A Study of Second-Order Linear Recurrence Sequences via Continuants

Hongshen Chua Puchong 47170 Malaysia

                    hschua0622@hotmail.com

Abstract

This paper presents a reinterpretation of a second-order linear recurrence sequence as a sequence of continuants derived from the convergents to a continued fraction. As a result, we are able to derive the generating function and Binet formula for continuants. Using this result, we provide a continuant-based formulation for well-known identities associated with Lucas sequences.

1 Introduction

Consider the d-periodic Fibonacci numbers, defined by the second-order linear recurrence

f_{ν} = a_{r}f_{ν−1} + b_{r}f_{ν−2}, where ν ≡ r (mod d),

with initial values f_{0} = 0 and f_{1} = 1. Such a sequence was studied by Carson [3], who, using matrix manipulation, managed to rewrite the sequence in the form:

f_{ν} = af_{ν−d} + bf_{ν−2d},

for some constants a and b. This extends an earlier partial result by Lehmer [11]. An alternative method via continuants was provided by Panario et al. [15]. In the same paper, they also derived the generating function and Binet formula for d-periodic Fibonacci numbers, supplementing earlier work by Edson et al. [6]. Our paper will follow this approach closely. Other related studies include the papers by Tan [19] and Yayenie [21], which focus on the special case of d = 2.

The concept of continuants was initially introduced by Euler during the development of a theory for continued fractions. In particular, continuants appear in the numerators and denominators of the convergents to a continued fraction. Subsequently, Muir [14] studied continuants as determinants of tridiagonal matrices. A major contribution by Muir is the notation that we employ in this paper. Another perspective on continuants is through a combinatorial approach, where they are viewed as a tiling problem, as discussed by Benjamin et al. [2]. However, the main inspiration for the content of this paper stems from Perron’s book [16] on continued fractions.

The goal of this paper is to highlight the similarities between continuants and secondorder linear recurrence sequences, with a focus on d-periodic Fibonacci numbers. Additionally, we address a problem posed by Panario et al. [15], seeking an application for generalized continuants. We hope that this will give continuants the attention they deserve.

While not a prerequisite to understand this paper, we assume here that readers are familiar with Lucas sequences, especially with the Fibonacci numbers, as most of our results are analogous to classical theorems on Fibonacci numbers. This familiarity may help the readers to grasp the motivation behind the results. Some selected articles that may be useful include [4, 5, 7, 9, 17, 20].

2 Generalized continuants

Suppose that (a_{ν}) and (b_{ν}) are sequences of positive integers. Consider the generalized continued fraction

ξ_{λ} = b_{λ} + \frac{a_{λ+1}}{b_{λ+1}} + \frac{a_{λ+2}}{b_{λ+2}} + ⋯ = b_{λ} + \frac{a_{λ+1}}{b_{λ+1} + \frac{a_{λ+2}}{b_{λ+2} + ⋯}}.

The convergents to the continued fraction are the numbers obtained by truncating the continued fraction. For example, the ν-th convergent of ξ_{λ} is exactly

ξ_{ν,λ} = b_{λ} + \frac{a_{λ+1}}{b_{λ+1}} + ⋯ + \frac{a_{λ+ν}}{b_{λ+ν}} .

Define the ν-th generalized continuants

A_{ν,λ} = K\binom{a_{λ+1}, …, a_{λ+ν}}{b_{λ}, b_{λ+1}, …, b_{λ+ν}},
B_{ν,λ} = K\binom{a_{λ+2}, …, a_{λ+ν}}{b_{λ+1}, b_{λ+2}, …, b_{λ+ν}},

as the numerator and denominator, respectively, of the ν-th convergent. In other words,

\frac{A_{ν,λ}}{B_{ν,λ}} = b_{λ} + \frac{a_{λ+1}}{b_{λ+1}} + ⋯ + \frac{a_{λ+ν}}{b_{λ+ν}} .

By the choice of sequences (a_{ν}) and (b_{ν}), both A_{ν,λ} and B_{ν,λ} are integers. We can slightly extend ν to negative numbers by setting the initial values

A_{−1,λ} = 1, A_{0,λ} = b_{λ},
B_{−1,λ} = 0, B_{0,λ} = 1, B_{1,λ} = b_{λ+1}.

The ordinary continuants correspond to the case when a_{ν} = 1 is constant. To simplify our notation, we set A_{ν,0} = A_{ν} and B_{ν,0} = B_{ν}. Observe that A_{ν,λ} and B_{ν,λ} are related via the identity B_{ν,λ} = A_{ν−1,λ+1}. Furthermore, there are more general identities connecting A_{ν,λ} and B_{ν,λ}, notably an analogue of Cassini’s identity for Fibonacci numbers.

Theorem 1 (Cassini’s Identity). For non-negative integers λ, ν and μ, we have

A_{ν+λ+μ−1}B_{ν−1,λ} = A_{ν+λ−1}B_{ν+μ−1,λ} + (−1)^{ν−1}a_{λ+1} ⋯ a_{λ+ν}A_{λ−1}B_{μ−1,ν+λ},
B_{ν+λ+μ−1}B_{ν−1,λ} = B_{ν+λ−1}B_{ν+μ−1,λ} + (−1)^{ν−1}a_{λ+1} ⋯ a_{λ+ν}B_{λ−1}B_{μ−1,ν+λ}.

Corollary 2 (Catalan’s Identity). For non-negative integers λ and ν, we have

A_{ν+λ} = A_{λ}B_{ν,λ} + a_{λ+1}A_{λ−1}B_{ν−1,λ+1},
B_{ν+λ} = B_{λ}B_{ν,λ} + a_{λ+1}B_{λ−1}B_{ν−1,λ+1}.

Corollary 3 (d’Ocagne’s Identity). For non-negative integers λ and ν, we have

A_{λ}B_{ν−1,λ−ν} = A_{λ−1}B_{ν,λ−ν} + (−1)^{ν−1}a_{λ} ⋯ a_{λ−ν+1}A_{λ−ν−1},
B_{λ}B_{ν−1,λ−ν} = B_{λ−1}B_{ν,λ−ν} + (−1)^{ν−1}a_{λ} ⋯ a_{λ−ν+1}B_{λ−ν−1}.

Corollary 4. For integer ν ≥ −1, we have

A_{ν+2} = b_{ν+2}A_{ν+1} + a_{ν+2}A_{ν},
B_{ν+2} = b_{ν+2}B_{ν+1} + a_{ν+2}B_{ν}.

Corollary 5. For non-negative integers λ and ν, we have

A_{ν,λ} = b_{λ}A_{ν−1,λ+1} + a_{λ+1}A_{ν−2,λ+2},
B_{ν,λ} = b_{λ+1}B_{ν−1,λ+1} + a_{λ+2}B_{ν−2,λ+2}.

From Corollary 4, we can express A_{ν} and B_{ν} as a determinant of tridiagonal matrices.

Corollary 6. For non-negative integer ν, we have

A_{ν} = |\begin{matrix}b_{0} & −1 & 0 & … & 0 & 0 \\ a_{1} & b_{1} & −1 & … & 0 & 0 \\ 0 & a_{2} & b_{2} & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & 0 & … & b_{ν−1} & −1 \\ 0 & 0 & 0 & … & a_{ν} & b_{ν} \end{matrix}| , B_{ν} = |\begin{matrix}b_{1} & −1 & 0 & … & 0 & 0 \\ a_{2} & b_{2} & −1 & … & 0 & 0 \\ 0 & a_{3} & b_{3} & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & 0 & … & b_{ν−1} & −1 \\ 0 & 0 & 0 & … & a_{ν} & b_{ν} \end{matrix}| .

Alternatively, we can write A_{ν} and B_{ν} as a product of matrices.

Corollary 7. For non-negative integer ν, we have

(\begin{matrix}A_{ν} & A_{ν−1} \\ B_{ν} & B_{ν−1} \end{matrix}) = (\begin{matrix}1 & 1 \\ 1 & 0 \end{matrix}) ∏_{j=1}^{ν} (\begin{matrix}b_{j} & 1 \\ a_{j} & 0 \end{matrix}) .

Proof. The proofs of all the results above can be found in Perron’s book [16], so we omit them here. It is worth noting that it is easier to prove the corollaries before proving Theorem 1.

Remark 8. The results presented here exhibit similarities to those of Fibonacci numbers. This connection is not surprising since choosing (a_{ν}) = (b_{ν}) = (1) leads to (B_{ν}) = (F_{ν+1}), which is just the usual Fibonacci sequence.

