INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A08

CONJUGATE SEQUENCES IN A FIBONACCI-LUCAS SENSE AND SOME IDENTITIES FOR SUMS OF POWERS OF THEIR ELEMENTS

Roman Wituła

Institute of Mathematics, Silesian University of Technology, Kaszubska 23, 44-100 Gliwice, Poland

Damian Słota

Institute of Mathematics, Silesian University of Technology, Kaszubska 23, 44-100 Gliwice, Poland

                             d.slota@polsl.pl

Received: 5/30/06, Revised: 11/16/06, Accepted: 2/3/07, Published: 2/9/07

Abstract

The aim of this paper is to introduce the notion of a pair of conjugate sequences in a Fibonacci-Lucas sense. New identities are generated by means of such sequences (including separate derivations for Fibonacci and Lucas numbers) and by means of Lucas and Vieta-Lucas polynomials.

1. Introduction

The paper, which is a follow-up of [17], is still focused on analyzing the notion of a pair of sequences {x_{n}} and {y_{n}} that are conjugate in a Fibonacci-Lucas sense (Sections 2 and 4). A number of identities are derived for such pairs of conjugate sequences (see Lemma 7).

In Section 3, by means of Lucas and Vieta-Lucas polynomials, some identities (Lemma 9) are presented for sequences {z_{n}} defined by recurrence formula (z_{0}, z_{1} ∈ ℂ):

z_{n+2} = a z_{n+1} + z_{n}, n ∈ ℕ_{0} := ℕ ∪ {0},

and for auxiliary sequences (ξ_{0,n} = z_{n}):

ξ_{k,n} = z_{n+k} + (−1)^{k} z_{n−k}, k = 1, …, n, n ∈ ℕ.

In Section 5 the decompositions of some special symmetric polynomials of three, four, and five variables are given on the grounds of Vieta-Lucas polynomials. Finally, in Section 6 the decompositions of the polynomials discussed in Section 5 are applied to generate the corresponding identities for the pairs of conjugate sequences in a Fibonacci-Lucas sense. Special cases of such identities for Fibonacci and Lucas numbers are discussed separately ((6.42)– (6.45)).

Several publications were concerned with identities for the sums of the powers of Fibonacci and Lucas numbers (see [7, 8, 9, 10, 11, 17] and references therein). However, it should be emphasized that the research tasks undertaken in this paper are substantively different from those already discussed elsewhere.

2. Conjugate Sequences

Let us assume that the elements of sequences {x_{n}} and {y_{n}} are determined by means of the same recurrence relation:

x_{n+1} = a x_{n} + b x_{n−1},  y_{n+1} = a y_{n} + b y_{n−1}, n ∈ ℕ,  (2.1)

where b = ±1, a ≠ 0, a^{2} + 4b > 0. First our objective is to find some adjoint-conditions – connecting elements of sequences {x_{n}} and {y_{n}}. These conditions are derived from the following identities for Fibonacci and Lucas numbers:

F_{n−1} + F_{n+1} = L_{n},
L_{n−1} + L_{n+1} = 5 F_{n}.

Lemma 1 If there exists A ∈ ℝ and n_{0} ∈ ℕ such that x_{n_{0}} ≠ 0,

y_{n} = x_{n−1} + x_{n+1}, for n = n_{0} − 1, n_{0} + 1,

and

Ax_{n} = y_{n−1} + y_{n+1}, for n = n_{0},

then, A = 2 + 2b + a^{2}.

Proof. We have

Ax_{n_{0}} = y_{n_{0}−1} + y_{n_{0}+1} = x_{n_{0}−2} + 2x_{n_{0}} + x_{n_{0}+2}
= x_{n_{0}−2} + abx_{n_{0}−1} + (2 + b + a^{2})x_{n_{0}} = (2 + 2b + a^{2})x_{n_{0}}

so A = 2 + 2b + a^{2}. □

Lemma 2 Let α, β ∈ ℂ, αβ ≠ 0, x_{n} = uα^{n} + vβ^{n}, y_{n} = ζα^{n} + ξβ^{n}, n ∈ ℕ_{0}. If |α| ≠ |β| and identities:

\{ \begin{matrix}y_{n} = x_{n−1} + x_{n+1} \\ (2 + 2b + a^{2})x_{n} = y_{n−1} + y_{n+1} \end{matrix}  (2.2)

hold for every n ∈ ℕ, then the following identities are satisfied:

\begin{aligned}y_{n} & = (α^{−1} + α) u α^{n} + (β^{−1} + β) v β^{n} \\ x_{n} & = \frac{α + α^{−1}}{2 + 2b + a^{2}} ζ α^{n} + \frac{β + β^{−1}}{2 + 2b + a^{2}} ξ β^{n}, \end{aligned}  (2.3)

for every n ∈ ℕ.

Proof. We have

y_{n} = x_{n−1} + x_{n+1} = uα^{n−1} + uα^{n+1} + vβ^{n−1} + vβ^{n+1} = u(α^{−1} + α)α^{n} + v(β^{−1} + β)β^{n}.

□

Corollary 3 If b = −1 then by (2.2) we obtain: ax_{n} = y_{n}, for all n ∈ ℕ.

Definition 4 Sequences {x_{n}}_{n=1}^{∞} and {y_{n}}_{n=1}^{∞}, defined by the same relation

x_{n+1} = ax_{n} + x_{n−1} and y_{n+1} = ay_{n} + y_{n−1},

are called conjugate sequences in a Fibonacci-Lucas sense (with parameter a) if the following conditions are satisfied:

y_{n} = x_{n−1} + x_{n+1},
(4 + a^{2})x_{n} = y_{n−1} + y_{n+1},

for every n ∈ ℕ.

Remark 5 If a = 1, then α = (1+\sqrt{5})/2, β = (1−\sqrt{5})/2. Additionally, if x_{n} = uα^{n} +vβ^{n}, y_{n} = ζα^{n} + ξβ^{n}, n ∈ ℕ_{0} and identities (2.2) hold, then

y_{n} = \sqrt{5}uα^{n} − \sqrt{5}vβ^{n} and x_{n} = \frac{\sqrt{5}}{5}uα^{n} − \frac{\sqrt{5}}{5}vβ^{n}, n ∈ ℕ_{0}.

In the language of conjugate sequences, we may recapitulate the discussion to this point as follows:

Lemma 6 If elements of two sequences {x_{n}} and {y_{n}} are conjugates in a Fibonacci-Lucas sense with parameter a and x_{n} = uα^{n}+vβ^{n}, n ∈ ℕ_{0}, where α, β are roots of the characteristic equation 𝕏^{2} − a𝕏 − 1 = 0, then

αβ = −1,
α + α^{−1} = −(β + β^{−1}) = α − β = ±\sqrt{a^{2} + 4}

and, consequently, we obtain

y_{n} = ±\sqrt{a^{2} + 4}(uα^{n} − vβ^{n}), n ∈ ℕ,

or

y_{n} = ±\frac{1}{\sqrt{a^{2} + 4}}(uα^{n} − vβ^{n}), n ∈ ℕ.

Proof. We note that

(α − β)^{2} + 4αβ = (α + β)^{2}

i.e.,

α − β = ±\sqrt{a^{2} + 4}

and

α + α^{−1} + β + β^{−1} = (α + β)(αβ + 1)(αβ)^{−1} = 0.
□

Lemma 7 Let us assume that the elements of sequences {x_{n}} and {y_{n}} are conjugates in a Fibonacci-Lucas sense with parameter a, or, more precisely, let us assume that the following conditions are satisfied:

x_{n+1} = ax_{n} + x_{n−1},  y_{n+1} = ay_{n} + y_{n−1},  (2.4)

y_{n} = x_{n−1} + x_{n+1}  (2.5)

and

(a^{2} + 4)x_{n} = y_{n−1} + y_{n+1}.  (2.6)

Then, the following identities hold:

z_{n} + az_{n+3} = (a^{2} + 1)z_{n+2},  (2.7)

z_{n+4} + cz_{n+2} + z_{n} = az_{n+3} + (1 + c)z_{n+2} + z_{n}
= (a^{2} + 1 + c)z_{n+2} + az_{n+1} + z_{n}
= (a^{2} + 2 + c)z_{n+2}  (2.8)

for every c ∈ ℂ and z ∈ {x, y},

(a^{2} + 4)x_{n+2} + y_{n−1} = y_{n+3} + y_{n+1} + y_{n−1} = (a^{2} + 3)y_{n+1},  (2.9)

y_{n+2} + x_{n−1} = x_{n+3} + x_{n+1} + x_{n−1} = (a^{2} + 3)x_{n+1},  (2.10)

x_{n} + x_{n+4} + ax_{n+5} = (a^{2} + 1)x_{n+4} + ax_{n+3} + x_{n}
= (a^{2} + 1)x_{n+4} + (a^{2} + 1)x_{n+2}
= (a^{2} + 1)y_{n+3},  (2.11)

x_{n+4} − x_{n} = x_{n+4} + x_{n+2} − (x_{n+2} + x_{n}) = y_{n+3} − y_{n+1} = ay_{n+2},  (2.12)

x_{n} + ax_{n+5} = a^{2}y_{n+3} + x_{n+2},  (2.13)

y_{n} + y_{n+4} + ay_{n+5} = (a^{2} + 1)y_{n+4} + ay_{n+3} + y_{n}
= (a^{2} + 1)y_{n+4} + (a^{2} + 1)y_{n+2}
= (a^{2} + 1)(a^{2} + 4)x_{n+3},  (2.14)

y_{n+4} − y_{n} = y_{n+4} + y_{n+2} − (y_{n+2} + y_{n}) =
= (a^{2} + 4)(x_{n+3} − x_{n+1}) = a(a^{2} + 4)x_{n+2}  (2.15)

y_{n} + ay_{n+5} = a^{2}(a^{2} + 4)x_{n+3} + y_{n+2}.  (2.16)

However, we have

z_{n} + z_{n+2} + z_{n+4} + z_{n+6} + a(z_{n+5} + z_{n+7}) = (a^{2} + 1)(a^{2} + 4)z_{n+4}  (2.17)

and

z_{n+8} − 2z_{n+4} + z_{n} = a^{2}(a^{2} + 4)z_{n+4}  (2.18)

for every z ∈ {x, y}.

