Some Finite Sums Involving Generalized Fibonacci and Lucas Numbers

Abstract

By considering Melham's sums (Melham, 2004), we compute various more general nonalternating sums, alternating sums, and sums that alternate according to (−1)^2𝑛+1 involving the generalized Fibonacci and Lucas numbers.

1. Introduction

Let 𝑎, 𝑏, and 𝑝 be assumed to be arbitrary nonzero complex numbers with 𝑝(𝑝²+2)(𝑝²+4)≠0. Define second-order linear recursion {𝑊_𝑛} by𝑊_𝑛=𝑝𝑊_𝑛−1+𝑊_𝑛−2, (1.1) with 𝑊₀=𝑎,𝑊₁=𝑏 for all integers 𝑛. Since Δ=𝑝²+4≠0, the roots 𝛼 and 𝛽 of 𝑥²−𝑝𝑥−1=0 are distinct.

Also define the sequence {𝑋_𝑛} via the terms of sequence {𝑊_𝑛} as 𝑋_𝑛=𝑊_𝑛+1+𝑊_𝑛−1.

The Binet formulas for the sequences {𝑊_𝑛} and {𝑋_𝑛} are𝑊_𝑛=𝐴𝛼^𝑛−𝐵𝛽^𝑛𝛼−𝛽,𝑋_𝑛=𝐴𝛼^𝑛+𝐵𝛽^𝑛, (1.2) where 𝐴=𝑏−𝑎𝛽 and 𝐵=𝑏−𝑎𝛼.

For 𝑎=0, 𝑏=1, we denote 𝑊_𝑛=𝑈_𝑛 and so 𝑋_𝑛=𝑉_𝑛, respectively. When 𝑝=1, 𝑈_𝑛=𝐹_𝑛 (𝑛th Fibonacci number) and 𝑉_𝑛=𝐿_𝑛 (𝑛th Lucas number).

Inspired by the well-known identity^𝑗_𝑛=1𝐹²_𝑛=𝐹_𝑗𝐹_𝑗+1, (1.3) Clary and Hemenway [1] obtained factored closed-form expressions for all sums of the form ∑^𝑗_𝑛=1𝐹³_𝑚𝑛, where 𝑚 is an integer. Motivated by the results in [1], Melham [2] computed all sums of the form ∑^𝑗_𝑛=1(−1)^𝑛𝐹⁴_𝑚𝑛 and ∑^𝑗_𝑛=1(−1)^𝑛𝐿⁴_𝑚𝑛. In [3], Melham computed various nonalternating sums, alternating sums, and sums that alternate according to (−1)^2𝑛+1 for sequences {𝑊_𝑛} and {𝑋_𝑛}. The author gathers his sums in three sets. Here we recall one example from each set for the reader's convenience:^𝑗_𝑛=𝑖𝑊_𝑛=1𝑝𝑉_{(𝑗−𝑖+1)/2}𝑊_{(𝑗+𝑖+1)/2}+𝑊_{(𝑗+𝑖−1)/2}(),1if𝑗−𝑖≡1mod4𝑝𝑈_{(𝑗−𝑖+1)/2}𝑋_{(𝑗+𝑖+1)/2}+𝑋_{(𝑗+𝑖−1)/2}if𝑗−𝑖≡3(mod4_{4𝑗+3𝑛=4𝑖}),(−1)^2𝑛+1𝑊_2𝑛=𝑝𝑈Δ−2^{4𝑗−4𝑖+4}𝑋_{4𝑗+4𝑖+3},_{4𝑗+3𝑛=4𝑖+2}(−1)^2𝑛+1𝑈_𝑛𝑋_𝑛=𝑝𝑉Δ−2^{4𝑗−4𝑖+5}𝑊_{4𝑗+4𝑖+3}+2𝑊₀. (1.4)

We refer to [4] for general expansion formulas for sums of powers of Fibonacci and Lucas numbers, as considered by Melham, as well as some extensions such that ^𝑛_𝑘=0𝐹^2𝑚+𝜖_2𝑘+𝛿,^𝑛_𝑘=0𝐿^2𝑚+𝜖_2𝑘+𝛿, (1.5) where 𝛿,𝜖∈{0,1}.

For alternating analogues of the results given by Prodinger, that is, ^𝑛_𝑘=0(−1)^𝑘𝐹^2𝑚+𝜖_2𝑘+𝛿,^𝑛_𝑘=0(−1)^𝑘𝐿^2𝑚+𝜖_2𝑘+𝛿, (1.6) we refer to [5].