3 Periodic continuants

In this section and the remainder of this paper, we consider the case when (a_{ν}) and (b_{ν}) are both periodic with the same period d for ν ≥ 1, i.e., a_{ν+d} = a_{ν} and b_{ν+d} = b_{ν} for all ν ≥ 1. This type of continued fraction arises frequently in various contexts. For instance, when N is a positive integer, the continued fraction expansion of \sqrt{N} is periodic. Under this assumption,

K\binom{a_{d+2}, …, a_{d+ν}}{b_{d+1}, b_{d+2}, …, b_{d+ν}} = K\binom{a_{2}, …, a_{ν}}{b_{1}, b_{2}, …, b_{ν}},

so B_{ν,d} = B_{ν} (in fact, A_{ν,λ} and B_{ν,λ} are periodic with period d in the second argument). Using this fact and Corollary 3, we can derive an identity that will be useful later on.

Corollary 9. For non-negative integers λ and ν with d | (λ − ν), we have

A_{λ−1}B_{ν} − A_{λ}B_{ν−1} = (−1)^{ν}a_{λ} ⋯ a_{λ−ν+1}A_{λ−ν−1},
B_{λ−1}B_{ν} − B_{λ}B_{ν−1} = (−1)^{ν}a_{λ} ⋯ a_{λ−ν+1}B_{λ−ν−1}.

3.1 An equivalent recurrence relation

When (a_{ν}) and (b_{ν}) are periodic with period d, we can write (B_{ν}) as a second-order linear recurrence relation,

B_{ν} = b_{r}B_{ν−1} + a_{r}B_{ν−2}, where ν ≡ r (mod d),

with the appropriate initial values. Motivated by Carson’s result [3], we seek a recurrence relation of the form

B_{ν+2d} = C_{d}B_{ν+d} + D_{d}B_{ν},

for some integers C_{d} and D_{d}. The next theorem provides the desired values.

Theorem 10. If (a_{ν}) and (b_{ν}) are periodic with period d, then

B_{ν+2d} = \frac{B_{2d−1}}{B_{d−1}} B_{ν+d} + (−1)^{d−1}a_{1} ⋯ a_{d}B_{ν}.

Proof. Using Corollary 2 with λ = d and 2d, we get

B_{ν+d} = B_{d}B_{ν} + a_{1}B_{d−1}B_{ν−1,1},
B_{ν+2d} = B_{2d}B_{ν} + a_{1}B_{2d−1}B_{ν−1,1}.

By canceling out a_{1}B_{ν−1,1}, we obtain

B_{ν+2d} = B_{2d}B_{ν} + B_{2d−1}(\frac{B_{ν+d} − B_{d}B_{ν}}{B_{d−1}} ).

= \frac{B_{2d−1}}{B_{d−1}} B_{ν+d} + (\frac{B_{d−1}B_{2d} − B_{2d−1}B_{d}}{B_{d−1}} )B_{ν}.

Again using Corollary 2 with λ = d and ν = d − 1, we get

B_{2d−1} = B_{d}B_{d−1} + a_{1}B_{d−1}B_{d−2,1},

so C_{d} = B_{2d−1}/B_{d−1} = B_{d} + a_{1}B_{d−2,1} is a non-negative integer. On the other hand, using Corollary 9 with λ = 2d and ν = d, we get

B_{2d}B_{d−1} − B_{2d−1}B_{d} = (−1)^{d−1}a_{1} ⋯ a_{d}B_{d−1}.

Thus, D_{d} = (−1)^{d−1}a_{1} ⋯ a_{d} is an integer.

One can check that the constants are indeed the same as those found by Carson [3]. From Theorem 10, it is immediate to deduce the next corollary.

Corollary 11. Let Δ = C_{d}^{2} + 4D_{d}. For n ≥ 1, we have

B_{(n+1)d−1} = \frac{C_{d}B_{nd−1} + \sqrt{ΔB_{nd−1}^{2} + 4(−D_{d})^{n}B_{d−1}^{2}}}{2} .

Proof. If we choose λ = μ = d and ν = nd with n ≥ 1 in Theorem 1, then

B_{(n+1)d−1}^{2} − B_{nd−1}B_{(n+2)d−1} = B_{(n+1)d−1}^{2} − C_{d}B_{nd−1}B_{(n+1)d−1} − D_{d}B_{nd−1}^{2} = (−D_{d})^{n}B_{d−1}^{2}.

Solving this quadratic equation for B_{(n+1)d−1} gives the desired equation.

Remark 12. For the corresponding identities for Lucas sequences, refer to the book by Andrica and Bagdasar [1].

3.2 Generating function and Binet’s formula

A benefit of having a single recurrence relation for (B_{ν}) is that we can compute its generating function and Binet formula easily. We follow the proof by Panario et al. [15] for the next two theorems. Firstly, let

G(x) = ∑_{n=0}^{∞} B_{n}x^{n},

be the required generating function. Then

(1 − C_{d}x^{d} − D_{d}x^{2d})G(x) = ∑_{n=0}^{∞} B_{n}x^{n} − C_{d} ∑_{n=d}^{∞} B_{n−d}x^{n} − D_{d} ∑_{n=2d}^{∞} D_{d}B_{n−2d}x^{n}

= ∑_{n=0}^{2d−1} B_{n}x^{n} − C_{d} ∑_{n=0}^{d−1} B_{n}x^{n+d}.

Therefore, we arrive at our generating function.

Theorem 13. The generating function of B_{n} is

G(x) = \frac{∑_{n=0}^{2d−1} B_{n}x^{n} − C_{d} ∑_{n=0}^{d−1} B_{n}x^{n+d}}{1 − C_{d}x^{d} − D_{d}x^{2d}} .

Using this generating function, we can then calculate the Binet formula for B_{ν}. Since the continued fraction has period d, it is perhaps not surprising that the Binet formula depends on the residue modulo d. Let Δ = C_{d}^{2} + 4D_{d} be the discriminant of the equation D_{d}z^{2} + C_{d}z − 1 = 0. We assume throughout this paper that Δ ≠ 0. The next theorem gives the Binet formula for (B_{ν}).

Theorem 14 (Binet’s formula). Suppose that Δ ≠ 0. Let α and β be the positive and negative roots of D_{d}z^{2} + C_{d}z − 1 = 0, respectively. For integers n and r such that n ≥ 0 and r ≥ −1, we have

B_{nd+r} = (−D_{d})^{n−1}(\frac{α^{n} − β^{n}}{α − β} B_{d+r} − \frac{α^{n−1} − β^{n−1}}{α − β} B_{r}).

Proof. Consider the generating function of the subsequence (B_{nd+r}) given by

G_{r}(x) = \frac{B_{r}x^{r} + B_{d+r}x^{d+r} − C_{d}B_{r}x^{d+r}}{1 − C_{d}x^{d} − D_{d}x^{2d}} .

Let α and β be the positive and negative roots of D_{d}z^{2} + C_{d}z − 1 = 0, respectively, or equivalently,

α = \frac{−C_{d} + \sqrt{Δ}}{2D_{d}}, β = \frac{−C_{d} − \sqrt{Δ}}{2D_{d}}.

Note that, α + β = −C_{d}/D_{d} and αβ = −1/D_{d}. By partial fraction decomposition, we have

\frac{1}{1 − C_{d}x^{d} − D_{d}x^{2d}} = \frac{1}{α − β} (\frac{α}{1 + D_{d}αx^{d}} − \frac{β}{1 + D_{d}βx^{d}} ).

Substituting this into G_{r}(x), we obtain

G_{r}(x) = \frac{B_{r}x^{r} + B_{d+r}x^{d+r} − C_{d}B_{r}x^{d+r}}{α − β} ∑_{n=0}^{∞}(−1)^{n}D_{d}^{n}(α^{n+1} − β^{n+1})x^{nd}

= ∑_{n=0}^{∞}(−1)^{n} \frac{D_{d}^{n}(α^{n+1} − β^{n+1})B_{r}}{α − β} x^{nd+r} + ∑_{n=0}^{∞}(−1)^{n} \frac{D_{d}^{n}(α^{n+1} − β^{n+1})(B_{d+r} − C_{d}B_{r})}{α − β}x^{(n+1)d+r}.

