Fibonacci Identities via Fibonacci Functions

Kunle Adegoke Department of Physics and Engineering Physics Obafemi Awolowo University 220005 Ile-Ife Nigeria

                  adegoke00@gmail.com

Abstract

We present a differential-calculus-based method which allows one to derive more identities from a given Fibonacci-Lucas identity containing a finite number of terms and having at least one free index. The method has two independent components. The first component allows new identities to be obtained directly from an existing identity while the second yields a generalization of the existing identity. The strength of the first component is that no additional information is required about the given original identity. We illustrate the method by providing new generalizations of some well-known identities such as d’Ocagne’s identity, Candido’s identity, the Gelin-Cesàro identity, and Catalan’s identity. The method readily extends to a generalized Fibonacci sequence.

1 Introduction

Let F_{j} and L_{j} be the jth Fibonacci and Lucas numbers, defined for all integers by

F_{j} = \frac{α^{j} − β^{j}}{α − β} , L_{j} = α^{j} + β^{j},  (1)

where α = (1 + \sqrt{5})/2, the golden ratio, and β = (1 − \sqrt{5})/2 = −1/α. Of course, α + β = 1, αβ = −1 and α−β = \sqrt{5}. Let (G_{j})_{j∈ℤ} be the gibonacci sequence having the same recurrence relation as the Fibonacci sequence but starting with arbitrary initial values; that is, let

G_{j} = G_{j−1} + G_{j−2}, (j ≥ 2),

with G_{0} and G_{1} arbitrary numbers (usually integers) not both zero; and

G_{−j} = G_{−(j−2)} − G_{−(j−1)}.

If, inspired by (1), we introduce infinitely differentiable, complex-valued Fibonacci and Lucas functions, f(x) and l(x), defined by

f(x) = \frac{α^{x} − β^{x}}{α − β} , l(x) = α^{x} + β^{x}, x ∈ ℝ;  (2)

then, clearly,

f(x)|_{x=j∈ℤ} = F_{j}, l(x)|_{x=j∈ℤ} = L_{j};

and we will show that (see §4.2 and §5)

ℜ ( \frac{d}{dx}f(x)|_{x=j∈ℤ}) = \frac{L_{j}}{\sqrt{5}} ln α, ℜ ( \frac{d}{dx}l(x)|_{x=j∈ℤ}) = F_{j}\sqrt{5} ln α,  (3)

and

ℑ ( \frac{d}{dx}f(x)|_{x=j∈ℤ}) = −\frac{πβ^{j}}{\sqrt{5}} , ℑ ( \frac{d}{dx}l(x)|_{x=j∈ℤ}) = πβ^{j};  (4)

where, here and throughout this paper, ℜ(X) or ℜX denotes the real part of X and ℑ(X) or ℑX stands for the imaginary part of X. Many authors have studied various Fibonacci and Lucas functions in the past; we mention Halsey [5], Parker [20], Spickerman [22], Horadam and Shannon [11], and Han et al. [6]. The main difference between the approach in this paper and that in previous work by other authors is that the latter focused on seeking real-valued Fibonacci and Lucas functions. It is, precisely, the complex-valued nature of the Fibonacci and Lucas functions defined in (2) and their derivatives that motivated the method developed in this paper.

Our goal is to present a two-component method, based on (2)–(4) and their extensions, which allows the discovery of more identities from any known Fibonacci-Lucas identity or any gibonacci identity consisting of a finite number of terms and having at least one free index; that is an index that is not being summed over.

To illustrate what we mean, consider the identity

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}F_{j+k}^{4} = 25^{n} (F_{2n+k+1}^{4} − F_{2n+k}^{4}) ,  (5)

derived, among other similar results, by Hoggatt and Bicknell [7]. This identity has a free index, k. Working only with the knowledge of (5), our method (first component) allows us to derive the following presumably new identity:

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}F_{j+k}^{3}L_{j+k} = 25^{n} (F_{2n+k+1}^{3}L_{2n+k+1} − F_{2n+k}^{3}L_{2n+k});  (6)

which, in turn, implies the identity

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}F_{2j+k}^{2} = 25^{n}F_{2(4n+k+1)}.  (7)

We are not done yet, as (7) implies

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}F_{4j+2k} = 25^{n}L_{2(4n+k+1)};  (8)

which finally implies

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}L_{4j+2k} = 5^{2n+1}F_{2(4n+k+1)}.  (9)

Thus, the four identities (6), (7), (8) and (9) all follow from a knowledge of (5).

Our method (second component) provides the following generalization of (5) to (identity (131)):

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}F_{j+k}^{3}G_{j+r} = 25^{n} (F_{2n+k+1}^{3}G_{2n+r+1} − F_{2n+k}^{3}G_{2n+r}) .

A further generalization is derived in Proposition 32 on page 39.

As another example, consider the following well-known identity (see, for example, Hoggatt and Ruggles [8, Theorem 4])

tan^{−1} \frac{1}{F_{2k+1}} = tan^{−1} \frac{1}{F_{2k}} − tan^{−1} \frac{1}{F_{2k+2}} .  (10)

Our method (first component) shows that (10) implies the following apparently new identity:

\frac{L_{2k+1}}{F_{2k+1}^{2} + 1} = \frac{L_{2k}}{F_{2k}^{2} + 1} − \frac{L_{2k+2}}{F_{2k+2}^{2} + 1};

and the method (second component) yields a generalization:

\frac{G_{2k+r+3} + G_{2k+r+1}}{F_{2k+1}(F_{2k+1}^{2} + 1)} − \frac{G_{r+1}}{F_{2k+1}^{2} + 1} = \frac{G_{r+2}}{F_{2k}^{2} + 1} − \frac{G_{r}}{F_{2k+2}^{2} + 1}.

Yet another example, our method (first component) shows that the following identity of Howard [12, Corollary 3.5]:

F_{s}G_{k+r} + (−1)^{r−1}F_{s−r}G_{k} = F_{r}G_{k+s},  (11)

having three free indices r, s, and k, implies the following identities:

L_{s}G_{k+r} + (−1)^{r−1}L_{s−r}G_{k} = F_{r}(G_{k+s+1} + G_{k+s−1}),  (12)
F_{s} (G_{k+r+1} + G_{k+r−1}) + (−1)^{r}L_{s−r}G_{k} = L_{r}G_{k+s},  (13)
L_{s} (G_{k+r+1} + G_{k+r−1}) + (−1)^{r}5F_{s−r}G_{k} = L_{r} (G_{k+s+1} + G_{k+s−1}) .  (14)

The method (second component) provides the following generalization of (11):

H_{s}G_{k+r} + (−1)^{r−1}H_{s−r}G_{k} = F_{r} (G_{0}H_{k+s−1} + G_{1}H_{k+s}) ;  (15)

where here, and throughout this paper, (H_{j})_{j∈ℤ} is a gibonacci sequence with seeds H_{0} and H_{1}. Another generalization of (11) is given in Proposition 36 on page 41.

In Section 5 we will apply the method (second component) to provide a generalization of Candido’s identity

2 (F_{k}^{4} + F_{k+1}^{4} + F_{k+2}^{4}) = (F_{k}^{2} + F_{k+1}^{2} + F_{k+2}^{2})^{2} ,

to the following (identity (139)):

\begin{matrix}2 (H_{r}G_{k}^{3} + H_{r+1}G_{k+1}^{3} + H_{r+2}G_{k+2}^{3}) \\ = (G_{k}^{2} + G_{k+1}^{2} + G_{k+2}^{2}) (H_{r}G_{k} + H_{r+1}G_{k+1} + H_{r+2}G_{k+2}) , \end{matrix}

with a further generalization given in Proposition 38 on page 41; a particular case of which is

6 (F_{k}F_{r}F_{s}F_{t} + F_{k+1}F_{r+1}F_{s+1}F_{t+1} + F_{k+2}F_{r+2}F_{s+2}F_{t+2})
= (F_{k}F_{s} + F_{k+1}F_{s+1} + F_{k+2}F_{s+2}) (F_{r}F_{t} + F_{r+1}F_{t+1} + F_{r+2}F_{t+2})
+ (F_{k}F_{r} + F_{k+1}F_{r+1} + F_{k+2}F_{r+2}) (F_{s}F_{t} + F_{s+1}F_{t+1} + F_{s+2}F_{t+2})
+ (F_{k}F_{t} + F_{k+1}F_{t+1} + F_{k+2}F_{t+2}) (F_{r}F_{s} + F_{r+1}F_{s+1} + F_{r+2}F_{s+2}) .

The method (second component) extends the d’Ocagne identity

F_{r+1}F_{k} − F_{r}F_{k+1} = (−1)^{r}F_{k−r},

to the gibonacci sequence as

G_{r+1}G_{k} − G_{r}G_{k+1} = (−1)^{r} (G_{1}G_{k−r} − G_{0}G_{k−r+1}) ;

and extends the well-known formula for the sum of the squares of two consecutive Fibonacci numbers, namely,

F_{k+1}^{2} + F_{k}^{2} = F_{2k+1},

to the gibonacci sequence as

G_{k+1}^{2} + G_{k}^{2} = G_{0}G_{2k} + G_{1}G_{2k+1}.

We will also establish the following generalization of Catalan’s identity (identity (147)):

F_{k+r}G_{k−r+s} + F_{k−r}G_{k+r+s} = 2F_{k}G_{k+s} + (−1)^{k+r+1}F_{r}^{2} (G_{s+1} + G_{s−1}) ,

and the Gelin-Cesàro identity

H_{k+r−2}G_{k−1}G_{k+1}G_{k+2} + G_{k−2}H_{k+r−1}G_{k+1}G_{k+2}
+ G_{k−2}G_{k−1}H_{k+r+1}G_{k+2} + G_{k−2}G_{k−1}G_{k+1}H_{k+r+2}
= 4H_{k+r}G_{k}^{3} − 2e_{G}G_{0} (H_{r+2} + H_{r}) + 2e_{G}G_{1} (H_{r+1} + H_{r−1}) ,

where e_{G} = G_{0}^{2} − G_{1}^{2} + G_{0}G_{1}.

The method (second component) extends the fundamental identity of Fibonacci and Lucas numbers,

5F_{k}^{2} − L_{k}^{2} = (−1)^{k−1}4,

to the gibonacci sequence as

5G_{k}^{2} − (G_{k+1} + G_{k−1})^{2} = (−1)^{k}4e_{G};

and offers an extension of the triple-angle formula of Lucas

F_{3k} = F_{k+1}^{3} + F_{k}^{3} − F_{k−1}^{3},

to

G_{0}^{2}G_{3k−2} + 2G_{0}G_{1}G_{3k−1} + G_{1}^{2}G_{3k} = G_{k+1}^{3} + G_{k}^{3} − G_{k−1}^{3}.