Hendel [6] gave the factorization theorem which exhibits factorizations of sums of the form ∑^{𝑛+𝑖−1}_𝑗=𝑖𝐹_{𝑎𝑗−𝑏}. The author also introduced a unified proof method based on formulae for the factorizations of 𝐹_𝑞−𝑑+𝐹_𝑞+𝑑.

In [7], Curtin et al. derived formulae for the shifted summations^𝑑−1_𝑗=0𝐹_𝑛+𝑗𝐹_𝑚+𝑗,^𝑑−1_𝑗=0𝐿_𝑛+𝑗𝐿_𝑚+𝑗,^𝑑−1_𝑗=0𝐹_𝑛+𝑗𝐿_𝑚+𝑗, (1.7) and the shifted convolutions^𝑑−1_𝑗=0𝐹_𝑛+𝑗𝐹_{𝑑−𝑚−𝑗},^𝑑−1_𝑗=0𝐿_𝑛+𝑗𝐿_{𝑑−𝑚−𝑗},^𝑑−1_𝑗=0𝐹_𝑛+𝑗𝐿_{𝑑−𝑚−𝑗} (1.8) for positive integers 𝑑 and arbitrary integers 𝑛 and 𝑚.

In this paper, our main purpose is to consider Melham's sums involving double products of terms of {𝑊_𝑛},{𝑋_𝑛},{𝑈_𝑛}, and {𝑉_𝑛} given in [3] and then compute several more general nonalternating sums, alternating sums, and sums that alternate according to (−1)^2𝑛+1.

2. Certain Finite Sums of Double Products of Terms

In this section, we will investigate certain sums consisting of products of at most two terms of {𝑊_𝑛}: nonalternating sums, alternating sums and sums that alternate according to (−1)^2𝑛+1. From the Binet forms of {𝑊_𝑛} and {𝑋_𝑛}, we give the following lemma for further use without proof.

Lemma 2.1. Let 𝑎, 𝑏, and 𝑝 be as in Section 1, and let 𝑟=𝑎𝑊₂−𝑏𝑊₁. Then for all integers 𝑘, 𝑏²𝑈_2𝑘+2𝑎𝑏𝑈_2𝑘−1+𝑎²𝑈_2𝑘−2=𝑊_𝑘𝑋_𝑘,𝑏²𝑈_2𝑘+1+2𝑎𝑏𝑈_2𝑘+𝑎²𝑈_2𝑘−1=𝑊_𝑘+1𝑋_𝑘+(−1)^𝑘𝑏𝑟,₂𝑉_2𝑘+2𝑎𝑏𝑉_2𝑘−1+𝑎²𝑉_2𝑘−2=𝑋²_𝑘+(−1)^𝑘𝑏2𝑟,₂𝑉_2𝑘+1+2𝑎𝑏𝑉_2𝑘+𝑎²𝑉_2𝑘−1=𝑋_𝑘𝑋_𝑘+1+(−1)^𝑘𝑝𝑟. (2.1)

Theorem 2.2. Fix integers 𝑐,𝑑, and 𝑚.(i)If 𝑚 is even, then for all integers 𝑗>𝑖,^𝑗_𝑛=𝑖𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}=𝑈_{𝑚(𝑗−𝑖+1)}𝑋_{𝑚(𝑗+𝑖)+𝑐+𝑑}Δ𝑈_𝑚−(−1)^𝑐𝑗−𝑖+1𝑋_𝑑−𝑐Δ. (2.2)(i)If 𝑚 is odd, then for all integers 𝑗>𝑖,^𝑗_𝑛=𝑖𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}=𝑈_{𝑚(𝑗−𝑖+1)}𝑊_{𝑚(𝑗+𝑖)+𝑐+𝑑}𝑉_𝑚(),𝑉if𝑗−𝑖≡1mod2_{𝑚(𝑗−𝑖+1)}𝑋_{𝑚(𝑗+𝑖)+𝑐+𝑑}Δ𝑉_𝑚−(−1)^𝑐+𝑗𝑋_𝑑−𝑐Δif𝑗−𝑖≡0(mod2). (2.3)