By replacing n with n − 1 in the second sum, it becomes

∑_{n=0}^{∞}(−1)^{n−1} \frac{D_{d}^{n−1}(α^{n} − β^{n})(B_{d+r} − C_{d}B_{r})}{α − β}x^{nd+r}.

This combined with the first sum gives the expression

G_{r}(x) = ∑_{n=0}^{∞}(−1)^{n}(\frac{D_{d}^{n}(α^{n+1} − β^{n+1})B_{r} − D_{d}^{n−1}(α^{n} − β^{n})(B_{d+r} − C_{d}B_{r})}{α − β})x^{nd+r}.

The terms containing B_{r} simplifies to

D_{d}^{n}(α^{n+1} − β^{n+1}) + C_{d}D_{d}^{n−1}(α^{n} − β^{n}) = D_{d}^{n−1}(α^{n−1} − β^{n−1}).

Therefore, we arrive at

G_{r}(x) = ∑_{n=0}^{∞}(−D_{d})^{n−1}(\frac{α^{n} − β^{n}}{α − β} B_{d+r} − \frac{α^{n−1} − β^{n−1}}{α − β} B_{r})x^{nd+r}.

The theorem holds by comparing the coefficient of x^{nd+r}.

In particular, when r = −1, one observes that B_{nd−1}/B_{d−1} and B_{2nd−1}/B_{nd−1} behave like the Lucas sequence of the first kind and second kind, respectively. Hence, new identities can be discovered by utilizing known identities involving Fibonacci and Lucas numbers. For example, the identity L_{n}^{2} − 5F_{n}^{2} = 4(−1)^{n} yields the following identity, which can be verified directly:

\frac{B_{2nd−1}^{2}}{B_{nd−1}^{2}} − \frac{ΔB_{nd−1}^{2}}{B_{d−1}^{2}} − = 4(−D_{d})^{n}.

Remark 15. We also have another related identity

\frac{B_{2nd−1}^{2}}{B_{nd−1}^{2}} + \frac{ΔB_{nd−1}^{2}}{B_{d−1}^{2}} = \frac{2B_{4nd−1}^{2}}{B_{2nd−1}^{2}} .

3.2.1 Example: bi-periodic Fibonacci numbers

When d = 2, the recurrence relation for (B_{ν}) can be written as

B_{ν} = \begin{cases}b_{1}B_{ν−1} + a_{1}B_{ν−2}, if ν is odd; \\ b_{2}B_{ν−1} + a_{2}B_{ν−2}, if ν is even, \end{cases}

with initial values B_{−1} = 0 and B_{0} = 1. The first few terms of B_{ν} are

B_{1} = b_{1}, B_{2} = b_{1}b_{2} + a_{2}, B_{3} = b_{1}^{2}b_{2} + a_{1}b_{1} + a_{2}b_{1},
B_{4} = b_{1}^{2}b_{2}^{2} + a_{2}^{2} + 2a_{2}b_{1}b_{2} + a_{1}b_{1}b_{2}.

We calculate

C_{2} = \frac{B_{3}}{B_{1}} = b_{1}b_{2} + a_{1} + a_{2}, D_{2} = −a_{1}a_{2}.

Hence (B_{ν}) can be defined using a single recurrence relation

B_{ν} = (b_{1}b_{2} + a_{1} + a_{2})B_{ν−2} − a_{1}a_{2}B_{ν−4},

with initial values as stated above. Its generating function is

G(x) = \frac{1 + b_{1}x − a_{2}x^{2}}{1 − (b_{1}b_{2} + a_{1} + a_{2})x^{2} + a_{1}a_{2}x^{4}} ,

and its Binet formula is

B_{2n+r} = (a_{1}a_{2})^{n−1}(\frac{α^{n} − β^{n}}{α − β} B_{r+2} − \frac{α^{n−1} − β^{n−1}}{α − β} B_{r}).

where α and β are the positive and negative roots of a_{1}a_{2}z^{2} − (b_{1}b_{2} + a_{1} + a_{2})z + 1 = 0, respectively.

Consider the convergents to the continued fraction of \sqrt{8} = [2, \overline{1, 4}]. The denominators of these convergents are given by the sequence 1, 1, 5, 6, 29, 35, 169, 204, 985, … (see \underline{A041011} in the On-Line Encyclopedia of Integer Sequences (OEIS)). This sequence is generated by the following second-order linear recurrence relation:

B_{ν} = 6B_{ν−2} − B_{ν−4}.

Its generating function is

G(x) = \frac{1 + x − x^{2}}{1 − 6x^{2} + x^{4}} .

The roots of 1 − 6x + x^{2} = 0 are 3 ± 2\sqrt{2}. Thus, the Binet formula is given by

B_{2n+r} = \frac{(3 + 2\sqrt{2})^{n} − (3 − 2\sqrt{2})^{n}}{4\sqrt{2}} B_{r+2} − \frac{(3 + 2\sqrt{2})^{n−1} − (3 − 2\sqrt{2})^{n−1}}{4\sqrt{2}} B_{r}.

We will return to this example a few more times throughout this paper.

3.3 Limits

We use Theorem 14 to compute the limit of consecutive term, which we need for later. Observe that, |α| < |β|. Hence we deduce the next two theorems. Theorem 16. Let α and β be as defined in Theorem 14. For integer r ≥ 0, we have

lim_{n→∞} \frac{B_{nd+r}}{B_{nd+r−1}} = \frac{βB_{d+r} − B_{r}}{βB_{d+r−1} − B_{r−1}} .

Proof. Take n ⟶ ∞ in the expression below

\frac{B_{nd+r}}{B_{nd+r−1}} = \frac{(α(α/β)^{n−1} − β)B_{d+r} − ((α/β)^{n−1} − 1)B_{r}}{(α(α/β)^{n−1} − β)B_{d+r−1} − ((α/β)^{n−1} − 1)B_{r−1}} ,

and use the fact that (α/β)^{n} ⟶ 0. Theorem 17. Let α and β be as defined in Theorem 14. For integer r ≥ −1, we have

lim_{n→∞} \frac{B_{(n+1)d+r}}{B_{nd+r}} = −D_{d}β.

Proof. Take n ⟶ ∞ in the expression below

\frac{B_{(n+1)d+r}}{B_{nd+r}} = \frac{−D_{d}(α^{2}(α/β)^{n−1} − β^{2})B_{d+r} − (α(α/β)^{n−1} − β)B_{r}}{(α(α/β)^{n−1} − β)B_{d+r} − ((α/β)^{n−1} − 1)B_{r}} ,

and use the fact that (α/β)^{n} ⟶ 0.

3.4 Extending to negative indices

Replacing n with −n in Binet’s formula (Theorem 14) yields the next theorem. Theorem 18. For integers n and r such that n ≥ 0 and r ≥ −1, we have

B_{−nd+r} = −(−D_{d})^{n−1}(\frac{α^{n} − β^{n}}{(−D_{d})^{n}(α − β)}B_{d+r} − \frac{α^{n+1} − β^{n+1}}{(−D_{d})^{n−1}(α − β)}B_{r}).

In particular, when r = −1, (−D_{d})^{n}B_{−nd−1} = −B_{nd−1}.

3.5 Some curious telescoping sums

Many interesting sums involving Fibonacci numbers arise from telescoping sums. We demonstrate here some of the analogous sums based on continuants.

3.5.1 Telescoping sum 1 When d = 2, we have encountered this sequence before. It has the form

B_{ν} = \begin{cases}b_{1}B_{ν−1} + a_{1}B_{ν−2}, if ν is odd; \\ b_{2}B_{ν−1} + a_{2}B_{ν−2}, if ν is even, \end{cases}

with initial values B_{−1} = 0 and B_{0} = 1. The next theorem is a continuant version of the Millin series [13] for Fibonacci numbers. Theorem 19. If d = 2, then

∑_{n=1}^{∞} \frac{(a_{1}a_{2})^{2^{n−1}}}{B_{2^{n+1}−1}} = \frac{1}{b_{1}β} .

Proof. By choosing λ = 2^{n+1} and ν = 2^{n} with n ≥ 1 in Corollary 9, we get

\frac{B_{2^{n}}}{B_{2^{n}−1}} − \frac{B_{2^{n+1}}}{B_{2^{n+1}−1}} = \frac{(a_{1}a_{2})^{2^{n−1}}}{B_{2^{n+1}−1}} .

