωsLs6)=ωsn−1FsnFs, Cn(ωsLs6)=ωsnLsn2,$

where $ωs=1$, if s is even, and $ωs=i=−1$, if s is odd. These links will be used to prove our results.

### 3. Relations Between Euler and Balancing (Lucas-Balancing) Polynomials

We start with the following result involving even indexed balancing and Lucas-balancing polynomials.

Theorem 3.1. For each n≥ 1 and $x∈ℂ$, we have

$∑k=1n/2n−12k−1C2(n−2k)(x)(144x2(9x2−1))kE2k−1(0) =12x(1−9x2)B2n−2*(x).$

Proof. Since $tanhz=1−2e2z+1$, from (1.8) we get

$I(0,12x9x2−1z)=1−tanh(6x9x2−1z)$

and, by (2.2) and (2.4),

$∑n=0∞(∑k=0nnkC2(n−k)(12x9x2−1)kEk(0))znn! =∑n=0∞(∑k=0n−1nkC2k(12x9x2−1)n−kEn−k(0)+C2n(x))znn! =c2(x,z)I(0,12x9x2−1z) =e(18x2−1)z(cosh(6x9x2−1z)−sinh(6x9x2−1z)) =c2(x,z)−9x2−1b2(x,z) =∑n=0∞(C2n(x)−9x2−1B2n*(x))znn!.$

Thus,

$∑ k=0nnkC2(n−k)(12x9x2−1)kEk(0)=C2n(x)−9x2−1B2n*(x).$

Since $E2n−1=0$ for n ≥ 1, after some algebra we have (3.1).

Corollary 3.2. For each n ≥ 1 and j ≥ 1,

$∑ k=0 n/2n−12k−15k−1F2j2k−1L2j(n−2k)E2k−1(0)=−F2j(n−1).$

Proof. Evaluate (3.1) at the $x=ωjLj/6$ and use the links from (2.5). To simplify recall that $Ln2−5Fn2=(−1)n4$ and $F2n=FnLn$.

Using (1.10), we can write (3.2) as

$∑ k=0 n/2n−12k−120k−5kkF2j2k−1L2j(n−2k)B2k=5F2j(n−1),$

which is easily reduced to (1.6).

We also have the following interesting identity.

Theorem 3.3. For each $n≥0$ and $x∈ℂ$, we have the relation

$∑ k=0 n/2n2kC2(n−2k)(x)(36x2(9x2−1))kE2k=(18x2−1)n.$

Proof. The result is a consequence of the fact that

$c2(x,z)I(1/2,12x9x2−1z)=e(18x2−1)z.$

Corollary 3.4. For each n ≥ 0 and j ≥ 1,

$∑ k=0 n/2n2k(54)kF2j2kL2j(n−2k)E2k=21−nL2jn.$

Proof. Evaluate (3.3) at the point $x=ωjLj/6$ and use the links from (2.5). When simplifying you will also need the formula $Ln2−L2n=(−1)n2$.

Interestingly, if j=1/2 from (3.4) we obtain Byrd's result (1.4). Also, when j=1 and j=2, from (3.4) we obtain the following Lucas-Euler relations:

$∑k=0n/2n2k(54)kL2(n−2k)E2k=2(32)n,∑k=0n/2n2k(454)kL4(n−2k)E2k=2(72)n,$

respectively. The first example appears as equation (31) in [5].

A different expression for the sum on the left of (3.3) is stated next.

Theorem 3.5. For each $n≥0$ and $x∈ℂ$, we have

$∑k=0n/2n2kC2(n−2k)(x)(36x2(9x2−1))kE2k =∑k=0nnk(C2k(x)−9x2−1B2k*(x))(6x9x2−1)n−k.$

Proof. We use the identity

$I(1/2,2z)=ez(1−tanhz),$

from which the functional equation follows

$c2(x,z)I(1/2,12x9x2−1z)=e(6x9x2−1)z(c2(x,z)−9x2−1b2(x,z)).$

Thus,

$∑k=0nnkC2k(x)(6x9x2−1)n−kEn−k =∑k=0nnk(C2k(x)−9x2−1B2k*(x))(6x9x2−1)n−k,$

that is equivalent to (3.6).