In the next section some generalizations of some of the identities from Lemma 7 will be presented.

3. Lucas and Vieta-Lucas Polynomials

Let us set

Ω_{n}(x) := 2T_{n}(x/2)

and

W_{n}(x) := (−i)^{n}Ω_{n}(ix)

for every n = 0, 1, 2, …, where T_{n}(x) means the n-th Chebyshev polynomial of the first kind. We note that the polynomials Ω_{n}(x), called Vieta-Lucas polynomials (see [6, 15]), satisfy the following recurrence relations:

Ω_{0}(x) = 2, Ω_{1}(x) = x,
Ω_{k+2}(x) = x Ω_{k+1}(x) − Ω_{k}(x), k ∈ ℕ.

Table 1: Triangle of the coefficients of polynomials W_{n}(x)

\begin{aligned}& 1 & x & x^{2} & x^{3} & x^{4} & x^{5} & x^{6} & x^{7} & x^{8} & x^{9} \\ W_{0}(x) & 2 \\ W_{1}(x) & & 1 \\ W_{2}(x) & 2 [figure] & & 1 \\ W_{3}(x) & & 3 [figure] & & 1 \\ W_{4}(x) & 2 [figure] & & 4 [figure] & & 1 \\ W_{5}(x) & & 5 [figure] & & 5 [figure] & & 1 \\ W_{6}(x) & 2 [figure] & & 9 [figure] & & 6 [figure] & & 1 \\ W_{7}(x) & & 7 [figure] & & 14[figure] & & 7 [figure] & & 1 \\ W_{8}(x) & 2 [figure] & & 16 [figure] & & 20[figure] & & 8 [figure] & & 1 \\ W_{9}(x) & & 9 [figure] & & 30[figure] & & 27[figure] & & 9 [figure] & & 1 \end{aligned} Legende: b a [figure] a+b

Hence, we obtain

Ω_{2}(x) = x^{2} − 2, Ω_{3}(x) = x^{3} − 3x, Ω_{4}(x) = x^{4} − 4x^{2} + 2,
Ω_{5}(x) = x^{5} − 5x^{3} + 5x,  Ω_{6}(x) = x^{6} − 6x^{4} + 9x^{2} − 2,
Ω_{7}(x) = x^{7} − 7x^{5} + 14x^{3} − 7x,  etc.

Similarly, we have

W_{0}(x) = 2, W_{1}(x) = x,
W_{n+2}(x) = x W_{n+1}(x) + W_{n}(x),

which generate the so-called Lucas polynomials [1, 2, 3, 4, 6]):

W_{2}(x) = x^{2} + 2, W_{3}(x) = x^{3} + 3x, W_{4}(x) = x^{4} + 4x^{2} + 2,
W_{5}(x) = x^{5} + 5x^{3} + 5x,  W_{6}(x) = x^{6} + 6x^{4} + 9x^{2} + 2,
W_{7}(x) = x^{7} + 7x^{5} + 14x^{3} + 7x,  etc.

Lemma 8 We have

coeff[x^{2}; W_{2k}(x)] = k^{2}, coeff[x^{3}; W_{2k+1}(x)] = ∑_{l=1}^{k} l^{2} = \frac{1}{6} k (k + 1) (2k + 1),

coeff[x^{4}; W_{2k}(x)] = ∑_{l=1}^{k}(k − l)l^{2} = \frac{1}{12} k^{2} (k^{2} − 1),

W_{n}(x) = ∑_{k=0}^{⌊n/2⌋} \frac{n}{n − k} \binom{n − k}{k} x^{n−2k},  (3.19)

x ∑_{r=1}^{s} W_{2r−1}(x) = W_{2s}(x) − 2,  (3.20)

and

x ∑_{r=0}^{s} W_{2r}(x) = W_{2s+1}(x).  (3.21)

The polynomials W_{n}(x), n ∈ ℕ, enable the description of many fundamental identities, involving the elements of the sequence {z_{n}}_{n=1}^{∞} determined by the following one-parameter recurrence relation:

z_{0}, z_{1} ∈ ℂ  and  z_{n+2} = a z_{n+1} + z_{n},  n ∈ ℕ_{0}.  (3.22)

Additionally, let us assume that ξ_{0,n} := z_{n} and

ξ_{k,n} := z_{n+k} + (−1)^{k} z_{n−k}

for every k = 1, 2, …, n − 1.

Some of the identities are generalized forms of the identities from Lemma 7 (especially identities (2.4)–(2.8) and (2.17)–(2.18)). All the identities presented below seem to be original.

Lemma 9 The following identities hold true

ξ_{k+1,n} = aξ_{k,n} + ξ_{k−1,n},  (3.23)
ξ_{k+1,n} = W_{k−s}(a)ξ_{s+1,n} + W_{k−s−1}(a)ξ_{s,n}, s = 1 ≤ k,  (3.24)
ξ_{k,n} = W_{k}(a)z_{n},  (3.25)
∑_{r=−s}^{s} z_{n−2r} = ( ∑_{r=0}^{s} W_{2r}(a))z_{n} = \frac{1}{a}W_{2s+1}(a)z_{n},  (3.26)

∑_{r=1}^{s} (z_{n+2r−1} − z_{n−2r+1}) = ( ∑_{r=1}^{s} z_{n+2r−1}) − ( ∑_{t=1}^{s} z_{n−2r+1}) =

= ( ∑_{r=1}^{s} W_{2r−1}(a))z_{n} = \frac{1}{a}(W_{2s}(a) − 2).  (3.27)

Proof. First, we note that

z_{n+2k+1} − z_{n−2k−1} = az_{n+2k} + z_{n+2k−1} − (z_{n−2k+1} − az_{n−2k})
= a(z_{n+2k} + z_{n−2k}) + z_{n+2k−1} − z_{n−2k+1},
z_{n+2k} + z_{n−2k} = a(z_{n+2k−1} − z_{n−2k+1}) + z_{n+2k−2} + z_{n−2k+2},

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A08 8

i.e., that

ξ_{k+1,n} = aξ_{k,n} + ξ_{k−1,n} = W_{1}(a)ξ_{k,n} + W_{0}(a)ξ_{k−1,n}.

Now, suppose that for some s ∈ ℕ_{0}, s ⩽ k we have

ξ_{k+1,n} = W_{k−s}(a)ξ_{s+1,n} + W_{k−s−1}(a)ξ_{s,n}.

Hence, we get

ξ_{k+1,n} = W_{k−s}(a)(aξ_{s,n} + ξ_{s−1,n}) + W_{k−s−1}(a)ξ_{s,n}
= W_{k−s−1}(a)ξ_{s,n} + W_{k−s}(a)ξ_{s−1,n}.

This leads to

ξ_{k+1,n} = W_{k}(a)ξ_{1,n} + W_{k−1}(a)ξ_{0,n}
= (aW_{k}(a) + W_{k−1}(a))z_{n} = W_{k+1}(a)z_{n}.
□

Corollary 10 We have

z_{n+8} − 2z_{n+4} + 2z_{n} − 2z_{n−4} + z_{n−8} = a^{2}(a^{2} + 4)W_{4}(a)z_{n},

z_{2k} + (−1)^{k}z_{0} = W_{k}(a)z_{k},  (3.28)
az_{2k+1} + 2(−1)^{k+1}z_{0} = W_{k+1}(a)z_{k+1} − W_{k}(a)z_{k},  (3.29)
a ∑_{k=0}^{2s} z_{2k} = W_{2s+1}(a)z_{2s},  (3.30)
a ∑_{k=0}^{s} z_{2k+1} = W_{s+1}(a)z_{s+1} + ((−1)^{s} − 1)z_{0},  (3.31)
a( ∑_{k=s}^{2s−1} z_{2k} − ∑_{l=0}^{s−1} z_{2l}) = (W_{2s}(a) − 2)z_{2s−1},  (3.32)
a ∑_{k=0}^{4s} z_{k} = W_{2s+1}(a)z_{2s} + W_{2s−1}(a)z_{2s−1},  (3.33)
a ∑_{k=1}^{s} W_{k}(a)z_{k} = a ∑_{k=1}^{s} z_{2k} + \frac{a}{2}((−1)^{s} − 1)z_{0}
= \{ \begin{matrix}W_{s+1}(a)z_{s} − az_{0} & for s ∈ 2ℕ, \\ W_{s}(a)z_{s−1} − az_{s} − 2az_{0} & for s ∈ 2ℕ − 1. \end{matrix}  (3.34)

4. Some Generalizations of Conjugate Sequences Concept

Let us assume that the sequences {x_{n}} and {y_{n}} are determined by the same recurrence equations (2.1). The following result is a response to an attempt to substitute conditions (2.2) by some other conditions connecting the elements of the sequences {x_{n}} and {y_{n}}.

