Ramanujan Type Trigonometric Formulas:

The General Form for the Argument \frac{2 π}{7}

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

                     roman.witula@polsl.pl

Abstract In this paper, we present many general identities connected with the classical Ramanujan equality. Moreover, we give Binet formulas for an accelerator sequence for Catalan’s constant.

1 Introduction

The main objective of this paper is to obtain some general trigonometric formulas related to the known Ramanujan equality [1, 2, 3, 5, 7]:

\sqrt[3]{cos α} + \sqrt[3]{cos 2α} + \sqrt[3]{cos 4α} = \sqrt[3]{\frac{1}{2}(5 − 3 \sqrt[3]{7})},  (1)

where α = 2π/7. Other formulas of this type, referring to the ninth and eleventh primitive roots of unity, etc., will be published in separate papers. The present paper, in a way, is a continuation of previous papers [11, 12, 13] and I will take advantage of some results from those papers. The quasi-Fibonacci sequences of the seventh order discussed in the above mentioned papers are applied here for describing some attractive formulas involving radicals.

The paper is divided into five parts:

– Section 2 – where some striking equalities related to equality (1) are presented. Furthermore, Binet formulas for two new sequences {S_{n}} and {S_{n}^{∗}} are derived, which, at the same time, resolves the problem of an algebraic description of the zeros of polynomials x^{3} − \sqrt[3]{7}x − 1 and x^{3} − \sqrt[3]{49}x − 1 (see Remark 1).

– Section 3 – where the fundamental formula (10) for a sum of the cube roots of the three roots of a cubic polynomial is given.

– Section 4 – where many basic sequences of integers, reals and complex numbers, introduced and discussed earlier by the authors in papers [11, 12, 13], are presented. In addition, some new relations between the elements of the sequences are discussed. Furthermore, a new description of Binet’s formula is introduced for an accelerator sequence for Catalan’s constant, which, naturally, makes it possible to extend this formula for all integers (see Remark 7).

– Section 5 – where applications of the formula (10) to many special kinds of polynomial of degree three are given. This section contains many Ramanujan type trigonometric formulas. Moreover in Remark 10 the nontrivial theoretical discussion of some numerical case is presented.

2 Delicious

Now we are going to prove the following three interesting identities:

\sqrt[3]{\frac{cos α}{cos 2α}} (2 cos α)^{k} + \sqrt[3]{\frac{cos 2α}{cos 4α}} (2 cos 2α)^{k} + \sqrt[3]{\frac{cos 4α}{cos α}} (2 cos 4α)^{k} =

= \sqrt[3]{\frac{cos α}{cos 2α}} (2 cos 2α)^{k+1} + \sqrt[3]{\frac{cos 2α}{cos 4α}} (2 cos 4α)^{k+1} +

+ \sqrt[3]{\frac{cos 4α}{cos α}} (2 cos α)^{k+1} = \sqrt[3]{7} ψ_{k}, (2)

where ψ_{0} = −1, ψ_{1} = 0, ψ_{2} = −3 and

ψ_{k+3} + ψ_{k+2} − 2 ψ_{k+1} − ψ_{k} = 0, k ∈ ℤ;

\sqrt[3]{\frac{cos α}{cos 4α}} (2 cos α)^{k} + \sqrt[3]{\frac{cos 2α}{cos α}} (2 cos 2α)^{k} + \sqrt[3]{\frac{cos 4α}{cos 2α}} (2 cos 4α)^{k} =

= \sqrt[3]{\frac{cos 2α}{cos α}} (2 cos α)^{k+1} + \sqrt[3]{\frac{cos 4α}{cos 2α}} (2 cos 2α)^{k+1} +

+ \sqrt[3]{\frac{cos α}{cos 4α}} (2 cos 4α)^{k+1} = \sqrt[3]{49} φ_{k}, (3)

where φ_{0} = 0, φ_{1} = −1, φ_{2} = 1 and

φ_{k+3} + φ_{k+2} − 2 φ_{k+1} − φ_{k} = 0, k ∈ ℤ;

\sqrt[3]{sec 2α} (2 cos α)^{k} + \sqrt[3]{sec 4α} (2 cos 2α)^{k} + \sqrt[3]{sec α} (2 cos 4α)^{k} =

= δ_{k} \sqrt[3]{8 − 6 \sqrt[3]{7}} + σ_{k} \sqrt[3]{6 (1 + \sqrt[3]{7})^{2}} + ξ_{k} \sqrt[3]{2 (5 − 3 \sqrt[3]{7})^{2}} =

= \sqrt[3]{f_{3k+1} + 6 − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{3k+1,8} + \sqrt{T_{3k+1,8}}} + \sqrt[3]{S_{3k+1,8} − \sqrt{T_{3k+1,8}}} )}, (4)

where

δ_{0} = 1,  δ_{1} = 0,  δ_{2} = 0,
σ_{0} = 0,  σ_{1} = −1,  σ_{2} = 0,
ξ_{0} = 0,  ξ_{1} = 0,  ξ_{2} = 1,
𝕏_{k+3} + 𝕏_{k+2} − 2 𝕏_{k+1} − 𝕏_{k} = 0, k = 0, 1, 2, …,

for every 𝕏 ∈ {δ, σ, ξ}, whereas the sequences f_{3k+1}, S_{3k+1,8} and T_{3k+1,8} are defined by formulas (21), (99) and (100) (in Section 4, other sequences occurring in the definition of sequences S_{3k+1,8} and T_{3k+1,8} are defined as well). The first twelve values of numbers ψ_{n} and φ_{n} are presented in Table 1.

Moreover an interesting numerical link to the formula (2) are the considerations from Remark 10.

Proof of formulas (2)–(4).

For k = 0, 1, 2, the formulas (2)–(4) follow from (34), (37), (40), (41) and (44), and (or) from the following equalities (in both cases, equality (5) below for 𝕏 = 0 is needed):

\sqrt[3]{\frac{cos α}{cos 2α}} (2 cos α)^{2} + \sqrt[3]{\frac{cos 2α}{cos 4α}} (2 cos 2α)^{2} + \sqrt[3]{\frac{cos 4α}{cos α}} (2 cos 4α)^{2} =

= \sqrt[3]{\frac{cos α}{cos 2α}} (2 + 2 cos 2α) + \sqrt[3]{\frac{cos 2α}{cos 4α}} (2 + 2 cos 4α) + \sqrt[3]{\frac{cos 4α}{cos α}} (2 + 2 cos α) =
= 2 \sqrt[3]{7} ψ_{0} + 2 (\sqrt[3]{cos α cos^{2} 2α} + \sqrt[3]{cos 2α cos^{2} 4α} + \sqrt[3]{cos 4α cos^{2} α} ) =
= −2 \sqrt[3]{7} + \sqrt[3]{\frac{cos 2α}{cos 4α}} + \sqrt[3]{\frac{cos 4α}{cos α}} + \sqrt[3]{\frac{cos α}{cos 2α}} = −2 \sqrt[3]{7} + \sqrt[3]{7} ψ_{0} = −3 \sqrt[3]{7},

and

\sqrt[3]{\frac{cos α}{cos 4α}} (2 cos α)^{2} + \sqrt[3]{\frac{cos 2α}{cos α}} (2 cos 2α)^{2} + \sqrt[3]{\frac{cos 4α}{cos 2α}} (2 cos 4α)^{2} =

= \sqrt[3]{\frac{cos α}{cos 4α}} (2 + 2 cos 2α) + \sqrt[3]{\frac{cos 2α}{cos α}} (2 + 2 cos 4α) + \sqrt[3]{\frac{cos 4α}{cos 2α}} (2 + 2 cos α) =

= 2 \sqrt[3]{49} φ_{0} + (\sqrt[3]{\frac{cos 2α}{cos 4α}})^{2} + (\sqrt[3]{\frac{cos 4α}{cos α}})^{2} + (\sqrt[3]{\frac{cos α}{cos 2α}})^{2} =

=(\sqrt[3]{\frac{cos 2α}{cos 4α}} + \sqrt[3]{\frac{cos 4α}{cos α}} + \sqrt[3]{\frac{cos α}{cos 2α}})^{2} − 2(\sqrt[3]{\frac{cos 2α}{cos α}} + \sqrt[3]{\frac{cos α}{cos 4α}} + \sqrt[3]{\frac{cos 4α}{cos 2α}})=

= (\sqrt[3]{7} ψ_{0})^{2} − 2 \sqrt[3]{49} φ_{0} = \sqrt[3]{49}.

