A short remark on Horadam identities with binomial coefficients

A short remark on Horadam identities with binomial coefficients

Robert Frontczak

Landesbank Baden-Württemberg (LBBW), Stuttgart, Germany robert.frontczak@lbbw.de

Submitted: February 17, 2021 Accepted: March 17, 2021 Published online: April 27, 2021

Abstract

In this note, we introduce a very simple approach to obtain Horadam identities with binomial coefficients including an additional parameter. Many known Fibonacci identities (as well as polynomial identities) will follow im- mediately as special cases.

Keywords:Horadam number, Fibonacci number, binomial transform AMS Subject Classification:11B37, 11B39

1. Introduction and motivation

Layman [15] recalled the Fibonacci identities

𝐹2𝑛=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

𝐹𝑘, 2^𝑛𝐹𝑛=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

𝐹2𝑘, 3^𝑛𝐹𝑛=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

𝐹4𝑘,

and attributed them to Hoggatt [9]. Here, as usual,𝐹𝑛is the𝑛th Fibonacci number, defined by 𝐹0 = 0, 𝐹1 = 1and 𝐹𝑛+2 =𝐹𝑛+1+𝐹𝑛, 𝑛≥0. Layman proved more such identities, in particular, the following alternating sums:

(−1)^𝑛𝐹3𝑛=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−2)^𝑘𝐹2𝑘, (−5)^𝑛𝐹3𝑛=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−2)^𝑘𝐹5𝑘,

∗Statements and conclusions made in this article are entirely those of the author. They do not necessarily reflect the views of LBBW.

doi: https://doi.org/10.33039/ami.2021.03.016 url: https://ami.uni-eszterhazy.hu

5

and

(−4)^𝑛𝐹3𝑛 =

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−1)^𝑘𝐹6𝑘.

Several additional sums of this kind and generalizations were derived by Carlitz and Ferns [5], Carlitz [4], Haukkanen [7, 8] and Prodinger [16]. More recently, some authors also worked on generalizations and derived expressions for sums with weighted binomial sums, sums with polynomials and sums where only half of the binomial coefficients are used. We refer to [2] and [10–14]. Adegoke [1] generalized many of the above results and derived summation identities involving Horadam numbers and binomial coefficients, some of which we will encounter below.

In this note, we give another type of generalization of some Horadam binomial sums. More precisely, we introduce a very simple approach to obtain Horadam identities with an additional parameter. All results are derived completely routinely using standard methods.

Let 𝑤𝑛 = 𝑤𝑛(𝑎, 𝑏;𝑝, 𝑞) be a general Horadam sequence, i.e., a second order recurrence

𝑤𝑛=𝑝𝑤_𝑛−1−𝑞𝑤_𝑛−2, 𝑛≥2,

with nonzero constant𝑝,𝑞and initial values𝑤0=𝑎,𝑤1=𝑏. We mention the fol- lowing instances: 𝑤𝑛(0,1; 1,−1) =𝐹𝑛 is the Fibonacci sequence,𝑤𝑛(0,1; 2,−1) = 𝑃𝑛is the Pell sequence,𝑤𝑛(0,1; 1,−2) =𝐽𝑛is the Jacobsthal sequence,𝑤𝑛(0,1; 3,2)

= 𝑀𝑛 is the Mersenne sequence, 𝑤𝑛(0,1; 6,1) = 𝐵𝑛 is the balancing number sequence, 𝑤𝑛(2,1; 1,−1) = 𝐿𝑛 is the Lucas sequence, 𝑤𝑛(2,2; 2,−1) = 𝑄𝑛 is the Pell-Lucas sequence, 𝑤𝑛(2,1; 1,−2) = 𝑗𝑛 is the Jacobsthal-Lucas sequence, and 𝑤𝑛(1,3; 6,1) = 𝐶𝑛 is Lucas-balancing number sequence. All sequences are listed in OEIS [17] where additional information and references are available. We also note that the sequence 𝑤𝑛 also contains important sequences of polynomials:

𝑤𝑛(0,1;𝑥,−1) = 𝐹𝑛(𝑥) are the Fibonacci polynomials, 𝑤𝑛(0,1; 2𝑥,−1) = 𝑃𝑛(𝑥) are the Pell polynomials, 𝑤𝑛(0,1; 1,−2𝑥) =𝐽𝑛(𝑥)are the Jacobsthal polynomials, and𝑤𝑛(0,1; 6𝑥,1) =𝐵𝑛(𝑥)are the balancing polynomials, respectively.