Lemma 11 Let A, B, C, D, α, β ∈ ℂ, |α| ≠ |β|. Let us assume that there exist u, v ∈ ℂ so that for every n ∈ ℕ the following identities are satisfied:

x_{n} = Aα^{n} + Bβ^{n}  (4.35)

y_{n} = Cα^{n} + Dβ^{n}  (4.36)

x_{n} + x_{n+2} + x_{n+4} + x_{n+6} = u · y_{n+3}  (4.37)

and

y_{n} + y_{n+2} + y_{n+4} + y_{n+6} = v · x_{n+3}.  (4.38)

If ABCD ≠ 0 then −b = αβ = ±1. Moreover, if AC ≠ 0 (resp. BD ≠ 0) then

u = \frac{A}{C}(α + α^{−1})[(α + α^{−1})^{2} − 2]  and  v = \frac{C}{A}(α + α^{−1})[(α + α^{−1})^{2} − 2]

(resp., u = \frac{B}{D}(β + β^{−1})[(β + β^{−1})^{2} − 2] and v = \frac{D}{B}(β + β^{−1})[(β + β^{−1})^{2} − 2]).

Proof. The following four identities from (4.37) and (4.38) can be generated:

\{ \begin{matrix}A\frac{α^{8}−1}{α^{3}(α^{2}−1)} = A\frac{α^{4}−α^{−4}}{α−α^{−1}} = uC, \\ B \frac{β^{8}−1}{β^{3}(β^{2}−1)} = B \frac{β^{4}−β^{−4}}{β−β^{−1}} = uD, \\ C \frac{α^{8}−1}{α^{3}(α^{2}−1)} = C \frac{α^{4}−α^{−4}}{α−α^{−1}} = vA, \\ D\frac{β^{8}−1}{β^{3}(β^{2}−1)} = D\frac{β^{4}−β^{−4}}{β−β^{−1}} = uB. \end{matrix}  (4.39)

Hence

A(\frac{α^{4} − α^{−4}}{α − α^{−1}} )^{2} = uvA and A(\frac{β^{4} − β^{−4}}{β − β^{−1}} )^{2} = uvB.

If ABCD ≠ 0 then

(\frac{α^{4} − α^{−4}}{α − α^{−1}} )^{2} = (\frac{β^{4} − β^{−4}}{β − β^{−1}} )^{2},

i.e.,

α^{3} + α + α^{−1} + α^{−3} = ±(β^{3} + β + β^{−1} + β^{−3})

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A08 10

or

(α + α^{−1})^{3} − 2(α + α^{−1}) ± ((β + β^{−1})^{3} − 2(β + β^{−1})) =

= ((α + α^{−1}) ± (β + β^{−1}))((α + α^{−1})^{2} ∓ (β + β^{−1})(α + α^{−1}) + (β + β^{−1})^{2} − 2), while

△_{α+α^{−1}} = (β + β^{−1})^{2} − 4(β + β^{−1})^{2} + 8 = −3β^{−2}[β^{4} − \frac{2}{3}β^{2} + 1] < 0.

Therefore

α + α^{−1} = ±(β + β^{−1}),

i.e.,

α^{2} ± (β + β^{−1})α + 1 = 0,

which implies for plus sign

(α + β)(α + β^{−1}) = 0, i.e., αβ = −1

and for minus sign

(α − β)(α − β^{−1}) = 0, i.e., αβ = 1

(since |α| ≠ |β|). □ Corollary 12 If ABCD ≠ 0 and αβ = 1 then

α + α^{−1} = α + β = β^{−1} + β

and

AD = BC.

So, sequences \{x_{n}\} and \{y_{n}\} are linearly dependent; or, more precisely, we have x_{n} = \frac{A}{D}y_{n}, n ∈ ℕ. Corollary 13 If ABCD ≠ 0 and αβ = −1 then

α + α^{−1} = α − β = −(β^{−1} + β)

and

AD + BC = 0.

Moreover, we have

x_{n} + x_{n+2} = \frac{B}{D}(β − α)y_{n+1} and y_{n} + y_{n+2} = \frac{D}{B}(β − α)x_{n+1};

or

x_{n}^{∗} + x_{n+2}^{∗} = (β − α)^{2}y_{n+1} and y_{n} + y_{n+2} = x_{n+1}^{∗}

where x_{n}^{∗} := \frac{D}{B}(β − α)x_{n}, n ∈ ℕ.

In the sequel if B(β − α) = D or D(β − α) = B then sequences {x_{n}} and {y_{n}} are conjugates in a Fibonacci-Lucas sense with parameter a = α + β.

Proof. We have

x_{n} + x_{n+2} = A(1 + α^{2})α^{n} + B(1 + β^{2})β^{n}

= −\frac{BC}{D} α(α + α^{−1})α^{n} + Bβ(β + β^{−1})β^{n}

= \frac{B}{D}(β − α)[Cα^{n+1} + Dβ^{n+1}] = \frac{B}{D}(β − α)y_{n+1}.

□

5. Decompositions of Some Special Polynomials of Many Variables

In this section an attempt is made at decomposing some symmetric polynomials of 3, 4, and 5 variables into factors (such decompositions may easily be generalized to polynomials of 6 or even more variables). The decompositions are derived by means of modified Chebyshev polynomials Ω_{n}(x) (Vieta-Lucas polynomials).

Lemma 14 The following identities hold

p_{n}^{(3)}(x, y, z):= (x+ y+ z)^{n}− (x+ y)^{n}− (x+ z)^{n}− (y+ z)^{n}+ x^{n}+ y^{n}+ z^{n}

= nxyz[\frac{(x+ y+ z)^{n−1}}{xy+ xz+ yz} − (xy+ xz+ yz)^{(n−3)/2}Ω_{n−1}(\frac{x+ y+ z}{(xy+ xz+ yz)^{1/2}} )

+ F_{3,n}(xyz, x+ y+ z, xy, x+ y, xz, x+ z, yz, y+ z)]  (5.40)

(in the following formulas, the compact form of the respective identities is given)

= \{ \begin{matrix}0, & for & n ⩽ 2, \\ 6xyz, & for & n = 3, \\ 12xyz(x+ y+ z), & for & n = 4, \\ 10xyz[2(x+ y+ z)^{2}− xy− xz− yz], & for & n = 5, \\ 30xyz(x+ y+ z)[(x+ y+ z)^{2}− xy− xz− yz], & for & n = 6, \\ 7xyz[6(x+ y+ z)^{4}− 9(xy+ xz+ yz)(x+ y+ z)^{2} \\ +2(xy+ xz+ yz)^{2}− xyz(x+ y+ z)], & for & n = 7, \\ 56xyz(x+ y+ z)[((x+ y+ z)^{2}− xy− xz− yz)^{2} \\ −\frac{1}{2}xyz(x+ y+ z)], & for & n = 8, \\ 18xyz[4(x+ y+ z)^{6}− 10(xy+ xz+ yz)(x+ y+ z)^{4} \\ +8(xy+ xz+ yz)^{2}(x+ y+ z)^{2}− (xy+ xz+ yz)^{3} \\ −4xyz(x+ y+ z)^{3} + xyz(xy(x+ y)+ xz(x+ z) \\ +yz(y+ z))+ \frac{10}{3}(xyz)^{2}], & for & n = 9. \end{matrix}

Moreover, we have F_{3,n}(xyz, x+ y+ z, xy, x+ y, xz, x+ z, yz, y+ z) =

= \{ \begin{matrix}0, & for & n ⩽ 6, \\ −xyz(x+ y+ z), & for & n = 7, \\ −\frac{7}{2}xyz(x+ y+ z)^{2}, & for & n = 8, \\ −8xyz(x+ y+ z)^{3}+ 2xyz(xy(x+ y) \\ +xz(x+ z)+ yz(y+ z))+ \frac{20}{3}(xyz)^{2}, & for & n = 9, \\ −15xyz(x+ y+ z)^{4}+ 9xyz(xy(x+ y)^{2} \\ +xz(x+ z)^{2}+ yz(y+ z)^{2}) + 48x^{2}y^{2}z^{2}(x+ y+ z), & for & n = 10, \\ −25xyz(x+ y+ z)^{5}+ 25xyz(xy(x+ y)^{3} \\ +xz(x+ z)^{3}+ yz(y+ z)^{3}) − 3xyz(x^{3}(y+ z)^{2} \\ +y^{3}(x+ z)^{2}+ z^{3}(x+ y)^{2})+ 92(xyz)^{2}× \\ ×((x+ y)^{2}+ (x+ z)^{2}+ (y+ z)^{2}) \\ +117(xyz)^{3}(x+ y+ z), & for & n = 11, \end{matrix}

and

F_{3,12} = −\frac{77}{2}x^{2}y^{2}z^{2}(x + y + z)^{6}

+ 55x^{2}y^{2}z^{2}(xy(x + y)^{4} + xz(x + z)^{4} + yz(y + z)^{4})

− \frac{33}{2}x^{2}y^{2}z^{2}(x^{2}y^{2}(x + y)^{2} + x^{2}z^{2}(x + z)^{2} + y^{2}z^{2}(y + z)^{2})

+ 484x^{3}y^{3}z^{3}(x + y + z)^{3} − 319x^{3}y^{3}z^{3}(xy + xz + yz)(x + y + z) + \frac{6699}{2}x^{4}y^{4}z^{4}.