Now let us set

𝔅_{n} := ∑_{k=0}^{2} x_{k} ( cos (2^{k}α))^{n},

where x_{k} ∈ ℝ, k = 1, 2, 3, are given. Then, from Newton’s formula we obtain

𝔅_{n+3} + 𝔅_{n+2} − 2 𝔅_{n+1} − 𝔅_{n} = 0

since [5, 11, 12]:

∏_{k=0}^{2} (𝕏 − 2 cos (2^{k}α)) = 𝕏^{3} + 𝕏^{2} − 2 𝕏 − 1.  (5)

Hence, on account of the definitions of sequences φ_{k} and ψ_{k}, k ∈ ℕ, by applied induction arguments the formulas (2) and (3) follow. Similarly, by applying (33), (36) and (39) we deduce the first part of (4). The second part of (4) from (98) follows.

Remark 1. Let a, b, c ∈ ℂ and a + b + c = 0. Put

s_{k} := a^{k} + b^{k} + c^{k}, k ∈ ℕ.

Then the following relations hold [5, 6]:

2 s_{4} = s_{2}^{2};  6 s_{5} = 5 s_{2} s_{3};
6 s_{7} = 7 s_{3} s_{4};  10 s_{7} = 7 s_{2} s_{5};
25 s_{3} s_{7} = 21s_{5}^{2};  50 s_{7}^{2} = 49 s_{4} s_{5}^{2}

and the respective Newton formula has the form

s_{n+3} = a b c s_{n} + \frac{1}{2} s_{2} s_{n+1}, n ∈ ℕ.

Hence, and from (3) for k = 0, we get

S_{n+3} = \sqrt[3]{7} S_{n+1} + S_{n},

where S_{0} = 3, S_{1} = 0, S_{2} = 2 \sqrt[3]{7},

S_{n} := (\frac{cos α}{cos 4α})^{n/3} + (\frac{cos 2α}{cos α})^{n/3} + (\frac{cos 4α}{cos 2α})^{n/3}

which implies the following formula

S_{n} = â_{n} + b̂_{n} \sqrt[3]{7} + ĉ_{n} \sqrt[3]{49},

where

â_{n+3} = â_{n} + 7 ĉ_{n},
b̂_{n+3} = b̂_{n} + â_{n+1},
ĉ_{n+3} = ĉ_{n} + b̂_{n+1},
â_{0} = 3, â_{1} = 0, â_{2} = 0,
b̂_{0} = 0, b̂_{1} = 0, b̂_{2} = 1,
ĉ_{0} = 0, ĉ_{1} = 0, ĉ_{2} = 0.

On the other hand, from (2) for k = 1, we obtain

S_{n+3}^{∗} = \frac{1}{2} S_{2}^{∗} S_{n+1}^{∗} + S_{n}^{∗}, n ∈ ℕ_{0},

where S_{0}^{∗} = 3, S_{1}^{∗} = 0,

S_{n}^{∗} := (2 cos α \sqrt[3]{\frac{cos α}{cos 2α}} )^{n} + (2 cos 2α \sqrt[3]{\frac{cos 2α}{cos 4α}} )^{n} + (2 cos 4α \sqrt[3]{\frac{cos 4α}{cos α}} )^{n}, n ∈ ℕ, and by (5) for 𝕏 = 0:

S_{2}^{∗} := (\frac{cos α}{cos 2α})^{2/3} (2 + 2 cos 2α) + (\frac{cos 2α}{cos 4α})^{2/3}(2 + 2 cos 4α) + (\frac{cos 4α}{cos α})^{2/3}(2 + 2 cos α) =

= 2 ((\frac{cos α}{cos 2α})^{1/3} + (\frac{cos 2α}{cos 4α})^{1/3} + (\frac{cos 4α}{cos α})^{1/3} )^{2} −

− 4 ((\frac{cos α}{cos 4α})^{1/3} + (\frac{cos 4α}{cos 2α})^{1/3} + (\frac{cos 2α}{cos α})^{1/3} ) +

+ (2 cos α)^{1/3} (4 cos α cos 2α)^{1/3} + (2 cos 2α)^{1/3} (4 cos 2α cos 4α)^{1/3} +

+ (2 cos 4α)^{1/3} (4 cos α cos 4α)^{1/3} = 2 (\sqrt[3]{7} ψ_{0})^{2} − 4 S_{1} + S_{1} = 2 \sqrt[3]{49}.

So, we have

S_{n+3}^{∗} = \sqrt[3]{49} S_{n+1}^{∗} + S_{n}^{∗}, n ∈ ℕ_{0},

which implies

S_{n}^{∗} = a_{n}^{∗} + b_{n}^{∗} \sqrt[3]{7} + c_{n}^{∗} \sqrt[3]{49},

where

a_{n+3}^{∗} = a_{n}^{∗} + 7 b_{n+1}^{∗},
b_{n+3}^{∗} = b_{n}^{∗} + 7 c_{n+1}^{∗},
c_{n+3}^{∗} = c_{n}^{∗} + a_{n+1}^{∗},

a_{0}^{∗} = 3, a_{1}^{∗} = 0, a_{2}^{∗} = 0,
b_{0}^{∗} = 0, b_{1}^{∗} = 0, b_{2}^{∗} = 0,
c_{0}^{∗} = 0, c_{1}^{∗} = 0, c_{2}^{∗} = 2.

We note that the elements S_{3n} = â_{n} and S_{3n}^{∗} = a_{n}^{∗}, n = 0, 1, …, are all integers.

3 Some theoretical deliberations

Let us assume that ξ_{1}, ξ_{2}, ξ_{3} are complex roots of the following polynomial with complex coefficients

f(z) := z^{3} + p z^{2} + q z + r.

The symbols \sqrt[3]{ξ_{1}}, \sqrt[3]{ξ_{2}}, \sqrt[3]{ξ_{3}} will denote any of the third complex roots of the numbers ξ_{1}, ξ_{2} and ξ_{3}, respectively (only in the case where ξ_{1}, ξ_{2} and ξ_{3} are real numbers, we will assume that \sqrt[3]{ξ_{1}}, \sqrt[3]{ξ_{2}} and \sqrt[3]{ξ_{3}} also denote the respective real roots).

Let us assume that

A := (\sqrt[3]{ξ_{1}} + \sqrt[3]{ξ_{2}} + \sqrt[3]{ξ_{3}})^{3}

and

B := (\sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{2}} + \sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{3}} + \sqrt[3]{ξ_{2}} \sqrt[3]{ξ_{3}})^{3} .

Thus, the numbers

\sqrt[3]{ξ_{1}} + \sqrt[3]{ξ_{2}} + \sqrt[3]{ξ_{3}}  and  \sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{2}} + \sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{3}} + \sqrt[3]{ξ_{2}} \sqrt[3]{ξ_{3}}

belong to the sets of the third complex roots of A and B, respectively, which, for conciseness of notation, will be denoted by the symbols \sqrt[3]{A} and \sqrt[3]{B}, respectively. In other words, we have

\sqrt[3]{ξ_{1}} + \sqrt[3]{ξ_{2}} + \sqrt[3]{ξ_{3}} ∈ \sqrt[3]{A}

and

\sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{2}} + \sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{3}} + \sqrt[3]{ξ_{2}} \sqrt[3]{ξ_{3}} ∈ \sqrt[3]{B}.

Hence, after two-sided raising of the numbers to the third power, we obtain the following formulas:

A = ξ_{1} + ξ_{2} + ξ_{3} + 3 ∑_{k≠l} (\sqrt[3]{ξ_{k}})^{2} \sqrt[3]{ξ_{l}} + 6 \sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{2}} \sqrt[3]{ξ_{3}},

and

B = ξ_{1} ξ_{2} + ξ_{1} ξ_{3} + ξ_{2} ξ_{3} + 3∑_{k≠l} (\sqrt[3]{ξ_{k}})^{2}\sqrt[3]{ξ_{l}} \sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{2}} \sqrt[3]{ξ_{3}} + 6(\sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{2}} \sqrt[3]{ξ_{3}})^{2} ,

where

\sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{2}} \sqrt[3]{ξ_{3}} ∈ \sqrt[3]{ξ_{1} ξ_{2} ξ_{3}} = \sqrt[3]{−r} = −\sqrt[3]{r}.

Also here, for abbreviation, the product (−1) \sqrt[3]{ξ_{1}} \sqrt[3]{ξ_{2}} \sqrt[3]{ξ_{3}} will be denoted by the symbol \sqrt[3]{r}.

