Quasi-Fibonacci Numbers of Order 11

Roman Wituła and Damian Słota Institute of Mathematics Silesian University of Technology Kaszubska 23, Gliwice 44-100 POLAND

                       r.witula@polsl.pl
                       d.slota@polsl.pl

Abstract In this paper we introduce and investigate the so-called quasi-Fibonacci numbers of order 11. These numbers are defined by five conjugate recurrence equations of order five. We study some relations and identities concerning these numbers. We present some applications to the decomposition of some polynomials. Many of the identities presented here are the generalizations of the identities characteristic for general recurrence sequences of order three given by Rabinowitz.

1 Introduction

Wituła, Słota and Warzyński [9] analyzed the relationships between the so-called quasi-Fibonacci numbers of order 7. In this paper we are focused on the generalization of the identities and some facts derived in [9] to the quasi-Fibonacci numbers of order 11 A_{n}(δ), B_{n}(δ), C_{n}(δ), D_{n}(δ), and E_{n}(δ), n ∈ ℕ, δ ∈ ℂ, defined by the relations (see also Section 3 in this paper for more details):

(1 + δ (ξ^{m} + ξ^{10m}))^{n} = A_{n}(δ) + B_{n}(δ) (ξ^{m} + ξ^{10m}) + C_{n}(δ) (ξ^{2m} + ξ^{9m}) + + D_{n}(δ) (ξ^{3m} + ξ^{8m}) + E_{n}(δ) (ξ^{4m} + ξ^{7m})

for m, n ∈ ℕ, m ≤ 5, δ ∈ ℂ, where ξ ∈ ℂ is a primitive root of unity of order 11. Additionally, the following important auxiliary sequence is considered:

A_{n}(δ) := 5A_{n}(δ) − B_{n}(δ) − C_{n}(δ) − D_{n}(δ) − E_{n}(δ).

Moreover, in order to make shorter expressions and formulas we introduce the following notation:

k_{m} := cos (\frac{m π}{11} ),  s_{m} := sin (\frac{m π}{11} ),  ξ := exp (\frac{i 2 π}{11} ),
σ_{m} := ξ^{m} + ξ^{−m}  and  τ_{m}(δ) := 1 + δ σ_{m}

for m ∈ ℤ and δ ∈ ℂ. We note that

σ_{m} = 2 k_{2m}  and  τ_{m}(δ) = 1 + 2 δ k_{2m}

and

σ_{m} σ_{n} = σ_{m+n} + σ_{|m−n|},  (1.1)
σ_{m n} = σ_{(11−m)n},  (1.2)
σ_{m n} = σ_{|m|n},  (1.3)
1 + σ_{1} + σ_{2} + σ_{3} + σ_{4} + σ_{5} = 0.  (1.4)

Moreover we have

τ_{m}(\frac{1}{2}) = 2 k_{m}^{2},  τ_{m}(−\frac{1}{2}) = 2 s_{m}^{2},  (1.5)

and

τ_{m}(\frac{T_{r}(k_{2s})}{2k_{2m}}) = τ_{m}(\frac{T_{2r}(k_{s})}{2k_{2m}}) = 1 + k_{2rs} = 2 k_{rs}^{2},  (1.6)

where T_{r}(x) denote the r-th Chebyshev polynomial of the first kind. For example, from the last formula we get

τ_{m}(2 k_{2m}^{2} − \frac{3}{2}) = τ_{m}(\frac{T_{3}(k_{2m})}{2k_{2m}}) = 2 k_{3m}^{2},  (1.7)
τ_{m}(8 k_{2m}^{4} − 10 k_{2m}^{2} + \frac{5}{2}) = τ_{m}(\frac{T_{5}(k_{2m})}{2k_{2m}}) = 2 k_{5m}^{2},  (1.8)

for every m ∈ ℕ.

2 Minimal polynomials, linear independence over ℚ

Let Ψ_{n}(x) be the minimal polynomial of cos(2π/n) for every n ∈ ℕ. W. Watkins and J. Zeitlin described in [8] the following identities

T_{s+1}(x) − T_{s}(x) = 2^{s} ∏_{d|n} Ψ_{d}(x)

if n = 2s + 1 and

T_{s+1}(x) − T_{s−1}(x) = 2^{s} ∏_{d|n} Ψ_{d}(x)

if n = 2s. In the sequel if n = 2s + 1 is a prime number, then we get

T_{s+1}(x) − T_{s}(x) = 2^{s}Ψ_{1}(x)Ψ_{n}(x),

which for example implies the formula

Ψ_{11}(x) = \frac{1}{32(x − 1)}(T_{6}(x) − T_{5}(x))

= \frac{1}{32(x − 1)}(32x^{6} − 16x^{5} − 48x^{4} + 20x^{3} + 18x^{2} − 5x − 1)

= x^{5} + \frac{1}{2}x^{4} − x^{3} − \frac{3}{8}x^{2} + \frac{3}{16}x + \frac{1}{32}.

Lemma 1. Let ξ = exp(i2π/11). Then

p_{11}(x) = ∏_{m=1}^{5}(x − σ_{m}) = x^{5} + x^{4} − 4x^{3} − 3x^{2} + 3x + 1

is a minimal polynomial of the numbers σ_{m}, m = 1, 2, …, 5. Moreover, we have the identity

Ψ_{11}(x) = \frac{1}{32}p_{11}(2x).

Corollary 2. The numbers k_{2m}, m = 0, 1, …, 4 are linearly independent over ℚ. Proof. If we suppose that

a + b cos(2π/11) + c cos(4π/11) + d cos(6π/11) + e cos(8π/11) = 0

for some a, b, c, d, e ∈ ℚ, then by the formulas cos 2α = 2 cos^{2} α − 1,

cos 3α = 4 cos^{3} α − 3 cos α,
cos 4α = 8 cos^{4} α − 8 cos^{2} α + 1,

the number cos(2π/11) would be a root of some polynomial ∈ ℚ[x] having degree ⩽ 4, which by Lemma 1 is impossible. Corollary 3. Every five numbers which belong to the set {1, σ_{1}, σ_{2}, σ_{3}, σ_{4}, σ_{5}} are linearly independent over ℚ. Proof. It follows from the identity (1.4).

The following observation will play a central role in our arguments. Lemma 4. Let a_{1}, a_{2}, …, a_{n} ∈ ℝ be linearly independent over ℚ and let f_{k}, g_{k} ∈ ℚ[δ], k = 1, 2, …, n. If the identity holds

∑_{k=1}^{n} f_{k}(δ) a_{k} = ∑_{k=1}^{n} g_{k}(δ) a_{k}  (2.1)

for every δ ∈ ℚ, then f_{k}(δ) = g_{k}(δ) for every k = 1, 2, …, n and δ ∈ ℂ. Proof. Since f_{k}(δ) ∈ ℚ and g_{k}(δ) ∈ ℚ for any δ ∈ ℚ we get from (2.1) that f_{k}(δ) = g_{k}(δ) for every k = 1, 2, …, n and δ ∈ ℚ. The last equality implies that all, respective coefficients of polynomials f_{k} and g_{k} are the same. So f_{k}(δ) = g_{k}(δ) for all δ ∈ ℂ and k = 1, 2, …, n.

3 Quasi-Fibonacci numbers of order 11

Let us start from the following basic result. Lemma 5. Let δ ∈ ℂ and n ∈ ℕ. Then the following identities hold τ_{m}^{n}(δ) = a_{n}(δ) + b_{n}(δ) σ_{m} + c_{n}(δ) σ_{2m} + d_{n}(δ) σ_{3m} + e_{n}(δ) σ_{4m} + f_{n}(δ) σ_{5m} (3.2m − 1) τ_{m}^{n}(δ) = A_{n}(δ) + B_{n}(δ) σ_{m} + C_{n}(δ) σ_{2m} + D_{n}(δ) σ_{3m} + E_{n}(δ) σ_{4m} (3.2m) for every m = 1, 2, …, 5, i.e.,

τ_{1}^{n}(δ) =a_{n}(δ) + b_{n}(δ) σ_{1} + c_{n}(δ) σ_{2} + d_{n}(δ) σ_{3} + e_{n}(δ) σ_{4} + f_{n}(δ) σ_{5}  (3.1)
=A_{n}(δ) + B_{n}(δ) σ_{1} + C_{n}(δ) σ_{2} + D_{n}(δ) σ_{3} + E_{n}(δ) σ_{4},  (3.2)
τ_{2}^{n}(δ) =a_{n}(δ) + b_{n}(δ) σ_{2} + c_{n}(δ) σ_{4} + d_{n}(δ) σ_{5} + e_{n}(δ) σ_{3} + f_{n}(δ) σ_{1}  (3.3)
=A_{n}(δ) + B_{n}(δ) σ_{2} + C_{n}(δ) σ_{4} + D_{n}(δ) σ_{5} + E_{n}(δ) σ_{3},  (3.4)
τ_{3}^{n}(δ) =a_{n}(δ) + b_{n}(δ) σ_{3} + c_{n}(δ) σ_{5} + d_{n}(δ) σ_{2} + e_{n}(δ) σ_{1} + f_{n}(δ) σ_{4}  (3.5)
=A_{n}(δ) + B_{n}(δ) σ_{3} + C_{n}(δ) σ_{5} + D_{n}(δ) σ_{2} + E_{n}(δ) σ_{1},  (3.6)
τ_{4}^{n}(δ) =a_{n}(δ) + b_{n}(δ) σ_{4} + c_{n}(δ) σ_{3} + d_{n}(δ) σ_{1} + e_{n}(δ) σ_{5} + f_{n}(δ) σ_{2}  (3.7)
=A_{n}(δ) + B_{n}(δ) σ_{4} + C_{n}(δ) σ_{3} + D_{n}(δ) σ_{1} + E_{n}(δ) σ_{5},  (3.8)
τ_{5}^{n}(δ) =a_{n}(δ) + b_{n}(δ) σ_{5} + c_{n}(δ) σ_{1} + d_{n}(δ) σ_{4} + e_{n}(δ) σ_{2} + f_{n}(δ) σ_{3}  (3.9)
=A_{n}(δ) + B_{n}(δ) σ_{5} + C_{n}(δ) σ_{1} + D_{n}(δ) σ_{4} + E_{n}(δ) σ_{2},  (3.10)

where

a_{0}(δ) = 1, b_{0}(δ) = … = f_{0}(δ) = 0,

\{ \begin{matrix}a_{n+1}(δ) & = & a_{n}(δ) + 2δb_{n}(δ), \\ b_{n+1}(δ) & = & δa_{n}(δ) + b_{n}(δ) + δc_{n}(δ), \\ c_{n+1}(δ) & = & δb_{n}(δ) + c_{n}(δ) + δd_{n}(δ), \\ d_{n+1}(δ) & = & δc_{n}(δ) + d_{n}(δ) + δe_{n}(δ), \\ e_{n+1}(δ) & = & δd_{n}(δ) + e_{n}(δ) + δf_{n}(δ), \\ f_{n+1}(δ) & = & δe_{n}(δ) + (1 + δ)f_{n}(δ) \end{matrix}  (3.11)

and

A_{n}(δ) = a_{n}(δ) − f_{n}(δ), B_{n}(δ) = b_{n}(δ) − f_{n}(δ),
C_{n}(δ) = c_{n}(δ) − f_{n}(δ), D_{n}(δ) = d_{n}(δ) − f_{n}(δ), E_{n}(δ) = e_{n}(δ) − f_{n}(δ),

for every n = 0, 1, 2, . . .. Hence, the following identities are derived:

A_{n+1}(δ) = a_{n}(δ) + 2δb_{n}(δ) − δe_{n}(δ) − (1 + δ)f_{n}(δ)
= A_{n}(δ) + 2δB_{n}(δ) − δE_{n}(δ),

B_{n+1}(δ) = δa_{n}(δ) + b_{n}(δ) + δc_{n}(δ) − δe_{n}(δ) − (1 + δ)f_{n}(δ)
= δA_{n}(δ) + B_{n}(δ) + δC_{n}(δ) − δE_{n}(δ),
C_{n+1}(δ) = δb_{n}(δ) + c_{n}(δ) + δd_{n}(δ) − δe_{n}(δ) − (1 + δ)f_{n}(δ)
= δB_{n}(δ) + C_{n}(δ) + δD_{n}(δ) − δE_{n}(δ),
D_{n+1}(δ) = δc_{n}(δ) + d_{n}(δ) − (1 + δ)f_{n}(δ)
= δC_{n}(δ) + D_{n}(δ),
E_{n+1}(δ) = δd_{n}(δ) + e_{n}(δ) − δe_{n}(δ) − f_{n}(δ)
= δD_{n}(δ) + (1 − δ)E_{n}(δ),

i.e., the following system of linear equations holds

\{ \begin{matrix}A_{n+1}(δ) = A_{n}(δ) + 2δB_{n}(δ) − δE_{n}(δ), \\ B_{n+1}(δ) = δA_{n}(δ) + B_{n}(δ) + δC_{n}(δ) − δE_{n}(δ), \\ C_{n+1}(δ) = δB_{n}(δ) + C_{n}(δ) + δD_{n}(δ) − δE_{n}(δ), \\ D_{n+1}(δ) = δC_{n}(δ) + D_{n}(δ), \\ E_{n+1}(δ) = δD_{n}(δ) + (1 − δ)E_{n}(δ), \end{matrix}  (3.12)

where A_{0}(δ) = 1, B_{0}(δ) = C_{0}(δ) = D_{0}(δ) = E_{0}(δ) = 0.