Remark 15 We note that xyz divides p_{n}^{(3)}(x, y, z) for every n ∈ ℕ. Lemma 16 We have

p_{n}^{(4)}(x, y, z, u) := (x + y + z + u)^{n} − (x + y + z)^{n} − (x + y + u)^{n} − − (x + z + u)^{n} − (y + z + u)^{n} + (x + y)^{n} + (x + z)^{n} + (x + u)^{n} + (y + z)^{n} + +(y + u)^{n} + (z + u)^{n} − x^{n} − y^{n} − z^{n} − u^{n} =

= n(n − 1)xyzu[\frac{(x + y + z + u)^{n−2}}{xy + xz + xu + yz + yu + zu} −

− (xy + xz + xu + yz + yu + zu)^{(n−4)/2}Ω_{n−2}(\frac{x + y + z + u}{(xy + xz + xu + yz + yu + zu)^{1/2}} ) +

+ F_{4,n}(xyzu, x + y + z + u, xyz, x + y + z, xyu, x + y + u, xzu,

x + z + u, yzu, y + z + u)] =  (5.41)

= \{ \begin{matrix}0, & for & n ⩽ 3, \\ 24xyzu, & for & n = 4, \\ 60xyzu(x+ y+ z+ u), & for & n = 5, \\ 60xyzu[2(x+ y+ z+ u)^{2} − \\ −(xy+ xz+ xu+ yz+ yu+ zu)], & for & n = 6, \\ 210xyzu(x+ y+ z+ u)[(x+ y+ z+ u)^{2} − \\ −(xy+ xz+ xu+ yz+ yu+ zu)], & for & n = 7, \\ 56xyzu[6(x+ y+ z+ u)^{4}− 9(xy+ xz+ xu+ \\ +yz+ yu+ zu)(x+ y+ z+ u)^{2} + 2(xy+ xz+ \\ +xu+ yz+ yu+ zu)^{2}− xyz(x+ y+ z)− \\ −xyu(x+ y+ u)− xuz(x+ u+ z)− \\ −yuz(y+ u+ z)− 3xyzu] = \\ = 28xyzu[5(x+ y+ z+ u)^{4}+ 5(x^{2}+ y^{2}+ z^{2}+ u^{2})× \\ ×(x+ y+ z+ u)^{2} + 2(x^{2}+ y^{2}+ z^{2}+ u^{2})^{2}− \\ −30xyzu− 10xyz(x+ y+ z) − 10xyu(x+ y+ u)− \\ −10xuz(x+ u+ z)− 10uyz(u+ y+ z)− \\ −4(x^{2}y^{2}+ x^{2}u^{2}+ x^{2}z^{2}+ y^{2}u^{2}+ y^{2}z^{2}+ u^{2}z^{2})], & for & n = 8, \end{matrix}

and for n = 9:

= 252xyzu(x+y +z +u)[2(x+y +z +u)^{4} −4(xy +xz +xu+yz +yu+zu)(x+y +z +u)^{2} + + 2(xy + xz + xu + yz + yu + zu)^{2} − xyz(x + y + z)^{2} − xuz(x + u + z)^{2} − xuy(x + u + y)^{2} −

− uyz(u + y + z)^{2} − 42xyuz(x + y + u + z)],

where F_{4,n}(xyzu, x+ y+ z+ u, xyz, x+ y+ z, xyu, x+ y+ u, xzu, x+ z+ u, yzu, y+ z+ u) =

= \{ \begin{matrix}0 & for & n ⩽ 7, \\ −xyz(x + y + z) − xyu(x + y + u)− \\ −xuz(x + u + z) − yuz(y + u + z) − 3xyzu, & for & n = 8, \\ −\frac{7}{2}xyz(x + y + z)^{2} − \frac{7}{2}xyu(x + y + u)^{2} \\ −\frac{7}{2}xuz(x + u + z)^{2} − \frac{7}{2}yuz(y + u + z)^{2}− \\ −147xyuz(x + y + u + z) & for & n = 9. \end{matrix}

Remark 17 We note that xyzu divides p_{n}^{(4)}(x, y, z, u) for every n ∈ ℕ.

Lemma 18 We have the following decompositions:

p_{n}^{(5)}(x, y, z, u, v) := (x + y + z + u + v)^{n} − (x + y + z + u)^{n} − (x + y + z + v)^{n} − − (x + y + u + v)^{n} − (x + z + u + v)^{n} − (z + y + u + v)^{n} + (x + y + z)^{n} + (x + y + u)^{n} + + (x + y + v)^{n} + (x + z + u)^{n} + (x + z + v)^{n} + (x + u + v)^{n} + (y + z + u)^{n} + (y + z + v)^{n} + + (y + u + v)^{n} + (z + u + v)^{n} − (x + y)^{n} − (x + z)^{n} − (x + u)^{n} − (x + v)^{n} − (y + z)^{n} −

− (y + u)^{n} − (y + v)^{n} − (z + u)^{n} − (z + v)^{n} − (u + v)^{n} + x^{n} + y^{n} + z^{n} + u^{n} + v^{n} =

= \{ \begin{matrix}0 & for n ⩽ 4, \\ 5!xyzuv & for n = 5, \\ \frac{6!}{2!}xyzuv(x + y + z + u + v) & for n = 6, \\ \frac{7!}{2!3!}xyzuv[3(x + y + z + u + v)^{2}− \\ −x^{2} − y^{2} − z^{2} − u^{2} − v^{2}] = \\ = \frac{7!}{3!}xyzuv[(x + y + z + u + v)^{2} + xy + xz+ \\ +xu + xv + yz + yu + yv + zu + zv + uv] & for n = 7. \end{matrix}

6. Applications

6.1 Identities for the Powers of Elements of Conjugate Sequences

We now present some identities obtained by applying identity (5.40) to the elements of sequences {x_{n}} and {y_{n}} satisfying (2.4)–(2.6).

a) For x = x_{n+4}, y = −x_{n}, z = −ay_{n+2} and odd powers only:

x_{n+4}^{k} − x_{n}^{k} − a^{k}y_{n+2}^{k} = \{ \begin{matrix}0 & for k = 1, \\ 3ax_{n}y_{n+2}x_{n+4} & for k = 3, \\ 5ax_{n}y_{n+2}x_{n+4}(x_{n}x_{n+4} + a^{2}y_{n+2}^{2}) & for k = 5, \\ 7ax_{n}y_{n+2}x_{n+4}(x_{n}x_{n+4} + a^{2}y_{n+2}^{2})^{2} & for k = 7, \\ 9ax_{n}y_{n+2}x_{n+4}[\frac{1}{3}(ax_{n}y_{n+2}x_{n+4})^{2}− \\ −(x_{n}x_{n+4} + a^{2}y_{n+2}^{2})^{3}] & for k = 9; \end{matrix}

b) for x = x_{n+4}, y = −x_{n}, z = ay_{n+2}:

(2a)^{k}y_{n+2}^{k} + x_{n+4}^{k} + (−1)^{k}x_{n}^{k} − (ay_{n+2} − x_{n})^{k} − (2ay_{n+2} + x_{n})^{k} =

= \{ \begin{matrix}0 & for k ⩽ 2, \\ −6ax_{n}y_{n+2}x_{n+4} & for k = 3, \\ −24a^{2}x_{n}y_{n+2}^{2}x_{n+4} & for k = 4, \\ −10ax_{n}y_{n+2}x_{n+4}[7a^{2}y_{n+2}^{2} + x_{n}x_{n+4}] & for k = 5, \\ −60a^{2}x_{n}y_{n+2}^{2}x_{n+4}[3a^{2}y_{n+2}^{2} + x_{n}x_{n+4}] & for k = 6, \\ −112a^{2}x_{n}y_{n+2}^{2}x_{n+4}[(3a^{2}y_{n+2}^{2} + x_{n}x_{n+4})^{2}+ \\ +a^{2}x_{n}y_{n+2}^{2}x_{n+4}] & for k = 8; \end{matrix}

c) for x = x_{n+4}, y = x_{n}, z = −ay_{n+2}:

x_{n+4}^{k} − (a^{2} + 2)^{k}x_{n+2}^{k} + 2^{k}x_{n}^{k} + (−a)^{k}y_{n+2}^{k} − (x_{n} − ay_{n+2})^{k} =