Taking into account Viète’s formulas for polynomial f(z), the expressions for A and B can be attributed the following form (from now on, symbols \sqrt[3]{A} and \sqrt[3]{B} will mean the properly selected elements from sets \sqrt[3]{A} and \sqrt[3]{B}, respectively):

A = −p + 3 \sqrt[3]{A} \sqrt[3]{B} + 3\sqrt[3]{r}  (6)

and

B = q − 3 \sqrt[3]{A} \sqrt[3]{B} \sqrt[3]{r} − 3 (\sqrt[3]{r})^{2} .  (7)

By multiplying the first of these equations by \sqrt[3]{r} and adding the equations side-by-side, we obtain

B = q − (A + p) \sqrt[3]{r}.  (8)

At the same time, the equation (6) yields

3 \sqrt[3]{A} \sqrt[3]{B} = A + p − 3 \sqrt[3]{r},

i.e.,

27 A B = (A + p − 3 \sqrt[3]{r})^{3} ,

hence, with respect to (8) we obtain

27 A (q − (A + p) \sqrt[3]{r}) = A^{3} + p^{3} − 27 r + + 3 (A^{2} (p − 3\sqrt[3]{r}) + A (p^{2} + 9 (\sqrt[3]{r})^{2}) − 3 p^{2} \sqrt[3]{r} + 9 p (\sqrt[3]{r})^{2}) − 18 A p \sqrt[3]{r},

and having rearrange the summands (with respect to the powers of A), we obtain the basic equality

A^{3} + 3 (p + 6 \sqrt[3]{r})A^{2} + 3 (p^{2} + 3 p \sqrt[3]{r} + 9 (\sqrt[3]{r})^{2} − 9 q)A + (p − 3 \sqrt[3]{r})^{3} = 0.  (9)

By applying Cardano’s formula to this polynomial, we get the following basic formula (the right side of the formula below means a properly selected third root of the number present in the formula):

\sqrt[3]{A} = \sqrt[3]{ξ_{1}} + \sqrt[3]{ξ_{2}} + \sqrt[3]{ξ_{3}} =

= \sqrt[3]{−p − 6 \sqrt[3]{r} − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S + \sqrt{T}} + \sqrt[3]{S − \sqrt{T}} )}, (10)

where

S := p q + 6 q \sqrt[3]{r} + 6 p \sqrt[3]{r^{2}} + 9 r,
T := p^{2} q^{2} − 4 q^{3} − 4 p^{3} r + 18 p q r − 27 r^{2}.

In the case when T ≥ 0, S ∈ ℝ, r ∈ ℝ, we assume that all the roots appearing here are real.

Remark 2. We note that, if in the formula (9) the following condition holds

(p + 6 \sqrt[3]{r})^{2} = p^{2} + 3 p \sqrt[3]{r} + 9 (\sqrt[3]{r})^{2} − 9 q,

i.e.,

p \sqrt[3]{r} + 3 (\sqrt[3]{r})^{2} + q = 0,  (11)

then the equation (9) could be given in the form

(A + p + 6 \sqrt[3]{r})^{3} = (p + 6 \sqrt[3]{r})^{3} − (p − 3 \sqrt[3]{r})^{3},

hence we get

A = −p − 6 \sqrt[3]{r} + \sqrt[3]{(p + 6 \sqrt[3]{r})^{3} − (p − 3 \sqrt[3]{r})^{3}}.  (12)

Remark 3. The analysis which enabled describing the value of A by means of coefficients of polynomial f(z) comes from the papers [7, 5] (see also [4]).

4 Basic sequences

We will now provide definitions of a dozen basic sequences (not only integer sequences) used further in the paper. For more information concerning these sequences (including the trigonometric relationships defining their terms), see the papers [11, 12].

The sequences {A_{n}(δ)}_{n=0}^{∞}, {B_{n}(δ)}_{n=0}^{∞} and {C_{n}(δ)}_{n=0}^{∞} are the so-called quasi-Fibonacci numbers of the seventh order described in [11] by means of relations

(1 + δ (ξ^{k} + ξ^{6k}))^{n} = A_{n}(δ) + B_{n}(δ) (ξ^{k} + ξ^{6k}) + C_{n}(δ) (ξ^{2k} + ξ^{5k})  (13)

for k = 1, 2, 3, where ξ ∈ ℂ is a primitive root of unity of the seventh order (ξ^{7} = 1 and ξ ≠ 1), δ ∈ ℂ, δ ≠ 0. These sequences satisfy the following recurrence relations

\{ \begin{matrix}A_{0}(δ) = 1, B_{0}(δ) = C_{0}(δ) = 0, \\ A_{n+1}(δ) = A_{n}(δ) + 2 δ B_{n}(δ) − δ C_{n}(δ), \\ B_{n+1}(δ) = δ A_{n}(δ) + B_{n}(δ), \\ C_{n+1}(δ) = δ B_{n}(δ) + (1 − δ) C_{n}(δ), \end{matrix}  (14)

for every n ∈ ℕ.

Two auxiliary sequences {A_{n}(δ)}_{n=0}^{∞} and {B_{n}(δ)}_{n=0}^{∞} connected with these ones are defined by the following relations:

A_{n}(δ) := 3 A_{n}(δ) − B_{n}(δ) − C_{n}(δ)  (15)

and

B_{n}(δ) := \frac{1}{2}((A_{n}(δ))^{2} − A_{2n}(δ)).  (16)

Furthermore, to simplify notation, we will write

A_{n} = A_{n}(1), B_{n} = B_{n}(1), A_{n} = A_{n}(1), B_{n} = B_{n}(1) and C_{n} = C_{n}(1),  (17)

for every n ∈ ℕ.

We note that the elements of the sequences {A_{n}}_{n=0}^{∞}, {A_{n}}_{n=0}^{∞}, {B_{n}}_{n=0}^{∞} and {C_{n}}_{n=0}^{∞} respectively, satisfy the following recurrence relation (see [12, eq. (3.20)]):

𝕏_{n+3} − 2 𝕏_{n+2} − 𝕏_{n+1} + 𝕏_{n} ≡ 0.

Simultaneously, the elements of sequence {B_{n}}_{n=0}^{∞} by [12, eqs. (3.18), (3.13)] satisfy the following relation

B_{n+3} + B_{n+2} − 2 B_{n+1} − B_{n} ≡ 0.  (18)

Remark 4. The sequence {B_{n}}_{n=0}^{∞} is an accelerator sequence for Catalan’s constant (see [10, sequence \underline{A094648}] and papers [11, 12]).

The elements of the sequences {a_{n}}_{n=0}^{∞}, {b_{n}}_{n=0}^{∞} and {c_{n}}_{n=0}^{∞} are defined by the following recurrence relations:

a_{0} = b_{0} = c_{0} = \sqrt{7}

and

\{ \begin{matrix}a_{n+1} = 2 a_{n} + b_{n}, \\ b_{n+1} = a_{n} + 2 b_{n} − c_{n}, \\ c_{n+1} = c_{n} − b_{n}, \end{matrix}  (19)

for n = 0, 1, 2, . . ..

Moreover, we will use the following sequences

\{ \begin{matrix}\overline{α}_{n} := c_{n+1}, \\ \overline{β}_{n} := −a_{n} − b_{n}, \\ \overline{γ}_{n} := a_{n}, \end{matrix}  (20)

Next, sequences {f_{n}}_{n=0}^{∞}, {g_{n}}_{n=0}^{∞}, {h_{n}}_{n=0}^{∞} and {H_{n}}_{n=0}^{∞} are defined in the following way

f_{0} = g_{0} = h_{0} = −1,

and

\{ \begin{matrix}f_{n+1} = f_{n} + g_{n}, & n ≥ 0, \\ g_{n+1} = f_{n} + h_{n}, & n ≥ 0, \\ h_{n} = B_{n+1}, & n ≥ 1, \\ 2 H_{n} = f_{n}^{2} + f_{2n} − g_{n}^{2} + h_{n}^{2} − h_{2n+1} − 2 h_{2n}, & n ≥ 0, \end{matrix}  (21)

(the numbers B_{n} are defined by the formula (17) above).

And at last, the elements of sequences {u_{n}}_{n=0}^{∞}, {v_{n}}_{n=0}^{∞}, {w_{n}}_{n=0}^{∞}, {x_{n}}_{n=0}^{∞}, {y_{n}}_{n=0}^{∞} and {z_{n}}_{n=0}^{∞} are defined by

\{ \begin{matrix}u_{n+1} = x_{n}, \\ v_{n+1} = −y_{n} − z_{n} = x_{n} − \sqrt{7} z_{n−1}, \\ w_{n+1} = y_{n} − x_{n}, \\ x_{n+1} = u_{n} − w_{n}, \\ y_{n+1} = w_{n} − v_{n}, \\ z_{n+1} = 2 z_{n−1} − v_{n}, \end{matrix}  (22)

for n = 0, 1, 2, …, where u_{0} = v_{0} = w_{0} = −1, x_{0} = y_{0} = z_{0} = \sqrt{7} and z_{1} = 7.