Proof. Proofs of the identities (3.2m − 1) for every m = 1, 2, …, 5 by induction follow. For n = 1 we have

τ_{m}(δ) = a_{1}(δ) + b_{1}(δ) σ_{m} + c_{1}(δ) σ_{2m} + d_{1}(δ) σ_{3m} + e_{1}(δ) σ_{4m} + f_{1}(δ) σ_{5m}.

Suppose that for some n ∈ ℕ the equalities (3.2m−1), m = 1, 2, …, 5, holds. Then by (1.1), (1.2) and (3.11) we get

τ_{m}^{n+1}(δ) = τ_{m}(δ) τ_{m}^{n}(δ) =

= (1 + δ σ_{m})(a_{n}(δ) + b_{n}(δ) σ_{m} + c_{n}(δ) σ_{2m} + d_{n}(δ) σ_{3m} + e_{n}(δ) σ_{4m} + f_{n}(δ) σ_{5m}) =

= (1 + δ σ_{m}) a_{n}(δ) + (σ_{m} + δ (σ_{2m} + 2)) b_{n}(δ) + (σ_{2m} + δ (σ_{3m} + σ_{m})) c_{n}(δ) +

+ (σ_{3m} + δ (σ_{4m} + σ_{2m})) d_{n}(δ) + (σ_{4m} + δ (σ_{5m} + σ_{3m})) e_{n}(δ) + (σ_{5m} + δ (σ_{5m} + σ_{4m})) f_{n}(δ) = = (a_{n}(δ) + 2 δ b_{n}(δ)) + (δ a_{n}(δ) + b_{n}(δ) + δ c_{n}(δ)) σ_{m} + (δ b_{n}(δ) + c_{n}(δ) + δ d_{n}(δ)) σ_{2m} +

+(δ c_{n}(δ)+d_{n}(δ)+δ e_{n}(δ)) σ_{3m}+(δ d_{n}(δ)+e_{n}(δ)+δ f_{n}(δ)) σ_{4m}+(δ e_{n}(δ)+(1+δ) f_{n}(δ)) σ_{5m} = = a_{n+1}(δ) + b_{n+1}(δ) σ_{m} + c_{n+1}(δ) σ_{2m} + d_{n+1}(δ) σ_{3m} + e_{n+1}(δ) σ_{4m} + f_{n+1}(δ) σ_{5m},

which by the principle of mathematical induction means that (3.2m−1) hold for every n ∈ ℕ and m = 1, 2, …, 5.

Formula (3.2m) from (3.2m − 1) and (1.4) follows for every m = 1, 2, …, 5.

Definition 6. We call all elements of the sequences {A_{n}(δ)}, {B_{n}(δ)}, …, {E_{n}(δ)} the quasi-Fibonacci numbers of order (11, δ) (see Table 1 at the end of the paper). To simplify notation we shall write {A_{n}}, {B_{n}}, …, {E_{n}} instead of {A_{n}(1)}, {B_{n}(1)}, …, {E_{n}(1)}, respectively. We call the elements of the sequences {A_{n}}, {B_{n}}, …, {E_{n}} the quasi-Fibonacci numbers of order 11 (see Table 2 at the end of the paper).

Corollary 7. In the sequel, for δ = 1 we obtain the following recurrence relations:

\{ \begin{matrix}A_{n+1} = A_{n} + 2B_{n} − E_{n}, \\ B_{n+1} = A_{n} + B_{n} + C_{n} − E_{n}, \\ C_{n+1} = B_{n} + C_{n} + D_{n} − E_{n}, \\ D_{n+1} = C_{n} + D_{n}, \\ E_{n+1} = D_{n}. \end{matrix}  (3.13)

Corollary 8. Adding identities (3.k) for odd and even k = 1, …, 10 respectively we obtain the identity:

τ_{1}^{n}(δ) + τ_{2}^{n}(δ) + τ_{3}^{n}(δ) + τ_{4}^{n}(δ) + τ_{5}^{n}(δ) =
= 5a_{n}(δ) − b_{n}(δ) − c_{n}(δ) − d_{n}(δ) − e_{n}(δ) − f_{n}(δ)
= 5A_{n}(δ) − B_{n}(δ) − C_{n}(δ) − D_{n}(δ) − E_{n}(δ)
:= A_{n}(δ).  (3.14)

Hence, by (1.5) the following identities holds:

2^{−n}A_{n}(\frac{1}{2}) = ∑_{m=1}^{5} k_{m}^{2n}  (3.15)

and

2^{−n}A_{n}(−\frac{1}{2}) = ∑_{m=1}^{5} s_{m}^{2n}.  (3.16)

Moreover, from (3.14) we obtain

∑_{m=1}^{5} coeff (τ_{m}^{n}(δ); δ^{n}) = ∑_{m=1}^{5} σ_{m}^{n} = 2^{n} ∑_{m=1}^{5} k_{2m}^{n} =

= 5 coeff (A_{n}(δ); δ^{n}) − coeff (B_{n}(δ); δ^{n}) − … − coeff (E_{n}(δ); δ^{n}) = coeff (A_{n}(δ); δ^{n}).

Hence, by (3.15) we get

coeff (A_{2n}(δ); δ^{2n}) = 2^{n} A_{n}(\frac{1}{2})

and

coeff (A_{2n−1}(δ); δ^{2n−1}) = 2^{2n−1} ∑_{m=1}^{5}(−1)^{m} k_{m}^{n}.

In Table 3 twelve initial values of the sequences \{A(1)\} (see also \underline{A062883} in [6]), \{A(1/2)\} and {A(−1/2)} are presented.

Corollary 9. By (1.7) and (3.2m) we have the identity

2^{n} k_{3m}^{2n} = A_{n}(δ) + B_{n}(δ) σ_{m} + C_{n}(δ) σ_{2m} + D_{n}(δ) σ_{3m} + E_{n}(δ) σ_{4m} =

= A_{n}(δ) + 2 (B_{n}(δ) k_{2m} + C_{n}(δ) k_{4m} + D_{n}(δ) k_{5m} + E_{n}(δ) k_{3m}) for δ := 2 k_{2m}^{2} − \frac{3}{2} and for every m = 1, 2, …, 5. Furthermore, by (1.6) and (3.2m) the following general formula hold

2^{n} k_{r m}^{2n} = A_{n}(δ) + B_{n}(δ) σ_{m} + C_{n}(δ) σ_{2m} + D_{n}(δ) σ_{3m} + E_{n}(δ) σ_{4m} =

= A_{n}(δ) + 2 (B_{n}(δ) k_{2m} + C_{n}(δ) k_{4m} + D_{n}(δ) k_{5m} + E_{n}(δ) k_{3m})

for δ := T_{r}(k_{2m})/(2 k_{2m}), m, r ∈ ℕ. We note that x divides T_{r}(x) iff r is an odd positive integer.

Corollary 10. Since deg_{δ} (τ_{m}^{n}) = n for every m = 1, 2, …, 5, by (3.2m) for m = 1, 2, …, 5 it can be easily deduced the following formula

max { deg (A_{n}(δ)), deg (B_{n}(δ)), deg (C_{n}(δ)), deg (D_{n}(δ)), deg (E_{n}(δ))} = n.

Hence, by (3.12) and Table 1, if deg(D_{n}(δ)) = n for n = 5, 6, … then also deg(C_{n}(δ)) = n for n = 5, 6, …, and deg(E_{n}(δ)) = n for infinite many n ∈ ℕ (see also the Section 6 in this paper).

Corollary 11. Taking into account the equations (from the last to the first one, respectively) of the recurrence system (3.12), we obtain:

\{ \begin{matrix}δD_{n}(δ) & = E_{n+1}(δ) − (1 − δ)E_{n}(δ), \\ δ^{2}C_{n}(δ) & = δD_{n+1}(δ) − δD_{n}(δ) \\ & = E_{n+2}(δ) + (δ − 2)E_{n+1}(δ) + (1 − δ)E_{n}(δ), \\ δ^{3}B_{n}(δ) & = δ^{2}C_{n+1}(δ) − δ^{2}C_{n}(δ) − δ^{3}D_{n}(δ) + δ^{3}E_{n}(δ) \\ & = E_{n+3}(δ) + (δ − 3)E_{n+2}(δ) + (3 − 2δ − δ^{2})E_{n+1}(δ) \\ & +(δ^{2} + δ − 1)E_{n}(δ), \\ δ^{4}A_{n}(δ) & = δ^{3}B_{n+1}(δ) − δ^{3}B_{n}(δ) − δ^{4}C_{n}(δ) + δ^{4}E_{n}(δ) \\ & = E_{n+4}(δ) + (δ − 4)E_{n+3}(δ) + (−2δ^{2} − 3δ + 6)E_{n+2}(δ) \\ & +(−δ^{3} + 4δ^{2} + 3δ − 4)E_{n+1}(δ) + (δ^{4} + δ^{3} − 2δ^{2} − δ + 1)E_{n}(δ) \end{matrix}  (3.17)

and, finally:

δ^{4}A_{n+1}(δ) − δ^{4}A_{n}(δ) − 2δ^{5}B_{n}(δ) + δ^{5}E_{n}(δ) = 0,  (3.18)

i.e.

E_{n+5}(δ) + (δ − 5)E_{n+4}(δ) + (−4δ^{2} − 4δ + 10)E_{n+3}(δ)+

+ (−3δ^{3} + 12δ^{2} + 6δ − 10)E_{n+2}(δ) + (3δ^{4} + 6δ^{3} − 12δ^{2} − 4δ + 5)E_{n+1}(δ)+ + (δ^{5} − 3δ^{4} − 3δ^{3} + 4δ^{2} + δ − 1)E_{n}(δ) = 0. (3.19)

Immediately from equations (3.13) or (3.17) and (3.19) for δ = 1 we obtain the following formulas:

\{ \begin{matrix}E_{n+1} = D_{n}, \\ C_{n} = D_{n+1} − D_{n}, \\ B_{n} = D_{n+2} − 2D_{n+1} + D_{n−1}, \\ A_{n} = D_{n+3} − 3D_{n+2} + D_{n+1} + 2D_{n}, \end{matrix}  (3.20)

and

D_{n+4} − 4D_{n+3} + 2D_{n+2} + 5D_{n+1} − 2D_{n} − D_{n−1} = 0.  (3.21)

The characteristic polynomial p_{11}(𝕏; δ) of the recurrence equation (3.19) has the following decomposition (see general formula (3.39) below):

p_{11}(𝕏; δ) := 𝕏^{5} + (δ − 5) 𝕏^{4} + (−4δ^{2} − 4δ + 10) 𝕏^{3} + (−3δ^{3} + 12δ^{2} + 6δ − 10) 𝕏^{2} + + (3δ^{4} + 6δ^{3} − 12δ^{2} − 4δ + 5) 𝕏 + (δ^{5} − 3δ^{4} − 3δ^{3} + 4δ^{2} + δ − 1) =

= ∏_{m=1}^{5} (𝕏 − τ_{m}(δ)). (3.22)

Sketch of the proof: We have

∑_{m=1}^{5} τ_{m}(δ) = ∑_{m=1}^{5} (1 + δ σ_{m}) = 5 + δ ∑_{m=1}^{5} σ_{m} =^{(1.4)} 5 − δ,

∑_{1≤m<n≤5} τ_{m}(δ) τ_{n}(δ) = \frac{1}{2} (( ∑_{m=1}^{5} τ_{m}(δ))^{2} − ∑_{m=1}^{5} τ_{m}^{2}(δ)) = \frac{1}{2} ((5 − δ)^{2} − ∑_{m=1}^{5} (1 + δ σ_{m})^{2}) =