= \{ \begin{matrix}0 & for k ⩽ 2, \\ −6ax_{n}y_{n+2}x_{n+4} & for k = 3, \\ −24ax_{n}^{2}y_{n+2}x_{n+4} & for k = 4, \\ −10ax_{n}y_{n+2}x_{n+4}[9x_{n}^{2} − ay_{n+2}x_{n+4}] & for k = 5, \\ −60ax_{n}^{2}y_{n+2}x_{n+4}[5x_{n}^{2} − ay_{n+2}x_{n+4}] & for k = 6, \\ −112ax_{n}^{2}y_{n+2}x_{n+4}[(5x_{n}^{2} − ay_{n+2}x_{n+4})^{2}+ \\ +ax_{n}^{2}y_{n+2}x_{n+4}] & for k = 8; \end{matrix}

d) for x = x_{n+4}, y = x_{n}, z = ay_{n+2}:

(2^{k} − 1)(ay_{n+2} + x_{n})^{k} − (2x_{n} + ay_{n+2})^{k} − (x_{n} + 2ay_{n+2})^{k} +

+ x_{n+4}^{k} + x_{n}^{k} + a^{k}y_{n+2}^{k} =

= \{ \begin{matrix}0 & for k ⩽ 2, \\ 6ax_{n}y_{n+2}x_{n+4} & for k = 3, \\ 24ax_{n}y_{n+2}x_{n+4}^{2} & for k = 4, \\ 10ax_{n}y_{n+2}x_{n+4}[7x_{n+4}^{2} − ax_{n}y_{n+2}] & for k = 5, \\ 60ax_{n}y_{n+2}x_{n+4}^{2}[3x_{n+4}^{2} − ax_{n}y_{n+2}] & for k = 6, \\ 112ax_{n}y_{n+2}x_{n+4}^{2}[(3x_{n+4}^{2} − ax_{n}y_{n+2})^{2}− \\ −ax_{n}y_{n+2}x_{n+4}^{2}] & for k = 8; \end{matrix}

e) for x = y_{n+4}, y = −y_{n}, z = −a(a^{2} + 4)x_{n+2} and odd powers only: y_{n+4}^{k} − y_{n}^{k} − (a(a^{2} + 4))^{k}x_{n+2}^{k} =

= \{ \begin{matrix}0 & for k = 1, \\ 3a(a^{2} + 4)y_{n}x_{n+2}y_{n+4} & for k = 3, \\ 5a(a^{2} + 4)y_{n}x_{n+2}y_{n+4}(a^{2}(a^{2} + 4)^{2}x_{n+2}^{2} + y_{n}y_{n+4}) & for k = 5, \\ 7a(a^{2} + 4)y_{n}x_{n+2}y_{n+4}(a^{2}(a^{2} + 4)^{2}x_{n+2}^{2} + y_{n}y_{n+4})^{2} & for k = 7, \\ 9a(a^{2} + 4)y_{n}x_{n+2}y_{n+4}[(a^{2}(a^{2} + 4)^{2}x_{n+2}^{2} + y_{n}y_{n+4})^{3}+ \\ +\frac{1}{3}(a(a^{2} + 4)y_{n}x_{n+2}y_{n+4})^{2}] & for k = 9; \end{matrix}

f) for x = y_{n+4}, y = −y_{n}, z = a(a^{2} + 4)x_{n+2}:

(2a(a^{2} + 4))^{k}x_{n+2}^{k} + y_{n+4}^{k} + (−1)^{k}y_{n}^{k} − (a(a^{2} + 4)x_{n+2} − y_{n})^{k} −

− (2a(a^{2} + 4)x_{n+2} + y_{n})^{k} =

= \{ \begin{matrix}0 & for k ⩽ 2, \\ −6a(a^{2} + 4)y_{n}x_{n+2}y_{n+4} & for k = 3, \\ −24a^{2}(a^{2} + 4)^{2}y_{n}x_{n+2}^{2}y_{n+4} & for k = 4, \\ −10a(a^{2} + 4)y_{n}x_{n+2}y_{n+4}× \\ ×[7a^{2}(a^{2} + 4)^{2}x_{n+2}^{2} + y_{n}y_{n+4}] & for k = 5, \\ −60a^{2}(a^{2} + 4)^{2}y_{n}x_{n+2}^{2}y_{n+4}× \\ ×[3a^{2}(a^{2} + 4)^{2}x_{n+2}^{2} + y_{n}y_{n+4}] & for k = 6, \\ −112a^{2}(a^{2} + 4)^{2}y_{n}x_{n+2}^{2}y_{n+4}× \\ ×[(3a^{2}(a^{2} + 4)^{2}x_{n+2}^{2} + y_{n}y_{n+4})^{2}+ \\ +a^{2}(a^{2} + 4)^{2}y_{n}x_{n+2}^{2}y_{n+4}] & for k = 8; \end{matrix}

g) for x = y_{n+4}, y = y_{n}, z = −a(a^{2} + 4)x_{n+2}:

y_{n+4}^{k} − (a^{2} + 2)^{k}y_{n+2}^{k} + 2^{k}y_{n}^{k} +

+ (−a(a^{2} + 4))^{k}x_{n+2}^{k} − (y_{n} − a(a^{2} + 4)x_{n+2})^{k} =

= \{ \begin{matrix}0 & for k ⩽ 2, \\ −6a(a^{2} + 4)y_{n}x_{n+2}y_{n+4} & for k = 3, \\ −24a(a^{2} + 4)y_{n}^{2}x_{n+2}y_{n+4} & for k = 4, \\ −10a(a^{2} + 4)y_{n}x_{n+2}y_{n+4}× \\ ×[7y_{n}^{2} + a(a^{2} + 4)x_{n+2}y_{n+4}] & for k = 5, \\ −60a(a^{2} + 4)y_{n}^{2}x_{n+2}y_{n+4}× \\ ×[3y_{n}^{2} + a(a^{2} + 4)x_{n+2}y_{n+4}] & for k = 6, \\ −112a(a^{2} + 4)y_{n}^{2}x_{n+2}y_{n+4}× \\ ×[(3y_{n}^{2} + a(a^{2} + 4)x_{n+2}y_{n+4})^{2}+ \\ +a(a^{2} + 4)y_{n}^{2}x_{n+2}y_{n+4}] & for k = 7; \end{matrix}

h) for x = x_{n}, y = ax_{n+1}, z = ax_{n+3}:

x_{n+4}^{k} + a^{k}x_{n+3}^{k} − [1 + (a^{2} + 1)^{k}]x_{n+2}^{k} − a^{k}y_{n+2}^{k} + a^{k}x_{n+1}^{k} + x_{n}^{k} =

= \{ \begin{matrix}0 & for k ⩽ 2, \\ 6a^{2}x_{n}x_{n+1}x_{n+3} & for k = 3, \\ 12a^{2}x_{n}x_{n+1}x_{n+3}x_{n+4} & for k = 4, \\ 10a^{2}x_{n}x_{n+1}x_{n+3}× \\ ×[2x_{n+4}^{2} − ax_{n}x_{n+1} − ax_{n+2}x_{n+3}] & for k = 5, \\ 30a^{2}x_{n}x_{n+1}x_{n+3}x_{n+4}× \\ ×[a^{2}x_{n+3}^{2} + x_{n+2}x_{n+4} − ax_{n}x_{n+1}] & for k = 6; \end{matrix}

i) for x = y_{n}, y = ay_{n+1}, z = ay_{n+3}:

y_{n+4}^{k} + a^{k}y_{n+3}^{k} − [(a^{2} + 1)^{k} + 1]y_{n+2}^{k} +

+ a^{k}y_{n+1}^{k} + y_{n}^{k} − a^{k}(a^{2} + 4)^{k}x_{n+2}^{k} =

= \{ \begin{matrix}0 & for k ⩽ 2, \\ 6a^{2}y_{n}y_{n+1}y_{n+3} & for k = 3, \\ 12a^{2}y_{n}y_{n+1}y_{n+3}y_{n+4}, & for k = 4, \\ 10a^{2}y_{n}y_{n+1}y_{n+3}× \\ ×[2y_{n+4}^{2} − a(y_{n}y_{n+1} + y_{n+2}y_{n+3})], & for k = 5, \\ 30a^{2}y_{n}y_{n+1}y_{n+3}y_{n+4}× \\ ×[y_{n+4}^{2} − a(y_{n}y_{n+1} + y_{n+2}y_{n+3})], & for k = 6, \\ 56a^{2}y_{n}y_{n+1}y_{n+3}y_{n+4}× \\ ×[(y_{n+4}^{2} − a(y_{n}y_{n+1} + y_{n+2}y_{n+3}))^{2}− \\ −\frac{1}{2}a^{2}y_{n}y_{n+1}y_{n+3}y_{n+4}] & for k = 8; \end{matrix}

j) for x = x_{n−1}, y = x_{n+1}, z = y_{n+2}:

y_{n+2}^{k} + [(4 + a^{2})^{k} − (3 + a^{2})^{k} + 1]x_{n+1}^{k} + x_{n−1}^{k} − y_{n}^{k} − (x_{n+1} + y_{n+2})^{k} =