The Binet formula of𝑤𝑛 in the non-degenerated case,𝑝²−4𝑞 >0, is 𝑤𝑛 =𝐴𝛼^𝑛+𝐵𝛽^𝑛,

with

𝐴=𝑏−𝑎𝛽

𝛼−𝛽, 𝐵=𝑎𝛼−𝑏 𝛼−𝛽,

and where 𝛼and𝛽 are roots of the equation𝑥²−𝑝𝑥+𝑞= 0, that is

𝛼= 𝑝+√︀

𝑝²−4𝑞

2 , 𝛽 =𝑝−√︀

𝑝²−4𝑞

2 .

In what follows we will need the following expressions:

𝛼+𝛽=𝑝, 𝛼𝛽=𝑞, 𝛼−𝛽=√︀

𝑝²−4𝑞,

as well as

𝛼²=𝑝𝛼−𝑞, 𝛽²=𝑝𝛽−𝑞, (1.1)

𝛼³= (𝑝²−𝑞)𝛼−𝑝𝑞, 𝛽²= (𝑝²−𝑞)𝛽−𝑝𝑞, (1.2) 𝛼⁴= (𝑝³−2𝑝𝑞)𝛼−𝑞(𝑝²−𝑞), 𝛽²= (𝑝³−2𝑝𝑞)𝛽−𝑞(𝑝²−𝑞),

and so on.

Finally, we mention the standard fact about sequences and their binomial trans- forms [3]: Let (𝑎𝑛)𝑛≥0 be a sequence of numbers and (𝑏𝑛)𝑛≥0 be its binomial transform. Then, we have the following relations:

𝑏𝑛=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

𝑎𝑘 ⇔ 𝑎𝑛 =

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−1)^𝑛−𝑘𝑏𝑘.

2. A simple generalization

The next lemma will be the key ingredient to derive our results.

Lemma 2.1. Let 𝑛and𝑗 be integers with0≤𝑗 ≤𝑛. Then, for each𝑎, 𝑥∈Cwe have the identity

(︂𝑛 𝑗 )︂

𝑥^𝑗(𝑎±𝑥)^𝑛−𝑗=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

(±1)^𝑘−𝑗𝑥^𝑘𝑎^𝑛−𝑘.

Proof. From the binomial theorem we have (︂𝑛

𝑗 )︂

𝑥^𝑗(𝑎±𝑥)^𝑛−𝑗 = (︂𝑛

𝑗 )︂^𝑛−𝑗

∑︁

𝑚=0

(︂𝑛−𝑗 𝑚

)︂

(±1)^𝑚𝑥^𝑚+𝑗𝑎^{𝑛−(𝑗+𝑚)}

= (︂𝑛

𝑗 )︂∑︁^𝑛

𝑘=𝑗

(︂𝑛−𝑗 𝑘−𝑗 )︂

(±1)^𝑘−𝑗𝑥^𝑘𝑎^𝑛−𝑘

=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

(±1)^𝑘−𝑗𝑥^𝑘𝑎^𝑛−𝑘,

where in the last step we have used the identity (︂𝑛

𝑗

)︂(︂𝑛−𝑗 𝑘−𝑗 )︂

= (︂𝑛

𝑘 )︂(︂𝑘

𝑗 )︂

, 0≤𝑗≤𝑘≤𝑛.

Example 2.2. Setting (𝑥;𝑎) = (𝛼²;−𝑞), (𝑥;𝑎) = (𝛽²;−𝑞), using (1.1) and the linearity of the Binet form gives the following identity valid for all 0≤𝑗≤𝑛

(︂𝑛 𝑗

)︂

𝑝^𝑛−𝑗𝑤𝑛+𝑗+𝑚=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

𝑞^𝑛−𝑘𝑤2𝑘+𝑚, 𝑚≥0.