= \frac{1}{2} ((5 − δ)^{2} − 5 − 2 δ ∑_{m=1}^{5} σ_{m} − δ^{2} ∑_{m=1}^{5} σ_{m}^{2}) =^{(1.4),(1.1)}

= \frac{1}{2} (25 − 10 δ + δ^{2} − 5 + 2 δ − δ^{2} ∑_{m=1}^{5} (σ_{2m} + 2)) =^{(1.4)} 10 − δ − 4 δ^{2},

etc. □ It follows from decomposition (3.22) that there exist numbers α, β, γ, ε, φ ∈ ℝ such that:

E_{n}(δ) = α τ_{1}^{n}(δ) + β τ_{2}^{n}(δ) + γ τ_{3}^{n}(δ) + ε τ_{4}^{n}(δ) + φ τ_{5}^{n}(δ)

for every n ∈ ℕ. Solving the respective system of linear equations:

\{ \begin{matrix}E_{n}(δ) & = & 0,  n = 1, 2, 3, \\ E_{4}(δ) & = & δ^{4}, \\ E_{5}(δ) & = & 5δ^{4} − δ^{5}, \end{matrix}

we obtain:

11 E_{n}(δ) = ∑_{m=1}^{5} (σ_{4m} − σ_{5m}) τ_{m}^{n}(δ)  (3.23)
= (σ_{4} − σ_{5}) τ_{1}^{n}(δ) + (σ_{3} − σ_{1}) τ_{2}^{n}(δ) + (σ_{1} − σ_{4}) τ_{3}^{n}(δ)
+ (σ_{5} − σ_{2}) τ_{4}^{n}(δ) + (σ_{2} − σ_{3}) τ_{5}^{n}(δ)

= 2(k_{1} − k_{3})(1 + 2 δ k_{2})^{n} + 2(k_{6} − k_{2})(1 + 2 δ k_{4})^{n} + 2(k_{2} + k_{3})(1 + 2 δ k_{6})^{n}

+ 2(−k_{1} − k_{4})(1 − 2 δ k_{3})^{n} + 2(k_{4} − k_{6})(1 − 2 δ k_{1})^{n},

which implies, the following identities for δ = \frac{1}{2}:

\frac{11}{2^{n+1}} E_{n}(\frac{1}{2}) = (k_{1} − k_{3}) k_{1}^{2n} + (k_{6} − k_{2}) k_{2}^{2n}  (3.24)
+ (k_{2} + k_{3}) k_{3}^{2n} − (k_{1} + k_{4}) k_{4}^{2n} + (k_{4} − k_{6}) k_{5}^{2n}
= 2 s_{1} s_{2} k_{1}^{2n} − 2 s_{2} s_{4} k_{2}^{2n} + 2 s_{3} s_{5} k_{3}^{2n} − 2 s_{3} s_{4} k_{4}^{2n} + 2 s_{1} s_{5} k_{5}^{2n},

and for δ = −\frac{1}{2} (by (1.5)):

\frac{11}{2^{n+2}} E_{n}(−\frac{1}{2}) = s_{2} s_{1}^{2n+1} − s_{4} s_{2}^{2n+1} + s_{5} s_{3}^{2n+1} − s_{3} s_{4}^{2n+1} + s_{1} s_{5}^{2n+1}.  (3.25)

Now, from (3.17), (3.23), (1.1) and (1.2) the following decomposition may be generated:

11 D_{n}(δ) = \frac{1}{δ}((11 E_{n+1}(δ)) − (1 − δ) (11 E_{n}(δ))) =  (3.26)

= \frac{1}{δ} ∑_{m=1}^{5} (σ_{4m} − σ_{5m})(τ_{m}(δ) − 1 + δ) τ_{m}^{n}(δ)

= ∑_{m=1}^{5} (σ_{4m} − σ_{5m})(σ_{m} + 1) τ_{m}^{n}(δ) = ∑_{m=1}^{5} (σ_{3m} − σ_{5m})τ_{m}^{n}(δ)

= (σ_{3} − σ_{5}) τ_{1}^{n}(δ) + (σ_{5} − σ_{1}) τ_{2}^{n}(δ) + (σ_{2} − σ_{4}) τ_{3}^{n}(δ)

+ (σ_{1} − σ_{2}) τ_{4}^{n}(δ) + (σ_{4} − σ_{3}) τ_{5}^{n}(δ)

= 2 (k_{1} + k_{6}) (1 + 2 δ k_{2})^{n} − 2 (k_{1} + k_{2}) (1 + 2 δ k_{4})^{n} + 2 (k_{3} + k_{4}) (1 + 2 δ k_{6})^{n}

+ 2 (k_{2} − k_{4}) (1 − 2 δ k_{3})^{n} − 2 (k_{3} + k_{6}) (1 − 2 δ k_{1})^{n},

which implies the following identity for δ = \frac{1}{2}:

\frac{11}{2^{n+1}} D_{n}(\frac{1}{2}) = (k_{1} − k_{5}) k_{1}^{2n} − (k_{1} + k_{2}) k_{2}^{2n} + (k_{3} + k_{4}) k_{3}^{2n}  (3.27)
+ (k_{2} − k_{4}) k_{4}^{2n} − (k_{3} + k_{6}) k_{5}^{2n}
= 2 s_{2} s_{3} k_{1}^{2n} − 2 s_{4} s_{6} k_{2}^{2n} + 2 s_{2} s_{6} k_{3}^{2n} + 2 s_{1} s_{3} k_{4}^{2n} − 2 s_{1} s_{4} k_{5}^{2n},

and for δ = −\frac{1}{2} (by (1.5)):

\frac{11}{2^{n+2}} D_{n}(−\frac{1}{2}) = s_{2} s_{3} s_{1}^{2n} − s_{4} s_{6} s_{2}^{2n} + s_{2} s_{6} s_{3}^{2n} + s_{1} s_{3} s_{4}^{2n} − s_{1} s_{4} s_{5}^{2n}.  (3.28)

Next, by (3.17), (3.26), (1.1) and (1.2), we obtain:

11 C_{n}(δ) = \frac{11}{δ}(D_{n+1}(δ) − D_{n}(δ)) =  (3.29)

= ∑_{m=1}^{5} (σ_{3m} − σ_{5m}) σ_{m} τ_{m}^{n}(δ) = ∑_{m=1}^{5} (σ_{2m} − σ_{5m})τ_{m}^{n}(δ)

= (σ_{2} − σ_{5}) τ_{1}^{n}(δ) + (σ_{4} − σ_{1}) τ_{2}^{n}(δ) + (σ_{5} − σ_{4}) τ_{3}^{n}(δ)

+ (σ_{3} − σ_{2}) τ_{4}^{n}(δ) + (σ_{1} − σ_{3}) τ_{5}^{n}(δ)

= 2 (k_{1} + k_{4}) (1 + 2 δ k_{2})^{n} − 2 (k_{2} + k_{3}) (1 + 2 δ k_{4})^{n} + 2 (k_{3} − k_{1}) (1 + 2 δ k_{6})^{n}

+ 2 (k_{6} − k_{4}) (1 − 2 δ k_{3})^{n} + 2 (k_{2} − k_{6}) (1 − 2 δ k_{1})^{n}, which implies the following identity for δ = \frac{1}{2}:

\frac{11}{2^{n+1}} C_{n}(\frac{1}{2}) = (k_{1} + k_{4}) k_{1}^{2n} − (k_{2} + k_{3}) k_{2}^{2n}  (3.30)
+ (k_{3} − k_{1}) k_{3}^{2n} + (k_{6} − k_{4}) k_{4}^{2n} + (k_{2} − k_{6}) k_{5}^{2n}

= 2 s_{3} s_{4} k_{1}^{2n} − 2 s_{3} s_{5} k_{2}^{2n} − 2 s_{1} s_{2} k_{3}^{2n} − 2 s_{1} s_{5} k_{4}^{2n} + 2 s_{2} s_{4} k_{5}^{2n},

and for δ = −\frac{1}{2} (by (1.5)):

\frac{11}{2^{n+2}} C_{n}(−\frac{1}{2}) = s_{3} s_{4} s_{1}^{2n} − s_{3} s_{5} s_{2}^{2n} − s_{1} s_{2} s_{3}^{2n} − s_{1} s_{5} s_{4}^{2n} + s_{2} s_{4} s_{5}^{2n}.  (3.31)

Now, by (3.17), (3.23), (3.26), (3.29), (1.1) and (1.2), we may generate the formula:

11 B_{n}(δ) = \frac{11}{δ}(C_{n+1}(δ) − C_{n}(δ)) + 11 (E_{n}(δ) − D_{n}(δ)) =  (3.32)

= ∑_{m=1}^{5} ((σ_{2m} − σ_{5m}) σ_{m} + σ_{4m} − σ_{5m} − σ_{3m} + σ_{5m}) τ_{m}^{n}(δ) = ∑_{m=1}^{5} (σ_{m} − σ_{5m})τ_{m}^{n}(δ)

= 2 (k_{1} + k_{2}) (1 + 2 δ k_{2})^{n} + 2 (k_{4} − k_{2}) (1 + 2 δ k_{4})^{n}

+ 2 (k_{3} + k_{6}) (1 + 2 δ k_{6})^{n} − 2 (k_{3} + k_{4}) (1 − 2 δ k_{3})^{n} − 2 (k_{1} + k_{6}) (1 − 2 δ k_{1})^{n},

hence, for δ = \frac{1}{2} we obtain:

\frac{11}{2^{n+1}} B_{n}(\frac{1}{2}) = (k_{1} + k_{2}) k_{1}^{2n} + (k_{4} − k_{2}) k_{2}^{2n}  (3.33)
+ (k_{3} + k_{6}) k_{3}^{2n} − (k_{3} + k_{4}) k_{4}^{2n} − (k_{1} + k_{6}) k_{5}^{2n},

and for δ = −\frac{1}{2} (by (1.5)):

\frac{11}{2^{n+2}} B_{n}(−\frac{1}{2}) = s_{4} s_{5} s_{1}^{2n} − s_{1} s_{3} s_{2}^{2n} + s_{1} s_{4} s_{3}^{2n} − s_{2} s_{5} s_{4}^{2n} − s_{2} s_{3} s_{5}^{2n}.  (3.34)

Immediately from Corollary 8 we derive the identity:

55 A_{n}(δ) = 11 ( ∑_{m=1}^{5} τ_{m}^{n}(δ) + B_{n}(δ) + C_{n}(δ) + D_{n}(δ) + E_{n}(δ))  (3.35)

= ∑_{m=1}^{5} (11 + σ_{m} + σ_{2m} + σ_{3m} + σ_{4m} − 4 σ_{5m}) τ_{m}^{n}(δ)

= ∑_{m=1}^{5} ( \underbrace{(1 + σ_{m} + σ_{2m} + σ_{3m} + σ_{4m} + σ_{5m})}_{=0} +10 − 5 σ_{5m}) τ_{m}^{n}(δ)

= 5 ∑_{m=1}^{5} (2 − σ_{5m})τ_{m}^{n}(δ),

and thus we obtain the following trigonometrical form of A_{n}(δ):

11 A_{n}(δ) = s_{5}^{2} (1 + 2 δ k_{2})^{n} + s_{1}^{2} (1 + 2 δ k_{4})^{n} + s_{4}^{2} (1 + 2 δ k_{6})^{n} + + s_{2}^{2} (1 − 2 δ k_{3})^{n} + s_{3}^{2} (1 − 2 δ k_{1})^{n}. (3.36)

Hence, for δ = \frac{1}{2} we obtain:

\frac{11}{2^{n+2}} A_{n}(\frac{1}{2}) = s_{5}^{2} k_{1}^{2n} + s_{1}^{2} k_{2}^{2n} + s_{4}^{2} k_{3}^{2n} + s_{2}^{2} k_{4}^{2n} + s_{3}^{2} k_{5}^{2n},  (3.37)

and for δ = −\frac{1}{2} (by (1.5)):

\frac{11}{2^{n+2}} A_{n}(−\frac{1}{2}) = s_{5}^{2} s_{1}^{2n} + s_{1}^{2} s_{2}^{2n} + s_{4}^{2} s_{3}^{2n} + s_{2}^{2} s_{4}^{2n} + s_{3}^{2} s_{5}^{2n}.  (3.38)

The next lemma is an attempt at generalizing the Lemma 3.13 from [9]. In view of an extensive form of the formulas, only some identities will be presented here.