Using the method, the golden ratio power reduction formula

α^{k} = αF_{k} + F_{k−1},

will be shown to imply

\begin{matrix}G_{0} (H_{k+r} + H_{k+r−2}) + G_{1} (H_{k+r+1} + H_{k+r−1}) \\ = G_{r} (H_{k} + H_{k−2}) + G_{r+1} (H_{k+1} + H_{k−1}) , \end{matrix}

which subsumes several Fibonacci-Lucas identities.

Consider a generalized Fibonacci sequence (W_{j}) = (W_{j}(W_{0}, W_{1}; P)) defined, for all integers and arbitrary real numbers W_{0}, W_{1}, and P ≠ 0, by the recurrence relation

W_{j} = P W_{j−1} + W_{j−2}, j ≥ 2,  (16)

with W_{−j} = W_{−j+2} − P W_{−j+1}.

Two important cases of (W_{j}) are the special Lucas sequences of the first kind, (U_{j}(P)) = (W_{j}(0, 1; P)), and the second kind, (V_{j}(P)) = (W_{j}(2, P; P)); so that

U_{0} = 0, U_{1} = 1, U_{j} = P U_{j−1} + U_{j−2}, j ≥ 2,  (17)

and

V_{0} = 2, V_{1} = P, V_{j} = P V_{j−1} + V_{j−2}, j ≥ 2,  (18)

with U_{−j} = U_{−j+2} − P U_{−j+1} and V_{−j} = V_{−j+2} − P V_{−j+1}.

We will show that the new method also applies to the generalized Fibonacci sequence. For example, the method (first component) shows that the identity [9, Equation (3.14), Q = −1]:

U_{r}W_{k+1} + U_{r−1}W_{k} = W_{k+r}

implies

V_{r}W_{k+1} + V_{r−1}W_{k} = W_{k+r+1} + W_{k+r−1}.  (19)

The new method presented in this paper complements some previous research (for example the work of Long [16], Dresel [4], and Melham [18]).

The rest of the paper is arranged as follows. In Section 2 we describe the method (first component) and give examples. We cast about for identities to apply the method (first component) in Section 3. Further justification of the method (first component) is addressed in Section 4. A description of the method (second component), with examples including various extensions and generalizations of some known identities, is presented in Section 5. Finally, an extension of the method to the general second order (Horadam) sequence is offered in Section 6.

2 The method, first component

Delaying further justification to Section 4, we present the method (first component) and give examples.

Here then is how to obtain more identities from any given Fibonacci-Lucas identity having a free index:

1. Let k be a free index in the known identity. Replace each Fibonacci number, say F_{h(k,...)}, with a certain differentiable function of k, namely, f(h(k, …)), with k now considered a variable; and replace each Lucas number, say L_{h(k,...)}, with a certain differentiable function l(h(k, …)). The subscript h will be considered a function of several variables; that is variable k and other parameters (if any) indicated by ellipsis: “…”. The explicit form of f(h(k, …)) or l(h(k, …)) will not enter into consideration.

2. By applying the usual rules of calculus, differentiate, with respect to k, through the identity obtained in step 1.

3. Simplify the equation obtained in step 2 and make the following replacements:

f(h(k, …)) → F_{h(k,...)},  (20)
l(h(k, …)) → L_{h(k,...)}.  (21)

4. Take the real part of the whole expression/equation obtained in step 3, using also the following prescription:

ℜ\frac{∂f}{∂k} (h(k, …)) → \frac{L_{h(k,...)}}{\sqrt{5}} ln α,  (22)
ℜ\frac{∂l}{∂k} (h(k, …)) → F_{h(k,...)}\sqrt{5} ln α.  (23)

Remark 1. Formally, the method (first component) of obtaining new identities from a known Fibonacci-Lucas identity proceeds in two quick steps:

(i) Treat the subscripts of Fibonacci and Lucas numbers as variables and differentiate through the given identity, with respect to the free index of interest, using the rules of differential calculus.

(ii) Make the following replacements:

\frac{∂}{∂k} F_{h(k,...)} → \frac{L_{h(k,...)}}{\sqrt{5}} \frac{∂}{∂k} h(k, …),  (24)
\frac{∂}{∂k} L_{h(k,...)} → F_{h(k,...)}\sqrt{5} \frac{∂}{∂k} h(k, …),  (25)
ln α → 1,  (26)
i → 0;  (27)

where i = \sqrt{−1} is the imaginary unit.

For example, given the double-angle identity

F_{2k} = L_{k}F_{k},

we have, by step (i),

\frac{d}{dk} F_{2k} = \frac{d}{dk} (L_{k}F_{k}) = L_{k} \frac{d}{dk} F_{k} + F_{k} \frac{d}{dk} L_{k};

so that, by step (ii), using (24) and (25), we get

\frac{L_{2k}}{\sqrt{5}} · \frac{d}{dk} (2k) = L_{k} · \frac{L_{k}}{\sqrt{5}} + F_{k} · F_{k}\sqrt{5};

and hence,

2L_{2k} = L_{k}^{2} + 5F_{k}^{2}.  (28)

2.1 Examples

We illustrate the method (first component) with some examples from familiar identities.

2.2 Example from a connecting formula between Fibonacci and

Lucas numbers

In this example we show that

L_{k} = F_{k+1} + F_{k−1} ⟹ 5F_{k} = L_{k+1} + L_{k−1}.

Following step 1 we write

l(k) = f(k + 1) + f(k − 1)

and (step 2) differentiate with respect to k, obtaining

\frac{d}{dk} l(k) = \frac{d}{dk} f(k + 1) + \frac{d}{dk} f(k − 1).

Steps 3 and 4 now give

ℜ\frac{d}{dk} l(k) = ℜ\frac{d}{dk} f(k + 1) + ℜ\frac{d}{dk} f(k − 1);

and by (22) and (23),

F_{k}\sqrt{5} ln α = \frac{L_{k+1}}{\sqrt{5}} ln α + \frac{L_{k−1}}{\sqrt{5}} ln α;

that is

5F_{k} = L_{k+1} + L_{k−1}.

2.2.1 Example from the double-angle identity of Fibonacci and Lucas numbers In this example we demonstrate that:

F_{2k} = L_{k}F_{k} ⟹ 2L_{2k} = L_{k}^{2} + 5F_{k}^{2}.  (29)

For the identity F_{2k} = L_{k}F_{k}, step 1 is

f(2k) = l(k)f(k);

where k is now considered a variable.

Following step 2, we differentiate with respect to k to obtain

2\frac{df}{dk} (2k) = l(k)\frac{df}{dk} (k) + f(k)\frac{dl}{dk} (k).

Steps 3 and 4 give

2ℜ\frac{df}{dk} (2k) = L_{k}ℜ\frac{df}{dk} (k) + F_{k}ℜ\frac{dl}{dk} (k).

Thus, using (22) and (23), we have

2\frac{L_{2k}}{\sqrt{5}} ln α = L_{k} \frac{L_{k}}{\sqrt{5}} ln α + F_{k}\sqrt{5}F_{k} ln α;

which, dropping ln α and multiplying through by \sqrt{5}, is

2L_{2k} = L_{k}^{2} + 5F_{k}^{2}.

The interested reader may wish to verify the converse of (29), that is 2L_{2k} = L_{k}^{2} + 5F_{k}^{2} ⟹ F_{2k} = L_{k}F_{k}.

2.2.2 Example from the multiplication formula of Fibonacci and Lucas numbers Here we show that the multiplication formula

F_{k+m} + (−1)^{m}F_{k−m} = L_{m}F_{k}

implies

L_{k+m} + (−1)^{m}L_{k−m} = L_{m}L_{k}  (30)

and

L_{k+m} − (−1)^{m}L_{k−m} = 5F_{m}F_{k}.  (31)

We write

f(k + m) + (−1)^{m}f(k − m) = l(m)f(k);  (32)

so that, treating k as the free index of interest gives

\frac{∂f}{∂k} (k + m) + (−1)^{m} \frac{∂f}{∂k} (k − m) = l(m)\frac{∂}{∂k} f(k).

Thus, by steps 3 and 4, we have

ℜ\frac{∂f}{∂k} (k + m) + (−1)^{m}ℜ\frac{∂f}{∂k} (k − m) = L_{m}ℜ\frac{∂f}{∂k} (k);

and hence, using (22) and (23), we obtain

\frac{L_{k+m}}{\sqrt{5}} ln α + (−1)^{m} \frac{L_{k−m}}{\sqrt{5}} ln α = L_{m} \frac{L_{k}}{\sqrt{5}} ln α;

from which we get (30).

Taking m as the index of interest and differentiating (32) with respect to m yields

\frac{∂f}{∂m}(k + m) − (−1)^{m} \frac{∂f}{∂m}(k − m) + (−1)^{m}iπf(k − m) = f(k)\frac{∂}{∂m}l(m),

so that

ℜ\frac{∂f}{∂m}(k + m) − (−1)^{m}ℜ\frac{∂f}{∂m}(k − m) = F_{k}ℜ\frac{∂}{∂m}l(m);

and hence

\frac{L_{k+m}}{\sqrt{5}} ln α − (−1)^{m} \frac{L_{k−m}}{\sqrt{5}} ln α = F_{k}F_{m}\sqrt{5} ln α;

from which (31) follows.

The reader may verify that the remaining multiplication formula can be discovered by differentiating (30) with respect to m.

2.2.3 Example from an inverse tangent Fibonacci number identity

Consider the following identity:

tan^{−1} \frac{F_{2m}}{F_{2k+2m−1}} = tan^{−1} \frac{L_{m}}{L_{2k+m−1}} − tan^{−1} \frac{L_{m}}{L_{2k+3m−1}} , m even,  (33)

which can be derived using the inverse tangent addition formula and basic Fibonacci-Lucas identities.

We now demonstrate that (33) implies

\frac{1}{5} \frac{F_{2m}L_{2k+2m−1}}{F_{2k+2m−1}^{2} + F_{2m}^{2}} = \frac{L_{m}F_{2k+m−1}}{L_{2k+m−1}^{2} + L_{m}^{2}} − \frac{L_{m}F_{2k+3m−1}}{L_{2k+3m−1}^{2} + L_{m}^{2}} , m even.  (34)

We treat k as the free index of interest. Step 1 gives the Fibonacci-Lucas function form of (33) as

tan^{−1} \frac{f(2m)}{f(2k + 2m − 1)} = tan^{−1} \frac{l(m)}{l(2k + m − 1)} − tan^{−1} \frac{l(m)}{l(2k + 3m − 1)};

so that step 2 yields

\frac{2f(2m)}{f(2k + 2m − 1)^{2} + f(2m)^{2}} \frac{∂f}{∂k} (2k + 2m − 1)

= \frac{2l(m)}{l(2k + m − 1)^{2} + l(m)^{2}} \frac{∂l}{∂k} (2k + m − 1)

− \frac{2l(m)}{l(2k + 3m − 1)^{2} + l(m)^{2}} \frac{∂l}{∂k} (2k + 3m − 1),

which, by step 3, results in

\begin{matrix}\frac{F_{2m}}{F_{2k+2m−1}^{2} + F_{2m}^{2}} \frac{∂f}{∂k} (2k + 2m − 1) \\ = \frac{L_{m}}{L_{2k+m−1}^{2} + L_{m}^{2}} \frac{∂l}{∂k} (2k + m − 1) − \frac{L_{m}}{L_{2k+3m−1}^{2} + L_{m}^{2}} \frac{∂l}{∂k} (2k + 3m − 1), \end{matrix}

whence taking the real part and replacing the derivatives using (22) and (23) gives (34).