The case 𝑗= 0 and the corresponding inverse binomial transform produce imme- diately

(︁𝑝 𝑞

)︁𝑛

𝑤𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

𝑞^−𝑘𝑤2𝑘+𝑚

as well as

𝑞^−𝑛𝑤2𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−1)^𝑛−𝑘(︁𝑝 𝑞

)︁𝑘

𝑤𝑘+𝑚.

Obviously, with𝑤𝑛 =𝐹𝑛 (or𝐿𝑛) we recover the classical results, which appeared in [5] and [15]. The balancing number counterparts were stated in [6].

Example 2.3. If we set ∆ =𝑝²−4𝑞, then a simple computation shows that 𝛼²−𝑞=√

∆𝛼 and 𝛽²−𝑞=−√

∆𝛼.

Thus, with(𝑥;𝑎) = (𝛼²;−𝑞), (𝑥;𝑎) = (𝛽²;−𝑞)and using again (1.1), we see that if𝑛and𝑗 have the same parity, then for all0≤𝑗≤𝑛

(︂𝑛 𝑗 )︂

∆^{(𝑛−𝑗)/2}𝑤𝑛+𝑗+𝑚=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

(−𝑞)^𝑛−𝑘𝑤2𝑘+𝑚, 𝑚≥0.

Especially, for𝑗= 0and𝑛even we get

𝑞^−𝑛∆^𝑛/2𝑤𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−𝑞)^−𝑘𝑤2𝑘+𝑚

and for𝑗= 1 and𝑛odd

−𝑞^−𝑛∆^(𝑛−1)/2𝑛𝑤𝑛+1+𝑚=

∑︁𝑛

𝑘=1

(︂𝑛 𝑘 )︂

𝑘(−𝑞)^−𝑘𝑤2𝑘+𝑚.

If𝑛and𝑗 are of unequal parity (𝑛odd and𝑗 even, for instance), then (︂𝑛

𝑗 )︂

∆^{(𝑛−1−𝑗)/2}𝑣𝑛+𝑗+𝑚=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

(−𝑞)^𝑛−𝑘𝑢2𝑘+𝑚, 𝑚≥0,

and (︂𝑛 𝑗 )︂

∆^{(𝑛+1−𝑗)/2}𝑢𝑛+𝑗+𝑚=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

(−𝑞)^𝑛−𝑘𝑣2𝑘+𝑚, 𝑚≥0,

with𝑢𝑛=𝑤𝑛(0,1;𝑝, 𝑞)and𝑣𝑛 =𝑤𝑛(2, 𝑝;𝑝, 𝑞).

Example 2.4. Setting(𝑥;𝑎) = (𝛼³;−𝑝𝑞),(𝑥;𝑎) = (𝛽³;−𝑝𝑞), using (1.2) yields for all0≤𝑗 ≤𝑛

(︂𝑛 𝑗 )︂

(𝑝²−𝑞)^𝑛−𝑗𝑤𝑛+2𝑗+𝑚=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

(𝑝𝑞)^𝑛−𝑘𝑤3𝑘+𝑚, 𝑚≥0.

The case 𝑗= 0in combination with the binomial transform produce (︁𝑝²−𝑞

𝑝𝑞 )︁^𝑛

𝑤𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(𝑝𝑞)^−𝑘𝑤3𝑘+𝑚

as well as

(𝑝𝑞)^−𝑛𝑤3𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−1)^𝑛−𝑘(︁𝑝²−𝑞 𝑝𝑞

)︁𝑘

𝑤𝑘+𝑚.

Example 2.5. Combining the values(𝑥;𝑎) = (𝑝𝛼³;𝑞²)and(𝑥;𝑎) = (𝑝𝛽³;𝑞²)with 𝑝𝛼³+𝑞²= (𝑝²−𝑞)𝛼², and 𝑝𝛽³+𝑞²= (𝑝²−𝑞)𝛽²,

Lemma 2.1 gives (︂𝑛

𝑗 )︂

𝑝^𝑗(𝑝²−𝑞)^𝑛−𝑗𝑤2𝑛+𝑗+𝑚=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

𝑝^𝑘𝑞^{2𝑛−2𝑘}𝑤3𝑘+𝑚, 𝑚≥0.