Lemma 12. The following identities hold:

a) (decomposition using Newton’s formulas for elementary symmetric polynomials):

∏_{m=1}^{5} (𝕏 − τ_{m}^{n}(δ)) = 𝕏^{5} − A_{n}(δ) 𝕏^{4} + \frac{1}{2} (A_{n}^{2}(δ) − A_{2n}(δ)) 𝕏^{3}  (3.39)

− \frac{1}{6} (A_{n}^{3}(δ) + 2 A_{3n}(δ) − 3 A_{2n}(δ) A_{n}(δ)) 𝕏^{2}

+ \frac{1}{24} (A_{n}^{4}(δ) − 6 A_{2n}(δ) A_{n}^{2}(δ) + 8 A_{3n}(δ) A_{n}(δ) − 6 A_{4n}(δ) + 3 A_{2n}^{2}(δ)) 𝕏

+ (δ^{5} − 3 δ^{4} − 3 δ^{3} + 4 δ^{2} + δ − 1)^{n}.

b)

2 A_{n}(δ) − 11 A_{n}(δ) = ∑_{m=1}^{5} σ_{5m}τ_{m}^{n}(δ),  (3.40)

11 (B_{n}(δ) − A_{n}(δ)) + 2 A_{n}(δ) = ∑_{m=1}^{5} σ_{m}τ_{m}^{n}(δ) = \frac{1}{δ} (A_{n+1}(δ) − A_{n}(δ)),  (3.41)

which implies also the recurrence formula:

A_{n+1}(δ) = (2 δ + 1) A_{n}(δ) + 11 δ (B_{n}(δ) − A_{n}(δ));  (3.42)

11 (C_{n}(δ) − A_{n}(δ)) + 2 A_{n}(δ) = ∑_{m=1}^{5} σ_{2m}τ_{m}^{n}(δ),  (3.43)

11 (D_{n}(δ) − A_{n}(δ)) + 2 A_{n}(δ) = ∑_{m=1}^{5} σ_{3m}τ_{m}^{n}(δ),  (3.44)

11 (E_{n}(δ) − A_{n}(δ)) + 2 A_{n}(δ) = ∑_{m=1}^{5} σ_{4m}τ_{m}^{n}(δ).  (3.45)

c) (we set here A_{n} ≡ A_{n}(1) and E_{n} ≡ E_{n}(1))

11(E_{n+1} A_{n} − E_{n} A_{n+1}) = ∑_{m=1}^{5} (5 σ_{5m} + 3 (σ_{m} + σ_{4m}))(−σ_{m})^{n}  (3.46)

+ ∑_{m=1}^{5} (2 + σ_{3m} − σ_{2m} − σ_{4m}))(σ_{m} − σ_{4m} − σ_{5m})^{n}

=2^{n+1} ((−5 k_{1} + 3 k_{2} − 3 k_{3}) k_{9}^{n} + (5 k_{2} + 3 k_{4} + 3 k_{6}) k_{7}^{n} + (−5 k_{3} + 3 k_{2} + 3 k_{6}) k_{5}^{n}

+ (5 k_{4} − 3 k_{3} − 3 k_{1}) k_{3}^{n} + (5 k_{6} + 3 k_{4} − 3 k_{1}) k_{1}^{n} + (1 + k_{6} (1 − 2 k_{2}))(k_{2} (1 + 2 k_{1}))^{n} + (1 − k_{1} (1 − 2 k_{4}))(k_{4} (1 − 2 k_{2}))^{n} + (1 + k_{4} (1 + 2 k_{5}))(k_{6} (1 + 2 k_{3}))^{n}

+ (1 + k_{2} (1 + 2 k_{3}))(k_{8} (1 − 2 k_{4}))^{n} + (1 − k_{3} (1 + 2 k_{1}))(k_{10} (1 − 2 k_{5}))^{n}).

Proof. (a) The decompositions (3.39) from Vieta’s formulas, formula (3.22) and the following (version) of Newton’s formula for elementary symmetric polynomials [2] follows:

χ_{m} = \frac{1}{m!} | \begin{matrix}η_{1} & 1 & 0 & … & 0 \\ η_{2} & η_{1} & 2 & … & 0 \\ η_{3} & η_{2} & η_{1} & … & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ \\ η_{m} & η_{m−1} & η_{m−2} & … & η_{1} \end{matrix} |  (3.47)

where η_{j} = x_{1}^{j} + x_{2}^{j} + … + x_{n}^{j}, χ_{j} = ∑_{1≤m_{1}<m_{2}<...<m_{j}≤n} x_{1}x_{2} … x_{j}. Indeed, for x_{m} = τ_{m}^{n}(δ),

m = 1, 2, …, 5, from (3.47) we get

χ_{m} = \frac{1}{m!} | \begin{matrix}A_{n}(δ) & 1 & 0 & 0 & … & 0 \\ A_{2n}(δ) & A_{n}(δ) & 2 & 0 & … & 0 \\ A_{3n}(δ) & A_{2n}(δ) & A_{n}(δ) & 3 & … & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ \\ A_{mn}(δ) & A_{(m−1)n}(δ) & A_{(m−2)n}(δ) & A_{(m−3)n}(δ) & … & A_{n}(δ) \end{matrix} | =

= \{ \begin{matrix}\frac{1}{2!} (A_{n}^{2}(δ) − A_{2n}(δ)), & for m = 2, \\ \frac{1}{3!} (A_{n}^{3}(δ) + 2 A_{3n}(δ) − 3 A_{n}(δ) A_{2n}(δ)), & for m = 3, \\ etc. \end{matrix}

We note that for this case we have

∏_{m=1}^{5} (𝕏 − τ_{m}^{n}(δ)) = 𝕏^{5} − χ_{1} 𝕏^{4} + χ_{2} 𝕏^{3} − χ_{3} 𝕏^{2} + χ_{4} 𝕏 − χ_{5}.

Moreover, from (3.22) we have

χ_{5} = ∏_{m=1}^{5} τ_{m}(δ) = −δ^{5} + 3 δ^{4} − 4 δ^{2} − δ + 1.

(b) (3.40): By (3.14) and (3.35) we have:

2 A_{n}(δ) − 11 A_{n}(δ) = 2 ∑_{m=1}^{5} τ_{m}^{n}(δ) − ∑_{m=1}^{5} (2 − σ_{m}) τ_{m}^{n}(δ) = ∑_{m=1}^{5} σ_{5m}τ_{m}^{n}(δ).

(3.41): From (3.14) we obtain

\frac{1}{δ}(A_{n+1}(δ) − A_{n}(δ)) = \frac{1}{δ} ∑_{m=1}^{5} (τ_{m}^{n+1}(δ) − τ_{m}^{n}(δ)) = ∑_{m=1}^{5} \frac{1}{δ} (τ_{m}(δ) − 1) τ_{m}^{n}(δ) = ∑_{m=1}^{5} σ_{m}τ_{m}^{n}(δ).

On the other hand, by (3.32) and (3.14) we get

∑_{m=1}^{5} σ_{m}τ_{m}^{n}(δ) = ∑_{m=1}^{5} σ_{5m}τ_{m}^{n}(δ) + ∑_{m=1}^{5} (σ_{m} − σ_{5m})τ_{m}^{n}(δ) = 11 (B_{n}(δ) − A_{n}(δ)) + 2 A_{n}(δ).

Formulas (3.43), (3.44) and (3.45) from (3.40) and formulas (3.2m − 1), m = 2, 3, 4 can be easily deduced.

(c) The following ten simple equalities form the technical base (by hand calculation) of the proof of identity (3.46):

(1 + σ_{1})(1 + σ_{3}) = −σ_{5},  (1 + σ_{1})(1 + σ_{2}) = σ_{1} − σ_{4} − σ_{5},
(1 + σ_{1})(1 + σ_{4}) = −σ_{2},  (1 + σ_{1})(1 + σ_{5}) = σ_{5} − σ_{2} − σ_{3},
(1 + σ_{2})(1 + σ_{3}) = −σ_{4},  (1 + σ_{2})(1 + σ_{4}) = σ_{2} − σ_{1} − σ_{3},
(1 + σ_{2})(1 + σ_{5}) = −σ_{1},  (1 + σ_{3})(1 + σ_{4}) = σ_{4} − σ_{2} − σ_{5},
(1 + σ_{4})(1 + σ_{5}) = −σ_{3},  (1 + σ_{3})(1 + σ_{5}) = σ_{3} − σ_{1} − σ_{4}.

4 Summation formulas

Lemma 13. We have:

δ ∑_{n=1}^{N} C_{n}(δ) = D_{N+1}(δ),  (4.1)

δ ∑_{n=1}^{N} B_{n}(δ) = C_{N+1}(δ) + δ ∑_{n=1}^{N} (E_{n}(δ) − D_{n}(δ)) = C_{N+1}(δ) − E_{N+1}(δ),  (4.2)

δ ∑_{n=1}^{N} E_{n}(δ) = A_{1}(δ) − A_{N+1}(δ) + 2δ ∑_{n=1}^{N} B_{n}(δ)  (4.3)
= 1 − A_{N+1}(δ) + 2C_{N+1}(δ) − 2E_{N+1}(δ),  (4.4)

and, from (3.19) we obtain:

δ^{5} ∑_{n=1}^{N} E_{n}(δ) = −E_{N+5}(δ) + (4 − δ)E_{N+4}(δ) + (4δ^{2} + 3δ − 6)E_{N+3}(δ)+

+ (3δ^{3} − 8δ^{2} − 3δ + 4)E_{N+2}(δ) + (−3δ^{4} − 3δ^{3} + 4δ^{2} + δ − 1)E_{N+1}(δ) + δ^{4}; (4.5)

δ ∑_{n=1}^{N} D_{n}(δ) = E_{N+1}(δ) + δ ∑_{n=1}^{N} E_{n}(δ),  (4.6)
δ ∑_{n=0}^{N−1}(1 − δ)^{n}D_{N−n}(δ) = E_{N+1}(δ)  (4.7)

or

\frac{E_{N}(δ)}{(1 − δ)^{N−1}} = δ ∑_{n=0}^{N−1} \frac{D_{n}(δ)}{(1 − δ)^{n}} ,  (4.8)
δ ∑_{n=1}^{N} A_{n}(δ) = −A_{N+1}(δ) + B_{N+1}(δ) + 2C_{N+1}(δ) − D_{N+1}(δ) − 2E_{N+1}(δ) + 1 − δ. (4.9)

and

δ^{2} ∑_{n=1}^{N}(N − n + 1)(2 B_{n}(δ) − E_{n}(δ)) = δ ∑_{n=1}^{N+1} A_{n}(δ) − (N + 1) δ.  (4.10)

Proof. (4.1): By (3.17) and Table 1 we get

δ ∑_{n=1}^{N} C_{n}(δ) = ∑_{n=1}^{N} (D_{n+1}(δ) − D_{n}(δ)) = D_{N+1}(δ) − D_{1}(δ) = D_{N+1}(δ).

(4.2): By (3.17) and Table 1 we get

δ ∑_{n=1}^{N} B_{n}(δ) = ∑_{n=1}^{N} (C_{n+1}(δ) − C_{n}(δ)) + δ ∑_{n=1}^{N} (E_{n}(δ) − D_{n}(δ)) =

= C_{N+1}(δ) − C_{1}(δ) + δ ∑_{n=1}^{N} (E_{n}(δ) − E_{n+1}(δ)) =

= C_{N+1}(δ) − C_{1}(δ) + E_{1}(δ) − E_{N+1}(δ) = C_{N+1}(δ) − E_{N+1}(δ).