By treating m as the free index, the interested reader can verify, using our method, that (33) also implies

\frac{2}{5} \frac{F_{2k−1}}{F_{2k+2m−1}^{2} + F_{2m}^{2}} = −\frac{F_{2k−1}}{L_{2k+m−1}^{2} + L_{m}^{2}} + \frac{F_{2k+3m−1}L_{m} + F_{2k+2m−1}}{L_{2k+3m−1}^{2} + L_{m}^{2}} , m even.

2.3 Extension to a generalized Fibonacci sequence

We now describe how the method (first component) for obtaining new identities from existing ones works for the generalized Fibonacci sequence (W_{j}(W_{0}, W_{1}; P)) whose terms are given in (16). The scheme is the following.

1. Let k be a free index in the known identity. Replace each generalized Fibonacci number, say W_{h(k,...)}, with a certain differentiable function of k, namely, w(h(k, …)), with k now considered a variable.

2. By applying the usual rules of calculus, differentiate, with respect to k, through the identity obtained in step 1.

3. Simplify the equation obtained in step 2 and make the following replacement:

w(h(k, …)) → W_{h(k,...)}.  (35)

4. Take the real part of the equation/expression obtained in step 3, using also the following prescription:

ℜ\frac{∂w}{∂k}(h(k, …)) → \frac{W_{h(k+1,···)} + W_{h(k−1,···)}}{δ} ln σ;  (36)

where σ = (P + δ)/2 and δ = \sqrt{P^{2} + 4}.

Note that, on account of (70) and (72), for the special Lucas sequences, (35) and (36) reduce to

u(h(k, …)) → U_{h(k,...)},  (37)
ℜ\frac{∂u}{∂k}(h(k, …)) → \frac{V_{h(k,...)}}{δ} ln σ  (38)

and

v(h(k, …)) → V_{h(k,...)},  (39)
ℜ\frac{∂v}{∂k} (h(k, …)) → U_{h(k,...)}δ ln σ;  (40)

of which the Fibonacci and Lucas relations (20)–(23) are particular cases.

For the gibonacci sequence, (35) and (36) reduce to

g(h(k, …)) → G_{h(k,...)},  (41)
ℜ\frac{∂g}{∂k} (h(k, …)) → \frac{G_{h(k+1,···)} + G_{h(k−1,···)}}{\sqrt{5}} ln α.  (42)

2.4 More examples

We give further examples involving the gibonacci sequence and the generalized Fibonacci sequence.

2.4.1 Examples from an identity of Howard

Consider the following identity, derived by Howard [12, Corollary 3.5]:

F_{s}G_{k+r} + (−1)^{r−1}F_{s−r}G_{k} = F_{r}G_{k+s},

Identity (11) has three free indices r, s, and k.

We write

f(s)g(k + r) + (−1)^{r−1}f(s − r)g(k) = f(r)g(k + s).  (43)

Treating s as the index of interest and differentiating (43) with respect to s gives

g(k + r)\frac{d}{ds}f(s) + (−1)^{r−1}g(k)\frac{∂f}{∂s} (s − r) = f(r)\frac{∂g}{∂s}(k + s);  (44)

so that, using (20) and (41) we get

G_{k+r}ℜ\frac{d}{ds}f(s) + (−1)^{r−1}G_{k}ℜ\frac{∂f}{∂s} (s − r) = F_{r}ℜ\frac{∂g}{∂s}(k + s).

We now use (22) to replace the derivatives on the left hand side and (42) to replace the derivative on the right hand side, obtaining

L_{s}G_{k+r} + (−1)^{r−1}L_{s−r}G_{k} = F_{r}(G_{k+s+1} + G_{k+s−1}).

On the other hand, treating r as the index of interest and differentiating (43) with respect to r yields

\begin{matrix}f(s)\frac{∂g}{∂r} (k + r) + (−1)^{r−1}iπf(s − r)g(k) − (−1)^{r−1}g(k)\frac{∂f}{∂r} (s − r) \\ = g(k + s)\frac{d}{dr}f(r); \end{matrix}  (45)

so that, taking the real part,

F_{s}ℜ\frac{∂g}{∂r} (k + r) − (−1)^{r−1}G_{k}ℜ\frac{∂f}{∂r} (s − r) = G_{k+s}ℜ\frac{d}{dr}f(r).

Use of (42) and (22) finally gives (identity 13):

F_{s} (G_{k+r+1} + G_{k+r−1}) + (−1)^{r}L_{s−r}G_{k} = L_{r}G_{k+s}.

The interested reader is invited to discover, by differentiating with respect to s, that (13) implies

L_{s} (G_{k+r+1} + G_{k+r−1}) + (−1)^{r}5F_{s−r}G_{k} = L_{r} (G_{k+s+1} + G_{k+s−1}) ;

and that differentiating (11) with respect to k does not produce a new result.

2.4.2 Example from a general recurrence relation Consider the following identity of Horadam [9, Equation (3.14), Q = −1]:

U_{r}W_{k+1} + U_{r−1}W_{k} = W_{k+r}.

We write

u(r)w(k + 1) + u(r − 1)w(k) = w(k + r);

and differentiate with respect to r, obtaining

\frac{d}{dr}u(r) · w(k + 1) + \frac{d}{dr}u(r − 1) · w(k) = \frac{∂w}{∂r}(k + r);

so that, taking the real part, we find

ℜ\frac{d}{dr}u(r) · W_{k+1} + ℜ\frac{d}{dr}u(r − 1) · W_{k} = ℜ\frac{∂w}{∂r}(k + r);

and hence, upon using (38) and (36) to replace the derivatives, we derive identity (19):

V_{r}W_{k+1} + V_{r−1}W_{k} = W_{k+r+1} + W_{k+r−1}.

In particular,

V_{r}U_{k+1} + V_{r−1}U_{k} = V_{k+r},
V_{r}V_{k+1} + V_{r−1}V_{k} = (P^{2} + 4)U_{k+r}.

2.4.3 Example from a multiplication formula Here we will demonstrate that the identity [9, Equation (3.16), Q = −1]:

W_{k+r} + (−1)^{r}W_{k−r} = V_{r}W_{k}

implies the identity

(W_{k+r+1} + W_{k+r−1}) − (−1)^{r}(W_{k−r+1} + W_{k−r−1}) = U_{r}W_{k}δ^{2}.  (46)

We write

w(k + r) + (−1)^{r}w(k − r) = v(r)w(k)

and differentiate through with respect to r to obtain

\frac{∂w}{∂r}(k + r) + (−1)^{r}πiw(k − r) − (−1)^{r} \frac{∂w}{∂r}(k − r) = w(k)\frac{d}{dr}v(r);

so that

ℜ\frac{∂w}{∂r}(k + r) − (−1)^{r}ℜ\frac{∂w}{∂r}(k − r) = w(k)ℜ\frac{d}{dr}v(r).

Using (36) and (40), we get

\frac{W_{k+r+1} + W_{k+r−1}}{δ} − (−1)^{r} \frac{(W_{k−r+1} + W_{k−r−1})}{δ} = W_{k}U_{r}δ;

and hence (46).

Identities

V_{k+r} − (−1)^{r}V_{k−r} = U_{k}U_{r}δ^{2}

and

U_{k+r} − (−1)^{r}U_{k−r} = U_{r}V_{k}

are special cases of (46).

3 Applications

In this section, we pick various known results from the literature and apply the method (first component) to discover new identities.

3.1 New identities from an identity of Long

Long [17, Equation (44)] showed that, for a non-negative integer n and integers k and r,

∑_{j=0}^{n} \binom{n}{j}F_{r+2kj} = L_{k}^{n}F_{r+nk}, if k is even.  (47)

Based on the knowledge of (47) alone, we will derive the results stated in the proposition.

Proposition 2. If n is a non-negative integer, k is an even integer and r is an integer, then

2 ∑_{j=0}^{n} j\binom{n}{j}L_{r+2kj} = 5nL_{k}^{n−1}F_{r+nk}F_{k} + nL_{k}^{n}L_{r+nk},  (48)

2 ∑_{j=0}^{n} j\binom{n}{j}F_{r+2kj} = nL_{k}^{n−1}L_{r+nk}F_{k} + nL_{k}^{n}F_{r+nk}.  (49)

Identity (47) contains two free indices r and k. Treating r as the index of interest immediately gives the Lucas version of (47), namely,

∑_{j=0}^{n} \binom{n}{j}L_{r+2kj} = L_{k}^{n}L_{r+nk}, if k is even;

coming from

∑_{j=0}^{n} \binom{n}{j}ℜ\frac{∂f}{∂r} (r + 2kj) = l(k)^{n}ℜ\frac{∂f}{∂r} (r + nk)

and prescription (22).

To derive (48), write (47) as

∑_{j=0}^{n} \binom{n}{j}f(r + 2kj) = l(k)^{n}f(r + nk);

treat k as the index of interest and differentiate with respect to k (step 2) to obtain

∑_{j=0}^{n} 2j\binom{n}{j}\frac{∂f}{∂k} (r + 2kj) = nl(k)^{n−1}f(r + nk)\frac{∂}{∂k} l(k) + nl(k)^{n} \frac{∂f}{∂k} (r + nk),

and, taking the real part,

∑_{j=0}^{n} 2j\binom{n}{j}ℜ\frac{∂f}{∂k} (r + 2kj) = nL_{k}^{n−1}F_{r+nk}ℜ\frac{∂}{∂k} l(k) + nL_{k}^{n}ℜ\frac{∂f}{∂k} (r + nk).  (50)

Thus (48) follows from step 4 of Section 2, after using (22) and (23) to replace the derivatives in (50).

To derive (49) treat r as the free index of interest in (48) and write

2 ∑_{j=0}^{n} j\binom{n}{j}\frac{∂l}{∂r}(r + 2kj) = 5nL_{k}^{n−1} \frac{∂f}{∂r} (r + nk)f(k) + nL_{k}^{n} \frac{∂l}{∂r}(r + nk).