The LHS is telescoping, hence summing from 1 to N, we have

\frac{B_{2}}{B_{1}} − \frac{B_{2^{N+1}}}{B_{2^{N+1}−1}} = ∑_{n=1}^{N} \frac{(a_{1}a_{2})^{2^{n−1}}}{B_{2^{n+1}−1}} .

When N ⟶ ∞, the second term on the LHS approaches a finite limit

\frac{βB_{2} − 1}{βB_{1}},

by Theorem 16. Therefore,

∑_{n=1}^{∞} \frac{(a_{1}a_{2})^{2^{n−1}}}{B_{2^{n+1}−1}} = \frac{B_{2}}{B_{1}} − \frac{βB_{2} − 1}{βB_{1}} = \frac{1}{b_{1}β} .

In the case of \sqrt{8}, this sum becomes

∑_{n=1}^{∞} \frac{1}{B_{2^{n+1}−1}} = \frac{1}{6} + \frac{1}{204} + \frac{1}{235416} + ⋯ = \frac{1}{3 + 2\sqrt{2}} = 3 − 2\sqrt{2}.

3.5.2 Telescoping sum 2

For general d, (B_{ν}) has the form

B_{ν} = b_{r}B_{ν−1} + a_{r}B_{ν−2}, where ν ≡ r (mod d),

for the appropriate initial values.

Theorem 20. If the period is d, then

∑_{n=1}^{∞} \frac{(−D_{d})^{n−1}}{B_{nd−1}B_{(n+1)d−1}} = \frac{α}{B_{d−1}^{2}} .

Proof. By choosing λ = μ = d and ν = nd with n ≥ 1, Theorem 1 becomes

\frac{B_{(n+1)d−1}}{B_{nd−1}} − \frac{B_{(n+2)d−1}}{B_{(n+1)d−1}} = \frac{(−1)^{nd}(a_{1} ⋯ a_{d})^{n}B_{d−1}^{2}}{B_{nd−1}B_{(n+1)d−1}} .

The LHS is telescoping, so summing from 1 to N, we get

\frac{B_{2d−1}}{B_{d−1}} − \frac{B_{(N+2)d−1}}{B_{(N+1)d−1}} = ∑_{n=1}^{N} \frac{(−1)^{nd}(a_{1} ⋯ a_{d})^{n}B_{d−1}^{2}}{B_{nd−1}B_{(n+1)d−1}} .

When N ⟶ ∞, the second term on the LHS approaches β by Theorem 17. Therefore,

∑_{n=1}^{∞} \frac{(−1)^{nd}(a_{1} ⋯ a_{d})^{n}B_{d−1}^{2}}{B_{nd−1}B_{(n+1)d−1}} = C_{d} + D_{d}β = −D_{d}α.

This proves the theorem after simplifications.

It is a known fact that, for a positive integer N, the continued fraction of \sqrt{N} is periodic (refer to Khrushchev’s book [10] for a proof), say with period d. When (B_{ν}) represents the denominators of the convergents to \sqrt{N}, we have C_{d} = B_{2d−1}/B_{d−1} = 2A_{d−1} and D_{d} = (−1)^{d−1}. This implies that α = (−1)^{d}A_{d−1} − (−1)^{d}\sqrt{A_{d−1}^{2} − (−1)^{d}}. Furthermore, there is an interesting interpretation of the number B_{nd−1} as a solution to the Pell’s equation x^{2} − Ny^{2} = 1, as stated in the next two theorems.

Theorem 21. The fundamental (minimal) solution to x^{2} − Ny^{2} = 1 is

(x_{1}, y_{1}) = \begin{cases}(A_{d−1}, B_{d−1}), & if d is even; \\ (A_{2d−1}, B_{2d−1}), & if d is odd. \end{cases}

Theorem 22. All the solutions to x^{2} − Ny^{2} = 1 are given by

(x_{n}, y_{n}) = \begin{cases}(A_{nd−1}, B_{nd−1}), & if d is even; \\ (A_{2nd−1}, B_{2nd−1}), & if d is odd, \end{cases}

where n is a positive integer.

Proof. See Khrushchev’s book [10] for proofs.

Hence our telescoping sum can be reinterpreted as the next corollary. Corollary 23. If d is even and (x_{n}, y_{n}) is the n-th solution to x^{2} − Ny^{2} = 1, then

∑_{n=1}^{∞} \frac{1}{y_{n}y_{n+1}} = \frac{x_{1} − \sqrt{x_{1}^{2} − 1}}{y_{1}^{2}}.

In the case when N = 8, the sum becomes

∑_{n=1}^{∞} \frac{1}{y_{n}y_{n+1}} = \frac{1}{6} + \frac{1}{210} + \frac{1}{7140} + ⋯ = 3 − \sqrt{8}.

3.5.3 Telescoping sum 3 We assume for this part that (B_{ν}) is the sequence of denominators of the convergents to \sqrt{N}. In this case, a_{n} = 1 is constant and D_{d} = (−1)^{d−1}. The next corollary is similar to the previous one. Corollary 24. If d is even and (x_{n}, y_{n}) is the n-th solution to x^{2} − Ny^{2} = 1, then

∑_{n=1}^{∞} \frac{1}{x_{n}x_{n+1}} = \frac{x_{1} − \sqrt{x_{1}^{2} − 1}}{x_{1}\sqrt{x_{1}^{2} − 1}}.

Proof. If we choose λ = μ = d and ν = (n − 1)d with n ≥ 1, then Theorem 1 becomes

\frac{B_{nd−1}}{A_{(n+1)d−1}} − \frac{B_{(n−1)d−1}}{A_{nd−1}} = \frac{(−1)^{(n−1)d}(a_{1} ⋯ a_{d})^{n−1}A_{d−1}B_{d−1}}{A_{nd−1}A_{(n+1)d−1}} .

The LHS is telescoping, so summing from 1 to N, we get

\frac{B_{Nd−1}}{A_{(N+1)d−1}} − \frac{B_{−1}}{A_{d−1}} = \frac{B_{Nd−1}}{A_{(N+1)d−1}} = ∑_{n=1}^{N} \frac{(−1)^{(n−1)d}(a_{1} ⋯ a_{d})^{n−1}A_{d−1}B_{d−1}}{A_{nd−1}A_{(n+1)d−1}} .

When N ⟶ ∞, using the same procedure as in Theorem 17, one checks that

\frac{B_{Nd−1}}{A_{(N+1)d−1}} → −\frac{2B_{d−1}}{D_{d}^{2}β(α − β)}.

Therefore,

∑_{n=1}^{∞} \frac{(−1)^{(n−1)d}(a_{1} ⋯ a_{d})^{n−1}A_{d−1}B_{d−1}}{A_{nd−1}A_{(n+1)d−1}} = −\frac{2B_{d−1}}{D_{d}^{2}β(α − β)},

which can be simplified to

∑_{n=1}^{∞} \frac{(−D_{d})^{n}}{A_{nd−1}A_{(n+1)d−1}} = −\frac{2α}{A_{d−1}(α − β)}.

We get our conclusion by noting that D_{d} = −1 and α − β = −2\sqrt{A_{d−1}^{2} − 1}.

In the case when N = 8, we have the sum

∑_{n=1}^{∞} \frac{1}{x_{n}x_{n+1}} = \frac{1}{51} + \frac{1}{1683} + \frac{1}{57123} + ⋯ = \frac{3\sqrt{2} − 4}{12}.

3.5.4 Telescoping sum 4 Let (x_{n}) and (y_{n}) be any sequences . Consider the identity, which can be verified directly,

\frac{4(x_{n+1}y_{n} − x_{n}y_{n+1})(x_{n}x_{n+1} − y_{n}y_{n+1})}{(x_{n} − y_{n})^{2}(x_{n+1} − y_{n+1})^{2}} = \frac{(x_{n} + y_{n})^{2}}{(x_{n} − y_{n})^{2}} − \frac{(x_{n+1} + y_{n+1})^{2}}{(x_{n+1} − y_{n+1})^{2}} .

We use this identity to prove the next theorem. Theorem 25. If the period is d, then

∑_{n=1}^{∞} \frac{(−D_{d})^{n−1}B_{(2n+1)d−1}}{B_{nd−1}^{2}B_{(n+1)d−1}^{2}} = \frac{1}{B_{d−1}^{3}} .