Again, from the case𝑗= 0and the binomial transform we get (︁𝑝²−𝑞

𝑞² )︁𝑛

𝑤2𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘

)︂(︁𝑝 𝑞²

)︁𝑘

𝑤3𝑘+𝑚

as well as

(︁𝑝 𝑞²

)︁𝑛

𝑤3𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−1)^𝑛−𝑘(︁𝑝²−𝑞 𝑞²

)︁𝑘

𝑤2𝑘+𝑚.

Example 2.6. In this example we combine the values(𝑥;𝑎) = (𝛼⁴;𝑞(𝑝²−𝑞))and (𝑥;𝑎) = (𝛽⁴;𝑞(𝑝²−𝑞))to get

(︂𝑛 𝑗 )︂

(𝑝(𝑝²−2𝑞))^𝑛−𝑗𝑤𝑛+3𝑗+𝑚=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

(𝑞(𝑝²−𝑞))^𝑛−𝑘𝑤4𝑘+𝑚, 𝑚≥0.

Hence,

(︁𝑝(𝑝²−2𝑞) 𝑞(𝑝²−𝑞)

)︁^𝑛

𝑤𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(𝑞(𝑝²−𝑞))^−𝑘𝑤4𝑘+𝑚

as well as

(𝑞(𝑝²−𝑞))^−𝑛𝑤4𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−1)^𝑛−𝑘(︁𝑝(𝑝²−2𝑞) 𝑞(𝑝²−𝑞)

)︁^𝑘 𝑤𝑘+𝑚.

Example 2.7. An application of Lemma 2.1 with (𝑥;𝑎) = (𝛼⁴;𝑞²) and (𝑥;𝑎) = (𝛽⁴;𝑞²)and noting that

𝛼⁴+𝑞²= (𝑝²−2𝑞)𝛼² and 𝛽⁴+𝑞²= (𝑝²−2𝑞)𝛽²,

proves the next identity:

(︂𝑛 𝑗 )︂

(𝑝²−2𝑞)^𝑛−𝑗𝑤2𝑛+2𝑗+𝑚=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

𝑞^{2𝑛−2𝑘}𝑤4𝑘+𝑚, 𝑚≥0.

The case 𝑗= 0in conjunction with the binomial transform yield (︁𝑝²−2𝑞

𝑞² )︁^𝑛

𝑤2𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

𝑞^−2𝑘𝑤4𝑘+𝑚

as well as

𝑞^−2𝑛𝑤4𝑛+𝑚=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−1)^𝑛−𝑘(︁𝑝²−2𝑞 𝑞²

)︁𝑘

𝑤2𝑘+𝑚.

3. Slightly more general identities

Lemma 3.1. For each 𝑛≥1 we have the relations

𝛼^𝑛=𝛼𝑢𝑛−𝑞𝑢_𝑛−1 and 𝛽^𝑛=𝛽𝑢𝑛−𝑞𝑢_𝑛−1 with 𝑢𝑛 =𝑤𝑛(0,1;𝑝, 𝑞).

Proof. We can prove the statements by induction on𝑛. Since,𝛼¹=𝛼𝑢1−𝑞𝑢0, the inductive step is

𝛼^𝑛+1=𝛼𝛼^𝑛

=𝛼²𝑢𝑛−𝑞𝛼𝑢_𝑛−1

= (𝛼𝑢2−𝑞𝑢1)𝑢𝑛−𝑞𝛼𝑢𝑛−1

=𝛼(𝑝𝑢𝑛−𝑞𝑢_𝑛−1)−𝑞𝑢𝑛 (𝑢2=𝑝)

=𝛼𝑢𝑛+1−𝑞𝑢𝑛.

The proof of the second statement is a copy of the first proof.

The next identity is stated as a proposition.

Proposition 3.2. For integers 𝑚≥2,𝑟≥0and 0≤𝑗≤𝑛it is true that (︂𝑛

𝑗 )︂

𝑢⁻_𝑚^𝑛₋₁𝑢^𝑛_𝑚^−𝑗𝑤𝑗(𝑚−1)+𝑛+𝑟=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

𝑢⁻_𝑚^𝑘₋₁𝑞^𝑛−𝑘𝑤𝑚𝑘+𝑟.