Proof. Using the Binet formulas, we compute ^𝑗_𝑛=𝑖𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}=^𝑗_𝑛=𝑖𝛼^{𝑚𝑛+𝑐}−𝛽^{𝑚𝑛+𝑐}𝛼−𝛽𝐴𝛼^{𝑚𝑛+𝑑}−𝐵𝛽^{𝑚𝑛+𝑑}=1𝛼−𝛽𝛼−𝛽^2𝑗_𝑛=𝑖𝐴𝛼^{2𝑚𝑛+𝑐+𝑑}+𝐵𝛽^{2𝑚𝑛+𝑐+𝑑}−(−1)^{𝑚𝑛+𝑐}𝛼−𝛽²𝐴𝛼^𝑑−𝑐+𝐵𝛽^𝑑−𝑐=1Δ^𝑗_𝑛=𝑖𝑋_{2𝑚𝑛+𝑐+𝑑}−(−1)^𝑐Δ𝑋^𝑗_{𝑑−𝑐𝑛=𝑖}(−1)^𝑚𝑛. (2.4) Since 𝑋_𝑛=𝑊_𝑛−1+𝑊_𝑛+1, we can obtain that for even 𝑚^𝑗_𝑛=𝑖𝑋_{2𝑚𝑛+𝑡}=𝑈_{𝑚(𝑗−𝑖+1)}𝑋_{𝑚(𝑗+𝑖)+𝑡}𝑈_𝑚. (2.5) The result follows.

For example, when 𝑖=2,𝑚=3,𝑎=0,𝑏=𝑐=𝑝=1, and 𝑑=5, we obtain^𝑗_𝑛=2𝐹_3𝑛+1𝐹_3𝑛+5=𝐹_3(𝑗−1)𝐹_3𝑗+44. (2.6)

Theorem 2.3. Fix integers 𝑐,𝑑, and 𝑚. Let 𝑆=∑^𝑗_𝑛=𝑖(−1)^𝑛𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}.(1)If 𝑚 is odd, then 𝑆 equals ()𝑆=−1^𝑗𝑈_{𝑚(𝑗−𝑖+1)}𝑋_{𝑚(𝑗+𝑖)+𝑐+𝑑}Δ𝑈_𝑚−(−1)^𝑐𝑗−𝑖+1𝑋_𝑑−𝑐Δ. (2.7)(2)If 𝑚 is odd and the parities of 𝑖 and 𝑗 are the same, then 𝑆 equals (−1)^𝑗𝑉_{𝑚(𝑗−𝑖+1)}𝑋_{𝑚(𝑗+𝑖)+𝑐+𝑑}Δ𝑉_𝑚−(−1)^𝑗+(−1)^𝑖(−1)^𝑐𝑋_𝑑−𝑐2Δ. (2.8)(3)If 𝑚 is odd and the parities of 𝑖 and 𝑗 are the different, then 𝑆 equals (−1)^𝑗𝑈_{𝑚(𝑗−𝑖+1)}𝑊_{𝑚(𝑗+𝑖)+𝑐+𝑑}𝑉_𝑚−(−1)^𝑗+(−1)^𝑖(−1)^𝑐𝑋_𝑑−𝑐2. (2.9)

Proof. Consider ^𝑗_𝑛=𝑖(−1)^𝑛𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}=^𝑗_𝑛=𝑖𝛼^{𝑚𝑛+𝑐}−𝛽^{𝑚𝑛+𝑐}𝛼−𝛽𝐴𝛼^{𝑚𝑛+𝑑}−𝐵𝛽^{𝑚𝑛+𝑑}=1𝛼−𝛽𝛼−𝛽^2𝑗_𝑛=𝑖(−1)^𝑛𝐴𝛼^{2𝑚𝑛+𝑐+𝑑}+𝐵𝛽^{2𝑚𝑛+𝑐+𝑑}−(−1)^{𝑚𝑛+𝑐}𝛼−𝛽²𝐴𝛼^𝑑−𝑐+𝐵𝛽^𝑑−𝑐=1Δ^𝑗_𝑛=𝑖(−1)^𝑛𝑋_{2𝑚𝑛+𝑐+𝑑}−1Δ𝑋^𝑗_{𝑑−𝑐𝑛=𝑖}(−1)^{(𝑚+1)𝑛+𝑐}. (2.10)
Since 𝑋_𝑛=𝑊_𝑛−1+𝑊_𝑛+1, for odd 𝑚, we find ^𝑗_𝑛=𝑖(−1)^𝑛𝑋_{2𝑚𝑛+𝑐}=(−1)^𝑗𝑈_{𝑚(𝑗−𝑖+1)}𝑋_{𝑚(𝑖+𝑗)+𝑐}𝑈_𝑚. (2.11) The result is now obtained by considering the values of ∑^𝑗_𝑛=𝑖(−1)^{(𝑚+1)𝑛+𝑐}.