Proof. Choose x_{n} = α^{n} and y_{n} = β^{n} in the identity above to get

\frac{4(α − β)(α^{2n+1} − β^{2n+1})}{(−D_{d})^{n}(α^{n} − β^{n})^{2}(α^{n+1} − β^{n+1})^{2}} = \frac{(α^{n} + β^{n})^{2}}{(α^{n} − β^{n})^{2}} − \frac{(α^{n+1} + β^{n+1})^{2}}{(α^{n+1} − β^{n+1})^{2}} .

The RHS is telescoping, so summing from n = 1 to N, we obtain

∑_{n=1}^{N} \frac{4(α − β)(α^{2n+1} − β^{2n+1})}{(−D_{d})^{n}(α^{n} − β^{n})^{2}(α^{n+1} − β^{n+1})^{2}} = \frac{(α + β)^{2}}{(α − β)^{2}} − \frac{(α^{N+1} + β^{N+1})^{2}}{(α^{N+1} − β^{N+1})^{2}} .

Taking N ⟶ ∞, the second term on RHS approaches 1. Therefore,

∑_{n=1}^{∞} \frac{4(α − β)(α^{2n+1} − β^{2n+1})}{(−D_{d})^{n}(α^{n} − β^{n})^{2}(α^{n+1} − β^{n+1})^{2}} = \frac{(α + β)^{2}}{(α − β)^{2}} − 1.

Rewriting this equation in terms of B_{ν}, we obtain

∑_{n=1}^{N} \frac{4(−D_{d})^{n−2}B_{(2n+1)d−1}B_{d−1}^{3}}{B_{nd−1}^{2}B_{(n+1)d−1}^{2}} = \frac{C_{d}^{2}}{D_{d}^{2}} − \frac{Δ}{D_{d}^{2}} .

We conclude our theorem upon simplification of this last equation.

Suppose that (B_{ν}) are the denominators of the convergents to \sqrt{N} for some positive integer N. The next corollary is immediate. Corollary 26. If d is even and (x_{n}, y_{n}) are the n-th solution to x^{2} − Ny^{2} = 1, then

∑_{n=1}^{∞} \frac{y_{2n+1}}{y_{n}^{2}y_{n+1}^{2}} = \frac{1}{y_{1}^{3}} .

When N = 8, we have

∑_{n=1}^{∞} \frac{y_{2n+1}}{y_{n}^{2}y_{n+1}^{2}} = \frac{35}{1^{2} · 6^{2}} + \frac{1189}{6^{2} · 35^{2}} + \frac{40391}{35^{2} · 204^{2}} + ⋯ = 1.

3.5.5 Telescoping sum 5 Suppose that D_{d} = 1, so

B_{(n+1)d−1} − B_{(n−1)d−1} = C_{d}B_{nd−1}.

Motivated by the identity F_{n}^{2} − F_{n+1}F_{n−1} = (−1)^{n−1} for Fibonacci numbers, we deduce

\frac{B_{nd−1}^{2}}{B_{d−1}^{2}} − \frac{B_{(n+1)d−1}}{B_{d−1}} · \frac{B_{(n−1)d−1}}{B_{d−1}} = (−1)^{n−1},

where n ≥ 1. The application of this identity yields the next result. Theorem 27. If D_{d} = 1, then

∑_{n=1}^{∞} tan^{−1} \frac{B_{2d−1}}{B_{(2n+1)d−1}} = tan^{−1} \frac{B_{d−1}}{B_{2d−1}} .

Proof. Using the addition formula for tan^{−1} (see §3.6.1),

tan^{−1} \frac{B_{d−1}}{B_{2nd−1}} − tan^{−1} \frac{B_{d−1}}{B_{2(n+1)d−1}} = tan^{−1} \frac{B_{d−1}(B_{2(n+1)d−1} − B_{2nd−1})}{B_{d−1}^{2} + B_{2(n+1)d−1}B_{2nd−1}}

= tan^{−1} \frac{B_{2d−1}}{B_{(2n+1)d−1}} .

The LHS is telescoping, so summing from 1 to N, we obtain

tan^{−1} \frac{B_{d−1}}{B_{2d−1}} − tan^{−1} \frac{B_{d−1}}{B_{2(N+1)d−1}} = ∑_{n=1}^{N} tan^{−1} \frac{B_{2d−1}}{B_{(2n+1)d−1}} .

Taking N ⟶ ∞, the second term on the LHS vanishes, so

∑_{n=1}^{∞} tan^{−1} \frac{B_{2d−1}}{B_{(2n+1)d−1}} = tan^{−1} \frac{B_{d−1}}{B_{2d−1}} .

Remark 28. This theorem generalizes an earlier theorem by Hoggatt and Bruggles [9, Theorem 4].

Using a similar construction, we can get a corresponding sum for tanh^{−1}.

Theorem 29. If D_{d} = 1, then

∑_{n=2}^{∞} tanh^{−1} \frac{B_{2d−1}}{B_{2nd−1}} = \frac{1}{2} ln \frac{B_{3d−1} + B_{d−1}}{B_{3d−1} − B_{d−1}} .

Proof. Using the addition formula for tanh^{−1} (see §3.6.2),

tanh^{−1} \frac{B_{d−1}}{B_{(2n−1)d−1}} − tanh^{−1} \frac{B_{d−1}}{B_{(2n+1)d−1}} = tanh^{−1} \frac{B_{d−1}(B_{(2n+1)d−1} − B_{(2n−1)d−1})}{B_{(2n+1)d−1}B_{(2n−1)d−1} − B_{d−1}^{2}}

= tanh^{−1} \frac{B_{2d−1}}{B_{2nd−1}} .

The LHS is telescoping, so summing from 2 to N (tanh^{−1} 1 is undefined), we obtain

tanh^{−1} \frac{B_{d−1}}{B_{3d−1}} − tanh^{−1} \frac{B_{d−1}}{B_{(2N+1)d−1}} = ∑_{n=2}^{N} tanh^{−1} \frac{B_{2d−1}}{B_{2nd−1}} .

Taking N ⟶ ∞, the second term on the LHS vanishes, so

∑_{n=2}^{∞} tanh^{−1} \frac{B_{2d−1}}{B_{2nd−1}} = tanh^{−1} \frac{B_{d−1}}{B_{3d−1}} = \frac{1}{2} ln \frac{B_{3d−1} + B_{d−1}}{B_{3d−1} − B_{d−1}} ,

where we used the fact that

tanh^{−1} x = \frac{1}{2} ln \frac{1 + x}{1 − x}.

Let us apply these theorems to the continued fraction of \sqrt{2} = [1, \overline{2}]. The denominators, (B_{ν}), of the convergents to \sqrt{2} are 1, 2, 5, 12, 29, 70, 169, 408, 985, . . .. Each B_{ν} corresponds to the (ν + 1)-th term of the Pell number (see \underline{A000129} in the OEIS). The theorems above imply that

∑_{n=1}^{∞} tan^{−1} \frac{2}{B_{2n}} = tan^{−1} \frac{2}{5} + tan^{−1} \frac{2}{29} + tan^{−1} \frac{2}{169} + ⋯ = tan^{−1} \frac{1}{2},

and

∑_{n=2}^{∞} tanh^{−1} \frac{2}{B_{2n−1}} = tanh^{−1} \frac{1}{6} + tanh^{−1} \frac{1}{35} + tanh^{−1} \frac{1}{204} + ⋯ = tanh^{−1} \frac{1}{5} = \frac{1}{2} ln \frac{3}{2}.

3.6 More series involving tan^{−1} and tanh^{−1}

Besides the telescoping sums in the last section, another set of series can be generated using tan^{−1} and tanh^{−1}. The results in this section are inspired by the papers of Castellanos [4, 5]. However, Castellanos’ original method using Chebyshev polynomials is cumbersome. Therefore, we provide a direct method that is applicable to his results, while generating new ones.

3.6.1 Series involving tan^{−1}

Let ζ be a number to be determined later. Using the addition formula for tan^{−1},

tan^{−1}(x − y) = \frac{tan^{−1} x − tan^{−1} y}{1 + tan^{−1} x tan^{−1} y} ,

and the Gregory series,

tan^{−1} z = ∑_{k=0}^{∞} \frac{(−1)^{k}z^{2k+1}}{2k + 1} ,

we get

tan^{−1} \frac{(α − β)ζ}{αβ + ζ^{2}} = tan^{−1} \frac{ζ}{β} − tan^{−1} \frac{ζ}{α} = ∑_{k=0}^{∞} \frac{(−1)^{k}(α^{2k+1} − β^{2k+1})}{(2k + 1)(αβ)^{2k+1}}ζ^{2k+1}.