Remark 5. All the recurrence sequences above have a third order; the selective identities, Binet formulas and generating functions, and some different identities for these numbers, are presented in papers [11, 12]. For example, the following equivalent recurrence relations hold

(19) ⟺ 𝕏_{n+3} − 5 𝕏_{n+2} + 6 𝕏_{n+1} − 𝕏_{n} = 0,

for n = 0, 1, 2, …, and 𝕏 ∈ \{a, b, c\}, and a_{0} = b_{0} = c_{0} = \sqrt{7}, a_{1} = 3 \sqrt{7}, b_{1} = 2 \sqrt{7}, c_{1} = 0, a_{2} = 8 \sqrt{7}, b_{2} = \sqrt{7^{3}}, c_{1} = −2 \sqrt{7}. We also have

a_{n} = 2^{2n+1} [ sin α (cos 4α)^{2n} + sin 2α (cos α)^{2n} + sin 4α (cos 2α)^{2n}],
b_{n} = 2^{2n+1} [ sin 2α (cos 4α)^{2n} + sin 4α (cos α)^{2n} + sin α (cos 2α)^{2n}],
c_{n} = 2^{2n+1} [ sin 4α (cos 4α)^{2n} + sin α (cos α)^{2n} + sin 2α (cos 2α)^{2n}],

etc.

Remark 6. Now we present some new identities for the above sequences (identities (23)– (26), (28)–(31)), which will be used in subsection 5.2. These identities significantly complete those obtained in paper [12]. By [12, Lemma 3.14 (a)], equality (5) and [12, eq. (3.21)], the following identity holds

(−1)^{n} A_{n} = H_{n+1}.  (23)

Hence, by [12, eq. (3.23)], we obtain

2 cos α (2 cos 2α)^{−n} + 2 cos 2α (2 cos 4α)^{−n} + 2 cos 4α (2 cos α)^{−n} = = (−1)^{n} (A_{n} + A_{n−1} − A_{n−2}) = (−1)^{n} (A_{n+1} − A_{n}), (24)

and next, by [12, eq. (3.22)], we obtain

2 cos α (2 cos 4α)^{−n} + 2 cos 2α (2 cos α)^{−n} + 2 cos 4α (2 cos 2α)^{−n} = = (−1)^{n} (A_{n+1} − A_{n−1} + A_{n−2} − 7 A_{n}) = (−1)^{n} (A_{n−1} − A_{n+1}), (25)

since, by [12, Remark 3.11], we have the identity

A_{n+3} = 2 A_{n+2} + A_{n+1} − A_{n}

and by [12, Remark 3.8], we have

4 A_{n} − A_{n−2} = 7 A_{n}.

Moreover, the following formula can be easily generated

B_{n} = g_{n+1} + h_{n+1}.  (26)

By [12, eq. (3.18)], we have

h_{n−1} = B_{n},  (27)

which by (26) implies

g_{n+1} = B_{n} − B_{n+2},  (28)

and next, by [12, eq. (3.12)] and by (18), we get

f_{n} = g_{n+1} − h_{n} = B_{n} − B_{n+2} − B_{n+1} = −B_{n−1} − B_{n}.  (29)

Moreover, we obtain

f_{n} + g_{n} = −B_{n} − B_{n+1}  (30)

and

f_{n} + h_{n} = B_{n} − B_{n+2}.  (31)

Remark 7. From (27) we get (see [12, eq. (3.11)]):

B_{n} = 2^{n} ( cos^{n} α + cos^{n} 2α + cos^{n} 4α),

from which the next form can be deduced

B_{n} = (\frac{sin 2α}{sin α})^{n} + (\frac{sin 4α}{sin 2α})^{n} + (\frac{sin α}{sin 4α})^{n}.  (32)

This forms of Binet’s formula for B_{n} are more attractive than the Sloane’s ones (see [10, sequence \underline{A094648}]). Moreover, the formula (32) makes it possible to extend the definition B_{n} for negative integers n. So {B_{n}}_{n=−∞}^{∞} is a two-sided sequence of integers which is defined for all integers either by recurrence formula (18), or equivalently by Binet’s formula (32).

5 Applications of the formula (10)

5.1 Some special formulas

By [12, eq. (3.30)]

for n = 1:

\sqrt[3]{sec α} + \sqrt[3]{sec 2α} + \sqrt[3]{sec 4α} = \sqrt[3]{8 − 6 \sqrt[3]{7}} ;  (33)

for n = 2:

\sqrt[3]{\frac{cos α}{cos 2α}} + \sqrt[3]{\frac{cos 2α}{cos 4α}} + \sqrt[3]{\frac{cos 4α}{cos α}} = −\sqrt[3]{7} ;  (34)

for n = 3:

cos 2α \sqrt[3]{2 cos α} + cos 4α \sqrt[3]{2 cos 2α} + cos α \sqrt[3]{2 cos 4α} = \frac{1}{2} \sqrt[3]{5 + 3 \sqrt[3]{7} − 3 \sqrt[3]{49}} ; (35)

for n = 4:

cos 2α \sqrt[3]{sec 4α} + cos 4α \sqrt[3]{sec α} + cos α \sqrt[3]{sec 2α} = −\sqrt[3]{\frac{3}{4} (1 + \sqrt[3]{7})^{2}} ;  (36)

for n = 5:

cos 2α \sqrt[3]{\frac{cos 2α}{cos 4α}} + cos 4α \sqrt[3]{\frac{cos 4α}{cos α}} + cos α \sqrt[3]{\frac{cos α}{cos 2α}} = 0 ;  (37)

for n = 6:

cos^{2}(2α) \sqrt[3]{2 cos α} + cos^{2}(4α) \sqrt[3]{2 cos 2α} + cos^{2}(α) \sqrt[3]{2 cos 4α} =

= \frac{1}{4} \sqrt[3]{12 \sqrt[3]{7} − (4 + 3 \sqrt[3]{7} )^{2}} ; (38)

for n = 7:

cos^{2}(2α) \sqrt[3]{sec 4α} + cos^{2}(4α) \sqrt[3]{sec α} + cos^{2}(α) \sqrt[3]{sec 2α} = \frac{1}{4} \sqrt[3]{2 (5 − 3 \sqrt[3]{7} )^{2}} ;  (39)

for n = 8:

cos^{3}(2α) \sqrt[3]{\frac{cos α}{cos 2α}} + cos^{3}(4α) \sqrt[3]{\frac{cos 2α}{cos 4α}} + cos^{3}(α) \sqrt[3]{\frac{cos 4α}{cos α}} =

= \frac{1}{2} ( cos^{2}(2α) \sqrt[3]{\frac{cos 2α}{cos 4α}} + cos^{2}(4α) \sqrt[3]{\frac{cos 4α}{cos α}} + cos^{2}(α) \sqrt[3]{\frac{cos α}{cos 2α}} ) = −\frac{3}{8} \sqrt[3]{7} , (40) which also can be generated from (34). Remark 8. The formula (39) also follows from (1) and (5) for 𝕏 = 0:

2 cos^{2}(2α) \sqrt[3]{sec 4α} + 2 cos^{2}(4α) \sqrt[3]{sec α} + 2 cos^{2}(α) \sqrt[3]{sec 2α} =

= ∑_{k=0}^{2} \sqrt[3]{sec(2^{k}α)} + ∑_{k=0}^{2} (\sqrt[3]{cos(2^{k}α)} )^{2} =

= ∑_{k=0}^{2} \sqrt[3]{sec(2^{k}α)} + ( ∑_{k=0}^{2} \sqrt[3]{cos(2^{k}α)} )^{2} − 2 ∑_{k=0}^{2} \sqrt[3]{cos(2^{k}α) cos(2^{k+1}α)} =

= ( ∑_{k=0}^{2} \sqrt[3]{cos(2^{k}α)} )^{2}.