Theorem 2.4. Fix integers 𝑐,𝑑, and 𝑚. For all integers 𝑗>𝑖, ^4𝑗_𝑛=4𝑖+1(−1)^2𝑛+1𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}=𝑈_{4𝑚(𝑗−𝑖)}𝑉_2𝑚𝑉_𝑚𝑊_𝑠+𝑚𝑈if𝑚iseven,_𝑚𝑋_𝑠+𝑚if𝑚isodd, (2.12)^4𝑗+3_𝑛=4𝑖(−1)^2𝑛+1𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}=𝑈_{4𝑚(𝑗−𝑖+1)}𝑉_2𝑚𝑈_𝑚𝑋_𝑠+3𝑚𝑉if𝑚iseven,_𝑚𝑊_𝑠+3𝑚if𝑚isodd,_{4𝑗𝑛=4𝑖+3}(−1)^2𝑛+1𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}=𝑉_𝑚𝑉_{2𝑚(2(𝑗−𝑖)−1)}𝑋_𝑠+3𝑚Δ𝑉_2𝑚−2(−1)^𝑐𝑋_𝑑−𝑐Δ𝑈if𝑚iseven,_𝑚𝑉_{2𝑚(2(𝑗−𝑖)−1)}𝑊_𝑠+3𝑚𝑉_2𝑚if𝑚isodd,_{4𝑗+3𝑛=4𝑖+2}(−1)^2𝑛+1𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}=𝑈_𝑚𝑉_{2𝑚(2(𝑗−𝑖)+1)}𝑊_𝑠+5𝑚𝑉_2𝑚𝑉if𝑚iseven,_𝑚𝑉_{𝑚(4(𝑗−𝑖+1)−2)}𝑋_𝑠+5𝑚Δ𝑉_2𝑚+2(−1)^𝑐𝑋_𝑑−𝑐Δif𝑚isodd, (2.13) where 𝑠=𝑚(4(𝑗+𝑖))+𝑐+𝑑.

Proof. Consider ^4𝑗_𝑛=4𝑖+1(−1)^2𝑛+1𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}=^4𝑗_𝑛=4𝑖+1(−1)^2𝑛+1𝛼^{𝑚𝑛+𝑐}−𝛽^{𝑚𝑛+𝑐}𝛼−𝛽𝐴𝛼^{𝑚𝑛+𝑑}−𝐵𝛽^{𝑚𝑛+𝑑}=1𝛼−𝛽𝛼−𝛽^24𝑗_𝑛=4𝑖+1(−1)^2𝑛+1𝐴𝛼^{2𝑚𝑛+𝑐+𝑑}+𝐵𝛽^{2𝑚𝑛+𝑐+𝑑}−(−1)^{𝑚𝑛+𝑐}𝛼−𝛽²𝐴𝛼^𝑑−𝑐+𝐵𝛽^𝑑−𝑐=1Δ^4𝑗_𝑛=4𝑖+1(−1)^2𝑛+1𝑋_{2𝑚𝑛+𝑐+𝑑}−1Δ𝑋_𝑑−𝑐(−1)^𝑐4𝑗_𝑛=4𝑖+1(−1)^2𝑛+1. (2.14) Here we have that ∑^4𝑗_𝑛=4𝑖+1(−1)^2𝑛+1=0 and, by 𝑋_𝑛=𝑊_𝑛−1+𝑊_𝑛+1, ^4𝑗_𝑛=4𝑖+1(−1)^2𝑛+1𝑋_{2𝑚𝑛+𝑐+𝑑}=Δ𝑉_𝑚𝑈_{4𝑚(𝑗−𝑖)}𝑊_{𝑚(4(𝑗+𝑖)+1)+𝑐+𝑑}𝑉_2𝑚, (2.15) for even 𝑚. Now formula (2.12) follows. The remaining formulas are proven in a similar manner.

Notice that in (2.12)-(2.13), one limit of summation is even while the other is odd. Accordingly we have observed that each of (2.12)-(2.13) has a dual sum that is obtained with the use of the rule below. We highlight this rule since it also applies to get certain groups of sums in Section 2.