Since the series expansion for tan^{−1} z is only valid for |z| < 1, we require that |ζ| < min(|α|, |β|) = |α|. The next two theorems are direct consequences of the last equation.

Theorem 30. Let ζ be a root of D_{d}z^{2} − \sqrt{3Δ}z − 1 = 0 that satisfies |ζ| < α. Then

\frac{πB_{d−1}}{6\sqrt{Δ}} = ∑_{k=0}^{∞} \frac{(−1)^{k+1}B_{(2k+1)d−1}}{2k + 1} ζ^{2k+1}.

Proof. We choose ζ such that

\frac{(α − β)ζ}{αβ + ζ^{2}} = \frac{1}{\sqrt{3}}.

We obtain our theorem upon simplification and using the fact tan^{−1}(1/\sqrt{3}) = π/6.

Theorem 31. Let ζ be a root of D_{d}z^{2} − (\sqrt{2} − 1)\sqrt{Δ}z − 1 = 0 that satisfies |ζ| < α. Then

\frac{πB_{d−1}}{8\sqrt{Δ}} = ∑_{k=0}^{∞} \frac{(−1)^{k+1}B_{(2k+1)d−1}}{2k + 1} ζ^{2k+1}.

Proof. We choose ζ such that

\frac{(α − β)ζ}{αβ + ζ^{2}} = \sqrt{2} − 1.

We obtain our theorem upon simplification and using the fact tan^{−1}(\sqrt{2} − 1) = π/8.

3.6.2 Series involving tanh^{−1}

We also have similar addition formula for tanh^{−1},

tanh^{−1}(x − y) = \frac{tanh^{−1} x − tanh^{−1} y}{1 − tanh^{−1} x tanh^{−1} y} ,

and Taylor series expansion,

tanh^{−1} z = ∑_{k=0}^{∞} \frac{z^{2k+1}}{2k + 1}.

So one expects similar series using tanh^{−1}. Again, take ζ as a number to be determined later, then

tanh^{−1} \frac{(α − β)ζ}{αβ − ζ^{2}} = tanh^{−1} \frac{ζ}{β} − tanh^{−1} \frac{ζ}{α} = ∑_{k=0}^{∞} \frac{α^{2k+1} − β^{2k+1}}{(2k + 1)(αβ)^{2k+1}} ζ^{2k+1}.

The series expansion for tanh^{−1} z is also valid for |z| < 1, so we have the same inequality condition |ζ| < |α|. The previous equation gives the next two theorems.

Theorem 32. Let ζ be a root of D_{d}z^{2} + 2\sqrt{Δ}z + 1 = 0 that satisfies |ζ| < α. Then

\frac{B_{d−1}}{2\sqrt{Δ}} ln 3 = − ∑_{k=0}^{∞} \frac{B_{(2k+1)d−1}}{2k + 1} ζ^{2k+1}.

Proof. We choose ζ such that

\frac{(α − β)ζ}{αβ − ζ^{2}} = \frac{1}{2}.

We obtain our theorem upon simplification and using the fact 2 tanh^{−1}(1/2) = ln 3.

Theorem 33. Let ζ be a root of D_{d}z^{2} + 3\sqrt{Δ}z + 1 = 0 that satisfies |ζ| < α. Then

\frac{B_{d−1}}{2\sqrt{Δ}} ln 2 = − ∑_{k=0}^{∞} \frac{B_{(2k+1)d−1}}{2k + 1} ζ^{2k+1}.

Proof. We choose ζ such that

\frac{(α − β)ζ}{αβ − ζ^{2}} = \frac{1}{3}.

We obtain our theorem upon simplification and using the fact 2 tanh^{−1}(1/3) = ln 2.

3.6.3 Example (continued) For \sqrt{8}, the first two theorems give

\frac{π}{24\sqrt{2}} = ∑_{k=0}^{∞} \frac{(−1)^{k+1}B_{4k+1}}{2k + 1} (2\sqrt{6} − 5)^{2k+1},
\frac{π}{32\sqrt{2}} = ∑_{k=0}^{∞} \frac{(−1)^{k+1}B_{4k+1}}{2k + 1} (\sqrt{23 + 16\sqrt{2}} − 2\sqrt{2} − 4)^{2k+1}.

On the other hand, the latter two theorems give

\frac{ln 3}{8\sqrt{2}} = ∑_{k=0}^{∞} \frac{B_{4k+1}}{2k + 1}(\sqrt{33} − 4\sqrt{2})^{2k+1},
\frac{ln 2}{8\sqrt{2}} = ∑_{k=0}^{∞} \frac{B_{4k+1}}{2k + 1}(\sqrt{73} − 6\sqrt{2})^{2k+1}.

3.7 Sums involving continuants

The Binet formula enables us to calculate an intriguing sum of involving B_{ν}, as stated in the next theorem. Theorem 34. Let x ≠ α, β. For integer r ≥ −1, we have

∑_{n=1}^{N} x^{n}B_{nd+r} = \frac{x^{N+1}(B_{(N+1)d+r}/D_{d} + xB_{Nd+r}) − x(B_{d+r}/D_{d} + xB_{r})}{(x − α)(x − β)}.

Proof. It is straightforward to check that

∑_{n=1}^{N}(xD_{d})^{n} \frac{α^{n} − β^{n}}{α − β} = \frac{x(xD_{d})^{N+1}(α^{N} − β^{N}) + (xD_{d})^{N+1}(α^{N+1} − β^{N+1}) − xD_{d}(α − β)}{(α − β)(x^{2}D_{d} − xC_{d} − 1)},
∑_{n=0}^{N−1}(xD_{d})^{n} \frac{α^{n} − β^{n}}{α − β} = \frac{x(xD_{d})^{N+1}(α^{N−1} − β^{N−1}) + (xD_{d})^{N}(α^{N} − β^{N}) − xD_{d}(α − β)}{(α − β)(x^{2}D_{d} − xC_{d} − 1)}.

Hence, by applying Binet’s formula (Theorem 14), we get

∑_{n=1}^{N} x^{n}B_{nd+r} = ∑_{n=1}^{N} ((−D_{d})^{−1}(−xD_{d})^{n} \frac{α^{n} − β^{n}}{α − β} B_{d+r} − x(−xD_{d})^{n−1} \frac{α^{n−1} − β^{n−1}}{α − β} B_{r})

= \frac{−x^{N+1}B_{(N+1)d+r} − x^{N+2}D_{d}B_{Nd+r} + xB_{d+r} + x^{2}D_{d}B_{r}}{1 + xC_{d} − x^{2}D_{d}}

= \frac{x^{N+1}(B_{(N+1)d+r}/D_{d} + xB_{Nd+r}) − x(B_{d+r}/D_{d} + xB_{r})}{(x − α)(x − β)}.

Remark 35. One could also apply telescoping sum directly to the identity

(x − α)(x − α)B_{kd+r} = x^{k+1}(B_{(k+1)d+r}/D_{d} + xB_{kd+r}) − x^{k}(B_{kd+r}/D_{d} + xB_{(k−1)d+r}), although our derivation gives an explanation for this identity.

For the next series, we take r = −1. Theorem 36. We have

∑_{n=1}^{N} \binom{N}{n}(−1)^{n}(α + β)^{n}B_{nd−1} = \frac{B_{2Nd−1}}{D_{d}^{N}} .

Proof. Note that 1 − C_{d}z = Dz^{2} for z = α and β. Hence

∑_{n=1}^{N} \binom{N}{n}(−C_{d})^{n} \frac{α^{n} − β^{n}}{α − β} = \frac{(1 − C_{d}α)^{N} − (1 − C_{d}β)^{N}}{α − β} = D_{d}^{N} \frac{α^{2N} − β^{2N}}{α − β} .

The sum in question is then

∑_{n=1}^{N} \binom{N}{n}(−1)^{n}(α + β)^{n}B_{nd−1} = ∑_{n=1}^{N} \binom{N}{n}(\frac{C_{d}}{D_{d}} )^{n}B_{nd−1}

= −D_{d}^{N−1} \frac{α^{2N} − β^{2N}}{α − β} B_{d−1} = \frac{B_{2Nd−1}}{D_{d}^{N}} .