Proof. From Lemma 3.1 we see that

𝑞+ 𝛼^𝑛

𝑢_𝑛−1 =𝛼 𝑢𝑛

𝑢_𝑛−1 and 𝑞+ 𝛽^𝑛

𝑢_𝑛−1 =𝛽 𝑢𝑛

𝑢_𝑛−1. Using Lemma 2.1 with𝑎=𝑞and𝑥= _𝑢^𝛼_𝑛^𝑛

−1 and𝑥= _𝑢^𝛽_𝑛^𝑛

−1 completes the proof.

The next two sum identities follow immediately:

(︁ 𝑢𝑚

𝑞𝑢𝑚−1

)︁𝑛

𝑤𝑛+𝑟=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(𝑞𝑢𝑚−1)^−𝑘𝑤𝑚𝑘+𝑟

as well as

(𝑞𝑢𝑚−1)^−𝑛𝑤𝑚𝑛+𝑟=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−1)^𝑛−𝑘(︁ 𝑢𝑚

𝑞𝑢_𝑚−1 )︁𝑘

𝑤𝑘+𝑟.

We also mention the formula for𝑗= 1:

𝑛𝑢⁻_𝑚−1^𝑛 𝑢^𝑛_𝑚⁻¹𝑤𝑛+𝑚+𝑟−1=

∑︁𝑛

𝑘=1

(︂𝑛 𝑘 )︂

𝑘𝑢⁻_𝑚−1^𝑘 𝑞^𝑛−𝑘𝑤𝑚𝑘+𝑟.

Lemma 3.3. For each 𝑘, 𝑛≥1 we have the relations 𝛼^𝑘𝑛= 𝑢𝑘𝑛

𝑢𝑛

𝛼^𝑛−𝑞^𝑛𝑢_{(𝑘−1)𝑛}

𝑢𝑛 and 𝛽^𝑘𝑛= 𝑢𝑘𝑛

𝑢𝑛

𝛽^𝑛−𝑞^𝑛𝑢_{(𝑘−1)𝑛} 𝑢𝑛

with 𝑢𝑛 =𝑤𝑛(0,1;𝑝, 𝑞).

Proof. Both statements can be verified directly by computation working with𝑢𝑛𝛼^𝑘𝑛 (respectively 𝑢𝑛𝛽^𝑘𝑛) and𝑞=𝛼𝛽.

Proposition 3.4. For integers 𝑚≥2,𝑠≥1, 𝑟≥0, and 0≤𝑗≤𝑛 we have the identity

(︂𝑛 𝑗

)︂(︁ 𝑢𝑠

𝑢𝑚𝑠

)︁𝑗(︁ 𝑢𝑚𝑠

𝑢_{(𝑚−1)𝑠} )︁𝑛

𝑞^−𝑠𝑛𝑤𝑠𝑛+𝑠𝑗(𝑚−1)+𝑟

=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

𝑞^−𝑠𝑘(︁ 𝑢𝑠

𝑢(𝑚−1)𝑠

)︁^𝑘

𝑤𝑚𝑠𝑘+𝑟.

Proof. The identity follows upon combining Lemma 2.1 with Lemma 3.3 with𝑎= 𝑞^𝑠𝑢(𝑚−1)𝑠/𝑢𝑠and 𝑥=𝛼^𝑚𝑠 and𝑥=𝛽^𝑚𝑠, respectively.

The special identities for𝑗 = 0are (︁ 𝑢𝑚𝑠

𝑢_{(𝑚−1)𝑠} )︁^𝑛

𝑞^−𝑠𝑛𝑤𝑠𝑛+𝑟=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

𝑞^−𝑠𝑘(︁ 𝑢𝑠

𝑢_{(𝑚−1)𝑠} )︁^𝑘

𝑤𝑚𝑠𝑘+𝑟

and

(︁ 𝑢𝑠

𝑢(𝑚−1)𝑠

)︁𝑛

𝑞^−𝑠𝑛𝑤𝑚𝑠𝑛+𝑟=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−1)^𝑛−𝑘𝑞^−𝑠𝑘(︁ 𝑢𝑚𝑠

𝑢(𝑚−1)𝑠

)︁𝑘

𝑤𝑠𝑘+𝑟.