Theorem 3.4. For each n≥ 0 and $x∈ℂ$, it is true that

$∑ k=0nnkC2(n−k)(x)12x 9x2−1kEk(x)=​(18x2−1+6x(2x−1)9x2 −1)n.$

Proof. The functional relation $cosh(z/2)I(x,z)=e(x−1/2)z$ produces immediately

$c2(x,z)I(x,12x9x2−1z)=e18x2−1+6x(2x−1)9x2 −1z.$

Comparing the coefficients of z in the power series expansions on both sides gives the identity.

When x=1/2, then we recover (3.5), by (1.9).

The following result appears as Theorem 13 in [7]: For each n ≥ 0, j ≥ 1, and $x∈ℂ$, we have

$∑ k=0nnkFjk(5Fj)n−kBn−k(x)=nFj((5x+β)Fj+Fj−1)n−1,$ $∑ k=0nnkFjk(−5Fj)n−kBn−k(x)=nFj((α−5x)Fj+Fj−1)n−1,$

where $α=(1+5)/2$ is the golden ratio and $β=(1−5)/2=−1/α$.

Now, we present the analogue result for the Lucas-Euler pair:

Theorem 4.1. The following polynomial identity is valid for all n ≥ 0, j ≥ 1, and $x∈ℂ$:

$∑ k=0nnkLjk(5Fj)n−kEn−k(x)=2((5x+β)Fj+Fj−1)n,$ $∑ k=0nnkLjk(−5Fj)n−kEn−k(x)=2((α−5x)Fj+Fj−1)n.$

Proof. Let L(z) be the exponential generating function for $(Ljn)n≥0$, j ≥ 1. Then, using the Binet formula for Ln, we get

$L(z)=2e(1/2Fj+Fj−1)zcosh​(5Fj2z).$

Thus, it follows that

$∑n=0∞(∑k=0nnkLjk(5Fj)n−kEn−k(x))znn!=L(z)I(x,5Fjz) =2e((x−1/2)5Fj+1/2Fj+Fj−1)z =2e((5x+β)Fj+Fj−1)z.$

This proves the first equation. The second follows upon replacing x by 1-x and using $En(1−x)=(−1)nEn(x)$ and $α−β=5$.

Note that the relations (4.1) and (4.2) provide a generalization of (3.4).

To see this, notice that they can be written more compactly as

$∑ k=0nnkLjk(±5Fj)n−kEn−k(x)=21−n(Lj±5Fj(2x−1))n.$

Now, if x=1/2, we get

$∑ k=0nnk(±5Fj)n−k2kLjkEn−k=2Ljn,$

which is equivalent to (3.4). We also mention the nice and curious identities

$∑ k=0nnk(±​5Fj)n−kLjkEn−k(α)=2(±1)nLj±1n,$ $∑ k=0nnk(±​5Fj)n−kLjkEn−k(β)=2(∓1)nLj∓1n,$

which can be deduced from (3.4) and $5Fn=Ln+1+Ln−1$.

We conclude this presentation with the following interesting corollary.

Corollary 4.2. Let n, j and q be integers with n, j ≥ 1 and q odd. Then it holds that

$∑ k=0nnk(5Fj)n−k(q−(n−k)−1)LjkEn−k(0)=2q−n∑ r=1 q−1(−1)r(rαj+(q−r)βj)n.$

Proof. The known multiplication formula for Euler polynomials for odd q [3, Chapter 24]

$qn∑ r=0 q−1(−1)rEn(x+rq)=En(qx)$

yields

$∑ r=1 q−1(−1)rEn(rq)=(q−n−1)En(0).$

Therefore,

$∑k=0nnkLjk(5Fj)n−k(q−(n−k)−1)En−k(0) =2∑r=1q−1(−1)r((5rq+β)Fj+Fj−1)n =2q−n∑r=1q−1(−1)r(5rFj+q(βFj+Fj−1))n =2q−n∑r=1q−1(−1)r(rαj+(q−r)βj)n.□$

The special instances for j=1, and q=3 and q=5, respectively, take the form

$∑ k=0nnk(5)n−k(3−(n−k)−1)En−k(0)=25⋅3−nF2n$

and

$∑ k=0nnk(5)n−k(5−(n−k)−1)LkEn−k(0)=2⋅5 (1−n)/2 (F 2n −F n ), if n is even;2⋅5 −n/2 (L 2n −L n ), if n is odd.$

In this section, we derive some mixed identities involving Bernoulli (Euler) polynomials and Bernoulli, Fibonacci and Lucas numbers.