= \{ \begin{matrix}0 & for k ⩽ 2, \\ 6x_{n−1}x_{n+1}y_{n+2} & for k = 3, \\ 12(2 + 2b + a^{2})x_{n−1}x_{n+1}^{2}y_{n+2} & for k = 4, \\ 10x_{n−1}x_{n+1}y_{n+2}× \\ ×[(2(4 + a^{2})^{2} − (3 + a^{2})^{2})x_{n+1}^{2} − x_{n−1}y_{n+2}] & for k = 5, \\ 30(4 + a^{2})x_{n−1}x_{n+1}^{2}y_{n+2}× \\ ×[(2a^{2} + 7)x_{n+1}^{2} − x_{n−1}y_{n+2}] & for k = 6, \\ 56(a^{2} + 4)x_{n−1}x_{n+1}^{2}y_{n+2}× \\ ×[((2a^{2} + 7)x_{n+1}^{2} − x_{n−1}y_{n+2})^{2}− \\ −\frac{1}{2}(4 + a^{2})x_{n−1}x_{n+1}^{2}y_{n+2}] & for k = 8; \end{matrix}

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A08 18

k) for x = bx_{n}, y = y_{n−1}, z = y_{n+1}, b ∈ ℂ:

(then bx_{n} + y_{n+1} = (b + 1)x_{n} + x_{n+2} and bx_{n} + y_{n−1} = (b + 1)x_{n} + x_{n−2}) we obtain

[(4 + b + a^{2})^{k} − (4 + a^{2})^{k} + b^{k}]x_{n}^{k} + y_{n−1}^{k} + y_{n+1}^{k} −

− (bx_{n} + y_{n+1})^{k} − (bx_{n} + y_{n−1})^{k} =

= \{ \begin{matrix}0 & for k ⩽ 2, \\ 6bx_{n}y_{n−1}y_{n+1} & for k = 3, \\ 12b(4 + b + a^{2})x_{n}^{2}y_{n−1}y_{n+1} & for k = 4, \\ 10bx_{n}y_{n−1}y_{n+1}[(2(4 + b + a^{2})^{2}− \\ −b(4 + a^{2}))x_{n}^{2} − y_{n−1}y_{n+1}] & for k = 5, \\ 30b(4 + b + a^{2})x_{n}^{2}y_{n−1}y_{n+1}× \\ ×[((4 + b + a^{2})^{2} − b(4 + a^{2}))x_{n}^{2} − y_{n−1}y_{n+1}] & for k = 6, \\ 56b(4 + b + a^{2})x_{n}^{2}y_{n−1}y_{n+1}× \\ ×[(\{(4 + b + a^{2})^{2} − b(4 + a^{2})\}x_{n}^{2}− \\ −y_{n−1}y_{n+1})^{2} − \frac{1}{2}b(4 + b + a^{2})x_{n}^{2}y_{n−1}y_{n+1}] & for k = 8; \end{matrix}

l) for x = w_{n}, y = bw_{n+2}, z = w_{n+4}, where b ∈ ℂ and w ∈ {x, y}:

w_{n+4}^{k} + [(a^{2} + 2 + b)^{k} − (a^{2} + 2)^{k} + b^{k}]w_{n+2}^{k} + w_{n}^{k} − − (bw_{n+2} + w_{n})^{k} − ((a^{2} + 2 + b)w_{n+2} − w_{n})^{k} =

= \{ \begin{matrix}0, & for k ⩽ 2, \\ 6bw_{n}w_{n+2}w_{n+4}, & for k = 3, \\ 12(a^{2} + 2 + b)bw_{n}w_{n+2}^{2}w_{n+4}, & for k = 4, \\ 10bw_{n}w_{n+2}w_{n+4}[(2(a^{2} + 2 + b)^{2}− \\ −b(a^{2} + 2))w_{n+2}^{2} − w_{n}w_{n+4}], & for k = 5, \\ 30b(a^{2} + 2 + b)w_{n}w_{n+2}^{2}w_{n+4}× \\ ×[((a^{2} + 2 + b)^{2} − b(a^{2} + 2))w_{n+2}^{2} − w_{n}w_{n+4}], & for k = 6, \\ 56b(a^{2} + 2 + b)w_{n}w_{n+2}^{2}w_{n+4}× \\ ×[(((a^{2} + 2 + b)^{2} − b(a^{2} + 2))w_{n+2}^{2} − w_{n}w_{n+4})^{2}− \\ −\frac{1}{2}b(a^{2} + 2 + b)w_{n}w_{n+2}^{2}w_{n+4}] & for k = 8; \end{matrix}

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A08 19

m) for x = (a^{2} + 3)x_{n+5}, y = −y_{n+2}, z = −x_{n−1}:

(a^{2} + 3)^{k}[x_{n+5}^{k} + a^{k}y_{n+3}^{k} − (−1)^{k}x_{n+1}^{k}] + (−y_{n+2})^{k} + (−x_{n−1})^{k} −

− ((a^{2} + 3)x_{n+5} − y_{n+2})^{k} − ((a^{2} + 3)x_{n+5} − x_{n−1})^{k} =

= \{ \begin{matrix}0, & for k ⩽ 2, \\ 6(a^{2} + 3)x_{n−1}y_{n+2}x_{n+5}, & for k = 3, \\ 12a(a^{2} + 3)^{2}x_{n−1}y_{n+2}y_{n+3}x_{n+5}, & for k = 4, \\ 10(a^{2} + 3)x_{n−1}y_{n+2}x_{n+5}× \\ ×[2a^{2}(a^{2} + 3)^{2}y_{n+3}^{2} − x_{n−1}y_{n+2}+ \\ +(a^{2} + 3)^{2}x_{n+1}x_{n+5}], & for k = 5, \\ 30a(a^{2} + 3)^{2}x_{n−1}y_{n+2}y_{n+3}x_{n+5}× \\ ×[a^{2}(a^{2} + 3)^{2}y_{n+3}^{2} − x_{n−1}y_{n+2}+ \\ +(a^{2} + 3)^{2}x_{n+1}x_{n+5}], & for k = 6, \\ 56a(a^{2} + 3)^{2}x_{n−1}y_{n+2}y_{n+3}x_{n+5}× \\ ×[(a^{2}(a^{2} + 3)^{2}y_{n+3}^{2} − x_{n−1}y_{n+2}+ \\ +(a^{2} + 3)^{2}x_{n+1}x_{n+5})^{2}− \\ −\frac{1}{2}a(a^{2} + 3)^{2}x_{n−1}y_{n+2}y_{n+3}x_{n+5}] & for k = 8. \end{matrix}

Below, the identities generated by (5.41) and the elements of sequences {x_{n}} and {y_{n}} satisfying relations (2.4)–(2.6) will be given:

a) for x = x_{n}, y = x_{n+2}, z = x_{n+2}, u = x_{n+4}

[(a^{2} + 2b + 2)^{k} − 2(a^{2} + 2b + 1)^{k} + (a^{2} + 2b)^{k} + 2^{k}]x_{n+2}^{k} +

+ 2y_{n+1}^{k} + 2y_{n+3}^{k} − (y_{n+1} + x_{n+2})^{k} − (y_{n+3} + x_{n+2})^{k} =

= \{ \begin{matrix}0 & for k ⩽ 3, \\ 24x_{n}x_{n+2}^{2}x_{n+4} & for k = 4, \\ 60(a^{2} + 2b + 2)x_{n}x_{n+2}^{3}x_{n+4} & for k = 5, \\ 60x_{n}x_{n+2}^{2}x_{n+4}[2((a^{2} + 2b + 2)^{2}− \\ −(a^{2} + 2b + \frac{1}{2}))x_{n+2}^{2} − x_{n}x_{n+4}] & for k = 6, \\ 210(a^{2} + 2b + 2)x_{n}x_{n+2}^{3}x_{n+4}[((a^{2} + 2b + 2)^{2}− \\ −2(a^{2} + 2b + \frac{1}{2}))x_{n+2}^{2} − x_{n}x_{n+4}] & for k = 7; \end{matrix}

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A08 20

b) for x = w_{n}, y = w_{n+2}, z = w_{n+4}, u = w_{n+6}, w ∈ {x, y} respectively:

(a^{2} + 4)^{k}x_{n+5}^{k} + (a^{2} + 4)^{k}((a^{2} + 2)^{k} + 1)x_{n+3}^{k} + (a^{2} + 4)^{k}x_{n+1}^{k} − y_{n+6}^{k} −

− y_{n}^{k} + ((a^{2} + 2)^{k} − (a^{2} + 3)^{k} − 1)(y_{n+4}^{k} + y_{n+2}^{k}) + (y_{n} + y_{n+6})^{k} − − [(a^{2} + 2)(a^{2} + 4)x_{n+3} − y_{n+2}]^{k} − [(a^{2} + 2)(a^{2} + 4)x_{n+3} − y_{n+4}]^{k} =