(4.3): Follows from (3.18). (4.4): Can be derived from (4.3) and (4.2). The other proof from formulas (3.23), (3.29) and (3.35) follows: first, by (3.23) we get

δ ∑_{n=1}^{N} E_{n}(δ) = \frac{δ}{11} ∑_{n=1}^{N} ∑_{m=1}^{5} (σ_{4m} − σ_{5m}) τ_{m}^{n}(δ) = \frac{δ}{11} ∑_{m=1}^{5} ∑_{n=1}^{N} (σ_{4m} − σ_{5m}) τ_{m}^{n}(δ) =

= \frac{δ}{11} ∑_{m=1}^{5} (σ_{4m} − σ_{5m}) \frac{τ_{m}^{N+1}(δ) − τ_{m}(δ)}{τ_{m}(δ) − 1} = \frac{1}{11} ∑_{m=1}^{5} \frac{σ_{4m} − σ_{5m}}{σ_{m}} (τ_{m}^{N+1}(δ) − τ_{m}(δ)), (4.11)

but we have

\frac{σ_{4m} − σ_{5m}}{σ_{m}} = \frac{σ_{4m} + σ_{5m}}{σ_{m}} − 2 \frac{σ_{5m}}{σ_{m}} = σ_{5m} − 2 (σ_{4m} − σ_{2m} + 1) =

= 2 (σ_{2m} − σ_{5m}) − 2 (σ_{4m} − σ_{5m}) − (2 − σ_{5m}),

which, by (4.11), (3.29), (3.23) and (3.35) implies

δ ∑_{n=1}^{N} E_{n}(δ) = 2 (C_{N+1}(δ) − C_{1}(δ)) − 2 (E_{N+1}(δ) − E_{1}(δ)) − (A_{N+1}(δ) − A_{1}(δ)).

Hence, by Table 1 the formula (4.4) follows. (4.7): By (3.18) we obtain

δ D_{N}(δ) = E_{N+1}(δ) − (1 − δ) E_{N}(δ),
δ (1 − δ) D_{N−1}(δ) = (1 − δ) E_{N}(δ) − (1 − δ)^{2} E_{N−1}(δ),
δ (1 − δ)^{2} D_{N−2}(δ) = (1 − δ)^{2} E_{N−1}(δ) − (1 − δ)^{3} E_{N−2}(δ),
⋮
δ (1 − δ)^{N−1} D_{1}(δ) = (1 − δ)^{N−1} E_{2}(δ) − (1 − δ)^{N} E_{1}(δ),

which, after summation implies the formula (4.7). (4.9): By (3.18) we have

δ ∑_{n=1}^{N} A_{n}(δ) = ∑_{n=1}^{N} (B_{n+1}(δ) − B_{n}(δ)) + δ ∑_{n=1}^{N} E_{n}(δ) − δ ∑_{n=1}^{N} C_{n}(δ).

Hence, by (4.1) and (4.3) we get the desired formula

= (B_{N+1}(δ) − B_{1}(δ)) + (1 − A_{N+1}(δ) + 2 C_{N+1}(δ) − 2 E_{N+1}(δ)) − D_{N+1}(δ). (4.10): From (3.18) we deduce the following system of equalities

A_{N+1}(δ) − A_{N}(δ) = 2 δ B_{N}(δ) − δ E_{N}(δ),
2 A_{N}(δ) − 2 A_{N−1}(δ) = 4 δ B_{N−1}(δ) − 2 δ E_{N−1}(δ),
3 A_{N−1}(δ) − 3 A_{N−2}(δ) = 6 δ B_{N−2}(δ) − 3 δ E_{N−2}(δ),
⋮
N A_{2}(δ) − N A_{1}(δ) = 2 N δ B_{1}(δ) − N δ E_{1}(δ),

which, after summation implies the desired formulas. Corollary 14. (See also Corollary 23.) From (4.7) for δ = ±\frac{1}{2} we get

2^{N+1} E_{N+1}(\frac{1}{2}) = ∑_{n=0}^{N−1} 2^{n} D_{n}(\frac{1}{2})

and

−2 (\frac{2}{3})^{N} E_{N+1}(−\frac{1}{2}) = ∑_{n=0}^{N−1} (\frac{2}{3})^{n} D_{n}(−\frac{1}{2}).

5 Reduction formulas for indices

Lemma 15. The following identities hold:

\begin{cases}A_{m+n} = A_{m}A_{n} + 2B_{m}B_{n} + 2C_{m}C_{n} + 2D_{m}D_{n} + 2E_{m}E_{n}, \\ B_{m+n} = A_{m}B_{n} + B_{m}A_{n} + B_{m}C_{n} + C_{m}B_{n} + C_{m}D_{n} + D_{m}C_{n} + D_{m}E_{n} + E_{m}D_{n}, \\ C_{m+n} = A_{m}C_{n} + B_{m}B_{n} + B_{m}D_{n} + D_{m}B_{n} + C_{m}A_{n} + C_{m}E_{n} + E_{m}C_{n}, \\ D_{m+n} = A_{m}D_{n} + B_{m}C_{n} + B_{m}E_{n} + D_{m}A_{n} + C_{m}B_{n} + E_{m}b_{n} + E_{m}E_{n}, \\ E_{m+n} = A_{m}E_{n} + B_{m}D_{n} + E_{m}A_{n} + D_{m}B_{n} + C_{m}C_{n} + D_{m}E_{n} + E_{m}D_{n}, \end{cases} (5.1)

and (the special cases when m = n):

\{ \begin{matrix}A_{2n} = A_{n}^{2} + 2B_{n}^{2} + 2C_{n}^{2} + 2D_{n}^{2} + 2E_{n}^{2}, \\ B_{2n} = 2A_{n}B_{n} + 2B_{n}C_{n} + 2C_{n}D_{n} + 2D_{n}E_{n}, \\ C_{2n} = 2A_{n}C_{n} + 2C_{n}E_{n} + 2D_{n}B_{n} + B_{n}^{2}, \\ D_{2n} = 2A_{n}D_{n} + 2C_{n}B_{n} + 2B_{n}E_{n} + E_{n}^{2}, \\ E_{2n} = 2A_{n}E_{n} + 2E_{n}D_{n} + 2D_{n}B_{n} + C_{n}^{2}. \end{matrix}  (5.2)

The proof runs similar to the proof of Lemma 3.16 from [9]. Lemma 16. We have the identity:

Δ := | \begin{matrix}A_{k}(δ) & 2B_{k}(δ) − E_{k}(δ) & 2C_{k}(δ) − D_{k}(δ) − E_{k}(δ) \\ B_{k}(δ) & A_{k}(δ) + C_{k}(δ) − E_{k}(δ) & B_{k}(δ) − E_{k}(δ) \\ C_{k}(δ) & B_{k}(δ) + D_{k}(δ) − E_{k}(δ) & A_{k}(δ) − D_{k}(δ) \\ D_{k}(δ) & C_{k}(δ) & B_{k}(δ) − D_{k}(δ) − E_{k}(δ) \\ E_{k}(δ) & D_{k}(δ) − E_{k}(δ) & C_{k}(δ) − D_{k}(δ) − E_{k}(δ) \end{matrix}

\begin{matrix}−C_{k}(δ) + D_{k}(δ) & −B_{k}(δ) − C_{k}(δ) + 2E_{k}(δ) \\ −D_{k}(δ) + E_{k}(δ) & −B_{k}(δ) − C_{k}(δ) + 2E_{k}(δ) \\ B_{k}(δ) − C_{k}(δ) − D_{k}(δ) & −B_{k}(δ) \\ A_{k}(δ) − C_{k}(δ) − D_{k}(δ) & −C_{k}(δ) + E_{k}(δ) \\ B_{k}(δ) − C_{k}(δ) − D_{k}(δ) + E_{k}(δ) & A_{k}(δ) − B_{k}(δ) − C_{k}(δ) + D_{k}(δ) \end{matrix} | =

= ( − δ^{5} + 3 δ^{4} + 3 δ^{3} − 4 δ^{2} − δ + 1)^{k}. (5.3)

The proof of this lemma runs like the proof of Lemma 3.21 from [9]. Lemma 17. The following identities hold:

A_{n−k}(δ) = \frac{Δ_{1}}{Δ} , B_{n−k}(δ) = \frac{Δ_{2}}{Δ} , C_{n−k}(δ) = \frac{Δ_{3}}{Δ} , D_{n−k}(δ) = \frac{Δ_{4}}{Δ} , E_{n−k}(δ) = \frac{Δ_{5}}{Δ} ,

where Δ_{i} arise from the determinant Δ by replacing its i-th column by the vector (A_{n}, B_{n}, C_{n}, D_{n}, E_{n})^{T} (the Cramers rule for respective system of equations).

The proof is similar to the proof of Lemma 3.20 from [9].

Remark 18. After multiplication we obtain

(τ_{2}(1))^{2n}(τ_{5}(1))^{2n} = (2 + σ_{2})^{n} = 2^{n}(τ_{2}(\frac{1}{2}))^{n} =^{by (3.4)}

= 2^{n}(A_{n}(\frac{1}{2}) + B_{n}(\frac{1}{2}) σ_{2} + C_{n}(\frac{1}{2}) σ_{4} + D_{n}(\frac{1}{2}) σ_{5} + E_{n}(\frac{1}{2}) σ_{3}). (5.4)

On the other hand, by (3.4) and (3.10) we get:

(τ_{2}(1))^{2n}(τ_{5}(1))^{2n} = = (A_{2n} + B_{2n} σ_{2} + C_{2n} σ_{4} + D_{2n} σ_{5} + E_{2n} σ_{3})(A_{2n} + B_{2n} σ_{5} + C_{2n} σ_{1} + D_{2n} σ_{4} + E_{2n} σ_{2}) =

= σ_{2} (A_{2n} (B_{2n} − C_{2n} + E_{2n}) − B_{2n} (C_{2n} − E_{2n}) + E_{2n} (2 C_{2n} − D_{2n} − E_{2n})) + + σ_{3} (C_{2n} (C_{2n} − A_{2n} − B_{2n} + D_{2n}) + B_{2n} (B_{2n} − D_{2n}) + E_{2n} (A_{2n} + B_{2n}) − D_{2n}^{2} − E_{2n}^{2}) + + σ_{4} (B_{2n} (B_{2n} − 2 C_{2n} − D_{2n} + E_{2n}) + C_{2n} (D_{2n} + E_{2n}) + D_{2n} (A_{2n} − D_{2n} + E_{2n}) − E_{2n}^{2}) +

+ σ_{5} (A_{2n} (B_{2n} + D_{2n} − C_{2n}) + C_{2n} (C_{2n} − 2 B_{2n} + D_{2n} + E_{2n}) − D_{2n} (D_{2n} + E_{2n})) + + (A_{2n} (A_{2n} − C_{2n}) − B_{2n} (2 C_{2n} − D_{2n} − 2 E_{2n}) + D_{2n} (2 C_{2n} − D_{2n} − E_{2n}) − E_{2n}^{2}).

Hence, comparing the coefficients in the expression σ_{m}, m = 0, 2, 3, 4, 5 we obtain, by Lemma 4, the following interesting identities:

2^{n} A_{n}(\frac{1}{2}) = A_{2n} (A_{2n} − C_{2n}) − B_{2n} (2 C_{2n} − D_{2n} − 2 E_{2n})  (5.5)
+ D_{2n} (2 C_{2n} − D_{2n} − E_{2n}) − E_{2n}^{2},
2^{n} B_{n}(\frac{1}{2}) = A_{2n} (B_{2n} − C_{2n} + E_{2n}) − B_{2n} (C_{2n} − E_{2n})  (5.6)
+ E_{2n} (2 C_{2n} − D_{2n} − E_{2n}),
2^{n} C_{n}(\frac{1}{2}) = B_{2n} (B_{2n} − 2 C_{2n} − D_{2n} + E_{2n}) + C_{2n} (D_{2n} + E_{2n})  (5.7)
+ D_{2n} (A_{2n} − D_{2n} + E_{2n}) − E_{2n}^{2},
2^{n} D_{n}(\frac{1}{2}) = A_{2n} (B_{2n} + D_{2n} − C_{2n})  (5.8)
+ C_{2n} (C_{2n} − 2 B_{2n} + D_{2n} + E_{2n}) − D_{2n} (D_{2n} + E_{2n}),
2^{n} E_{n}(\frac{1}{2}) = C_{2n} (C_{2n} − A_{2n} − B_{2n} + D_{2n}) + B_{2n} (B_{2n} − D_{2n})  (5.9)
+ E_{2n} (A_{2n} + B_{2n}) − D_{2n}^{2} − E_{2n}^{2}.

6 Some new recurrence formulas

The following recurrence formulas determined the coefficients of the polynomials A_{n}(Δ), B_{n}(Δ), …, E_{n}(Δ).