3.2 New identities arising from an identity of Hoggatt and Bicknell

Based on Hoggatt and Bicknell’s result [7, Identity 2′]:

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}F_{j+k}^{4} = 25^{n} (F_{2n+k+1}^{4} − F_{2n+k}^{4}) ,

we wish to derive the four identities (6), (7), (8) and (9) stated in the Introduction section. Write (5) as

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}f(j + k)^{4} = 25^{n} (f(2n + k + 1)^{4} − f(2n + k)^{4});

and differentiate through, with respect to k, to obtain

\begin{aligned}∑_{j=0}^{4n+1} & (−1)^{j−1}\binom{4n + 1}{j}4f(j + k)^{3} \frac{∂f}{∂k} (j + k) \\ & = 25^{n} (4f(2n + k + 1)^{3} \frac{∂f}{∂k} (2n + k + 1) − 4f(2n + k)^{3} \frac{∂f}{∂k} (2n + k)); \end{aligned}  (51)

and taking the real part:

\begin{aligned}∑_{j=0}^{4n+1} & (−1)^{j−1}\binom{4n + 1}{j}4F_{j+k}^{3}ℜ\frac{∂f}{∂k} (j + k) \\ & = 25^{n} (4F_{2n+k+1}^{3}ℜ\frac{∂f}{∂k} (2n + k + 1) − 4F_{2n+k}^{3}ℜ\frac{∂f}{∂k} (2n + k)) . \end{aligned}

Thus,

\begin{aligned}∑_{j=0}^{4n+1} & (−1)^{j−1}\binom{4n + 1}{j}F_{j+k}^{3} \frac{L_{j+k}}{\sqrt{5}} \\ & = 25^{n} (F_{2n+k+1}^{3} \frac{L_{2n+k+1}}{\sqrt{5}} − F_{2n+k}^{3} \frac{L_{2n+k}}{\sqrt{5}} ); \end{aligned}

and hence (6). Identities (7), (8) and (9) are derived in the same manner; (7) is obtained from (6), etc.

3.3 New identities from an inverse tangent identity

Proposition 3. If k is an integer, then

\frac{L_{2k+1}}{F_{2k+1}^{2} + 1} = \frac{L_{2k}}{F_{2k}^{2} + 1} − \frac{L_{2k+2}}{F_{2k+2}^{2} + 1},  (52)

\frac{L_{2k+1}}{L_{2k}L_{2k+2}} \frac{(F_{2k}^{2} + 1)(F_{2k+2}^{2} + 1)}{(F_{2k+1}^{2} + 1)} = \frac{(F_{2k+2}^{2} + 1)}{L_{2k+2}} − \frac{(F_{2k}^{2} + 1)}{L_{2k}}.  (53)

Recall identity (10):

tan^{−1} \frac{1}{F_{2k+1}} = tan^{−1} \frac{1}{F_{2k}} − tan^{−1} \frac{1}{F_{2k+2}} .

To derive (52), write (10) as

tan^{−1} \frac{1}{f(2k + 1)} = tan^{−1} \frac{1}{f(2k)} − tan^{−1} \frac{1}{f(2k + 2)},  (54)

and differentiate with respect to k to obtain

\begin{matrix}\frac{1}{f(2k + 1)^{2} + 1} \frac{df}{dk} (2k + 1) \\ = \frac{1}{f(2k)^{2} + 1} \frac{df}{dk} (2k) − \frac{1}{f(2k + 2)^{2} + 1} \frac{df}{dk} (2k + 2), \end{matrix}

so that, by (20),

\begin{matrix}\frac{1}{F_{2k+1}^{2} + 1} ℜ\frac{df}{dk} (2k + 1) \\ = \frac{1}{F_{2k}^{2} + 1} ℜ\frac{df}{dk} (2k) − \frac{1}{F_{2k+2}^{2} + 1} ℜ\frac{df}{dk} (2k + 2), \end{matrix}

and hence (52), upon using (22). Identity (53) is a rearrangement of (52). Simple telescoping of (52) and (53) produces the results stated in the next proposition.

Proposition 4. If n is an integer, then

∑_{k=1}^{n} \frac{L_{2k+1}}{F_{2k+1}^{2} + 1} = \frac{3}{2} − \frac{L_{2(n+1)}}{F_{2(n+1)}^{2} + 1},

∑_{k=1}^{n} \frac{L_{2k+1}}{L_{2k}L_{2k+2}} \begin{matrix}(F_{2k}^{2} + 1) & (F_{2k+2}^{2} + 1) \\ ( & F_{2k+1}^{2} + 1 & ) \end{matrix} = \frac{F_{2(n+1)}^{2} + 1}{L_{2n+2}} − \frac{2}{3};

with the limiting case:

∑_{k=1}^{∞} \frac{L_{2k+1}}{F_{2k+1}^{2} + 1} = \frac{3}{2}.

3.4 New identities from an identity of Jennings

Jennings [13, Theorem 2] showed, among results of a similar nature, that

F_{k} ∑_{j=0}^{n} (−1)^{(k+1)(n+j)}\binom{n + j}{2j}L_{k}^{2j} = F_{(2n+1)k}.

Writing

∑_{j=0}^{n} (−1)^{(k+1)(n+j)}\binom{n + j}{2j}l(k)^{2j} = \frac{f((2n + 1)k)}{f(k)}

and differentiating with respect to k gives

\begin{matrix}∑_{j=0}^{n} (−1)^{(k+1)(n+j)}(n + j)πi\binom{n + j}{2j}l(k)^{2j} + ∑_{j=0}^{n} (−1)^{(k+1)(n+j)}2j\binom{n + j}{2j}l(k)^{2j−1} \frac{dl}{dk} (k) \\ = \frac{2n + 1}{f(k)} \frac{df}{dk} ((2n + 1)k) − \frac{f((2n + 1)k)}{f(k)^{2}} \frac{df}{dk} (k), \end{matrix}

and taking the real part,

\begin{aligned}∑_{j=0}^{n} & (−1)^{(k+1)(n+j)}2j\binom{n + j}{2j}L_{k}^{2j−1}ℜ\frac{dl}{dk} (k) \\ & = \frac{2n + 1}{F_{k}}ℜ\frac{df}{dk} ((2n + 1)k) − \frac{F_{(2n+1)k}}{F_{k}^{2}} ℜ\frac{df}{dk} (k), \end{aligned}

which, by (22) and (23) gives

∑_{j=0}^{n} (−1)^{(k+1)(n+j)}2j\binom{n + j}{2j}L_{k}^{2j−1}F_{k}\sqrt{5} = \frac{2n + 1}{F_{k}} \frac{L_{(2n+1)k}}{\sqrt{5}} − \frac{F_{(2n+1)k}}{F_{k}^{2}} \frac{L_{k}}{\sqrt{5}};

and hence the result stated in the next proposition. Proposition 5. For non-negative integers k and n, we have

F_{k}^{3} ∑_{j=0}^{n} (−1)^{(k+1)(n+j)}j\binom{n + j}{2j}L_{k}^{2j} = \frac{1}{10} ((2n + 1)F_{2k}L_{(2n+1)k} − F_{(2n+1)k}L_{k}^{2}) .

We also have the following divisibility property. Proposition 6. If n and k are non-negative integers, then

10F_{k}^{3} divides (2n + 1)F_{2k}L_{(2n+1)k} − F_{(2n+1)k}L_{k}^{2}.

3.5 New identities from Candido’s identity

Setting x = G_{k}, y = G_{k+1} in the algebraic identity

2 (x^{4} + y^{4} + (x + y)^{4}) = (x^{2} + y^{2} + (x + y)^{2})^{2} ,

gives the following generalization of Candido’s identity:

2 (G_{k}^{4} + G_{k+1}^{4} + G_{k+2}^{4}) = (G_{k}^{2} + G_{k+1}^{2} + G_{k+2}^{2})^{2} .

Writing

2(g(k)^{4} + g(k + 1)^{4} + g(k + 2)^{4}) = (g(k)^{2} + g(k + 1)^{2} + g(k + 2)^{2})^{2} ,

and differentiating with respect to k gives

\begin{matrix}2 (g(k)^{3} \frac{dg}{dk} (k) + g(k + 1)^{3} \frac{dg}{dk} (k + 1) + g(k + 2)^{3} \frac{dg}{dk} (k + 2)) \\ = (g(k)^{2} + g(k + 1)^{2} + g(k + 2)^{2}) · \\ (g(k)\frac{dg}{dk} (k) + g(k + 1)\frac{dg}{dk} (k + 1) + g(k + 2)\frac{dg}{dk} (k + 2)); \end{matrix}  (55)

so that applying the prescription (41) and (42) yields

2 (G_{k}^{3}(G_{k+1} + G_{k−1}) + G_{k+1}^{3}(G_{k+2} + G_{k}) + G_{k+2}^{3}(G_{k+3} + G_{k+1}))

= (G_{k}^{2} + G_{k+1}^{2} + G_{k+2}^{2}) (G_{k}(G_{k+1} + G_{k−1})+

G_{k+1}(G_{k+2} + G_{k}) + G_{k+2}(G_{k+3} + G_{k+1})) ,

which can be arranged as stated in the next proposition.

Proposition 7. For every integer k, we have

G_{k}^{2} (G_{k+1}(G_{k+2} + G_{k}) + G_{k+2}(G_{k+3} + G_{k+1}) − G_{k}(G_{k+1} + G_{k−1}))
+ G_{k+1}^{2} (G_{k}(G_{k+1} + G_{k−1}) + G_{k+2}(G_{k+3} + G_{k+1}) − G_{k+1}(G_{k+2} + G_{k}))
+ G_{k+2}^{2} (G_{k}(G_{k+1} + G_{k−1}) + G_{k+1}(G_{k+2} + G_{k}) − G_{k+2}(G_{k+3} + G_{k+1}))
= 0.

In particular,

F_{k}^{2}F_{2k+3} + F_{k+1}^{2}F_{2k+2} = F_{k+2}^{2}F_{2k+1},  (56)
L_{k}^{2}F_{2k+3} + L_{k+1}^{2}F_{2k+2} = L_{k+2}^{2}F_{2k+1}.  (57)

Subtraction of (56) from (57) gives

F_{k−1}F_{k+1}F_{2k+3} + F_{k}F_{k+2}F_{2k+2} = F_{k+1}F_{k+3}F_{2k+1},

while their addition yields

(F_{k+1}^{2} + F_{k−1}^{2})F_{2k+3} + (F_{k+2}^{2} + F_{k}^{2})F_{2k+2} = (F_{k+3}^{2} + F_{k+1}^{2})F_{2k+1}.