Theorem 5.1. For each $n,j≥0$ and $x∈ℂ$,

$∑k=0nnk(±5Fj)kLj(n−k)Bk(x) =21−n∑k=0n/2n2k20kFj2k(​±​5Fj(2x−1)+Lj)n−2kB2k.$

Proof. The result follows from the functional relation

$H(x,±5Fjz)L(z)=5Fje(±(x−1/2)5Fj+1/2Lj)zzcoth(5Fj2z)$

and the well-known power series $cothz=∑ n=0∞4nB 2n(2n)!z2n−1. □$

If x=1/2, from (5.1) using $Bn(1/2)=(21−n−1)Bn$, we obtain the following result.

Corollary 5.2. For each $n,j≥0$,

$∑ k=0 n/2n2k5k(21−2k−1)Fj2kLj(n−2k)B2k=21−nLjn∑ k=0 n/2n2k20k( Fj Lj )2kB2k.$

For example,

$∑ k=0 n/2n2k(21−2k−1)5kB2kLn−2k=21−n∑ k=0 n/2n2k20kB2k$

and

$∑ k=0 n/2n2k(21−2k−1)20kB2kL3(n−2k)=21+n∑ k=0 n/2n2k5kB2k.$

If x = α and x=β, where $α=(1+5)/2$ is the golden ratio and β=-1/α, from (5.1) we have the following corollary.

Corollary 5.3. For each $n,j≥0$,

$∑ k=0nnk(5Fj)kLj(n−k)Bk(α)=2Lj+1n∑ k=0 n/2 n2k5k(FjL j+1)2kB2k,$ $∑ k=0nnk(−5Fj)kLj(n−k)Bk(α)=2(−1)nLj−1n∑ k=0 n/2 n2k5k(FjL j−1)2kB2k$

and

$∑ k=0nnk(5Fj)kLj(n−k)Bk(β)=2(−1)nLj−1n∑ k=0 n/2 n2k5k(FjL j−1)2kB2k,[2pt]$ $∑ k=0nnk(−5Fj)kLj(n−k)Bk(β)=2Lj+1n∑ k=0 n/2 n2k5k(FjL j+1)2kB2k.$

Finally, we present the theorem for Fibonacci-Euler pair.

Theorem 5.4. or each n ≥ 0 and j ≥ 1,

$(n+1)∑k=0nnk(5Fj)kFj(n−k)Ek(x)=4∑k=0n+12​n+12k5k−1(4k−1)Fj2k−1​Fj2+Fj−1+5Fj(x−12)n+1−2k​​B2k$

and

$(−1)n(n+1)∑k=0nnk(−5Fj)kFj(n−k)Ek(x)=4∑k=0n+12​n+12k5k−1(4k−1)Fj2k−1​Fj2+Fj−1−5Fj(x−12)n+1−2k​B2k.$

Proof. Let F(z) be the exponential generating function for $(Fjn)n≥0$ with j ≥1. Then, using the Binet formula for Fn, we get

$F(z)=25e1/2Fj+Fj−1zsinh​(5Fj2z).$

The relations follows from

$I(x,±5Fjz)F(z)=25eFj/2+Fj−1+(x−1/2)5Fjztanh(5Fj2z)$

and power series

In particularly, from Theorem 5.4 we have the following Euler-Bernoulli-Fibonacci-Lucas identity:

$(n+1)∑k=1nnk(±5Fj2)kFj(n−k)Ek=Ljn2n−2∑k=0n+12​n+12k5k−1(4k−1)2FjLj2k−1​B2k.$

In this paper, we have documented identities relating Euler numbers (polynomials) to balancing and Lucas-balancing polynomials. We have also derived a general identity involving Euler polynomials and Lucas numbers in arithmetic progression. All results must be seen as companion results to the Fibonacci-Bernoulli pair from [7]. In the future, we will work on more identities connecting Bernoulli/Euler numbers (polynomials) with Fibonacci/Lucas numbers (polynomials).

We would like to thank the referee for valuable suggestions.

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