Lemma 19. Let Δ ∈ ℂ \ {0}, n ∈ ℕ. Then the following recurrence formulas hold

(−1)^{n} A_{n}(Δ) = Δ^{n} (A_{n} (A_{n} − B_{n} − D_{n}) + B_{n} (D_{n} + 2 E_{n}) + C_{n} (D_{n} − C_{n} − E_{n}) − E_{n}^{2})
+ ∑_{k=0}^{n−1} \binom{n}{k} (−1)^{k+1} A_{k}(Δ);  (6.1)

(−1)^{n} B_{n}(Δ) = Δ^{n} (C_{n} (A_{n} + 2 B_{n} − C_{n} − D_{n} − E_{n}) + D_{n} (D_{n} + E_{n} − A_{n}) − A_{n} B_{n})
+ ∑_{k=0}^{n−1} \binom{n}{k} (−1)^{k+1} B_{k}(Δ);  (6.2)
(−1)^{n} C_{n}(Δ) = Δ^{n} (C_{n} (B_{n} − C_{n} − D_{n} + E_{n}) + D_{n} (D_{n} − A_{n}) + E_{n} (A_{n} + B_{n} − E_{n}))
+ ∑_{k=0}^{n−1} \binom{n}{k} (−1)^{k+1} C_{k}(Δ);  (6.3)
(−1)^{n} D_{n}(Δ) = Δ^{n} (B_{n} (B_{n} + C_{n} + E_{n} − A_{n} − D_{n}) + E_{n} (A_{n} + D_{n} − C_{n} − E_{n}) − A_{n} D_{n})
+ ∑_{k=0}^{n−1} \binom{n}{k} (−1)^{k+1} D_{k}(Δ);  (6.4)
(−1)^{n} E_{n}(Δ) = Δ^{n} (B_{n} (B_{n} + E_{n} − A_{n} − D_{n}) + C_{n} (A_{n} − C_{n}) + E_{n} (2 D_{n} − E_{n}))
+ ∑_{k=0}^{n−1} \binom{n}{k} (−1)^{k+1} E_{k}(Δ).  (6.5)

Proof. We have the equality:

τ_{1}(1) τ_{3}(1) = −σ_{5}.

Hence we get (δ ∈ ℂ \ {0}):

τ_{1}^{n}(1) τ_{3}^{n}(1) = (δ − (δ + σ_{5}))^{n} = ∑_{k=0}^{n} \binom{n}{k} (−1)^{k} δ^{n−k} (δ + σ_{5})^{k} = δ^{n} ∑_{k=0}^{n} \binom{n}{k} (−1)^{k} τ_{5}^{k}(\frac{1}{δ}),

which by (3.2), (3.6) and (3.10), implies the identity

(A_{n} + B_{n} σ_{1} + C_{n} σ_{2} + D_{n} σ_{3} + E_{n} σ_{4})(A_{n} + B_{n} σ_{3} + C_{n} σ_{5} + D_{n} σ_{2} + E_{n} σ_{1}) =

= δ^{n} ∑_{k=0}^{n} \binom{n}{k}(−1)^{k}(A_{n}(\frac{1}{δ}) + B_{n}(\frac{1}{δ}) σ_{5} + C_{n}(\frac{1}{δ}) σ_{1} + D_{n}(\frac{1}{δ}) σ_{4} + E_{n}(\frac{1}{δ}) σ_{2}).

After the calculation of the left side of the above identity and the following substitution:

σ_{3} = −1 − σ_{1} − σ_{2} − σ_{4} − σ_{5}

and making use of the linear independence over ℚ of the numbers 1, σ_{1}, σ_{2}, σ_{4}, σ_{5} (see Corollary 3), we obtain:

δ^{n} ∑_{k=0}^{n} \binom{n}{k} (−1)^{k} A_{k}(\frac{1}{δ}) = A_{n} (A_{n} − B_{n} − D_{n}) + B_{n} (D_{n} + 2 E_{n})+

+ C_{n} (D_{n} − C_{n} − E_{n}) − E_{n}^{2}

etc., which, after the substitution of Δ := 1/δ and minor transformations, implies identities (6.1)–(6.5).

Corollary 20. If deg (A_{n}(Δ)) = n then coeff (A_{n}(Δ); Δ^{n}) = (−1)^{n}(A_{n} (A_{n} − B_{n} − D_{n}) + B_{n} (D_{n} + 2 E_{n}) + C_{n} (D_{n} − C_{n} − E_{n}) − E_{n}^{2}) and coeff (B_{n}(Δ); Δ^{n}) = (−1)^{n}(C_{n} (A_{n} + 2 B_{n} − C_{n} − D_{n} − E_{n}) + D_{n} (D_{n} + E_{n} − A_{n}) − A_{n} B_{n}).

Problem. Is it true that

deg (A_{n}(Δ)) = deg (B_{n}(Δ)) = deg (C_{n}(Δ)) = deg (D_{n}(Δ)) = deg (E_{n}(Δ)) = n for every n = 5, 6, …?

7 Some convolution type identities

Lemma 21. We have

τ_{1}(1) τ_{2}(1) τ_{3}(1) = 1 + σ_{1} + σ_{2}.

Hence, for δ ∈ ℂ, δ (1 − δ) ≠ 0, we obtain

τ_{1}^{n}(1) τ_{2}^{n}(1) τ_{3}^{n}(1) = ((1 − δ + σ_{1}) + (δ + σ_{2}))^{n} =

= ∑_{k=0}^{n} \binom{n}{k}(1 − δ)^{k} δ^{n−k} τ_{1}^{k}(1/(1 − δ)) τ_{2}^{n−k}(1/δ),

which by (3.2), (3.4) and (3.6), implies the identity:

(A_{n} + B_{n} σ_{1} + C_{n} σ_{2} + D_{n} σ_{3} + E_{n} σ_{4}) (A_{n} + B_{n} σ_{2} + C_{n} σ_{4} + D_{n} σ_{5} + E_{n} σ_{3}) ×

× (A_{n} + B_{n} σ_{3} + C_{n} σ_{5} + D_{n} σ_{2} + E_{n} σ_{1}) =

= ∑_{k=0}^{n} \binom{n}{k}(1 − δ)^{k} δ^{n−k} (A_{k}(\frac{1}{1−δ}) + B_{k}(\frac{1}{1−δ}) σ_{1} + C_{k}(\frac{1}{1−δ}) σ_{2} + D_{k}(\frac{1}{1−δ}) σ_{3}

+ E_{k}(\frac{1}{1−δ}) σ_{4}) (A_{n−k}(\frac{1}{δ}) + B_{n−k}(\frac{1}{δ}) σ_{2} + C_{n−k}(\frac{1}{δ}) σ_{4} + D_{n−k}(\frac{1}{δ}) σ_{5} + E_{n−k}(\frac{1}{δ}) σ_{3}). Hence, after some calculation, by comparing the absolute terms and the coefficients in the expressions σ_{1}, σ_{2}, σ_{3} and σ_{4} we get the following five independent identities:

A_{n}^{3} + B_{n}^{3} + 2 C_{n}^{3} + D_{n}^{3} + E_{n}^{3} − A_{n}^{2} (C_{n} + D_{n}) − B_{n}^{2} (A_{n} + D_{n}) + C_{n}^{2} (D_{n} − A_{n} − B_{n} − 2 E_{n}) +

+ D_{n}^{2} (2 E_{n} − A_{n} − B_{n} − 3 C_{n}) − E_{n}^{2} (A_{n} + 2 B_{n}) + A_{n} B_{n} (D_{n} + 2 E_{n} − C_{n}) +

+ A_{n} D_{n} (3 C_{n} − 2 E_{n}) + B_{n} C_{n} (2 D_{n} + E_{n}) + D_{n} E_{n} (B_{n} − 2 C_{n}) =

= ∑_{k=0}^{n} \binom{n}{k}(1 − δ)^{k} δ^{n−k} [A_{k}(\frac{1}{1−δ}) (A_{n−k}(\frac{1}{δ}) − D_{n−k}(\frac{1}{δ})) +

+ C_{k}(\frac{1}{1−δ}) (2 B_{n−k}(\frac{1}{δ}) − C_{n−k}(\frac{1}{δ}) − E_{n−k}(\frac{1}{δ})) − B_{k}(\frac{1}{1−δ}) (D_{n−k}(\frac{1}{δ}) + C_{n−k}(\frac{1}{δ})) + + D_{k}(\frac{1}{1−δ}) (E_{n−k}(\frac{1}{δ}) − B_{n−k}(\frac{1}{δ})) + E_{k}(\frac{1}{1−δ}) (2 C_{n−k}(\frac{1}{δ}) − B_{n−k}(\frac{1}{δ}))], (7.1)

B_{n}^{3} − C_{n}^{3} − D_{n}^{3} + E_{n}^{3} + A_{n}^{2} (C_{n} + D_{n} − E_{n}) + B_{n}^{2} (C_{n} − A_{n}) + C_{n}^{2} (A_{n} + D_{n} − E_{n}) +

+ D_{n}^{2} (C_{n} − B_{n} + E_{n}) − E_{n}^{2} (2 B_{n} + 2 D_{n}) + A_{n} B_{n} (E_{n} − 2 C_{n} − D_{n}) +

+ D_{n} E_{n} (A_{n} + 3 B_{n} + C_{n}) − C_{n} (A_{n} E_{n} + B_{n} D_{n}) =

= ∑_{k=0}^{n} \binom{n}{k}(1 − δ)^{k} δ^{n−k} [A_{k}(\frac{1}{1−δ}) B_{n−k}(\frac{1}{δ}) + B_{k}(\frac{1}{1−δ}) (E_{n−k}(\frac{1}{δ}) − A_{n−k}(\frac{1}{δ}) − B_{n−k}(\frac{1}{δ})) +

+ C_{k}(\frac{1}{1−δ}) (A_{n−k}(\frac{1}{δ}) + C_{n−k}(\frac{1}{δ}) − E_{n−k}(\frac{1}{δ})) + D_{k}(\frac{1}{1−δ}) (D_{n−k}(\frac{1}{δ}) − B_{n−k}(\frac{1}{δ}) − C_{n−k}(\frac{1}{δ})) + + E_{k}(\frac{1}{1−δ}) (B_{n−k}(\frac{1}{δ}) − E_{n−k}(\frac{1}{δ}))], (7.2)

E_{n}^{3} − D_{n}^{3} + A_{n}^{2} D_{n} + B_{n}^{2} (3 C_{n} − A_{n} − 2 E_{n}) − C_{n}^{2} B_{n} + D_{n}^{2} (A_{n} + B_{n} + C_{n} + E_{n}) + +A_{n} E_{n} (C_{n} −2 D_{n})−E_{n}^{2} (3 C_{n} +2 D_{n})+B_{n} E_{n} (2 C_{n} +4 D_{n} −A_{n})+C_{n} D_{n} (2 E_{n} −5 B_{n}) =

= ∑_{k=0}^{n} \binom{n}{k}(1 − δ)^{k} δ^{n−k} [A_{k}(\frac{1}{1−δ}) E_{n−k}(\frac{1}{δ}) + B_{k}(\frac{1}{1−δ}) (C_{n−k}(\frac{1}{δ}) − A_{n−k}(\frac{1}{δ})) +

+ C_{k}(\frac{1}{1−δ}) (D_{n−k}(\frac{1}{δ}) − E_{n−k}(\frac{1}{δ})) + D_{k}(\frac{1}{1−δ}) (A_{n−k}(\frac{1}{δ}) − B_{n−k}(\frac{1}{δ}) − C_{n−k}(\frac{1}{δ}) + D_{n−k}(\frac{1}{δ})) + + E_{k}(\frac{1}{1−δ}) (C_{n−k}(\frac{1}{δ}) − D_{n−k}(\frac{1}{δ}) − E_{n−k}(\frac{1}{δ}))], (7.3)

B_{n}^{3} + A_{n}^{2} (C_{n} − B_{n}) + B_{n}^{2} (C_{n} − A_{n} − D_{n} − E_{n}) + C_{n}^{2} (D_{n} − 3 B_{n}) + D_{n}^{2} (E_{n} − C_{n}) + + E_{n}^{2} (A_{n} − 3 D_{n}) + A_{n} C_{n} (2 B_{n} + D_{n} − 3 E_{n}) + B_{n} C_{n} (E_{n} − D_{n}) +

+ D_{n} E_{n} (5 B_{n} + C_{n}) = ∑_{k=0}^{n} \binom{n}{k}(1 − δ)^{k} δ^{n−k} [A_{k}(\frac{1}{1−δ}) C_{n−k}(\frac{1}{δ}) +

+ B_{k}(\frac{1}{1−δ}) (D_{n−k}(\frac{1}{δ}) − A_{n−k}(\frac{1}{δ}) − B_{n−k}(\frac{1}{δ}) + E_{n−k}(\frac{1}{δ})) + + C_{k}(\frac{1}{1−δ}) (B_{n−k}(\frac{1}{δ}) + D_{n−k}(\frac{1}{δ}) − E_{n−k}(\frac{1}{δ})) − D_{k}(\frac{1}{1−δ}) B_{n−k}(\frac{1}{δ}) +

+ E_{k}(\frac{1}{1−δ}) (A_{n−k}(\frac{1}{δ}) − D_{n−k}(\frac{1}{δ}))], (7.4)

etc.