By [12, eq. (3.31)] for n = 2:

\sqrt[3]{\frac{cos 2α}{cos α}} + \sqrt[3]{\frac{cos 4α}{cos 2α}} + \sqrt[3]{\frac{cos α}{cos 4α}} = 0 ;  (41)

for n = 3:

cos α \sqrt[3]{cos 2α} + cos 2α \sqrt[3]{cos 4α} + cos 4α \sqrt[3]{cos α} = −\frac{1}{2} \sqrt[3]{1 + \frac{3}{2} \sqrt[3]{49}} ;  (42)

for n = 4:

cos α \sqrt[3]{sec 4α} + cos 4α \sqrt[3]{sec 2α} + cos 2α \sqrt[3]{sec α} = \sqrt[3]{\frac{9}{4} (2 − \sqrt[3]{7})};  (43)

for n = 5:

cos^{2}(2α) \sqrt[3]{\frac{cos 4α}{cos 2α}} + cos^{2}(4α) \sqrt[3]{\frac{cos α}{cos 4α}} + cos^{2}(α) \sqrt[3]{\frac{cos 2α}{cos α}} = −\frac{1}{4} \sqrt[3]{49};  (44)

for n = 6:

cos^{2}(2α) \sqrt[3]{cos 4α} + cos^{2}(α) \sqrt[3]{cos 2α} + cos^{2}(4α) \sqrt[3]{cos α} =

= 2^{−7/3} \sqrt[3]{47 + 3 \sqrt[3]{7} − 12 \sqrt[3]{49}} ; (45)

for n = 7:

cos^{2}(α) \sqrt[3]{sec 4α} + cos^{2}(4α) \sqrt[3]{sec 2α} + cos^{2}(2α) \sqrt[3]{sec α} =

= −2^{5/3} \sqrt[3]{73 + 36 \sqrt[3]{7} + 3 \sqrt[3]{49}} ; (46)

Remark 9. We note that

2 cos^{2}(2α) \sqrt[3]{2 cos α} + 2 cos^{2}(4α) \sqrt[3]{2 cos 2α} + 2 cos^{2}(α) \sqrt[3]{2 cos 4α} =

= (\sqrt[3]{2 cos α} + \sqrt[3]{2 cos 2α} + \sqrt[3]{2 cos 4α}) +

+ cos 4α \sqrt[3]{2 cos α} + cos α \sqrt[3]{2 cos 2α} + cos 2α \sqrt[3]{2 cos 4α} =

=^{(1),(42)} \sqrt[3]{5 − 3 \sqrt[3]{7}} − \frac{1}{2} \sqrt[3]{2 + 3 \sqrt[3]{49}} =^{(38)} \frac{1}{2} \sqrt[3]{12 \sqrt[3]{7} − (4 + 3 \sqrt[3]{7})^{2}} ,

which implies the identity

\sqrt[3]{16 + 12 \sqrt[3]{7} + 9 \sqrt[3]{49}} = 2 \sqrt[3]{3 \sqrt[3]{7} − 5} + \sqrt[3]{2 + 3 \sqrt[3]{49}}.

By [12, eq. (3.32)] for n = 2:

\sqrt[3]{\frac{cos α}{cos^{2}(2α)}} + \sqrt[3]{\frac{cos 2α}{cos^{2}(4α)}} + \sqrt[3]{\frac{cos 4α}{cos^{2}(α)}} = \sqrt[3]{2 (11 − 3 \sqrt[3]{49})};  (47)

for n = 3:

\frac{\sqrt[3]{cos α}}{cos 2α} + \frac{\sqrt[3]{cos 2α}}{cos 4α} + \frac{\sqrt[3]{cos 4α}}{cos α} = −\sqrt[3]{36 (1 + \sqrt[3]{7})};  (48)

By [12, eq. (3.33)] for n = 2:

\sqrt[3]{\frac{cos α}{cos^{2}(4α)}} + \sqrt[3]{\frac{cos 2α}{cos^{2}(α)}} + \sqrt[3]{\frac{cos 4α}{cos^{2}(2α)}} = \sqrt[3]{6 ( − 1 + \sqrt[3]{7} − \sqrt[3]{49})} = −2 \sqrt[3]{\frac{6}{1 + \sqrt[3]{7}}} ; (49)

for n = 3:

\frac{\sqrt[3]{cos α}}{cos 4α} + \frac{\sqrt[3]{cos 2α}}{cos α} + \frac{\sqrt[3]{cos 4α}}{cos 2α} = \sqrt[3]{4 (26 − 6 \sqrt[3]{7} − 3 \sqrt[3]{49})};  (50)

By [12, eq. (3.34)] for n = 2 we get (33); for n = 3:

(2 cos α)^{−2/3} + (2 cos 2α)^{−2/3} + (2 cos 4α)^{−2/3} = \sqrt[3]{12 + 6 \sqrt[3]{7} + 3 \sqrt[3]{49}} ;  (51)

By [12, eq. (4.30)] for n = 1:

\sqrt[18]{7} (\sqrt[3]{cot α} + \sqrt[3]{cot 2α} + \sqrt[3]{cot 4α} ) =

= \sqrt[3]{\sqrt[3]{49} − 6 + 3 \sqrt[3]{3 (1 − \sqrt[3]{7} + \sqrt[3]{49})} − 3 \sqrt[3]{5 + 3 \sqrt[3]{7} − 3 \sqrt[3]{49}}} ; (52)

for n = 2:

\sqrt[9]{\frac{7}{8}} (\sqrt[3]{cot α csc α} + \sqrt[3]{cot 2α csc 2α} + \sqrt[3]{cot 4α csc 4α} ) =

= \sqrt[3]{6 − 2 \sqrt[3]{7} − 3 \sqrt[3]{3 (1 + \sqrt[3]{7})^{2}} − 3 \sqrt[3]{−26 + 6 \sqrt[3]{7} + 3 \sqrt[3]{49}}} , (53)

i.e,

∑_{k=0}^{2} \sqrt[3]{cot 2^{k}α csc 2^{k}α} = \sqrt[3]{\frac{6}{\sqrt[3]{7}} (2 − \frac{2}{3} \sqrt[3]{7} − \sqrt[3]{3(1 + \sqrt[3]{7})^{2}} − \sqrt[3]{3(1 + \sqrt[3]{7})^{2} − 27} )} . (54)

Moreover from (117) below we have for n = 1:

\sqrt[3]{tan α} + \sqrt[3]{tan 2α} + \sqrt[3]{tan 4α} =

= \sqrt[18]{7} \sqrt[3]{3 \sqrt[3]{3 (1 − \sqrt[3]{7} + \sqrt[3]{49})} − 3 \sqrt[3]{5 + 3 \sqrt[3]{7} − 3 \sqrt[3]{49}} − 6 − \sqrt[3]{7}} ; (55)

5.2 Some general formulas

1. By [12, eq. (2.1)] we obtain (for n = 0, 1, 2, …):

(2 cos α)^{n/3} + (2 cos 2α)^{n/3} + (2 cos 4α)^{n/3} =

= \sqrt[3]{B_{n} + 6 − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n,1} + \sqrt{T_{n,1}}} + \sqrt[3]{S_{n,1} − \sqrt{T_{n,1}}} )}, (56)

where

S_{n,1} := (−1)^{n−1} A_{n} (B_{n} + 6) − 6 B_{n} − 9,  (57)
T_{n,1} := A_{n}^{2} B_{n}^{2} + 4 (−1)^{n−1} A_{n}^{3} − 4 B_{n}^{3} + 18 (−1)^{n} A_{n} B_{n} − 27.  (58)

We note that

B_{n} = (2 cos α)^{n} + (2 cos 2α)^{n} + (2 cos 4α)^{n},  (59)

so we get the following interesting identity

\frac{1}{3} (6 + B_{3n} − B_{n}^{3}) = \sqrt[3]{\frac{1}{2}(S_{3n,1} + \sqrt{T_{3n,1}} )} + \sqrt[3]{\frac{1}{2}(S_{3n,1} − \sqrt{T_{3n,1}} )}.  (60)

Moreover, if n ∈ ℕ then by [12, eq. (3.34)] and by (23) and (26) we obtain

(2 cos α)^{−n/3} + (2 cos 2α)^{−n/3} + (2 cos 4α)^{−n/3} =

= \sqrt[3]{(−1)^{n} A_{n} + 6 − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n,1} + \sqrt{T_{n,1}}} + \sqrt[3]{S_{n,1} − \sqrt{T_{n,1}}} )}. (61)

Hence, for example for n = 2 the formula (33) follows.