Before closing this section, we bring forth a Candido-type identity of Melham and discover new identities from it. Melham [19, Theorem 1] has shown that

6 (∑_{j=0}^{2n−1} G_{k+j}^{2})^{2} = F_{2n}^{2} (G_{k+n−2}^{4} + 4G_{k+n−1}^{4} + 4G_{k+n}^{4} + G_{k+n+1}^{4});  (58)

from which, writing f(2n) for F_{2n}, g(k + n − 2) for G_{k+n−2}, etc. , and differentiating with respect to k, we have

(12 ∑_{j=0}^{2n−1} g(k + j)^{2}) ∑_{j=0}^{2n−1} 2g(k + j)\frac{∂g}{∂k} (k + j)

= f(2n)^{2} (4g(k + n − 2)^{3} \frac{∂g}{∂k} (k + n − 2) + 16g(k + n − 1)^{3} \frac{∂g}{∂k} (k + n − 1)

+16g(k + n)^{3} \frac{∂g}{∂k} (k + n) + 4g(k + n + 1)^{3} \frac{∂g}{∂k} (k + n + 1)) .

Taking the real part according to the prescription of steps 2, 3, and 4 of Section 2.3, using (20), (41) and (42) to replace the Fibonacci and gibonacci functions and derivatives, we obtain the result stated in the next proposition.

Proposition 8. If n is a non-negative integer and k is an integer, then

6∑_{j=0}^{2n−1} G_{j+k}^{2} ∑_{j=0}^{2n−1} G_{j+k}(G_{j+k+1} + G_{j+k−1})

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

In particular,

\begin{matrix}6 ∑_{j=0}^{2n−1} F_{j+k}^{2} ∑_{j=0}^{2n−1} F_{2j+2k} & = F_{2n}^{2} (F_{k+n−2}^{2}F_{2(k+n−2)} + 4F_{k+n−1}^{2}F_{2(k+n−1)} \\ +4F_{k+n}^{2}F_{2(k+n)} + F_{k+n+1}^{2}F_{2(k+n+1)} & ) \end{matrix}  (59)

and

\begin{aligned}6 ∑_{j=0}^{2n−1} L_{j+k}^{2} ∑_{j=0}^{2n−1} F_{2j+2k} & = F_{2n}^{2} (L_{k+n−2}^{2}F_{2(k+n−2)} + 4L_{k+n−1}^{2}F_{2(k+n−1)} \\ & +4L_{k+n}^{2}F_{2(k+n)} + L_{k+n+1}^{2}F_{2(k+n+1)}) . \end{aligned}  (60)

Subtraction of (59) from (60) gives

6 ∑_{j=0}^{2n−1} F_{j+k+1}F_{j+k−1} ∑_{j=0}^{2n−1} F_{2j+2k}

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

3.6 New identities from the Gelin-Cesàro identity

The Gelin-Cesàro identity

F_{k−2}F_{k−1}F_{k+1}F_{k+2} = F_{k}^{4} − 1

has the following generalization (Horadam and Shannon [10, Identity (2.5), Q = −1]):

W_{k−2}W_{k−1}W_{k+1}W_{k+2} = W_{k}^{4} + (−1)^{k}γ_{W} W_{k}^{2} − h_{W}^{2} ;

where γ_{W} = e_{W} (P^{2} − 1), h_{W} = e_{W} P, and e_{W} = P W_{0}W_{1} + W_{0}^{2} − W_{1}^{2}.

For the sequence of Lucas numbers, we have γ_{L} = 0 and e_{L} = 5 = h_{L}, so that

L_{k−2}L_{k−1}L_{k+1}L_{k+2} = L_{k}^{4} − 25;

while for the gibonacci sequence, γ_{G} = 0, h_{G} = e_{G} = G_{0}G_{1} + G_{0}^{2} − G_{1}^{2} and

G_{k−2}G_{k−1}G_{k+1}G_{k+2} = G_{k}^{4} − e_{G}^{2}.

Writing

w(k − 2)w(k − 1)w(k + 1)w(k + 2) = w(k)^{4} + (−1)^{k}γw(k)^{2} − h_{W}^{2}

and differentiating with respect to k and making use of (35) and (36) from Section 2.3 yields the result stated in the next proposition.

Proposition 9. For every integer k,

(W_{k−1} + W_{k−3})W_{k−1}W_{k+1}W_{k+2} + W_{k−2}(W_{k} + W_{k−2})W_{k+1}W_{k+2}

+ W_{k−2}W_{k−1}(W_{k} + W_{k+2})W_{k+2} + W_{k−2}W_{k−1}W_{k+1}(W_{k+3} + W_{k+1})  (61)
= 2W_{k}(W_{k+1} + W_{k−1})(2W_{k}^{2} + (−1)^{k}γ_{W}).

In particular,

(G_{k−1} + G_{k−3})G_{k−1}G_{k+1}G_{k+2} + G_{k−2}(G_{k} + G_{k−2})G_{k+1}G_{k+2}

+ G_{k−2}G_{k−1}(G_{k} + G_{k+2})G_{k+2} + G_{k−2}G_{k−1}G_{k+1}(G_{k+3} + G_{k+1})
= 4G_{k}^{3}(G_{k+1} + G_{k−1});

with the special cases

F_{k+1}F_{k+2}F_{2k−3} + F_{k−1}F_{k−2}F_{2k+3} = 2F_{k}^{3}L_{k} = 2F_{k}^{2}F_{2k}  (62)

and

L_{k+1}L_{k+2}F_{2k−3} + L_{k−1}L_{k−2}F_{2k+3} = 2L_{k}^{3}F_{k} = 2L_{k}^{2}F_{2k};  (63)

where, to arrive at (62) and (63), we used

F_{k+1} + F_{k−1} = L_{k}, L_{k+1} + L_{k−1} = 5F_{k},

and [23, Identity (16a)]

L_{m}F_{n} + L_{n}F_{m} = 2F_{m+n}.

Substituting k + 2 for k and arranging (62) and (63) as

\frac{F_{2k+1}}{F_{k}F_{k+1}} + \frac{F_{2k+7}}{F_{k+3}F_{k+4}} = \frac{2F_{k+2}^{2}F_{2k+4}}{F_{k+2}^{4} − 1}

and

\frac{F_{2k+1}}{L_{k}L_{k+1}} + \frac{F_{2k+7}}{L_{k+3}L_{k+4}} = \frac{2L_{k+2}^{2}F_{2k+4}}{L_{k+2}^{4} − 25} ;

and the use of the telescoping summation formula

∑_{k=1}^{n} (−1)^{k−1} (f_{k} + (−1)^{m−1}f_{k+m}) = ∑_{k=1}^{m} (−1)^{k−1}f_{k} + (−1)^{n−1} ∑_{k=1}^{m} (−1)^{k−1}f_{k+n}

yields the summation identities stated in the next proposition. Proposition 10. If n is a non-negative integer, then

∑_{k=1}^{n} \frac{(−1)^{k−1}F_{k+2}^{2}F_{2k+4}}{F_{k+2}^{4} − 1} = \frac{5}{6} + \frac{(−1)^{n−1}}{2} (\frac{F_{2n+3}}{F_{n+1}F_{n+2}} − \frac{F_{2n+5}}{F_{n+2}F_{n+3}} + \frac{F_{2n+7}}{F_{n+3}F_{n+4}} )

∑_{k=1}^{n} \frac{(−1)^{k−1}L_{k+2}^{2}F_{2k+4}}{L_{k+2}^{4} − 25} = \frac{5}{14} + \frac{(−1)^{n−1}}{2} (\frac{F_{2n+3}}{L_{n+1}L_{n+2}} − \frac{F_{2n+5}}{L_{n+2}L_{n+3}} + \frac{F_{2n+7}}{L_{n+3}L_{n+4}} );

with

∑_{k=1}^{∞} \frac{(−1)^{k−1}F_{k+2}^{2}F_{2k+4}}{F_{k+2}^{4} − 1} = \frac{5}{6},

∑_{k=1}^{∞} \frac{(−1)^{k−1}L_{k+2}^{2}F_{2k+4}}{L_{k+2}^{4} − 25} = \frac{5}{14}.

Arranging (61) as

\begin{matrix}\frac{W_{j−1} + W_{j−3}}{W_{j−2}} + \frac{W_{j} + W_{j−2}}{W_{j−1}} + \frac{W_{j+2} + W_{j}}{W_{j+1}} + \frac{W_{j+3} + W_{j+1}}{W_{j+2}} \\ = \frac{2W_{j}(W_{j+1} + W_{j−1})(2W_{j}^{3} + (−1)^{j}γ_{W})}{W_{j−2}W_{j−1}W_{j+1}W_{j+2}} \end{matrix}

and summing produces the next result. Proposition 11. If n and k are integers then,

∑_{j=1}^{n} \frac{(−1)^{j−1}2W_{j+k}(W_{j+k+1} + W_{j+k−1})(2W_{j+k}^{2} + (−1)^{j+k}γ_{W})}{W_{j+k−2}W_{j+k−1}W_{j+k+1}W_{j+k+2}}

= (−1)^{n+1} \frac{W_{n+k} + W_{n+k−2}}{W_{n+k−1}} + \frac{W_{k} + W_{k−2}}{W_{k−1}}

+ (−1)^{n+1} \frac{W_{n+k+3} + W_{n+k+1}}{W_{n+k+2}} + \frac{W_{k+3} + W_{k+1}}{W_{k+2}} ;

provided none of the denominators vanishes.

3.7 New identities from a reciprocal series of Fibonacci numbers

with subscripts k2^{j}

In this section we apply our method (first component) to discover new results associated with the following identity of Rabinowitz [21, Equation (4)]:

∑_{j=0}^{n} \frac{1}{U_{k2^{j}}} = \frac{1 + U_{k−1}}{U_{k}} + \frac{1 − (−1)^{k}}{U_{2k}} − \frac{U_{k2^{n}−1}}{U_{k2^{n}}} .

Writing

∑_{j=0}^{n} \frac{1}{u(k2^{j})} = \frac{1 + u(k − 1)}{u(k)} + \frac{1 − (−1)^{k}}{u(2k)} − \frac{u(k2^{n} − 1)}{u(k2^{n})},

and differentiating with respect to k gives

∑_{j=0}^{n} \frac{−2^{j}}{u(k2^{j})^{2}} \frac{du}{dk}(k2^{j}) = \frac{1}{u(k)} \frac{du}{dk}(k − 1) − \frac{(1 + u(k − 1))}{u(k)^{2}} \frac{du}{dk}(k)

− \frac{(−1)^{k}πi}{u(2k)} − \frac{2(1 − (−1)^{k})}{u(2k)^{2}} \frac{du}{dk}(2k)

− \frac{2^{n}}{u(k2^{n})} \frac{∂u}{∂k}(k2^{n} − 1) + \frac{2^{n}u(k2^{n} − 1)}{u(k2^{n})^{2}} \frac{∂u}{∂k}(k2^{n}).

Taking the real part while using (37)–(40), we have the next result.