3.7.1 Example (continued) Returning to the example of \sqrt{8}, we get, for x ≠ 3 ± 2\sqrt{2}, the sums

∑_{n=1}^{N} x^{n}B_{2n+r} = \frac{x^{N+1}(−B_{2(N+1)+r} + xB_{2N+r}) − x(−B_{2+r} + xB_{r})}{(x − 3 + 2\sqrt{2})(x − 3 − 2\sqrt{2})},

and

∑_{n=1}^{N} \binom{N}{n}(−6)^{n}B_{2n−1} = (−1)^{N}B_{4N−1}.

3.8 Divisibility properties

In this section, we explore some of the divisibility behaviors of the continuants. For similar results associated with Lucas sequences, refer to Ribenboim’s book [17].

3.8.1 Divisibility within the sequence

We saw in the proof of Theorem 10 that B_{d−1} | B_{2d−1}. This can be further extended to the following theorem.

Theorem 37. If m | n, then B_{md−1} | B_{nd−1}.

Proof. Using Corollary 2, when ν = nd − 1, λ = md, then

B_{(m+n)d−1} = B_{d}B_{nd−1} + a_{1}B_{md−1}B_{md−2,1}.

We get our conclusion by an inductive argument, which is omitted here.

Stated differently, this theorem implies that the sequence (B′_{n}) = (B_{nd−1}) is a divisibility sequence. In fact, it is a strong divisibility sequence, i.e., B′_{gcd(m,n)} = gcd(B′_{m}, B′_{n}). One can prove this using the same technique for Fibonacci sequence, so we leave the proof as an exercise.

3.8.2 Divisibility by primes

Let p be an odd prime. If p | C_{d} and p | D_{d}, then p | B_{nd+r} (this also holds for 2). For the case of p | C_{d} and p ∤ D_{d}, the result is given in the next theorem.

Theorem 38. If p | C_{d}, p ∤ D_{d}, and n ≥ 2, then

B_{nd+r} = \begin{cases}(−1/2)^{n−2}D_{d}Δ^{(n−2)/2}B_{r}, & if n is even; \\ (−1/2)^{n−1}Δ^{(n−1)/2}B_{d+r}, & if n is odd, \end{cases} (mod p).

In particular, B_{nd−1} ≡ 0 (mod p) when n is even.

Proof. We expand the term (α^{n} − β^{n})/(α − β) in Binet’s formula (Theorem 14),

\frac{α^{n} − β^{n}}{α − β} = \frac{1}{(2D_{d})^{n−1}} ∑_{k=0}^{⌊(n−1)/2⌋} \binom{n}{2k + 1}(−C_{d})^{n−2k−1}Δ^{k}.

If p | C_{d} and p ∤ D_{d}, then

(−D_{d})^{n−1} \frac{α^{n} − β^{n}}{α − β} = \begin{cases}0, & if n is even; \\ Δ^{(n−1)/2}/2^{n−1}, & if n is odd, \end{cases} (mod p).

Hence the theorem follows.

If p ∤ C_{d} and p | D_{d}, then Δ ≡ C_{d}^{2} (mod p). Using this fact, we conclude the following theorem.

Theorem 39. If p ∤ C_{d} and p | D_{d}, then

B_{(p−1)d+r} ≡ B_{r}, B_{(p−1)d−1} ≡ 0 (mod p)

Proof. When p | D_{d}, the term involving B_{r} vanishes modulo p. Consider again the expansion of (α^{n} − β^{n})/(α − β),

2^{p−1}B_{pd+r} ≡ ∑_{k=0}^{(p−1)/2} \binom{p}{2k + 1}(−C_{d})^{p−1}B_{d+r} (mod p).

Since \binom{p}{k} ≡ 0 (mod p) for 1 ≤ k ≤ p − 1, this simplifies to

B_{pd+r} ≡ B_{d+r} (mod p),

and the theorem follows.

Suppose now that p ∤ C_{d}D_{d}. By setting n = p + 1 in the expansion of (α^{n} − β^{n})/(α − β), we get

2^{p}B_{(p+1)d+r} = ∑_{k=0}^{(p−1)/2} \binom{p + 1}{2k + 1}C_{d}^{p−2k}Δ^{k}B_{d+r} + 2D_{d} ∑_{k=0}^{(p−1)/2} \binom{p}{2k + 1}C_{d}^{p−2k−1}Δ^{k}B_{r}.

Taking modulo p on both sides,

2B_{(p+1)d+r} ≡ C_{d}B_{d+r} + C_{d}Δ^{(p−1)/2}B_{d+r} + 2D_{d}Δ^{(p−1)/2}B_{r} (mod p).

If p | Δ, then

2B_{(p+1)d+r} ≡ C_{d}B_{d+r} (mod p).

This is equivalent to our next theorem.

Theorem 40. If p | Δ and p ∤ C_{d}D_{d}, then

2B_{pd+r} ≡ C_{d}B_{r}, B_{pd−1} ≡ 0 (mod p).

Otherwise, Δ^{(p−1)/2} ≡ (Δ | p) (mod p), where (· | p) denotes the Legendre symbol modulo p. Hence we obtain our next result.

Theorem 41. For an odd prime p ∤ C_{d}D_{d}, if Δ is a quadratic residue modulo p, then

B_{(p+1)d+r} ≡ C_{d}B_{d+r} + D_{d}B_{r}, B_{(p+1)d−1} ≡ C_{d}B_{d−1} (mod p).

If Δ is a quadratic non-residue modulo p, then

B_{(p+1)d+r} ≡ −D_{d}B_{r}, B_{(p+1)d−1} ≡ 0 (mod p).

On the other hand, if we take n = p with p ∤ C_{d}D_{d}, then

2^{p−1}B_{pd+r} = ∑_{k=0}^{(p−1)/2} \binom{p}{2k + 1}C_{d}^{p−2k−1}Δ^{k}B_{d+r} + 2D_{d} ∑_{k=0}^{(p−3)/2} \binom{p − 1}{2k + 1}C_{d}^{p−2k−2}Δ^{k}B_{r}.

Taking modulo p on both sides and using the fact that \binom{p−1}{k} ≡ (−1)^{k} (mod p), we get

B_{pd+r} ≡ Δ^{(p−1)/2}B_{d+r} − \frac{2D_{d}}{C_{d}} ∑_{k=0}^{(p−3)/2} (1 + \frac{4D_{d}}{C_{d}^{2}} )^{k}B_{r} (mod p)

≡ Δ^{(p−1)/2}B_{d+r} + \frac{C_{d}}{2} (1 − Δ^{(p−1)/2})B_{r} (mod p).

Thus, we can deduce the following theorem.

Theorem 42. For an odd prime p ∤ C_{d}D_{d}, if Δ is a quadratic residue modulo p, then

B_{(p−1)d+r} ≡ B_{r}, B_{(p−1)d−1} ≡ 0 (mod p).

If Δ is a quadratic non-residue modulo p, then

B_{pd+r} ≡ C_{d}B_{r} − B_{d+r}, B_{pd−1} ≡ −B_{d−1} (mod p).

Remark 43. In particular, if d = 1, C_{d} = (a + 1) and D_{d} = −a for some integer a, then Δ = (a + 1)^{2} − 4a = (a − 1)^{2}. For any prime p ∤ C_{d}D_{d}Δ, then (Δ | p) = 1 and

p | B_{p−2} = a^{p−2} \frac{a^{−(p−1)} − 1}{a^{−1} − 1} = \frac{a^{p−1} − 1}{a − 1}.

Therefore, p | a^{p−1} − 1 (this is trivial for p | C_{d}D_{d}Δ), which is Fermat’s little theorem.

Theorems 40 to 42 suggest a certain periodicity of B_{ν} modulo p. This period is commonly referred to as the Pisano period modulo p. While we do not delve into the full theory here, we will demonstrate a minor consequence of these theorems. Set σ_{p}(n) as the order of n modulo p, i.e., the smallest positive integer k such that n^{k} ≡ 1 (mod p). The next corollary gives the Pisano period modulo p.

Corollary 44. The Pisano period of (B_{ν}) modulo p, where p ∤ C_{d}D_{d}, divides

pdσ_{p}(C_{d}/2),  if p | Δ;
(p − 1)d,  if Δ is a quadratic residue modulo p;
(p + 1)dσ_{p}(−D_{d}),  if Δ is a quadratic non-residue modulo p.