2. By [12, eq. (3.27)] we obtain \sqrt[3]{2 sin α (2 cos 4α)^{n}} + \sqrt[3]{2 sin 2α (2 cos α)^{n}} + \sqrt[3]{2 sin 4α (2 cos 2α)^{n}} =

= \sqrt[3]{p_{n,2} − 6 \sqrt[6]{7} − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n,2} + \sqrt{T_{n,2}}} + \sqrt[3]{S_{n,2} − \sqrt{T_{n,2}}} )}, (62)

where

S_{n,2} = 7 (−1)^{n} B_{n} (6 \sqrt[6]{7} − p_{n,2}) − 6 \sqrt[3]{7} p_{n,2} + 9 \sqrt{7},  (63)
T_{n,2} = 49 B_{n}^{2} (p_{n,2}^{2} − 28 (−1)^{n} B_{n}) +
+ 2 \sqrt{7} p_{n,2} (2 p_{n,2}^{2} − 63 (−1)^{n}B_{n}) − 189,  (64)
p_{n,2} = \{ \begin{matrix}a_{n/2} & if n is even, \\ \overline{α}_{(n−1)/2} & if n is odd, \end{matrix}  (65)
q_{n,2} = 7 (−1)^{n} B_{n},  (66)
r_{n,2} ≡ \sqrt{7}.  (67)

3. By [12, eq. (3.28)] we obtain \sqrt[3]{2 sin α (2 cos 2α)^{n}} + \sqrt[3]{2 sin 2α (2 cos 4α)^{n}} + \sqrt[3]{2 sin 4α (2 cos α)^{n}} =

= \sqrt[3]{p_{n,3} − 6 \sqrt[6]{7} − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n,3} + \sqrt{T_{n,3}}} + \sqrt[3]{S_{n,3} − \sqrt{T_{n,3}}} )}, (68)

where

S_{n,3} = 7 (−1)^{n} (B_{n} − C_{n}) (p_{n,3} − 6 \sqrt[6]{7}) − 6 \sqrt[3]{7} p_{n,3} + 9 \sqrt{7},  (69)
T_{n,3} = 49 (B_{n} − C_{n})^{2} (p_{n,3}^{2} + 28 (−1)^{n} (B_{n} − C_{n})) +
+ 2 \sqrt{7} p_{n,3} (2 p_{n,3}^{2} + 63 (−1)^{n}(B_{n} − C_{n})) − 189,  (70)
p_{n,3} = \{ \begin{matrix}b_{n/2} & if n is even, \\ \overline{β}_{(n−1)/2} & if n is odd, \end{matrix}  (71)
q_{n,3} = 7 (−1)^{n−1} (B_{n} − C_{n}),  (72)
r_{n,3} ≡ \sqrt{7}.  (73)

4. By [12, eq. (3.29)] we obtain \sqrt[3]{2 sin α (2 cos α)^{n}} + \sqrt[3]{2 sin 2α (2 cos 2α)^{n}} + \sqrt[3]{2 sin 4α (2 cos 4α)^{n}} =

= \sqrt[3]{p_{n,4} − 6 \sqrt[6]{7} − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n,4} + \sqrt{T_{n,4}}} + \sqrt[3]{S_{n,4} − \sqrt{T_{n,4}}} )}, (74)

where

S_{n,4} = 7 (−1)^{n} C_{n} (p_{n,4} − 6 \sqrt[6]{7}) − 6 \sqrt[3]{7}p_{n,4} + 9 \sqrt{7},  (75)
T_{n,4} = (7 p_{n,4} C_{n} + 9 (−1)^{n} \sqrt{7})^{2} + 4 (7^{3} (−1)^{n} C_{n}^{3} + \sqrt{7} p_{n,4}^{3}) − 756,  (76)
p_{n,4} = \{ \begin{matrix}c_{n/2} & if n is even, \\ \overline{γ}_{(n−1)/2} & if n is odd, \end{matrix}  (77)
q_{n,4} = 7 (−1)^{n−1} C_{n},  (78)
r_{n,4} ≡ \sqrt{7}.  (79)

5. By [12, eq. (3.31)] we obtain \sqrt[3]{2 cos α (2 cos 4α)^{n}} + \sqrt[3]{2 cos 2α (2 cos α)^{n}} + \sqrt[3]{2 cos 4α (2 cos 2α)^{n}} =

= \sqrt[3]{g_{n} + 6 − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n,5} + \sqrt{T_{n,5}}} + \sqrt[3]{S_{n,5} − \sqrt{T_{n,5}}} )}, (80)

where

S_{n,5} = −g_{n} q_{n,5} − 6 (g_{n} + q_{n,5}) − 9,  (81)
T_{n,5} = (g_{n} q_{n,5} + 9)^{2} − 4 (g_{n}^{3} + q_{n,5}^{3}) − 108,  (82)

p_{n,5} = −g_{n},  (83)
q_{n,5} = (−1)^{n} (A_{n} + A_{n+1} − 7 A_{n}),  (84)
r_{n,5} ≡ −1.  (85)

6. By [12, eq. (3.32)] and (24) we obtain \sqrt[3]{cos α ( sec 2α)^{n}} + \sqrt[3]{cos 2α ( sec 4α)^{n}} + \sqrt[3]{cos 4α ( sec α)^{n}} =

= \sqrt[3]{2^{n−1} (−p_{n,6} + 6 − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n,6} + \sqrt{T_{n,6}}} + \sqrt[3]{S_{n,6} − \sqrt{T_{n,6}}} ))}, (86)

where

S_{n,6} = (−1)^{n} (A_{n+1} − A_{n}) (B_{n+2} − B_{n} − 6) + 6 (B_{n+2} − B_{n}) − 9,  (87)
T_{n,6} = (B_{n+2} − B_{n})^{2} ((A_{n} − A_{n+1})^{2} + 4 (B_{n+2} − B_{n})) +
+ 2 (−1)^{n} (A_{n} − A_{n+1}) (2 (A_{n} − A_{n+1})^{2} + 9 (B_{n+2} − B_{n})) − 27,  (88)
p_{n,6} = (−1)^{n} (A_{n} − A_{n+1}),  (89)
q_{n,6} = f_{n} + h_{n} = B_{n} − B_{n+2},  (90)
r_{n,6} ≡ −1.  (91)

7. By [12, eq. (3.33)] and (25) we obtain \sqrt[3]{cos α ( sec 4α)^{n}} + \sqrt[3]{cos 2α ( sec α)^{n}} + \sqrt[3]{cos 4α ( sec 2α)^{n}} =

= \sqrt[3]{2^{n−1} (−p_{n,7} + 6 − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n,7} + \sqrt{T_{n,7}}} + \sqrt[3]{S_{n,7} − \sqrt{T_{n,7}}} ))}, (92)

where

S_{n,7} = (−1)^{n} (A_{n+1} − A_{n−1}) (6 − B_{n+1} − B_{n}) + 6 (B_{n+1} + B_{n}) − 9,  (93)
T_{n,7} = (B_{n+1} + B_{n})^{2} ((A_{n+1} − A_{n−1})^{2} + 4 (B_{n+1} + B_{n})) +
+ 2 (−1)^{n} (A_{n+1} − A_{n−1}) (2 (A_{n+1} − A_{n−1})^{2} + 9 (B_{n+1} + B_{n})) − 27,  (94)
p_{n,7} = (−1)^{n} (A_{n+1} − A_{n−1}),  (95)
q_{n,7} = f_{n} + g_{n} = −B_{n} − B_{n+1},  (96)
r_{n,7} ≡ −1.  (97)

8. By [12, eq. (3.30)], we obtain \sqrt[3]{2 cos α (2 cos 2α)^{n}} + \sqrt[3]{2 cos 2α (2 cos 4α)^{n}} + \sqrt[3]{2 cos 4α (2 cos α)^{n}} =

= \sqrt[3]{f_{n} + 6 − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n,8} + \sqrt{T_{n,8}}} + \sqrt[3]{S_{n,8} − \sqrt{T_{n,8}}} )}, (98)

where

S_{n,8} = −f_{n} q_{n,8} − 6 (f_{n} + q_{n,8}) − 9,  (99)
T_{n,8} = (f_{n} q_{n,8} + 9)^{2} − 4 (f_{n}^{3} + q_{n,8}^{3}) − 108,  (100)
p_{n,8} = −f_{n},  (101)
q_{n,8} = (−1)^{n} (7 A_{n} − 3 A_{n}) = (−1)^{n}(A_{n} − A_{n−2}),  (102)
r_{n,8} ≡ −1.  (103)

Remark 10. As results from direct observation of the value of expression

\sqrt[3]{\frac{1}{2} (S_{n,8} − \sqrt{T_{n,8}})}

for n = 0, 1, …, 2000, for the indicated index values, the following equalities hold

\sqrt[3]{\frac{1}{2} (S_{n,8} − \sqrt{T_{n,8}})} = \{ \begin{matrix}(−1)^{k−1} x̂_{k}, & for n = 3 k − 1 ≥ 5, \\ \sqrt[3]{7} (−1)^{k} ŷ_{k}, & for n = 3 k, \\ \sqrt[3]{49} (−1)^{k+1} ẑ_{k}, & for n = 3 k + 1, \end{matrix}

for k = 1, 2, …, where

x̂_{1} = 2,  x̂_{2} = 5,  x̂_{3} = 16,
ŷ_{1} = 1,  ŷ_{2} = 4,  ŷ_{3} = 12,
ẑ_{1} = 1,  ẑ_{2} = 3,  ẑ_{3} = 9,

and the elements of any of the sequences: {x̂_{k}}_{k=1}^{∞}, {ŷ_{k}}_{k=1}^{∞} and {ẑ_{k}}_{k=1}^{∞}, satisfy the following recurrent relation

𝕏_{n+3} − 4 𝕏_{n+2} + 3 𝕏_{n+1} + 𝕏_{n} = 0.