Proposition 12. If n and k are positive integers, then

∑_{j=0}^{n} \frac{2^{j}V_{k2^{j}}}{U_{k2^{j}}^{2}} = \frac{(−1)^{k}2 + V_{k}}{U_{k}^{2}} + \frac{2(1 − (−1)^{k})V_{2k}}{U_{2k}^{2}} − \frac{2^{n+1}}{U_{k2^{n}}^{2}} ,

∑_{j=0}^{∞} \frac{2^{j}V_{k2^{j}}}{U_{k2^{j}}^{2}} = \frac{(−1)^{k}2 + V_{k}}{U_{k}^{2}} + \frac{2(1 − (−1)^{k})V_{2k}}{U_{2k}^{2}} .

Note that in arriving at the final form of the first expression in Proposition 12, we used

U_{r}V_{s} − V_{r}U_{s} = (−1)^{s}2U_{r−s}.

In particular, we have

∑_{j=0}^{n} \frac{2^{j}V_{2^{j}}}{U_{2^{j}}^{2}} = P + \frac{2δ^{2}}{P^{2}} − \frac{2^{n+1}}{U_{2^{n}}^{2}}

and

∑_{j=0}^{n} \frac{2^{j}V_{2^{j+1}}}{U_{2^{j+1}}^{2}} = \frac{δ^{2}}{P^{2}} − \frac{2^{n+1}}{U_{2^{n+1}}^{2}} ;

with the special cases

∑_{j=0}^{n} \frac{2^{j}L_{2^{j}}}{F_{2^{j}}^{2}} = 11 − \frac{2^{n+1}}{F_{2^{n}}^{2}}

and

∑_{j=0}^{n} \frac{2^{j}L_{2^{j+1}}}{F_{2^{j+1}}^{2}} = 5 − \frac{2^{n+1}}{F_{2^{n+1}}^{2}} .

4 Justification of the method

In this section we provide the rationale behind the method that was described in Section 2. To facilitate the discussion, we need the closed formula for the generalized Fibonacci sequence (W_{j}).

4.1 Closed formula for the generalized Fibonacci sequence

Standard methods for solving difference equations give the closed (Binet) formula of the generalized Fibonacci sequence (W_{j}) defined by the recurrence relation (16), in the nondegenerate case, P^{2} + 4 > 0, as

W_{j} = \frac{Aσ^{j} − Bτ^{j}}{σ − τ} = \frac{Aσ^{j} − Bτ^{j}}{δ} ,  (64)

where

A = W_{1} − W_{0}τ, B = W_{1} − W_{0}σ,  (65)

with

σ = \frac{P + \sqrt{P^{2} + 4}}{2},  τ = \frac{P − \sqrt{P^{2} + 4}}{2};  (66)

so that

σ + τ = P, σ − τ = \sqrt{P^{2} + 4} = δ, and στ = −1.  (67)

In particular,

U_{j} = \frac{σ^{j} − τ^{j}}{σ − τ} , V_{j} = σ^{j} + τ^{j}.  (68)

Using the Binet formulas, it is readily established that

U_{−j} = (−1)^{j−1}U_{j}, V_{−j} = (−1)^{j}V_{j}.  (69)

It is also straightforward to establish the following:

U_{j+1} + U_{j−1} = V_{j},  (70)
U_{j+1} − U_{j−1} = P U_{j},  (71)
V_{j+1} + V_{j−1} = U_{j}δ^{2},  (72)

and

V_{j+1} − V_{j−1} = P V_{j}.  (73)

As for the gibonacci sequence, we have

G_{j} = \frac{(G_{1} − G_{0}β) α^{j} − (G_{1} − G_{0}α) β^{j}}{\sqrt{5}} .  (74)

Lemma 13. For an integer j,

Aσ^{j} + Bτ^{j} = W_{j+1} + W_{j−1},  (75)

where A and B are as defined in (65).

Proof. Let

R_{j} = Aσ^{j} + Bτ^{j}.  (76)

Then,

σR_{j} = Aσ^{j+1} − Bτ^{j−1},
τR_{j} = −Aσ^{j−1} + Bτ^{j+1}.

Thus,

R_{j} · (σ − τ) = (Aσ^{j+1} − Bτ^{j+1}) + (Aσ^{j−1} − Bτ^{j−1});

that is

R_{j}δ = W_{j+1}δ + W_{j−1}δ,

or

R_{j} = W_{j+1} + W_{j−1}.  (77)

Identity (75) now follows by equating (76) and (77).

Identity (75) is at the heart of the justification of the calculus-based method of obtaining Fibonacci identities.

4.2 Justification of the method

We first state a required lemma. Lemma 14. If λ is a non-zero real number and x is real, then

\frac{d}{dx}λ^{x} = \begin{cases}λ^{x} ln λ, & if λ > 0; \\ λ^{x} (iπ(2m + 1) + ln (−λ)) , & if λ < 0, \end{cases}

where m is an integer.

Thus, if λ is a negative number, then \frac{d}{dx}λ^{x} is complex multi-valued with the principal value being

\frac{d}{dx}λ^{x} = λ^{x} (iπ + ln (−λ)) .  (78)

Proof. If λ is a negative number, then

λ^{x} = exp (x ln λ) = exp (x (ln(−1) + ln (−λ))) ,  (79)

where ln(−1), the complex logarithm of −1, is evaluated as

ln(−1) = ln (exp(iπ(2m + 1))) = iπ(2m + 1), m ∈ ℤ;

so that (79) can now be written as

λ^{x} = (−λ)^{x} exp (iπ(2m + 1)x) ,

from which the second result in (14) now follows by differentiation.

Consider a generalized Fibonacci function w(x) defined by

w(x) = \frac{Aσ^{x} − Bτ^{x}}{σ − τ} = \frac{Aσ^{x} − Bτ^{x}}{δ} , x ∈ ℝ,  (80)

where A and B are as defined in (65) and σ and τ are as given in (66).

Corresponding to (1), (68) and (74), we have the following special cases of (80):

f(x) = \frac{α^{x} − β^{x}}{\sqrt{5}} , l(x) = α^{x} + β^{x},

u(x) = \frac{σ^{x} − τ^{x}}{σ − τ} , v(x) = σ^{x} + τ^{x},  (81)

and

g(x) = \frac{A_{G}α^{x} − B_{G}β^{x}}{α − β} = \frac{A_{G}α^{x} − B_{G}β^{x}}{\sqrt{5}} ,  (82)

where

A_{G} = G_{1} − G_{0}β, B_{G} = G_{1} − G_{0}α.  (83)

Clearly,

w(j) = W_{j}, j ∈ ℤ;  (84)

that is

u(j) = U_{j}, v(j) = V_{j}, g(j) = G_{j}, f(j) = F_{j}, l(j) = L_{j}, j ∈ ℤ.

Theorem 15. The following identity holds:

ℜ ( \frac{d}{dx}w(x)|_{x=j∈ℤ}) = \frac{W_{j+1} + W_{j−1}}{δ} ln σ,  (85)

where, as usual, ℜ(X) denotes the real part of X.

Proof. We have

\frac{d}{dx}w(x) = \frac{1}{δ} (A\frac{d}{dx}σ^{x} − B \frac{d}{dx}τ^{x}) .

From (66), it is clear that σ > 0 and τ < 0 for all real numbers P. Thus, employing Lemma 14, we find

\frac{d}{dx}w(x) = \frac{1}{δ} (Aσ^{x} ln σ − Bτ^{x} ln (−τ) − Bτ^{x}πi(2m + 1))

= \frac{1}{δ} (Aσ^{x} ln σ + Bτ^{x} ln σ − Bτ^{x} ln σ − Bτ^{x} ln (−τ) − Bτ^{x}πi(2m + 1))

= \frac{1}{δ} ((Aσ^{x} + Bτ^{x}) ln σ − Bτ^{x} ln (−στ) − Bτ^{x}πi(2m + 1)) .

Since στ = −1, we obtain

\frac{d}{dx}w(x) = \frac{1}{δ} ((Aσ^{x} + Bτ^{x}) ln σ − Bτ^{x}πi(2m + 1)) .  (86)

Evaluating (86) at x = j ∈ ℤ, we have

\frac{d}{dx}w(x)|_{x=j∈ℤ} = \frac{1}{δ} ((W_{j+1} + W_{j−1}) ln σ − Bτ^{j}πi(2m + 1)) , by (75),  (87)

from which, on taking the real part, (85) follows, since m, τ, δ, B, W_{j+1}, and W_{j−1} are real quantities and σ is a positive number.

Of course the derivatives given in (3) are particular cases of (85) with δ = \sqrt{5}, F_{j+1} + F_{j−1} = L_{j}, and L_{j+1} + L_{j−1} = 5F_{j}. Similarly, (36), (38), (40) and (42) are all consequences of (85).

Thus, given a (generalized) Fibonacci identity having a free index, on account of (80),

(84) and (85), we can replace (generalized) Fibonacci numbers with (generalized) Fibonacci functions, perform differentiation and evaluate at integer values to obtain a new (generalized) Fibonacci identity.

5 The method, second component

The imaginary part of (87) establishes a connection between powers of σ and τ and the (generalized) Fibonacci numbers; through which new (generalized) Fibonacci identities can be obtained. We have

ℑ ( \frac{d}{dx}w(x)|_{x=j∈ℤ}) = −\frac{B_{W}}{δ} τ^{j}π(2m + 1),

where m is some integer and

B_{W} = B = W_{1} − W_{0}σ;

the principal value being

ℑ ( \frac{d}{dx}w(x)|_{x=j∈ℤ}) = −\frac{B_{W}}{δ} τ^{j}π.  (88)

Specializing to the special Lucas sequences, we have

B_{U} = 1, B_{V} = P − 2σ = τ − σ = −δ;

so that

ℑ ( \frac{d}{dx}u(x)|_{x=j∈ℤ}) = −\frac{πτ^{j}}{δ}

and

ℑ ( \frac{d}{dx}v(x)|_{x=j∈ℤ}) = πτ^{j}.

For the gibonacci sequence, we have

B_{G} = G_{1} − G_{0}α;

so that

ℑ ( \frac{d}{dx}g(x)|_{x=j∈ℤ}) = \frac{G_{0}α − G_{1}}{\sqrt{5}} πβ^{j}.

For the Fibonacci and Lucas numbers, we have

B_{F} = 1, B_{L} = −\sqrt{5};

so that, in view of (88), the method described in Section 2 can now be applied to a Fibonacci-Lucas identity with the prescription in step 3 and step 4 of Section 2 replaced with the following:

3. Simplify the equation obtained in step 2 and make the following replacements:

f(h(k, …)) = F_{h(k,...)},  (89)
l(h(k, …)) = L_{h(k,...)}.  (90)

4. Take the imaginary part of the whole expression/equation obtained in step 3, using also the following prescription:

ℑ\frac{∂f}{∂k} (h(k, …)) = −\frac{πβ^{h(k,...)}}{\sqrt{5}} ,  (91)
ℑ\frac{∂l}{∂k} (h(k, …)) = πβ^{h(k,...)}.  (92)

Remark 16. Formally, the method (second component) of establishing a connection between the powers of β and Fibonacci and Lucas numbers in a given Fibonacci-Lucas identity proceeds in two quick steps:

(i) Treat the subscripts of Fibonacci and Lucas numbers as variables and differentiate through the given identity, with respect to the free index of interest, using the rules of differential calculus.