From [3], we recall the rule for the formation of the dual sum.(1)Replace the even limit by the even limit corresponding to the other residue class modulo 4 and the odd limit by the odd limit corresponding to the other residue class modulo 4.(2)Calculate the subscripts on the right in accordance with the paragraph following (2.13).(3)Multiply the right side by −1.

For example, for odd integer 𝑚, the dual of (2.13) is^4𝑗+1_𝑛=4𝑖(−1)^2𝑛+1𝑈_{𝑚𝑛+𝑐}𝑊_{𝑚𝑛+𝑑}1=−Δ𝑉_𝑚𝑉_{2𝑚(2(𝑗−𝑖)+1)}𝑋_𝑠+𝑚𝑉_2𝑚+2(−1)^𝑐𝑋_𝑑−𝑐, (2.16) where 𝑠 is defined as before.

Theorem 2.5. Fix integers 𝑐,𝑑, and 𝑚.(i)If 𝑐 and 𝑑 have the same parities, then ^4𝑗_𝑛=4𝑖+1(−1)^2𝑛+1𝑊_{(2𝑚+1)𝑛+𝑐}𝑊_{(2𝑚+1)𝑛+𝑑}=𝑈_2𝑚+1𝑈_{4(2𝑚+1)(𝑗−𝑖)}𝑉_2(2𝑚+1)×𝑋_{2(2𝑚+1)(𝑗+𝑖)+𝑡}𝑋_{2(2𝑚+1)(𝑗+𝑖)+𝑡+1}+𝑝𝑟(−1)^𝑡,_{4𝑗𝑛=4𝑖+1}(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=𝑉_2𝑚𝑈_{8𝑚(𝑗−𝑖)}𝑉_4𝑚×𝑊_{𝑚(4𝑗+4𝑖)+𝑡}𝑋_{𝑚(4𝑗+4𝑖)+𝑡},_{4𝑗𝑛=4𝑖+3}(−1)^2𝑛+1𝑊_{(2𝑚+1)𝑛+𝑐}𝑊_{(2𝑚+1)𝑛+𝑑}=𝑈_2𝑚+1𝑉_{2(2𝑚+1)(2(𝑗−𝑖)−1)}𝑉_2(2𝑚+1)×𝑊_{(2𝑚+1)(2(𝑗+𝑖)+1)+𝑡+1}𝑋_{(2𝑚+1)(2(𝑗+𝑖)+1)+𝑡}−𝑟(−1)^𝑡,_{4𝑗𝑛=4𝑖+3}(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=𝑉_2𝑚𝑉_{4𝑚(2(𝑗−𝑖)−1)}Δ𝑉_4𝑚×𝑋²_{𝑚(4(𝑗+𝑖)+2)+𝑡}+2𝑟(−1)^𝑡+2𝑟(−1)^𝑐𝑉_𝑑−𝑐Δ_{4𝑗+3𝑛=4𝑖},(−1)^2𝑛+1𝑊_{(2𝑚+1)𝑛+𝑐}𝑊_{(2𝑚+1)𝑛+𝑑}=𝑉_2𝑚+1𝑈_{4(2𝑚+1)(𝑗−𝑖+1)}𝑉_2(2𝑚+1)×𝑊_{(2𝑚+1)(2(𝑗+𝑖)+1)+𝑡+1}𝑋_{(2𝑚+1)(2(𝑗+𝑖)+1)+𝑡}−𝑟(−1)^𝑡,_{4𝑗+3𝑛=4𝑖}(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=𝑈_2𝑚𝑈_{8𝑚(𝑗−𝑖+1)}𝑉_4𝑚×𝑋²_{𝑚(4(𝑗+𝑖)+2)+𝑡}+2𝑟(−1)^𝑡,_{4𝑗+3𝑛=4𝑖+2}(−1)^2𝑛+1𝑊_{(2𝑚