Also, the following interesting relationships occur (see (14)):

x̂_{k} = A_{k}(−1) + 2 C_{k}(−1) − C_{k−2}(−1),  k ≥ 2,
ŷ_{k} = A_{k}(−1) + C_{k}(−1),
ẑ_{k} = C_{k}(−1) − B_{k}(−1).

Hence, by [11, eqs. (3.17), (3.18), (3.19)], we obtain, inter alia, the following Binet formulas:

x̂_{k} = (2 − 2 cos α + 4 cos 2α) (1 − 2 cos α)^{k−2} +

+ (2 − 2 cos 2α + 4 cos 4α) (1 − 2 cos 2α)^{k−2} +

+ (2 − 2 cos 4α + 4 cos α) (1 − 2 cos 4α)^{k−2},

ŷ_{k} = \frac{2}{7} (1 + cos 2α − 2 cos 4α) (1 − 2 cos α)^{k} +

+ \frac{2}{7} (1 + cos 4α − 2 cos α) (1 − 2 cos 2α)^{k} +

+ \frac{2}{7} (1 + cos α − 2 cos 2α) (1 − 2 cos 4α)^{k},

ẑ_{k} = \frac{2}{7} ( cos 2α − cos α) (1 − 2 cos α)^{k} +

+ \frac{2}{7} ( cos 4α − cos 2α) (1 − 2 cos 2α)^{k} +
+ \frac{2}{7} ( cos α − cos 4α) (1 − 2 cos 4α)^{k}.

Next, as results from direct observation of the value of expression

\sqrt[3]{\frac{1}{2} (S_{n,8} + \sqrt{T_{n,8}})}

for n = 0, 1, …, 2000, for the indicated index values, the following equations hold

\sqrt[3]{\frac{1}{2} (S_{n,8} + \sqrt{T_{n,8}})} = \{ \begin{matrix}x̃_{k}, & for n = 3 k + 2, \\ \sqrt[3]{7} ỹ_{k}, & for n = 3 k + 1, \\ \sqrt[3]{49} z̃_{k}, & for n = 3 k, \end{matrix}

for k = 1, 2, …, where x_{0} = 2 and

x̃_{1} = 8,  x̃_{2} = 29,  x̃_{3} = 120,
ỹ_{1} = 1,  ỹ_{2} = 2,  ỹ_{3} = 10,
z̃_{1} = 1,  z̃_{2} = 3,  z̃_{3} = 13,

and the elements of any of the sequences {x̃_{k}}_{k=2}^{∞}, {ỹ_{k}}_{k=1}^{∞} and {z̃_{k}}_{k=1}^{∞} satisfy the following recurrent relation:

𝕏_{n+3} − 3 𝕏_{n+2} − 4 𝕏_{n+1} − 𝕏_{n} = 0.

After substitution x ↦ (x − 1) in the respective characteristic polynomial of this relation, we obtain polynomial x^{3} − 7 x − 7, for which by [12, eq. (4.14)], for n = 1, we obtain

𝕏^{3} − 7 𝕏 − 7 = ∏_{k=0}^{2} (𝕏 + \frac{\sqrt{7}}{2} csc(2^{k}α)).

Hence, the following Binet formulas hold

p_{n} = a_{p} (1 − \frac{\sqrt{7}}{2} csc α)^{n} + b_{p} (1 − \frac{\sqrt{7}}{2} csc 2α)^{n} + c_{p} (1 − \frac{\sqrt{7}}{2} csc 4α)^{n},

for every p ∈ {x̃, ỹ, z̃}, and where

ax̃ ≈ −1.246979604,  bx̃ ≈ 0.4450418679,  cx̃ ≈ 1.801937736,
aỹ ≈ −1.064961507,  bỹ ≈ 0.9189943261,  cỹ ≈ 0.1459671806,
az̃ ≈ −0.4355596199,  bz̃ ≈ 0.2417173531,  cz̃ ≈ 0.1938422668.

Additionally, we note that

x̃_{0} = ax̃ + bx̃ + cx̃ = 1,
ỹ_{0} = aỹ + bỹ + cỹ = 0,
z̃_{0} = az̃ + bz̃ + cz̃ = 0.

By juxtaposing the obtained values of expressions

\sqrt[3]{\frac{1}{2} (S_{n,8} ± \sqrt{T_{n,8}})}

we can now invest formula (98) to the new interesting form (for cases n = 3k, 3k + 1, 3k + 2 respectively, and only within the indicated range of values n = 1, 2, …, 2000):

\sqrt[3]{2 cos α} (2 cos 2α)^{k} + \sqrt[3]{2 cos 2α} (2 cos 4α)^{k} + \sqrt[3]{2 cos 4α} (2 cos α)^{k} = = \sqrt[3]{f_{3k} + 6 + 3 (−1)^{k−1}ŷ_{k} \sqrt[3]{7} − 3 z̃_{k} \sqrt[3]{49}} ,

\sqrt[3]{sec 4α} (2 cos 2α)^{k} + \sqrt[3]{sec α} (2 cos 4α)^{k} + \sqrt[3]{sec 2α} (2 cos α)^{k} =

= \sqrt[3]{2 (f_{3k+1} + 6 − 3 ỹ_{k} \sqrt[3]{7} + 3 (−1)^{k} ẑ_{k} \sqrt[3]{49})} ,

and the formula which is equivalent to relation (2) and which generates the identity

7 ψ_{k}^{3} = f_{3k+2} + 6 − 3 (x̃_{k} + (−1)^{k} x̂_{k+1}).  (104)

9. By [12, eq. (4.15)] we obtain

(2 sin α)^{n/3} + (2 sin 2α)^{n/3} + (2 sin 4α)^{n/3} =

= \sqrt[3]{z_{n−1} + 6 (−1)^{n} 7^{n/6} − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n,9} + \sqrt{T_{n,9}}} + \sqrt[3]{S_{n,9} − \sqrt{T_{n,9}}} )}, (105)

where

S_{n,9} = (\frac{1}{2} z_{n−1} + 3 (−1)^{n} 7^{n/6}) (z_{2n−1} − z_{n−1}^{2})−
− 6 · 7^{n/3} z_{n−1} − 9 (−1)^{n} 7^{n/2},  (106)
T_{n,9} = (−1)^{n} 7^{n/2} z_{n−1} (5 z_{n−1}^{2} − 9 z_{2n−1})+
+ \frac{1}{4} (2 z_{2n−1} − z_{n−1}^{2}) (z_{n−1}^{2} − z_{2n−1})^{2} − 27 · 7^{n},  (107)
p_{n,9} = −z_{n−1}, q_{n,9} = \frac{1}{2} (z_{n−1}^{2} − z_{2n−1}), r_{n,9} = (−1)^{n−1} 7^{n/2}.  (108)

10. By [12, eq. (2.2)] we have (for δ ∈ ℝ):

(1 + 2 δ cos α)^{n/3} + (1 + 2 δ cos 2α)^{n/3} + (1 + 2 δ cos 4α)^{n/3} =

= [A_{n}(δ) + 6 δ̂^{n/3} − \frac{3}{\sqrt[3]{2}} (\sqrt[3]{S_{n}(δ) + \sqrt{T_{n}(δ)}} + \sqrt[3]{S_{n}(δ) − \sqrt{T_{n}(δ)}} )]^{1/3}, (109)

where

δ̂ = δ^{3} − 2 δ^{2} − δ + 1,  (110)
S_{n}(δ) = −A_{n}(δ) B_{n}(δ) − 6 B_{n}(δ) δ̂^{n/3} − 6 A_{n}(δ) δ̂^{2n/3} − 9 δ̂^{n},  (111)
T_{n}(δ) = A_{n}^{2}(δ) B_{n}^{2}(δ) − 4 A_{n}^{3}(δ) δ̂^{n} − 4 B_{n}^{3}(δ) + 18 A_{n}(δ) B_{n}(δ) δ̂^{n} − 27 δ̂^{2 n}  (112)
= (A_{n}(δ) B_{n}(δ) + 9 δ̂^{n})^{2} − 4 (A_{n}^{3}(δ) δ̂^{n} + B_{n}^{3}(δ)) − 108 δ̂^{2 n}.  (113)