(ii) Make the following replacements:

\frac{∂}{∂k} F_{h(k,...)} → \frac{−πβ^{h(k,...)}}{\sqrt{5}} \frac{∂}{∂k} h(k, …),  (93)
\frac{∂}{∂k} L_{h(k,...)} → πβ^{h(k,...)} \frac{∂}{∂k} h(k, …),  (94)
ln α → 0,  (95)
i → 1;  (96)

where i = \sqrt{−1} is the imaginary unit.

For example, given the double-angle formula:

F_{2k} = L_{k}F_{k},

we have, by step (i),

\frac{d}{dk} F_{2k} = \frac{d}{dk} (L_{k}F_{k}) = L_{k} \frac{d}{dk} F_{k} + F_{k} \frac{d}{dk} L_{k};

so that, by step (ii), using (93) and (94), we get

\frac{−πβ^{2k}}{\sqrt{5}} · \frac{d}{dk} (2k) = L_{k} · \frac{−πβ^{k}}{\sqrt{5}} + F_{k} · πβ^{k};

and hence,

2β^{k} = L_{k} − F_{k}\sqrt{5}.

For the special Lucas sequences, (89)–(92) read

u(h(k, …)) = U_{h(k,...)},  (97)
v(h(k, …)) = V_{h(k,...)},  (98)
ℑ\frac{∂u}{∂k}(h(k, …)) = −\frac{πτ^{h(k,...)}}{δ} ,  (99)
ℑ\frac{∂v}{∂k} (h(k, …)) = πτ^{h(k,...)};  (100)

while for the gibonacci sequence, we have

g(h(k, …)) = G_{h(k,...)},  (101)
ℑ\frac{∂g}{∂k} (h(k, …)) = \frac{(G_{0}α − G_{1})πβ^{h(k,...)}}{\sqrt{5}} = −\frac{B_{G}}{\sqrt{5}}πβ^{h(k,...)}.  (102)

5.1 Examples

We now give some examples to illustrate the use of (89)–(92) and (101) and (102) in obtaining new identities from known Fibonacci-Lucas identities.

Note that in the definitions in (1), F_{j} and L_{j} do not change when α and β are interchanged. Thus, α and β can be interchanged in a Fibonacci-Lucas identity involving Fibonacci numbers, Lucas numbers, α and β and no other irrational numbers. More generally, we have the observation stated in Proposition 17.

Proposition 17. A (generalized) Fibonacci identity involving (generalized) Fibonacci numbers as well as σ and τ and no other irrational numbers remains valid under the exchange of σ and τ.

Proof. From (64) and (65), we have

W_{j}(σ, τ) = \frac{W_{1} − W_{0}τ}{σ − τ} σ^{j} − \frac{W_{1} − W_{0}σ}{σ − τ} τ^{j}.

It is straightforward to verify that W_{j}(σ, τ) = W_{j}(τ, σ); and hence, the proposition.

The generalizations obtained in this section rest on Proposition 17.

5.1.1 Generalizations of the fundamental identity of Fibonacci and Lucas numbers

Differentiating the Fibonacci-Lucas function form

5f(k)^{2} − l(k)^{2} = (−1)^{k−1}4,

of the fundamental identity

5F_{k}^{2} − L_{k}^{2} = (−1)^{k−1}4,

and applying the prescription of (89)–(92) yields

5F_{k}β^{k+r} + L_{k}β^{k+r}\sqrt{5} = (−1)^{k}2β^{r}\sqrt{5},  (103)

where r is an arbitrary integer; and also, by Proposition 17,

5F_{k}α^{k+r} − L_{k}α^{k+r}\sqrt{5} = (−1)^{k−1}2α^{r}\sqrt{5}.  (104)

Combining (103) and (104) according to the Binet form (74) leads to the result stated in Proposition 18.

Proposition 18. If k and r are integers, then

5F_{k}G_{k+r} − L_{k} (G_{k+r+1} + G_{k+r−1}) = (−1)^{k−1}2 (G_{r+1} + G_{r−1}) .

Writing the function form of the identity of Proposition 18 as

\begin{matrix}5f(k − r)g(k) − l(k − r) (g(k + 1) + g(k − 1)) \\ = (−1)^{k−r−1}2 (g(r + 1) + g(r − 1)) , \end{matrix}

and differentiating with respect to r, using again the prescription (89)–(92) and (101) and (102), we find

\begin{matrix}5β^{k−r}G_{k} + β^{k−r}\sqrt{5} (G_{k+1} + G_{k−1}) \\ = (−1)^{k−r}2\sqrt{5} (G_{r+1} + G_{r−1}) + (−1)^{k−r}2\sqrt{5} (G_{0}β^{r−1} − G_{1}β^{r}) , \end{matrix}  (105)

and, on account of Proposition 17, also

\begin{matrix}5α^{k−r}G_{k} − α^{k−r}\sqrt{5} (G_{k+1} + G_{k−1}) \\ = (−1)^{k−r+1}2\sqrt{5} (G_{r+1} + G_{r−1}) + (−1)^{k−r+1}2\sqrt{5} (G_{0}α^{r−1} − G_{1}α^{r}) . \end{matrix}  (106)

Combining (105) and (106) gives the next result.

Proposition 19. If k, r, and s are integers, then

\begin{matrix}5G_{k}H_{k+s−r} − (G_{k+1} + G_{k−1}) (H_{k+s−r+1} + H_{k+s−r−1}) \\ = (−1)^{k−r+1}2 (G_{r+1} + G_{r−1}) (H_{s+1} + H_{s−1}) \\ + (−1)^{k−r}2 (G_{0} (H_{r+s} + H_{r+s−2}) + G_{1} (H_{r+s+1} + H_{r+s−1})) . \end{matrix}  (107)

Setting r = 0 in (107) gives

\begin{matrix}5G_{k}H_{k+s} − (G_{k+1} + G_{k−1}) (H_{k+s+1} + H_{k+s−1}) \\ = (−1)^{k−1}2 (G_{1} (H_{s+1} + H_{s−1}) − G_{0} (H_{s+2} + H_{s})) , \end{matrix}

which upon using s = 0 and (H_{j}) = (G_{j}) gives

5G_{k}^{2} − (G_{k+1} + G_{k−1})^{2} = (−1)^{k}4e_{G},

where, as usual, e_{G} = G_{0}^{2} − G_{1}^{2} + G_{0}G_{1}.

5.1.2 Generalizations of the formula for the sum of squares of two consecutive Fibonacci numbers

Differentiating the Fibonacci function form

f(k + 1)^{2} + f(k)^{2} = f(2k + 1),

of the identity

F_{k+1}^{2} + F_{k}^{2} = F_{2k+1},

we have

f(k + 1)\frac{df}{dk} (k + 1) + f(k)\frac{df}{dk} (k) = \frac{df}{dk} (2k + 1),

and taking the imaginary part, by (89),

F_{k+1}ℑ\frac{df}{dk} (k + 1) + F_{k}ℑ\frac{df}{dk} (k) = ℑ\frac{df}{dk} (2k + 1),

and by (91),

β^{k+1}F_{k+1} + β^{k}F_{k} = β^{2k+1}.

Thus

β^{r+k+1}F_{k+1} + β^{r+k}F_{k} = β^{2k+r+1},

or

β^{s+1}F_{k+1} + β^{s}F_{k} = β^{k+s+1},  (108)

where s is an arbitrary integer, and also

α^{s+1}F_{k+1} + α^{s}F_{k} = α^{k+s+1}.  (109)

Combining (108) and (109) according to the Binet formula, we have the next result, equivalent to Vajda [23, Identity (8)].

Proposition 20. If k and s are integers, then

F_{k+1}G_{s+1} + F_{k}G_{s} = G_{k+s+1}.  (110)

Writing k − 1 for k and setting s = 0 in (110) gives the well-known result:

F_{k}G_{1} + F_{k−1}G_{0} = G_{k}.  (111)

Differentiating the Fibonacci function form of (110), that is

f(k + 1)g(s + 1) + f(k)g(s) = g(k + s + 1),

with respect to k and using (91) and (102) gives

−β^{k+1}G_{s+1} − β^{k}G_{s} = β^{k+s+1} (G_{0}α − G_{1}) ,

or

β^{k+1}G_{s+1} + β^{k}G_{s} = G_{0}β^{k+s} + G_{1}β^{k+s+1}  (112)

and also

α^{k+1}G_{s+1} + α^{k}G_{s} = G_{0}α^{k+s} + G_{1}α^{k+s+1}.  (113)

Combining (112) and (113) produces the next result. Proposition 21. If k and s are integers, then

H_{k+1}G_{s+1} + H_{k}G_{s} = G_{0}H_{k+s} + G_{1}H_{k+s+1}.

In particular,

G_{k+1}^{2} + G_{k}^{2} = G_{0}G_{2k} + G_{1}G_{2k+1}.

5.1.3 Generalizations of the d’Ocagne identity Differentiating with respect to k, the Fibonacci function form

f(r + 1)f(k) − f(r)f(k + 1) = (−1)^{r}f(k − r),

of the d’Ocagne identity

F_{r+1}F_{k} − F_{r}F_{k+1} = (−1)^{r}F_{k−r},

gives, upon taking the imaginary part while using (89) and (91),

β^{k}F_{r+1} − β^{k+1}F_{r} = (−1)^{r}β^{k−r},

and also

α^{k}F_{r+1} − α^{k+1}F_{r} = (−1)^{r}α^{k−r},

and hence, the result stated in the next proposition.

Proposition 22. If r and k are integers, then

F_{r+1}G_{k} − F_{r}G_{k+1} = (−1)^{r}G_{k−r}.  (114)

Differentiating the Fibonacci function form of (114):

f(r + 1)g(k) − f(r)g(k + 1) = (−1)^{r}g(k − r),

with respect to r, we find

\begin{matrix}g(k)\frac{df}{dr}(r + 1) − g(k + 1)\frac{df}{dr}(r) \\ = (−1)^{r}πg(k − r)i − (−1)^{r} \frac{∂g}{∂r} (k − r); \end{matrix}

and consequent upon use of (91), (101) and (102),

\begin{matrix}β^{r+s}G_{k+1} − β^{r+s+1}G_{k} \\ = (−1)^{r}β^{s}G_{k−r}\sqrt{5} + (−1)^{r} (G_{0}β^{k+s−r−1} + G_{1}β^{k+s−r}) \end{matrix}

and also

\begin{matrix}α^{r+s}G_{k+1} − α^{r+s+1}G_{k} \\ = (−1)^{r−1}α^{s}G_{k−r}\sqrt{5} + (−1)^{r} (G_{0}α^{k+s−r−1} + G_{1}α^{k+s−r}) \end{matrix}

and hence, the identity stated in the next proposition. Proposition 23. If r, s, and k are integers, then H_{r+s}G_{k+1} − H_{r+s+1}G_{k} = (−1)^{r−1} (H_{s+1} + H_{s−1}) G_{k−r} + (−1)^{r} (G_{0}H_{k+s−r−1} + G_{1}H_{k+s−r}) .