Proof. If p | Δ, then Theorem 40 implies that

B_{pdσ_{p}(C_{d}/2)+r} ≡ (C_{d}/2)^{σ_{p}(C_{d}/2)}B_{r} ≡ B_{r} (mod p).

Thus, the Pisano period divides pdσ_{p}(C_{d}/2). The other cases are similar.

Remark 45. For a more detailed account of Pisano period involving Fibonacci numbers, refer to the papers by Falcón [7] and Wall [20].

3.8.3 Example (continued)

Using the example of \sqrt{8} as before, we calculate Δ = 32. We choose p = 3 and 7, since 32 is a quadratic non-residue modulo 3 and a quadratic residue modulo 7. Theorems 41 and 42 give us the following results, respectively:

B_{r+8} ≡ B_{r} (mod 3), B_{r+16} ≡ 6B_{r+2} − B_{r} (mod 7),

and

B_{r+6} ≡ 6B_{r} − B_{r+2} (mod 3), B_{r+12} ≡ B_{r} (mod 7).

3.9 Law of apparition and repetition

3.9.1 Law of apparition

The results in §3.8.2 resemble the law of apparition of Lucas sequences. For any prime p, define ω(p) as the rank of apparition of p, which is the minimum positive integer k such that B_{kd−1} ≡ 0 (mod p). The next theorem summarizes what we know about ω(p).

Theorem 46 (Law of apparition). For a prime p, we have

(i) If p | C_{d} and p | D_{d}, then ω(p) = 1.

(ii) When p = 2, the following results hold:

• If 2 ∤ C_{d} and 2 | D_{d}, then

ω(2) = \begin{cases}does not exist, & if B_{d−1} ≡ 1 (mod 2); \\ 1, & if B_{d−1} ≡ 0 (mod 2). \end{cases}

• If 2 | C_{d} (or 2 | Δ) and 2 ∤ D_{d}, then ω(2) = 1 or 2.

• If 2 ∤ C_{d} and 2 ∤ D_{d}, then ω(2) = 1 or 3.

(iii) When p is an odd prime, the following results hold:

• If p | C_{d} and p ∤ D_{d}, then ω(p) = 1 or 2.

• If p ∤ C_{d} and p | D_{d}, then ω(p) | p − 1.

• If p ∤ C_{d}, p ∤ D_{d}, and p | Δ, then ω(p) = 1 or p.

• If p ∤ C_{d}, p ∤ D_{d}, and p ∤ Δ, then ω(p) | p − (Δ | p).

Proof. We have already dealt with the case of odd primes, so we can suppose p = 2. Using the fact B_{2d−1} = C_{d}B_{d−1}, we see that if p | C_{d} (or p | Δ) and p ∤ D_{d}, then B_{2d−1} = 0. Otherwise, if p ∤ C_{d} and p ∤ D_{d}, then

B_{3d−1} ≡ B_{2d−1} + B_{d−1} ≡ 2B_{d−1} ≡ 0 (mod 2).

However, if p ∤ C_{d}, but p | D_{d}, then for n ≥ 1,

B_{(n+1)d−1} ≡ B_{nd−1} (mod 2).

An inductive argument shows that B_{nd−1} ≡ B_{d−1} (mod 2) for all n ≥ 1. Thus, it is possible that ω(2) does not exists if B_{d−1} ≡ 1 (mod 2).

3.10 Lucas pseudoprime

Theorem 46 provides an extension to the notion of Lucas pseudoprime. For any positive integer n, by an abuse of notation, set ϵ(n) = (Δ | n) as the Jacobi symbol modulo n. We say that n is a Lucas pseudoprime with respect to (B_{ν}), if n ∤ C_{d}D_{d}Δ and

B_{(n−ϵ(n))d−1} ≡ 0 (mod n).

If the congruence is false, we can conclude that n is composite. However, if the congruence holds, it does not necessarily imply that n is prime and hence the term ‘pseudoprime’. For example, we take (B_{ν}) as the sequence of denominators of the convergent to \sqrt{8} again. Choose n = 35, then the Jacobi symbol is (32 | 35) = −1. On the other hand,

B_{71} = 641614773393652358999201580 ≡ 0 (mod 35).

Hence 35 is a pseudoprime with respect to (B_{ν}).

3.10.1 Law of repetition

For the next part, we require an identity that can be traced back to Lagrange. Given any odd integer m, we have

X^{m} − Y^{m} = ∑_{k=0}^{(m−1)/2} \frac{m}{k}\binom{m − k − 1}{k − 1}(XY)^{k}(X − Y )^{m−2k},

This gives us an expansion for the term B_{mnd−1}/B_{d−1} as

\frac{B_{mnd−1}}{B_{d−1}} = ∑_{k=0}^{(m−1)/2} \frac{m}{k}\binom{m − k − 1}{k − 1}Δ^{(m−2k−1)/2}(−D_{d})^{nk}((−D_{d})^{n−1} \frac{α^{n} − β^{n}}{α − β} )^{m−2k}.

We write p^{e} || N if e is the highest power of p dividing N. The next theorem gives the law of repetition.

Theorem 47 (Law of repetition). If p^{e} || B_{nd−1}/B_{d−1} and p ∤ m, then p^{e+f} | B_{p^{f} mnd−1}/B_{d−1}. In addition, if p ∤ D_{d}, then the power is exact.

Proof. By induction, it suffices to establish the theorem for f = 0 and 1. The previous identity gives

\frac{B_{mnd−1}}{B_{d−1}} ≡ m(−D_{d})^{n(m−1)/2} \frac{B_{nd−1}}{B_{d−1}} (mod (B_{nd−1}/B_{d−1})^{3}),
\frac{B_{pmnd−1}}{B_{d−1}} ≡ pm(−D_{d})^{n(pm−1)/2} \frac{B_{nd−1}}{B_{d−1}} (mod (B_{nd−1}/B_{d−1})^{3}).

Hence if p^{e} || B_{nd−1}/B_{d−1}, then p^{e} | B_{mnd−1}/B_{d−1} and p^{e+1} | B_{pmnd−1}/B_{d−1}. The second part is clear.

3.11 Non-integer sequences

While all the theorems presented above assume that (a_{ν}) and (b_{ν}) are integer sequences, there is no reason not to extend them to real sequences as well. The only casualties are that C_{d}, D_{d}, and consequently B_{ν} are not necessarily integers. However, most of our results still apply except those involving divisibility properties in §3.8. We demonstrate here an example. Consider the sequence

B_{ν} = \begin{cases}\sqrt{3}B_{ν−1} + B_{ν−2}, & if ν is odd; \\ B_{ν−1} − B_{ν−2}, & if ν is even. \end{cases}

The first few values of B_{ν} are

B_{1} = \sqrt{3}, B_{2} = \sqrt{3} − 1, B_{3} = 3, B_{4} = 4 − 2\sqrt{3}.

We compute α = −(\sqrt{3}−\sqrt{7})/2 and β = −(\sqrt{3}+\sqrt{7})/2, so using Binet’s formula (Theorem 14), we have

B_{4k+1} = \sqrt{\frac{3}{7}}(α^{2k−1} − β^{2k−1} − \sqrt{3}(α^{2k} − β^{2k})).

Plugging this value into Theorem 27, we obtain

∑_{n=1}^{∞} tan^{−1} \frac{3}{B_{4k+1}} = tan^{−1} \frac{\sqrt{3}}{4} + tan^{−1} \frac{\sqrt{3}}{19} + tan^{−1} \frac{\sqrt{3}}{91} + ⋯ = tan^{−1} \frac{1}{\sqrt{3}} = \frac{π}{6} .

4 Further discussion

The results presented in this paper primarily focus on the denominator, B_{ν,λ}. However, it should be possible to give similar results involving the numerator, A_{ν,λ}. We leave that to the readers.

Apart from that, the Fibonacci numbers have a companion sequence called the Lucas numbers. By examining the initial values, we observe that (A_{ν,λ}) is not the desired companion sequence. Therefore, it is still an open problem to construct the corresponding Lucas sequence for (B_{ν,λ}).

5 Acknowledgments

The author would like to thank the editor and the anonymous referee for their helpful comments.

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2020 Mathematics Subject Classification: Primary 11A55; Secondary 11B39. Keywords: continued fraction, continuant, Lucas sequence, Fibonacci number.

(Concerned with sequences \underline{A000129} and \underline{A041011}.)

Received August 28 2023; revised versions received October 19 2023; October 20 2023. Published in Journal of Integer Sequences, October 31 2023.

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