+1)𝑛+𝑐}𝑊_{(2𝑚+1)𝑛+𝑑}=𝑉_2𝑚+1𝑉_{2(2𝑚+1)(2(𝑗−𝑖)+1)}Δ𝑉_2(2𝑚+1)×𝑋_{2(2𝑚+1)(𝑗+𝑖+1)+𝑡+1}𝑋_{2(2𝑚+1)(𝑗+𝑖+1)+𝑡}+𝑝𝑟(−1)^𝑡−2𝑟(−1)^𝑐𝑉_𝑑−𝑐Δ_{4𝑗+3𝑛=4𝑖+2},(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=1𝑉_4𝑚𝑈_2𝑚𝑉_{4𝑚(2(𝑗−𝑖)+1)}𝑊_{4𝑚(𝑗+𝑖+1)+𝑡}𝑋_{4𝑚(𝑗+𝑖+1)+𝑡}, (2.17) where 𝑡=(𝑐+𝑑)/2+𝑚.(ii)If 𝑐 and 𝑑 have different parities, then ^4𝑗_𝑛=4𝑖+1(−1)^2𝑛+1𝑊_{(2𝑚+1)𝑛+𝑐}𝑊_{(2𝑚+1)𝑛+𝑑}=𝑈_2𝑚+1𝑈_{4(2𝑚+1)(𝑗−𝑖)}𝑉_2(2𝑚+1)𝑋²_{2()2𝑚+1(𝑗+𝑖)+𝑣}+2𝑟(−1)^𝑣,_{4𝑗𝑛=4𝑖+1}(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=𝑉_2𝑚𝑈_{8𝑚(𝑗−𝑖)}𝑉_4𝑚𝑋_{4𝑚(𝑗+𝑖)+𝑣−1}𝑊_{4𝑚(𝑗+𝑖)+𝑣}−𝑟(−1)^𝑣,_{4𝑗𝑛=4𝑖+3}(−1)^2𝑛+1𝑊_{(2𝑚+1)𝑛+𝑐}𝑊_{(2𝑚+1)𝑛+𝑑}=1𝑉_2(2𝑚+1)×𝑈_2𝑚+1𝑉_{2(2𝑚+1)(2(𝑗−𝑖)−1)}𝑊_{(2𝑚+1)(2(𝑗+𝑖)+1)+𝑣}𝑋_{(2𝑚+1)(2(𝑗+𝑖)+1)+𝑣},_{4𝑗𝑛=4𝑖+3}(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=𝑉_2𝑚𝑉_{4𝑚(2(𝑗−𝑖)−1)}Δ𝑉_4𝑚×𝑋_{𝑚(4(𝑗+𝑖)+2)+𝑣}𝑋_{𝑚(4(𝑗+𝑖)+2)+𝑣−1}+𝑝𝑟(−1)^𝑣+2𝑟(−1)^𝑐𝑉_𝑑−𝑐Δ_{4𝑗+3𝑛=4𝑖},(−1)^2𝑛+1𝑊_{(2𝑚+1)𝑛+𝑐}𝑊_{(2𝑚+1)𝑛+𝑑}=𝑉_2𝑚+1𝑈_{4(2𝑚+1)(𝑗−𝑖+1)}𝑉_2(2𝑚+1)×𝑋_{(2𝑚+1)(2(𝑗+𝑖)+1)+𝑣}𝑊_{(2𝑚+1)(2(𝑗+𝑖)+1)+𝑣},_{4𝑗+3𝑛=4𝑖}(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=𝑈_2𝑚𝑈_{8𝑚(𝑗−𝑖+1)}𝑉_4𝑚×𝑋_{𝑚(4(𝑗+𝑖)+2)+𝑣}𝑋_{𝑚(4(𝑗+𝑖)+2)+𝑣−1}−𝑝𝑟(−1)^𝑣,_{4𝑗+3𝑛=4𝑖+2}(−1)^2𝑛+1𝑊_{(2𝑚+1)𝑛+𝑐}𝑊_{(2𝑚+1)𝑛+𝑑}()=−2𝑟−1^𝑐𝑉_𝑑−𝑐Δ+𝑉_2𝑚+1𝑉_{2(2𝑚+1)(2(𝑗−𝑖)+1)}Δ𝑉_2(2𝑚+1)𝑋²_{2()2𝑚+1(𝑗+𝑖+1)+𝑣}+2𝑟(−1)^𝑣,_{4𝑗+3𝑛=4𝑖+2}(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=𝑈_2𝑚𝑉_{4𝑚(2(𝑗−𝑖)+1)}𝑉_4𝑚×𝑊_{4𝑚(𝑗+𝑖+1)+𝑣}𝑋_{4𝑚(𝑗+𝑖+1)+𝑣−1}−𝑟(−1)^𝑣, (2.18) where 𝑟 is defined as before and 𝑣=(𝑐+𝑑+1)/2+𝑚.