8 Polynomials associated with quasi-Fibonacci numbers of order 11

By taking into account the Chebyshev polynomials T_{n}(x) again we obtain, for τ_{k} = τ_{k}(δ), k = 1, 2, …, 5:

\begin{cases}σ_{k} & = 2 cos \frac{2kπ}{11} = \frac{1}{δ}(τ_{k} − 1), \\ σ_{2k} & = 2 cos \frac{4kπ}{11} = 2T_{2}( cos \frac{2kπ}{11}) = 2T_{2}(\frac{1}{2δ}(τ_{k} − 1)) = \frac{1}{δ^{2}} (τ_{k}^{2} − 2τ_{k} + 1 − 2δ^{2}), \\ σ_{3k} & = 2 cos \frac{6kπ}{11} = 2T_{3}( cos \frac{2kπ}{11}) = 2T_{3}(\frac{1}{2δ}(τ_{k} − 1)) = \\ & = \frac{1}{δ^{3}} (τ_{k}^{3} − 3τ_{k}^{2} + (3 − 3δ^{2})τ_{k} − 1 + 3δ^{2}), \\ σ_{4k} & = 2 cos \frac{8kπ}{11} = 2T_{4}( cos \frac{2kπ}{11}) = 2T_{4}(\frac{1}{2δ}(τ_{k} − 1)) = \\ & = \frac{1}{δ^{4}} (τ_{k}^{4} − 4τ_{k}^{3} + (6 − 4δ^{2})τ_{k}^{2} + (−4 + 8δ^{2})τ_{k} + 1 − 4δ^{2} + 2δ^{4}), \\ σ_{5k} & = 2 cos \frac{10kπ}{11} = 2T_{5}( cos \frac{2kπ}{11}) = \frac{1}{δ^{5}} (τ_{k}^{5} − 5τ_{k}^{4} + (10 − 5δ^{2})τ_{k}^{3}+ \\ & + (−10 + 15δ^{2})τ_{k}^{2} + (10 − 30δ^{2} + 5δ^{4})τ_{k} − 6 + 20δ^{2} − 5δ^{4}). \end{cases}  (8.1)

Therefore, immediately from equation (3.1) we obtain the equality:

τ_{k}^{n} =A_{n}(δ) + \frac{1}{δ}B_{n}(δ)(τ_{k} − 1) + \frac{1}{δ^{2}} C_{n}(δ)(τ_{k}^{2} − 2τ_{k} + 1 − 2δ^{2})

+ \frac{1}{δ^{3}} D_{n}(δ)(τ_{k}^{3} − 3τ_{k}^{2} + (3 − 3δ^{2})τ_{k} − 1 + 3δ^{2})

+ \frac{1}{δ^{4}} E_{n}(δ)(τ_{k}^{4} − 4τ_{k}^{3} + (6 − 4δ^{2})τ_{k}^{2} + (−4 + 8δ^{2})τ_{k} + 1 − 4δ^{2} + 2δ^{4}),

i.e.,

W_{n,11}(τ_{k}; δ) := τ_{k}^{n} − \frac{1}{δ^{4}} E_{n}(δ)τ_{k}^{4} + [ − \frac{1}{δ^{3}} D_{n}(δ) + \frac{4}{δ^{4}} E_{n}(δ)]τ_{k}^{3}
+ [\frac{1}{δ^{4}} (−6 + 4δ^{2})E_{n}(δ) + \frac{3}{δ^{3}} D_{n}(δ) − \frac{1}{δ^{2}} C_{n}(δ)]τ_{k}^{2}
+ [\frac{1}{δ^{4}} (4 − 8δ^{2})E_{n}(δ) + \frac{1}{δ^{3}} (−3 + 3δ^{2})D_{n}(δ) + \frac{2}{δ^{2}} C_{n}(δ) − \frac{1}{δ}B_{n}(δ)]τ_{k}
+ [\frac{1}{δ^{4}} (−1 + 4δ^{2} − 2δ^{4})E_{n}(δ) + \frac{1}{δ^{3}} (1 − 3δ^{2})D_{n}(δ) + \frac{1}{δ^{2}} (−1 + 2δ^{2})C_{n}(δ)
+ \frac{1}{δ}B_{n}(δ) − A_{n}(δ)] = 0,

which means that:

p_{11}(x; δ)|W_{n,11}(x; δ), n ⩾ 5.

For example for δ = 1, we have the relation:

(x^{5} − 4x^{4} + 2x^{3} + 5x^{2} − 2x − 1)|[x^{n} − E_{n}x^{4} + (4E_{n} − D_{n})x^{3}+

+ (−2E_{n} + 3D_{n} − C_{n})x^{2} + (−4E_{n} + 2C_{n} − B_{n})x + E_{n} − 2D_{n} + C_{n} − A_{n}]

for all n ⩾ 5.

More precisely, the following decomposition can be deduced:

Lemma 22. We have:

p_{11}(x; δ)( ∑_{k=1}^{n−4} E_{k}(δ)x^{n−4−k}) = W_{n,11}(x; δ).  (8.2)

For δ = 1 we obtain special decomposition:

(x^{5} − 4x^{4} + 2x^{3} + 5x^{2} − 2x − 1)( ∑_{k=3}^{n−2} D_{k}x^{n−2−k}) =

= x^{n} − E_{n}x^{4} + (4E_{n} − D_{n})x^{3} + (−2E_{n} + 3D_{n} − C_{n})x^{2}+

+ (−4E_{n} + 2C_{n} − B_{n})x + E_{n} − 2D_{n} + C_{n} + B_{n} − A_{n}, (8.3)

and, after differentiating of (8.3):

(5x^{4} − 16x^{3} + 6x^{2} + 10x − 2)( ∑_{k=3}^{n−2} D_{k}x^{n−2−k})+

+ (x^{5} − 4x^{4} + 2x^{3} + 5x^{2} − 2x − 1)( ∑_{k=3}^{n−2} D_{k}x^{n−2−k})′ =

= n x^{n−1} − 4E_{n}x^{3} + 3(4E_{n} − D_{n})x^{2} + 2(−C_{n} + 3D_{n} − 2E_{n})x − B_{n} + 2C_{n} − 4E_{n}. (8.4) Hence, by applying (8.3), we get

(x^{5} − 4x^{4} + 2x^{3} + 5x^{2} − 2x − 1)^{2}( ∑_{k=3}^{n−3}(n − 2 − k)D_{k}x^{n−3−k}) =

= (x^{5} − 4x^{4} + 2x^{3} + 5x^{2} − 2x − 1)(n x^{n−1} − 4E_{n}x^{3} + 3(4E_{n} − D_{n})x^{2}+

+ 2(−C_{n} + 3D_{n} − 2E_{n})x − B_{n} + 2C_{n} − 4E_{n})+

+ (−5x^{4} + 16x^{3} − 6x^{2} − 10x + 2)(x^{n} − E_{n}x^{4} + (4E_{n} − D_{n})x^{3} + (−2E_{n} + 3D_{n} − C_{n})x^{2}+

+ (−4E_{n} + 2C_{n} − B_{n})x + E_{n} − 2D_{n} + C_{n} + B_{n} − A_{n}). (8.5)

Corollary 23. We have:

∑_{k=3}^{n−2} D_{k} = −A_{n} + 2C_{n} − 2E_{n} + 1,  (8.6)
(−1)^{n+1} ∑_{k=3}^{n−2}(−1)^{k}D_{k} = (−1)^{n} − A_{n} + 2B_{n} − 2C_{n} + 2D_{n} − 2E_{n},  (8.7)
∑_{k=3}^{n−3}(n − 2 − k)D_{k} = n − 3 + 3A_{n} − B_{n} − 6C_{n} + 3D_{n} + 6E_{n},  (8.8)
(−1)^{n+1} ∑_{k=3}^{n−3}(−1)^{k}(n − 2 − k)D_{k} = (−1)^{n}(n − 15) + 15A_{n} − 29B_{n}+  (8.9)
+ 26C_{n} − 21D_{n} + 14E_{n},

and

∑_{k=3}^{n−2} kD_{k} = 1 − (n + 1)A_{n} + B_{n} + 2(n + 1)C_{n} − 3D_{n} − 2(n + 1)E_{n}.  (8.10)

9 Some properties of zeros of polynomials D_{n}(x)

Let us set:

D_{n}(x) = ∑_{k=1}^{n} D_{k}x^{n−k}, n ∈ ℕ.

Then,

D_{1}(x) ≡ 0,  D_{2}(x) ≡ 0,  D_{3}(x) ≡ 1,
D_{4}(x) = x + 4,  D_{5}(x) = x^{2} + 4x + 14,

and we obtain the recurrence relations (see (3.21)):

D_{n+5}(x) = x^{n+2} + 4D_{n+4}(x) − 2D_{n+3}(x) − 5D_{n+2}(x)+
+ 2D_{n+1}(x) + D_{n}(x), n ⩾ 1.

All polynomials D_{n}(x) for n = 2k + 1, k ∈ ℕ, k ⩾ 2, have an even degree, for example:

D_{7}(x) = x^{4} + 4x^{3} + 14x^{2} + 43x + 126,
D_{9}(x) = x^{6} + 4x^{5} + 14x^{4} + 43x^{3} + 126x^{2} + 357x + 993.

As follows from calculation, these polynomials have no real roots for n ⩽ 200.

However, all polynomials D_{n}(x) for n = 2k, k ∈ ℕ, k ⩾ 2, have an odd degree, for example:

D_{6}(x) = x^{3} + 4x^{2} + 14x + 43,
D_{8}(x) = x^{5} + 4x^{4} + 14x^{3} + 43x^{2} + 126x + 357.

Remark 24. Polynomials D_{2k}(x) have exactly one real root s_{k}, which is less than zero (for k ⩽ 100). The roots s_{k} form an increasing sequence from s_{2} = −4 to s_{100} = −2.699746 (by numerical calculation). All numerical calculations were performed using Mathematica.^{1}

10 Jordan decomposition

For sequences A_{n}(δ), B_{n}(δ), C_{n}(δ), D_{n}(δ) and E_{n}(δ) we have the identity:

[ \begin{matrix}A_{n+1}(δ) \\ B_{n+1}(δ) \\ C_{n+1}(δ) \\ D_{n+1}(δ) \\ E_{n+1}(δ) \end{matrix} ] = W(δ) [ \begin{matrix}A_{n}(δ) \\ B_{n}(δ) \\ C_{n}(δ) \\ D_{n}(δ) \\ E_{n}(δ) \end{matrix} ] ,  n ∈ ℕ,  (10.1)

1Mathematica is registered trademark of Wolfram Research Inc.

where

W(δ) = [ \begin{matrix}1 & 2 δ & 0 & 0 & −δ \\ δ & 1 & δ & 0 & −δ \\ 0 & δ & 1 & δ & −δ \\ 0 & 0 & δ & 1 & 0 \\ 0 & 0 & 0 & δ & 1 − δ \end{matrix} ] .  (10.2)

Matrix W(δ) is diagonalizable, because of the following decomposition:

W(δ) = A · diag [τ_{1}(δ), τ_{2}(δ), …, τ_{5}(δ)] · A^{−1},  (10.3)

where:

A = [K_{1}, K_{2}, K_{3}, K_{4}, K_{5}]

and

K_{m} = [σ_{m}^{−4} + 2 σ_{m}^{−3} − 2 σ_{m}^{−2} − 2 σ_{m}^{−1}, σ_{m}^{−2} − σ_{m}^{−1} − 1, −σ_{m}^{−1} − σ_{m}^{−2}, −σ_{m}^{−3} − σ_{m}^{−2}, −σ_{m}^{−3}]^{T} ,

for every m = 1, 2, …, 5. We note, that:

A^{−1} = \frac{1}{11} [L_{1}, L_{2}, L_{3}, L_{4}, L_{5}] · [ \begin{matrix}11 & −3 & −20 & 8 & 16 \\ −2 & 2 & 7 & −7 & −11 \\ −13 & 7 & 22 & −15 & −17 \\ 1 & −1 & −3 & 3 & 4 \\ 3 & −2 & −5 & 4 & 4 \end{matrix} ] ,

where

L_{m} = [σ_{1}^{m+3}, σ_{2}^{m+3}, σ_{3}^{m+3}, σ_{4}^{m+3}, σ_{5}^{m+3}]^{T} ,

for m = 1, 2, …, 5.