11. By [12, eq. (6.14)] we have

7^{n/6} ((cot α)^{n/3} + (cot 2α)^{n/3} + (cot 4α)^{n/3}) =

= [3^{n} A_{n}(\frac{2}{3}) + 6 (−1)^{n} 7^{n/3} + \frac{3}{\sqrt[3]{2}} 7^{n/3} (\sqrt[3]{Sn^{′} + \sqrt{Tn^{′}}} + \sqrt[3]{Sn^{′} − \sqrt{Tn^{′}}} )]^{1/3}, (114)

where

Sn^{′} = (3^{n} A_{n}(\frac{2}{3}) + 6 (−1)^{n} 7^{n/3}) Ω_{n}(\frac{2 i}{\sqrt{7}}) + 6 ( \frac{3}{\sqrt[3]{7}})^{n} A_{n}(\frac{2}{3}) + 9 (−1)^{n},  (115)
Tn^{′} = ((−3)^{n} A_{n}(\frac{2}{3}) Ω_{n}(\frac{2 i}{\sqrt{7}}) + 9)^{2} − 4 (7^{n} Ω_{n}^{3}(\frac{2 i}{\sqrt{7}}) + ( − \frac{27}{7})^{n} A_{n}^{3}(\frac{2}{3})) − 108.  (116)

The numbers Ω_{n}(δ) are defined for n ∈ ℕ and δ ∈ ℂ, in the following way

Ω_{n}(δ) := (1 + 2 i δ sin α)^{n} + (1 + 2 i δ sin 2α)^{n} + (1 + 2 i δ sin 4α)^{n},

(see [12, Section 6] for more details). Remark 11. Moreover, we have

(𝕏 − (tan α)^{n})(𝕏 − (tan 2α)^{n})(𝕏 − (tan 4α)^{n}) =

= 𝕏^{3} − (−\sqrt{7})^{n} Ω_{n}(\frac{2 i}{\sqrt{7}}) 𝕏^{2} + (−3)^{n} A_{n}(\frac{2}{3}) 𝕏 − (−\sqrt{7})^{n}. (117)

This ”distribution” easily results from [12, eq. (6.14)]. Now we will present a direct proof of the relation (117), because the formula (6.14) in [12] is presented without a proof. For this purpose, let us suppose that ξ = exp(i 2 π/7). Then we have

(tan α)^{n} + (tan 2α)^{n} + (tan 4α)^{n} =

= (−i \frac{ξ − ξ^{6}}{ξ + ξ^{6}} )^{n} + (−i \frac{ξ^{2} − ξ^{5}}{ξ^{2} + ξ^{5}} )^{n} + (−i \frac{ξ^{4} − ξ^{3}}{ξ^{4} + ξ^{3}} )^{n} =

= (\frac{−i}{(ξ + ξ^{6})(ξ^{2} + ξ^{5})(ξ^{4} + ξ^{3})})^{n} [((ξ − ξ^{6})(ξ^{2} + ξ^{5})(ξ^{4} + ξ^{3}))^{n} +

+ ((ξ^{2} − ξ^{5})(ξ + ξ^{6})(ξ^{4} + ξ^{3}))^{n} + ((ξ^{4} − ξ^{3})(ξ + ξ^{6})(ξ^{2} + ξ^{5}))^{n}] =

{[12, eq. (1.4)] \atop =} (−i)^{n} [(2 (ξ^{2} − ξ^{5}) − (ξ − ξ^{6}) − (ξ^{2} − ξ^{5}) − (ξ^{4} − ξ^{3}))^{n} +

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

+ (2 (ξ − ξ^{6}) − (ξ − ξ^{6}) − (ξ^{2} − ξ^{5}) − (ξ^{4} − ξ^{3}))^{n}] =

{[12, eq. (1.1)] \atop =} [(−2 i (ξ^{2} − ξ^{5}) − \sqrt{7})^{n} + (−2 i (ξ^{4} − ξ^{3}) − \sqrt{7})^{n} +

+ (−2 i (ξ − ξ^{6}) − \sqrt{7})^{n}] = (−\sqrt{7})^{n} Ω_{n}(\frac{2 i}{\sqrt{7}}), and

( tan α tan 2α)^{n} + ( tan α tan 4α)^{n} + ( tan 2α tan 4α)^{n} =

= ((−i \frac{ξ − ξ^{6}}{ξ + ξ^{6}} ) (−i \frac{ξ^{2} − ξ^{5}}{ξ^{2} + ξ^{5}} ))^{n} + ((−i \frac{ξ − ξ^{6}}{ξ + ξ^{6}} ) (−i \frac{ξ^{4} − ξ^{3}}{ξ^{4} + ξ^{3}} ))^{n} +

+ ((−i \frac{ξ^{2} − ξ^{5}}{ξ^{2} + ξ^{5}} ) (−i \frac{ξ^{4} − ξ^{3}}{ξ^{4} + ξ^{3}} ))^{n} =

= ((ξ + ξ^{6})(ξ^{2} + ξ^{5})(ξ^{4} + ξ^{3}))^{−n} [((ξ^{6} − ξ)(ξ^{2} − ξ^{5})(ξ^{4} + ξ^{3}))^{n} +
+ ((ξ^{6} − ξ)(ξ^{4} − ξ^{3})(ξ^{2} + ξ^{5}))^{n} + ((ξ^{5} − ξ^{2})(ξ^{4} − ξ^{3})(ξ + ξ^{6}))^{n}] =

= (−3 − 2 (ξ + ξ^{6}))^{n} + (−3 − 2 (ξ^{4} + ξ^{3}))^{n} + (−3 − 2 (ξ^{2} + ξ^{5}))^{n} = (−3)^{n} A_{n}(\frac{2}{3}).

Final remark. I was only after I received the referee report on my paper that I learnt about two important publications in this field [8, 9]. Certainly, both papers supplement and enrich the contents of Section 3. As a spontaneous reaction to [8] and the report on the present paper, two more papers sprang up [14] and [15].

6 Acknowledgments

The author wish to express their gratitude to the Reviewer for several helpful comments concerning the first version of my paper.

Table 1: ; \begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ ψ_{n} & −1 & 0 & −3 & 2 & −8 & 9 & −23 & 33 & −70 & 113 & −220 & 376 \\ φ_{n} & 0 & −1 & 1 & −3 & 4 & −9 & 14 & −28 & 47 & −89 & 155 & −286 \end{matrix}

References

[1] B. C. Berndt, Ramanujan’s Notebooks, Part IV, Springer, 1994.

[2] B. C. Berndt, H. H. Chan and L.-Ch. Zhang, Radicals and units in Ramanujan’s work, Acta Arith. 87 (1998), 145–158.

[3] B. C. Berndt, Y.-S. Choi and S.-Y. Kang, The problems submitted by Ramanujan to the Journal of the Indian Mathematical Society, in B. C. Berndt and F. Gesztesy, eds., Continued Fractions: From Analytic Number Theory to Constructive Approximation, AMS, 1999, pp. 15–56.

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

[5] W. A. Kreczmar, A Collection of Problems in Algebra, Nauka, 1961.

[6] P. S. Modenov, Collection of Problems for a Special Course of Elementary Mathematics, Sovetskaja Nauka, 1957.

[7] V. S. Shevelev, Three Ramanujan’s formulas, Kvant 6 (1988), 52–55 (in Russian).

[8] V. Shevelev, On Ramanujan cubic polynomials, arXiv:0711.3420v1 , 2007.

[9] V. Shevelev, Gold ratio and a trigonometric identity, arXiv:0909.5072v1 , 2009.

[10] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, 2009.

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

[12] 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.

[13] R. Wituła and D. Słota, Quasi-Fibonacci numbers of order 11, J. Integer Seq. 10 (2007), Article 07.8.5.

[14] R. Wituła, Full description of Ramanujan cubic polynomials, submitted.

[15] R. Wituła, Ramanujan type trigonometric formulas for arguments \frac{2 π}{7} and \frac{2 π}{9} – supplement, submitted.

[16] A. M. Yaglom and I. M. Yaglom, An elementary derivation of the formulas of Wallis, Leibnitz and Euler for the number π, Uspehi Matem. Nauk 8 (1953), 181–187.

2000 Mathematics Subject Classification: Primary 11B37; Secondary 11B83, 11Y55, 33B10. Keywords: Ramanujan equalities, trigonometric recurrences.

(Concerned with sequences \underline{A094648} and \underline{A006053}.)

Received July 6 2009; Revised version received December 3 2009. Published in Journal of Integer Sequences, December 3 2009.

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