Proof. Suppose that 𝑐 and 𝑑 have the same parities. Consider ^4𝑗+3_𝑛=4𝑖+2(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=1Δ^4𝑗+3_𝑛=4𝑖+2(−1)^2𝑛+1𝐴²𝛼^{4𝑚𝑛+𝑐+𝑑}+𝐵²𝛽^{4𝑚𝑛+𝑐+𝑑}−𝐴𝐵𝛼^{2𝑚𝑛+𝑐}𝛽^{2𝑚𝑛+𝑑}−𝐴𝐵𝛽^{2𝑚𝑛+𝑐}𝛼^{2𝑚𝑛+𝑑}=1Δ𝑏^24𝑗+3_𝑛=4𝑖+2(−1)^2𝑛+1𝑉_{4𝑚𝑛+𝑐+𝑑}+2𝑎𝑏^4𝑗+3_𝑛=4𝑖+2(−1)^2𝑛+1𝑉_{4𝑚𝑛+𝑐+𝑑−1}+𝑎^24𝑗+3_𝑛=4𝑖+2(−1)^2𝑛+1𝑉_{4𝑚𝑛+𝑐+𝑑−2}+1Δ(−1)^𝑐𝑟𝑉_𝑑−𝑐^4𝑗+3_𝑛=4𝑖+2(−1)^2𝑛+1. (2.19) From the definition of {𝑉_𝑛}, we obtain ^4𝑗+3_𝑛=4𝑖+2(−1)^2𝑛+1𝑉_{4𝑚𝑛+𝑐}=Δ𝑈_2𝑚𝑉_{4𝑚(2(𝑗−𝑖)+1)}𝑈_{2𝑚(4(𝑗+𝑖+1)+1)+𝑐}𝑉_4𝑚. (2.20) Since ^4𝑗+3_𝑛=4𝑖+2(−1)^2𝑛+1=0, (2.21) we get ^4𝑗+3_𝑛=4𝑖+2(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=𝑈_2𝑚𝑉_{4𝑚(2(𝑗−𝑖)+1)}𝑉_4𝑚𝑏²𝑈_{2𝑚(4(𝑗+𝑖+1)+1)+𝑐+𝑑}+2𝑎𝑏𝑈_{2𝑚(4(𝑗+𝑖+1)+1)+𝑐+𝑑−1}+𝑎²𝑈_{2𝑚(4(𝑗+𝑖+1)+1)+𝑐+𝑑−2} (2.22) Taking 2𝑘=2𝑚(4(𝑗+𝑖+1)+1)+𝑐+𝑑 in Lemma 2.1, we write ^4𝑗+3_𝑛=4𝑖+2(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}=𝑈_2𝑚𝑉_{4𝑚(2(𝑗−𝑖)+1)}𝑊_{𝑚(4(𝑗+𝑖+1)+1)+(𝑑+𝑐)/2}𝑋_{𝑚(4(𝑗+𝑖+1)+1)+(𝑑+𝑐)/2}Δ𝑉_4𝑚. (2.23) Thus the result follows. Similar arguments yield the remaining formulas, where we must consider the parities of 𝑐,𝑑.

For example, the dual of (2.17) is given by if 𝑐 and 𝑑 have the same parities,^4𝑗+1_𝑛=4𝑖(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}𝑈=−^2𝑚𝑉_{4𝑚(2(𝑗−𝑖)+1)}𝑊_{4𝑚(𝑗+𝑖)+𝑡}𝑋_{4𝑚(𝑗+𝑖)+𝑡}𝑉_4𝑚, (2.24) and the dual of (2.18) is given by if 𝑐 and 𝑑 have different parities, ^4𝑗+1_𝑛=4𝑖(−1)^2𝑛+1𝑊_{2𝑚𝑛+𝑐}𝑊_{2𝑚𝑛+𝑑}𝑈=−^2𝑚𝑉_{4𝑚(2(𝑗−𝑖)+1)}𝑉_4𝑚×𝑊_{4𝑚(𝑗+𝑖)+𝑣}𝑋_{4𝑚(𝑗+𝑖)+𝑣−1}−𝑟(−1)^𝑣, (2.25) where 𝑡 and 𝑣 are defined as before.

Acknowledgment

Thanks are due to the anonymous referee who made suggestions towards a better presentation.