Notice, that the characteristic polynomial w(λ) of the matrix W(δ) has the form:

w(λ) = p_{11}(λ; δ).

11 Final remarks

Quasi-Fibonacci numbers of order (7, δ) and (11, δ) constitute a special case of the so-called quasi-Fibonacci numbers of order (k, δ), where k is an odd positive integer, k ≥ 5 and δ ∈ ℂ. These numbers are defined to be the elements of the following sequences of polynomials:

\{a_{n,i}(δ)\}_{n∈ℕ} ⊂ ℤ[δ], i = 1, 2, …, \frac{1}{2} φ(k),

where φ is the Euler’s totient function, which are determined by the following relations:

(1 + δ (ξ^{l} + ξ^{k−l}))^{n} = a_{n,1}(δ) + ∑_{{i∈\{2,3,...,(k−1)/2\} \atop (i,k)=1}} a_{n,f(i)}(δ) (ξ^{i l} + ξ^{i (k−l)})  (11.1)

for l = 1, 2, …, \frac{k−1}{2}, (l, k) = 1, n ∈ ℕ, where ξ ∈ ℂ is a primitive root of unity of order k and where f is the increasing bijection of the set {i ∈ ℕ : 2 ≤ i ≤ (k − 1)/2 and (i, k) = 1} onto the set {2, 3, …, φ(k)/2}. In the sequel, if k is a prime number, relations (11.1) have the following simpler form:

(1 + δ (ξ^{l} + ξ^{k−l}))^{n} = a_{n,1}(δ) + ∑_{i=1}^{(k−3)/2} a_{n,i+1}(δ) (ξ^{i l} + ξ^{i (k−l)})  (11.2)

for l = 1, 2, …, \frac{k−1}{2} and n ∈ ℕ.

Quasi-Fibonacci numbers of order (k, δ) possess many interesting properties and natural applications. We remark that these are for the most cases derived immediately from relations (11.1). For example the quasi-Fibonacci numbers enable us to describe the coefficients of the following polynomials in the pure algebraic language:

∏_{{l∈\{1,2,...,k\} \atop (l,2k+1)=1}} (𝕏 − cos^{n} (\frac{2 l π}{2 k+1})),  ∏_{{l∈\{1,2,...,k\} \atop (l,2k+1)=1}} (𝕏 − sin (\frac{2 l π}{2 k+1}) cos^{n} (\frac{2 l π}{2 k+1})),

etc. (see for example [9, 10]). It is possible also to do it for many other trigonometric sums, for example, in [10] the following identity was derived (plus twelve others like this one):

2 cos(\frac{2 π}{7}) \sqrt[3]{2 cos(\frac{4 π}{7})} + 2 cos(\frac{4 π}{7}) \sqrt[3]{2 cos(\frac{8 π}{7})} + 2 cos(\frac{8 π}{7}) \sqrt[3]{2 cos(\frac{2 π}{7})} = \sqrt[3]{−2 − 3 \sqrt[3]{49}};

which is a variation of the known identity of Ramanujan (see [1]):

\sqrt[3]{cos (\frac{2 π}{7})} + \sqrt[3]{cos (\frac{4 π}{7})} + \sqrt[3]{cos (\frac{8 π}{7})} = \sqrt[3]{\frac{1}{2} (5 − 3 \sqrt[3]{7})} .

It seems almost definite that some types of the identities will be characteristic of only a certain kind of the quasi-Fibonacci numbers of order (k, δ), with respect to the odd k ∈ ℕ, k ≥ 5. Accordingly, it is relevant to analyze these numbers separately and independently, for different values of k.

We note that quasi-Fibonacci numbers of order (5, 1) are equal to the classical Fibonacci numbers. Moreover, for simplicity of notation, the quasi-Fibonacci numbers of order (11, δ) discussed in our paper are denoted by (see the relations (11.2) above for k = 11 and the relations (3.2m) for m = 1, 2, 3, 4, 5 in Section 3):

\begin{matrix}A_{n}(δ) := a_{n,1}(δ) & B_{n}(δ) := a_{n,2}(δ) & C_{n}(δ) := a_{n,3}(δ) \\ D_{n}(δ) := a_{n,4}(δ) & E_{n}(δ) := a_{n,5}(δ). \end{matrix}

12 Acknowledgements

The authors wish to thank the reviewer for his valuable criticisms and suggestions, leading to the present improved version of our paper.

Table 1: ; \begin{matrix}n & A_{n}(δ) = A_{11,n}(δ) & D_{n}(δ) = D_{11,n}(δ) \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ 2 & 2δ^{2} + 1 & 0 \\ 3 & 6δ^{2} + 1 & δ^{3} \\ 4 & 6δ^{4} + 12δ^{2} + 1 & 4δ^{3} \\ 5 & −δ^{5} + 30δ^{4} + 20δ^{2} + 1 & 4δ^{5} + 10δ^{3} \\ 6 & 19δ^{6} − 6δ^{5} + 90δ^{4} + 30δ^{2} + 1 & −δ^{6} + 24δ^{5} + 20δ^{3} \\ 7 & −7δ^{7} + 133δ^{6} − 21δ^{5} + 210δ^{4} + 42δ^{2} + 1 & 14δ^{7} − 7δ^{6} + 84δ^{5} + 35δ^{3} \\ n & B_{n}(δ) = B_{11,n}(δ) & E_{n}(δ) = E_{11,n}(δ) \\ 0 & 0 & 0 \\ 1 & δ & 0 \\ 2 & 2δ & 0 \\ 3 & 3δ^{3} + 3δ & 0 \\ 4 & 12δ^{3} + 4δ & δ^{4} \\ 5 & 9δ^{5} + 30δ^{3} + 5δ & −δ^{5} + 5δ^{4} \\ 6 & −δ^{6} + 54δ^{5} + 60δ^{3} + 6δ & 5δ^{6} − 6δ^{5} + 15δ^{4} \\ 7 & 28δ^{7} − 7δ^{6} + 189δ^{5} + 105δ^{3} + 7δ & −6δ^{7} + 35δ^{6} − 21δ^{5} + 35δ^{4} \\ n & C_{n}(δ) = C_{11,n}(δ) \\ 0 & 0 \\ 1 & 0 \\ 2 & δ^{2} \\ 3 & 3δ^{2} \\ 4 & 4δ^{4} + 6δ^{2} \\ 5 & −δ^{5} + 20δ^{4} + 10δ^{2} \\ 6 & 14δ^{6} − 6δ^{5} + 60δ^{4} + 15δ^{2} \\ 7 & −7δ^{7} + 98δ^{6} − 21δ^{5} + 140δ^{4} + 21δ^{2} \end{matrix}

26

Table 2: ; \begin{matrix}n & A_{n} & B_{n} & C_{n} & D_{n} & E_{n} \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 \\ 2 & 3 & 2 & 1 & 0 & 0 \\ 3 & 7 & 6 & 3 & 1 & 0 \\ 4 & 19 & 16 & 10 & 4 & 1 \\ 5 & 50 & 44 & 29 & 14 & 4 \\ 6 & 134 & 119 & 83 & 43 & 14 \\ 7 & 358 & 322 & 231 & 126 & 43 \\ 8 & 959 & 868 & 636 & 357 & 126 \\ 9 & 2569 & 2337 & 1735 & 993 & 357 \\ 10 & 6886 & 6284 & 4708 & 2728 & 993 \\ 11 & 18461 & 16885 & 12727 & 7436 & 2728 \end{matrix}  Table 3: ; \begin{matrix}n & A_{n}(1) & A_{n}(1/2) & A_{n}(−1/2) \\ 0 & 5 & 5 & 5 \\ 1 & 4 & 9/2 & 11/2 \\ 2 & 12 & 25/4 & 33/4 \\ 3 & 25 & 39/4 & 55/4 \\ 4 & 64 & 257/16 & 385/16 \\ 5 & 159 & 437/16 & 693/16 \\ 6 & 411 & 1517/32 & 2541/32 \\ 7 & 1068 & 2671/32 & 4719/32 \\ 8 & 2808 & 38017/256 & 70785/256 \\ 9 & 7423 & 68169/256 & 133705/256 \\ 10 & 19717 & 245935/512 & 508079/512 \\ 11 & 52529 & 1782735/2048 & 3879865/2048 \end{matrix}

Table 4: ; \begin{matrix}n & A_{n}(1/2) & B_{n}(1/2) & C_{n}(1/2) & D_{n}(1/2) & E_{n}(1/2) \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1/2 & 0 & 0 & 0 \\ 2 & 3/2 & 1 & 1/4 & 0 & 0 \\ 3 & 5/2 & 15/8 & 3/4 & 1/8 & 0 \\ 4 & 35/8 & 7/2 & 7/4 & 1/2 & 1/16 \\ 5 & 251/32 & 209/32 & 119/32 & 11/8 & 9/32 \\ 6 & 911/64 & 779/64 & 241/32 & 207/64 & 53/64 \\ 7 & 3327/128 & 1449/64 & 1897/128 & 7 & 65/32 \\ 8 & 6095/128 & 1345/32 & 229/8 & 3689/256 & 289/64 \\ 9 & 5593/64 & 19941/256 & 27949/512 & 7353/256 & 4845/512 \\ 10 & 164407/1024 & 36903/256 & 105641/1024 & 57361/1024 & 19551/1024 \\ 11 & 604487/2048 & 545721/2048 & 12397/64 & 220363/2048 & 4807/128 \end{matrix}

Table 5: ; \begin{matrix}n & A_{n}(−1/2) & B_{n}(−1/2) & C_{n}(−1/2) & D_{n}(−1/2) & E_{n}(−1/2) \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & -1/2 & 0 & 0 & 0 \\ 2 & 3/2 & -1 & 1/4 & 0 & 0 \\ 3 & 5/2 & -15/8 & 3/4 & -1/8 & 0 \\ 4 & 35/8 & -7/2 & 7/4 & -1/2 & 1/16 \\ 5 & 253/32 & -209/32 & 121/32 & -11/8 & 11/32 \\ 6 & 935/64 & -781/64 & 253/32 & -209/64 & 77/64 \\ 7 & 3509/128 & -1463/64 & 2079/128 & -231/32 & 55/16 \\ 8 & 6655/128 & -1375/32 & 33 & -3927/256 & 561/64 \\ 9 & 3179/32 & -20757/256 & 34067/512 & -8151/256 & 10659/512 \\ 10 & 195415/1024 & -39325/256 & 136609/1024 & -66671/1024 & 48279/1024 \\ 11 & 753709/2048 & -598345/2048 & 136367/512 & -269951/2048 & 52877/512 \end{matrix}

References

[1] R. Grzymkowski and R. Wituła, Calculus Methods in Algebra, Part One, WPKJS, 2000 (in Polish).

[2] A. Mostowski and M. Stark, The Elements of Higher Algebra, PWN, 1974 (in Polish).

[3] S. Paszkowski, Numerical Applications of Chebyshef Polynomials and Series, PWN, 1975 (in Polish).

[4] S. Rabinowitz, Algorithmic manipulation of third-order linear recurrences, Fib. Quart. 34 (1996), 447–464.

[5] T. J. Rivlin, Chebyshef Polynomials from Approximation Theory to Algebra and Number Theory, 2nd ed., Wiley, 1990.

[6] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, 2007.

[7] D. Surowski and P. McCombs, Homogeneous polynomials and the minimal polynomial of cos(2π/n), Missouri J. Math. Sci. 15 (2003), 4–14.

[8] W. Watkins and J. Zeitlin, The minimal polynomials of cos(2π/n), Amer. Math. Monthly 100 (1993), 471–474.

[9] R. Wituła, D. Słota and A. Warzyński, Quasi-Fibonacci numbers of the seventh order, J. Integer Seq. 9 (2006), Article 06.4.3.

[10] R. Wituła and D. Słota, New Ramanujan-type formulas and quasi-Fibonacci numbers of order 7, J. Integer Seq. 10 (2007), Article 07.5.6.

2000 Mathematics Subject Classification: Primary 11B83, 11A07; Secondary 12D10, 39A10. Keywords: Fibonacci numbers, primitive roots of unity, recurrence relation.

(Concerned with sequence \underline{A062883}.)

Received December 9 2006. Revised version received August 3 2007. Published in Journal of Integer Sequences, August 5 2007.

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