Corollary 3.5. For integers 𝑠≥1,𝑟≥0 and0≤𝑗≤𝑛we have the identity (︂𝑛

𝑗 )︂

𝑣^𝑛−𝑗_𝑠 𝑞^−𝑠𝑛𝑤𝑠(𝑛+𝑗)+𝑟=

∑︁𝑛

𝑘=𝑗

(︂𝑘 𝑗

)︂(︂𝑛 𝑘 )︂

𝑞^−𝑠𝑘𝑤2𝑠𝑘+𝑟.

In particular,

(−1)^𝑛𝑞^−𝑠𝑛𝑤2𝑠𝑛+𝑟=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

(−1)^𝑘𝑞^−𝑠𝑘𝑣^𝑘_𝑠𝑤𝑠𝑘+𝑟

and

𝑞^−𝑠𝑛𝑣_𝑠^𝑛𝑤𝑠𝑛+𝑟=

∑︁𝑛

𝑘=0

(︂𝑛 𝑘 )︂

𝑞^−𝑠𝑘𝑤2𝑠𝑘+𝑟.

Proof. Set𝑚= 2in Proposition 3.4 and use𝑢2𝑛/𝑢𝑛=𝑣𝑛.

Some more examples could be stated, but we stop here, as the principle is clear.

Acknowledgements. The author thanks Kunle Adegoke for discussions. He is also grateful to the anonymous referee for a rapid review and for suggesting important references that are directly linked to this research project.

References

[1] K. Adegoke:A master identity for Horadam numbers, Preprint (2019), url:https://arxiv.org/abs/1903.11057.

[2] K. Adegoke:Weighted sums of some second-order sequences, Fibonacci Quart. 56.3 (2018), pp. 252–262.

[3] K. N. Boyadzhiev:Notes on the Binomial Transform: Theory and Table with Appendix on Stirling Transform, World Scientific, 2018.

[4] L. Carlitz:Some classes of Fibonacci sums, Fibonacci Quart. 16.5 (1978), pp. 411–425.

[5] L. Carlitz, H. H. Ferns: Some Fibonacci and Lucas identities, Fibonacci Quart. 8.1 (1970), pp. 61–73.

[6] R. Frontczak:Sums of balancing and Lucas-balancing numbers with binomial coefficients, Int. J. Math. Anal. 12.12 (2018), pp. 585–594.

[7] P. Haukkanen:Formal power series for binomial sums of sequences of numbers, Fibonacci Quart. 31.1 (1993), pp. 28–31.

[8] P. Haukkanen:On a binomial sum for the Fibonacci and related numbers, Fibonacci Quart.

34.4 (1996), pp. 326–331.

[9] V. E. Hoggatt,Jr.:Some special Fibonacci and Lucas generating functions, Fibonacci Quart. 9.2 (1971), pp. 121–133.

[10] E. Kilic,N. Irmak:Binomial identities involving the generalized Fibonacci type polynomi- als, Ars Combin. 98 (2011), pp. 129–134.

[11] E. Kiliç:Some classes of alternating weighted binomial sums, An. Ştiint. Univ. Al. I. Cuza Iaşi. Mat. 3.2 (2016), pp. 835–843.

[12] E. Kiliç, E. J. Ionascu: Certain binomial sums with recursive coefficients, Fibonacci Quart. 48.2 (2010), pp. 161–167.

[13] E. Kiliç,N. Ömür,Y. T. Ulutaş:Binomial sums whose coefficients are products of terms of binary sequences, Util. Math. 84 (2011), pp. 45–52.

[14] E. Kiliç,E. Tan:On binomial sums for the general second order linear recurrence, Integers 10 (2010), pp. 801–806,

doi:https://doi.org/10.1515/INTEGER.2010.057.

[15] J. W. Layman:Certain general binomial-Fibonacci sums, Fibonacci Quart. 15.3 (1977), pp. 362–366.

[16] H. Prodinger: Some information about the binomial transform, Fibonacci Quart. 32.5 (1994), pp. 412–415.

[17] N. J. A. Sloane:The On-Line Encyclopedia of Integer Sequences, Published electronically at https://oeis.org.