= \{ \begin{matrix}0 & for k ⩽ 3, \\ 24y_{n}y_{n+2}y_{n+4}y_{n+6} & for k = 4, \\ 60(a^{2} + 2)(a^{2} + 4)y_{n}y_{n+2}x_{n+3}y_{n+4}y_{n+6} & for k = 5, \\ 60y_{n}y_{n+2}y_{n+4}y_{n+6}[2(a^{2} + 2)^{2}(a^{2} + 4)^{2}x_{n+3}^{2}− \\ −(a^{2} + 3)y_{n+2}y_{n+6} − (a^{2} + 4)x_{n+1}y_{n+4} − y_{n}y_{n+2}] & for k = 6, \\ 210(a^{2} + 2)(a^{2} + 4)y_{n}y_{n+2}x_{n+3}y_{n+4}y_{n+6}× \\ ×[(a^{2} + 2)^{2}(a^{2} + 4)^{2}x_{n+3}^{3} − (a^{2} + 3)y_{n+2}y_{n+6} \\ −(a^{2} + 4)x_{n+1}y_{n+4} − y_{n}y_{n+2}] & for k = 7, \end{matrix}

y_{n+5}^{k} + ((a^{2} + 2)^{k} + 1)y_{n+3}^{k} + y_{n+1}^{k} − x_{n+6}^{k} − x_{n}^{k} +

+ ((a^{2} + 2)^{k} − (a^{2} + 3)^{k} − 1)(x_{n+4}^{k} + x_{n+2}^{k}) + (x_{n} + x_{n+6})^{k} − − ((a^{2} + 2)y_{n+3} − x_{n+4})^{k} − ((a^{2} + 2)y_{n+3} − x_{n+2})^{k} =

= \{ \begin{matrix}0 & for k ⩽ 3, \\ 24x_{n}x_{n+2}x_{n+4}x_{n+6} & for k = 4, \\ 60(a^{2} + 2)x_{n}x_{n+2}x_{n+4}x_{n+6}y_{n+3} & for k = 5, \\ 60x_{n}x_{n+2}x_{n+4}x_{n+6}[2(a^{2} + 2)^{2}y_{n+3}^{2} \\ −(a^{2} + 3)x_{n+2}x_{n+6} − y_{n+1}x_{n+4} − x_{n}x_{n+2}] & for k = 6, \\ 210(a^{2} + 2)x_{n}x_{n+2}x_{n+4}x_{n+6}y_{n+3}× \\ ×[(a^{2} + 2)^{2}y_{n+3}^{2} − (a^{2} + 3)x_{n+2}x_{n+6} \\ −y_{n+1}x_{n+4} − x_{n}x_{n+2}] & for k = 7. \end{matrix}

6.2 Identities for Powers of Fibonacci and Lucas Numbers Applications of identity (5.40) to generate some identities for Fibonacci and Lucas numbers will now be presented. a) For x = F_{n}, y = F_{n+1}, z = F_{n+2}:

2^{k}F_{n+2}^{k} − L_{n+1}^{k} − F_{n+3}^{k} + F_{n}^{k} + F_{n+1}^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, \\ 6F_{n}F_{n+1}F_{n+2}, & for k = 3, \\ 24F_{n}F_{n+1}F_{n+2}^{2}, & for k = 4, \\ 10F_{n}F_{n+1}F_{n+2}[7F_{n+2}^{2} − F_{n}F_{n+1}], & for k = 5, \\ 60F_{n}F_{n+1}F_{n+2}^{2}[3F_{n+2}^{2} − F_{n}F_{n+1}], & for k = 6, \\ 14F_{n}F_{n+1}F_{n+2}[(6F_{n+2}^{2} − F_{n}F_{n+1})^{2}− \\ −5F_{n+2}^{2}(F_{n+2}^{2} + F_{n}F_{n+1})], & for k = 7, \\ 112F_{n}F_{n+1}F_{n+2}^{2}[(3F_{n+2}^{2} − F_{n}F_{n+1})^{2}− \\ −F_{n}F_{n+1}F_{n+2}^{2}], & for k = 8; \end{matrix}

b) for x = F_{n}, y = F_{n+1}, z = F_{n+3}:

F_{n+4}^{k} − (2^{k} + 1)F_{n+2}^{k} − L_{n+2}^{k} + F_{n}^{k} + F_{n+1}^{k} + F_{n+3}^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, \\ 6F_{n}F_{n+1}F_{n+3}, & for k = 3, \\ 12F_{n}F_{n+1}F_{n+3}F_{n+4}, & for k = 4, \\ 10F_{n}F_{n+1}F_{n+3}× \\ ×[2F_{n+4}^{2} − F_{n+2}F_{n+3} − F_{n}F_{n+1}], & for k = 5, \\ 30F_{n}F_{n+1}F_{n+3}F_{n+4}× \\ ×[F_{n+4}^{2} − F_{n+2}F_{n+3} − F_{n}F_{n+1}], & for k = 6, \\ 56F_{n}F_{n+1}F_{n+3}F_{n+4}× \\ ×[(F_{n+4}^{2} − F_{n}F_{n+1} − F_{n+2}F_{n+3})^{2}− \\ −\frac{1}{2}F_{n}F_{n+1}F_{n+3}F_{n+4}], & for k = 8; \end{matrix}

c) for x = F_{n}, y = F_{n+1}, z = F_{n+4}: L_{n+3}^{k} + F_{n+4}^{k} − 2^{k}F_{n+3}^{k} − (1 + 3^{k})F_{n+2}^{k} + F_{n+1}^{k} + F_{n}^{k} =

= \begin{cases}0 & for k = 1, 2, \\ 6F_{n}F_{n+1}F_{n+4}, & for k = 3, \\ 12F_{n}F_{n+1}L_{n+3}F_{n+4}, & for k = 4, \\ 10F_{n}F_{n+1}F_{n+4}[2L_{n+3}^{2} − F_{n}F_{n+1} − F_{n+2}F_{n+4}] \\ = 2F_{n}F_{n+1}F_{n+4}[9L_{2n+6} − L_{2n+1} − 16(−1)^{n}], & for k = 5, \\ 30F_{n}F_{n+1}L_{n+3}F_{n+4}[L_{n+3}^{2} − F_{n}F_{n+1} − F_{n+2}F_{n+4}] \\ = 6F_{n}F_{n+1}L_{n+3}F_{n+4}[4L_{2n+6} − L_{2n+1} − 6(−1)^{n}], & for k = 6, \\ 56F_{n}F_{n+1}L_{n+3}F_{n+4}[(L_{n+3}^{2} − F_{n}F_{n+1} − F_{n+2}F_{n+4})^{2} \\ −\frac{1}{2}F_{n}F_{n+1}L_{n+3}F_{n+4}] & for k = 8; \end{cases}

d) for x = F_{n}, y = F_{n+2}, z = F_{n+4}:

F_{n+4}^{k} − L_{n+3}^{k} + (4^{k} − 3^{k} + 1)F_{n+2}^{k} − L_{n+1}^{k} + F_{n}^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, \\ 6F_{n}F_{n+2}F_{n+4}, & for k = 3, \\ 48F_{n}F_{n+2}^{2}F_{n+4}, & for k = 4, \\ 10F_{n}F_{n+2}F_{n+4}[29F_{n+2}^{2} − F_{n}F_{n+4}] \\ = 2F_{n}F_{n+2}F_{n+4}[28L_{2n+4} − 51(−1)^{n}] \\ = 10F_{n}F_{n+2}F_{n+4}[28F_{n+2}^{2} + (−1)^{n}], & for k = 5, \\ 120F_{n}F_{n+2}^{2}F_{n+4}[13F_{n+2}^{2} − F_{n}F_{n+4}] \\ = 24F_{n}F_{n+2}^{2}F_{n+4}[12L_{2n+4} − 19(−1)^{n}] \\ = 120F_{n}F_{n+2}^{2}F_{n+4}[12F_{n+2}^{2} + (−1)^{n}], & for k = 6, \\ 224F_{n}F_{n+2}^{2}F_{n+4}[(13F_{n+2}^{2} − F_{n}F_{n+4})^{2}− \\ −2F_{n}F_{n+2}^{2}F_{n+4}] & for k = 8; \end{matrix}

e) for x = 2F_{n}, y = F_{n+2}, z = F_{n+5}: F_{n+5}^{k} − 2^{k}F_{n+4}^{k} + (6^{k} − 5^{k} + 1)F_{n+2}^{k} − (F_{n+1} + 3F_{n})^{k} + 2^{k}F_{n}^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, \\ 12F_{n}F_{n+2}F_{n+5}, & for k = 3, \\ 144F_{n}F_{n+2}^{2}F_{n+5}, & for k = 4, \\ 20F_{n}F_{n+2}F_{n+5}[72F_{n+2}^{2} − 4F_{n}F_{n+4} − F_{n+2}F_{n+5}] \\ = 4F_{n}F_{n+2}F_{n+5}[71L_{2n+4} − 10F_{2n+5} − 112(−1)^{n}], & for k = 5, \\ 360F_{n}F_{n+2}^{2}F_{n+5}[36F_{n+2}^{2} − 4F_{n}F_{n+4} − F_{n+2}F_{n+5}] \\ = 360F_{n}F_{n+2}^{2}F_{n+5}[7L_{2n+4} − 2F_{2n+5} − 8(−1)^{n}], & for k = 6; \end{matrix}

f) for x = F_{n}, y = F_{n+3}, z = F_{n+4}:

(F_{n} + F_{n+5})^{k} − F_{n+5}^{k} + F_{n+4}^{k} + F_{n+3}^{k} − (2^{k} + 3^{k})F_{n+2}^{k} + F_{n}^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, \\ 6F_{n}F_{n+3}F_{n+4}, & for k = 3, \\ 12F_{n}F_{n+3}F_{n+4}(F_{n} + F_{n+5}), & for k = 4, \\ 10F_{n}F_{n+3}F_{n+4}[2(F_{n} + F_{n+5})^{2} − F_{2n+6} + 2(−1)^{n}], & for k = 5, \\ 30F_{n}F_{n+3}F_{n+4}(F_{n} + F_{n+5})× \\ ×[(F_{n} + F_{n+5})^{2} − F_{2n+6} + 2(−1)^{n}], & for k = 6; \end{matrix}

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A08 23

g) for x = 2F_{n}, y = F_{n+3}, z = F_{n+4}:

F_{n+4}^{k} + F_{n+3}^{k} + (5F_{n+2})^{k} + (2F_{n})^{k} − F_{n+5}^{k} −

− (F_{n} + 2F_{n+2})^{k} − (F_{n} + 3F_{n+2})^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, \\ 12F_{n}F_{n+3}F_{n+4}, & for k = 3, \\ 120F_{n}F_{n+2}F_{n+3}F_{n+4}, & for k = 4, \\ 20F_{n}F_{n+3}F_{n+4}× \\ ×[F_{n+2}^{2} − 2F_{n}F_{n+5} − F_{n+3}F_{n+4}] \\ = 4F_{n}F_{n+3}F_{n+4}× \\ ×[45L_{2n+4} − 4L_{2n+3} − 79(−1)^{n}], & for k = 5, \\ 300F_{n}F_{n+2}F_{n+3}F_{n+4}× \\ ×[25F_{n+2}^{2} − 2F_{n}F_{n+5} − F_{n+3}F_{n+4}] \\ = 60F_{n}F_{n+2}F_{n+3}F_{n+4}× \\ ×[20L_{2n+4} − 4L_{2n+3} − 29(−1)^{n}], & for k = 6; \end{matrix}

h) for x = L_{n}, y = L_{n+1}, z = L_{n+2}:

2^{k}L_{n+2}^{k} − 5^{k} · F_{n+1}^{k} − L_{n+3}^{k} + L_{n}^{k} + L_{n+1}^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, \\ 6L_{n}L_{n+1}L_{n+2}, & for k = 3, \\ 24L_{n}L_{n+1}L_{n+2}^{2}, & for k = 4, \\ 10L_{n}L_{n+1}L_{n+2}[7L_{n+2}^{2} − L_{n}L_{n+1}], & for k = 5, \\ 60L_{n}L_{n+1}L_{n+2}^{2}[3L_{n+2}^{2} − L_{n}L_{n+1}], & for k = 6; \end{matrix}

i) for x = L_{n}, y = L_{n+1}, z = L_{n+3}:

L_{n+4}^{k} − 5^{k} · F_{n+2}^{k} − (2^{k} + 1)L_{n+2}^{k} + L_{n}^{k} + L_{n+1}^{k} + L_{n+3}^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, \\ 6L_{n}L_{n+1}L_{n+3}, & for k = 3, \\ 12L_{n}L_{n+1}L_{n+3}L_{n+4}, & for k = 4, \\ 10L_{n}L_{n+1}L_{n+3}× \\ ×[2L_{n+4}^{2} − L_{n+2}L_{n+3} − L_{n}L_{n+1}], & for k = 5, \\ 30L_{n}L_{n+1}L_{n+3}L_{n+4}× \\ ×[L_{n+4}^{2} − L_{n+2}L_{n+3} − L_{n}L_{n+1}], & for k = 6. \end{matrix}

In the next step, identity (5.41) shall be applied to generate some selected identities for Fibonacci and Lucas numbers.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A08 24

a) For x = F_{n}, y = F_{n+1}, z = F_{n+3}, u = F_{n+5} we obtain the following identities:

F_{n+6}^{k} − F_{n+5}^{k} − (2^{k} + 1)F_{n+4}^{k} − (4^{k} − 3^{k} + 1)F_{n+3}^{k} + (2^{k} + 1)F_{n+2}^{k} − − F_{n+1}^{k} − F_{n}^{k} + L_{n+2}^{k} + L_{n+4}^{k} + (F_{n} + F_{n+5})^{k} − (F_{n} + L_{n+4})^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, 3, \\ 24F_{n}F_{n+1}F_{n+3}F_{n+5}, & for k = 4, \\ 60F_{n}F_{n+1}F_{n+3}F_{n+5}F_{n+6}, & for k = 5, \\ 60F_{n}F_{n+1}F_{n+3}F_{n+5}(2F_{n+6}^{2} − F_{n+5}F_{n+4}− \\ −F_{n+3}F_{n+2} − F_{n+1}F_{n}), & for k = 6, \\ 210F_{n}F_{n+1}F_{n+3}F_{n+5}F_{n+6}(F_{n+6}^{2}− \\ −F_{n+5}F_{n+4} − F_{n+3}F_{n+2} − F_{n+1}F_{n}), & for k = 7, \end{matrix} (6.42)

where the following identity was utilized: F_{n+1} + F_{n+5} = 3F_{n+3} and L_{n+1} + L_{n+5} = 3L_{n+3};

b) for x = L_{n}, y = L_{n+1}, z = L_{n+3}, u = L_{n+5}:

L_{n+6}^{k} − L_{n+5}^{k} − (2^{k} + 1)L_{n+4}^{k} − (4^{k} − 3^{k} + 1)L_{n+3}^{k} + (2^{k} + 1)L_{n+2}^{k} − − L_{n+1}^{k} − L_{n}^{k} + 5^{k}F_{n+2}^{k} + 5^{k}F_{n+4}^{k} + (L_{n} + L_{n+5})^{k} − (L_{n} + 5F_{n+4})^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, 3, \\ 24L_{n}L_{n+1}L_{n+3}L_{n+5} & for k = 4, \\ 60L_{n}L_{n+1}L_{n+3}L_{n+5}L_{n+6} & for k = 5, \\ 60L_{n}L_{n+1}L_{n+3}L_{n+5}[2L_{n+6}^{2} − L_{n+5}L_{n+4}− \\ −L_{n+3}L_{n+2} − L_{n+1}L_{n}], & for k = 6; \end{matrix} (6.43)

c) for x = F_{n}, y = F_{n+2}, z = F_{n+4}, u = F_{n+6}:

L_{n+5}^{k} + (3^{k} + 1)L_{n+3}^{k} + L_{n+1}^{k} − F_{n+6}^{k} − F_{n}^{k} + (3^{k} − 4^{k} − 1)(F_{n+4}^{k} + F_{n+2}^{k}) + + (F_{n} + F_{n+6})^{k} − (2F_{n+4} + 3F_{n+2})^{k} − (3F_{n+4} + 2F_{n+2})^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, 3, \\ 24F_{n}F_{n+2}F_{n+4}F_{n+6} & for k = 4, \\ 180F_{n}F_{n+2}F_{n+4}F_{n+6}L_{n+3} & for k = 5, \\ 60F_{n}F_{n+2}F_{n+4}F_{n+6}[18L_{n+3}^{2}− \\ −4F_{n+2}F_{n+6} − L_{n+1}F_{n+4} − F_{n}F_{n+2}] & for k = 6, \\ 630F_{n}F_{n+2}F_{n+4}F_{n+6}L_{n+3}[9L_{n+3}^{2}− \\ −4F_{n+2}F_{n+6} − L_{n+1}F_{n+4} − F_{n}F_{n+2}] & for k = 7; \end{matrix} (6.44)

d) for x = L_{n}, y = L_{n+2}, z = L_{n+4}, u = L_{n+6}:

(5F_{n+5})^{k} + (15^{k} + 5^{k})F_{n+3}^{k} + (5F_{n+1})^{k} − L_{n+6}^{k} − (4^{k} − 3^{k} + 1)L_{n+4}^{k} −

− (4^{k} − 3^{k} + 1)L_{n+2}^{k} − L_{n}^{k} + (6L_{n+2} + 2L_{n+1})^{k} −

− (4L_{n+2} + L_{n+5})^{k} − (4L_{n+4} − L_{n+1})^{k} =

= \{ \begin{matrix}0 & for k = 1, 2, 3, \\ 24L_{n}L_{n+2}L_{n+4}L_{n+6} & for k = 4, \\ 900L_{n}L_{n+2}L_{n+4}L_{n+6}F_{n+3} & for k = 5, \\ 60F_{n}F_{n+2}F_{n+4}F_{n+6}[18L_{n+3}^{2}− \\ −4F_{n+2}F_{n+6} − L_{n+1}F_{n+4} − F_{n}F_{n+2}] & for k = 6, \\ 630F_{n}F_{n+2}F_{n+4}F_{n+6}L_{n+3}[9L_{n+3}^{2}− \\ −4F_{n+2}F_{n+6} − L_{n+1}F_{n+4} − F_{n}F_{n+2}] & for k = 7. \end{matrix} (6.45)

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