(115)

Writing r − s for r in (115) and setting s = 0 gives

H_{r}G_{k+1} − H_{r+1}G_{k} = (−1)^{r−1} (2H_{1} − H_{0}) G_{k−r} + (−1)^{r} (G_{0}H_{k−r−1} + G_{1}H_{k−r}) ; and, in particular,

G_{r}G_{k+1} − G_{r+1}G_{k} = (−1)^{r} (G_{0}G_{k−r+1} − G_{1}G_{k−r}) .

5.1.4 Generalizations of Fibonacci power formulas The well-known identity

G_{k+1}^{2} + G_{k−2}^{2} = 2 (G_{k}^{2} + G_{k−1}^{2}) ,  (116)

has the gibonacci function form g(k + 1)^{2} + g(k − 2)^{2} = 2 (g(k)^{2} + g(k − 1)^{2}) , which, upon differentiation, gives

g(k + 1)\frac{∂g}{∂k} (k + 1) + g(k − 2)\frac{∂g}{∂k} (k − 2) = 2g(k)\frac{∂g}{∂k} (k) + 2g(k − 1)\frac{∂g}{∂k} (k − 1),

and by (101) and (102):

β^{k+1}G_{k+1} + β^{k−2}G_{k−2} = 2 (β^{k}G_{k} + β^{k−1}G_{k−1});

and multiplying through by β^{s−k}, s an arbitrary integer:

β^{s+1}G_{k+1} + β^{s−2}G_{k−2} = 2 (β^{s}G_{k} + β^{s−1}G_{k−1});  (117)

and also

α^{s+1}G_{k+1} + α^{s−2}G_{k−2} = 2 (α^{s}G_{k} + α^{s−1}G_{k−1}) .  (118)

Combining (117) and (118) according to the Binet formula yields the following generalization of (116).

Proposition 24. If s and k are integers, then

G_{k+1}H_{s+1} + G_{k−2}H_{s−2} = 2G_{k}H_{s} + 2G_{k−1}H_{s−1}.

Long’s identities [16, (31)–(35)] are all special cases of the above proposition.

The reader is invited to apply the method (second component) to verify that the identity

G_{k+1}^{3} = 3G_{k}^{3} + 6G_{k−1}^{3} − 3G_{k−2}^{3} − G_{k−3}^{3}, [2, Equation (3)],

has the following generalization.

Proposition 25. If k, r, and s are integers, then

G_{k+1}H_{r+1}I_{s+1} = 3G_{k}H_{r}I_{s} + 6G_{k−1}H_{r−1}I_{s−1} − 3G_{k−2}H_{r−2}I_{s−2} − G_{k−3}H_{r−3}I_{s−3},

where (G_{j}), (H_{j}), and (I_{j}) are gibonacci sequences.

5.1.5 Generalizations of a triple-angle identity of Lucas

Differentiating the Fibonacci function form

f(3k) = f(k + 1)^{3} + f(k)^{3} − f(k − 1)^{3},

of the identity (see Vorob’ev [24, p. 16]):

F_{3k} = F_{k+1}^{3} + F_{k}^{3} − F_{k−1}^{3},  (119)

gives, after using (89) and (91),

β^{3k} = F_{k+1}^{2}β^{k+1} + F_{k}^{2}β^{k} − F_{k−1}^{2}β^{k−1},

that is

β^{2k+s} = F_{k+1}^{2}β^{s+1} + F_{k}^{2}β^{s} − F_{k−1}^{2}β^{s−1},  (120)

where s is an arbitrary integer; and also

α^{2k+s} = F_{k+1}^{2}α^{s+1} + F_{k}^{2}α^{s} − F_{k−1}^{2}α^{s−1}.  (121)

Combining (120) and (121), we have the first generalization of (119). Proposition 26. If k and s are integers, then

G_{2k+s} = F_{k+1}^{2}G_{s+1} + F_{k}^{2}G_{s} − F_{k−1}^{2}G_{s−1}.

The reader is invited to check that application of the method (second component) two more times with respect to k gives the full generalization of (119), stated in Proposition 27. Proposition 27. If k, r, and s are integers, then

\begin{matrix}G_{0}H_{0}I_{k+r+s−2} + (G_{0}H_{1} + G_{1}H_{0}) I_{k+r+s−1} + G_{1}H_{1}I_{k+r+s} \\ = G_{s+1}H_{r+1}I_{k+1} + G_{s}H_{r}I_{k} − G_{s−1}H_{r−1}I_{k−1}, \end{matrix}

where (G_{j}), (H_{j}), and (I_{j}) are gibonacci sequences with initial terms G_{0}, G_{1}; H_{0}, H_{1} and I_{0}, I_{1}.

In particular,

G_{0}^{2}G_{3k−2} + 2G_{0}G_{1}G_{3k−1} + G_{1}^{2}G_{3k} = G_{k+1}^{3} + G_{k}^{3} − G_{k−1}^{3},

with the special cases:

F_{3k} = F_{k+1}^{3} + F_{k}^{3} − F_{k−1}^{3},
5L_{3k} = L_{k+1}^{3} + L_{k}^{3} − L_{k−1}^{3}.

Identities (42)–(45) of Long [16] are special cases of the identity stated in Proposition 27.

5.1.6 Identities from the golden ratio power reduction formula Differentiating the Fibonacci function form

α^{k} = αf(k) + f(k − 1),

of the golden ratio power reduction formula

α^{k} = αF_{n} + F_{n−1};

we obtain

α^{k} ln α = α\frac{∂}{∂k} f(k) + \frac{∂}{∂k} f(k − 1),

which, upon use of (22), gives

α^{k}\sqrt{5} = αL_{k} + L_{k−1},

and hence

α^{k+r}\sqrt{5} = α^{r+1}L_{k} + α^{r}L_{k−1},  (122)

where r is an arbitrary integer. Also,

−β^{k+r}\sqrt{5} = β^{r+1}L_{k} + β^{r}L_{k−1},  (123)

Combining (122) and (123) gives the next result. Proposition 28. If k and r are integers, then

G_{k+r+1} + G_{k+r−1} = L_{k}G_{r+1} + L_{k−1}G_{r}.

Differentiating the function form of the identity stated in Proposition 28 with respect to k, making use of (92) and (102) while taking the imaginary part gives

−\frac{B_{G}}{\sqrt{5}}β^{k+r+1} − \frac{B_{G}}{\sqrt{5}}β^{k+r−1} = β^{k}G_{r+1} + β^{k−1}G_{r};

that is

(G_{0}α − G_{1}) β^{k+r+1} + (G_{0}α − G_{1}) β^{k+r−1} = β^{k}G_{r+1}\sqrt{5} + β^{k−1}G_{r}\sqrt{5},

and hence

G_{0}β^{k+r} + G_{1}β^{k+r+1} + G_{0}β^{k+r−2} + G_{1}β^{k+r−1} = −β^{k}G_{r+1}\sqrt{5} − β^{k−1}G_{r}\sqrt{5}.  (124)

Also,

G_{0}α^{k+r} + G_{1}α^{k+r+1} + G_{0}α^{k+r−2} + G_{1}α^{k+r−1} = α^{k}G_{r+1}\sqrt{5} + α^{k−1}G_{r}\sqrt{5}.  (125)

Combining (124) and (125) using the Binet formula and the lemma gives the result stated in the next proposition. Proposition 29. If k and r are integers, then

G_{0} (H_{r+k} + H_{r+k−2}) + G_{1} (H_{r+k+1} + H_{r+k−1}) = G_{r} (H_{k} + H_{k−2}) + G_{r+1} (H_{k+1} + H_{k−1}) . In particular,

H_{k+r+1} + H_{k+r−1} = F_{r} (H_{k} + H_{k−2}) + F_{r+1} (H_{k+1} + H_{k−1})

and

5H_{k+r} = L_{r} (H_{k} + H_{k−2}) + L_{r+1} (H_{k+1} + H_{k−1}) ;

with the special cases

L_{r+k} = F_{r}L_{k−1} + F_{r+1}L_{k},
F_{r+k} = F_{r}F_{k−1} + F_{r+1}F_{k},
5F_{r+k} = L_{r}L_{k−1} + L_{r+1}L_{k},
L_{r+k} = L_{r}F_{k−1} + L_{r+1}F_{k}.

5.1.7 Sum of products of the terms of two gibonacci sequences Simple telescoping of the identity

G_{j+k}^{2} = G_{j+k+1}G_{j+k} − G_{j+k}G_{j+k−1},

gives

∑_{j=1}^{n} G_{j+k}^{2} = G_{n+k}G_{n+k+1} − G_{k}G_{k+1},

whose gibonacci function form is

∑_{j=1}^{n} g(j + k)^{2} = g(n + k)g(n + k + 1) − g(k)g(k + 1).

Differentiating:

\begin{aligned}2 ∑_{j=1}^{n} g(j + k)\frac{∂g}{∂k} (j + k) & = g(n + k)\frac{∂g}{∂k} (n + k + 1) + g(n + k + 1)\frac{∂g}{∂k} (n + k) \\ & − g(k)\frac{∂g}{∂k} (k + 1) − g(k + 1)\frac{∂g}{∂k} (k), \end{aligned}

and, upon use of (101) and (102), we have

2 ∑_{j=1}^{n} β^{j+k}G_{j+k} = β^{n+k+1}G_{n+k} + β^{n+k}G_{n+k+1} − β^{k}G_{k+1} − β^{k+1}G_{k};

which, multiplying through by β^{s−k}, s an arbitrary integer, gives

2 ∑_{j=1}^{n} β^{j+s}G_{j+k} = β^{n+s+1}G_{n+k} + β^{n+s}G_{n+k+1} − β^{s}G_{k+1} − β^{s+1}G_{k},  (126)

and hence also

2 ∑_{j=1}^{n} α^{j+s}G_{j+k} = α^{n+s+1}G_{n+k} + α^{n+s}G_{n+k+1} − α^{s}G_{k+1} − α^{s+1}G_{k}.  (127)

Combining (126) and (127) according to the Binet formula yields the result stated in the next proposition. Proposition 30. If k, s, and n are integers, then

∑_{j=1}^{n} G_{j+k}H_{j+s} = G_{n+k}H_{n+s+1} + G_{n+k+1}H_{n+s} − G_{k+1}H_{s} − G_{k}H_{s+1}.  (128)

Using generating function techniques, Berzsenyi [1] found alternative expressions for the special case

∑_{j=0}^{2n+s} G_{j}G_{j+2k+s}, s = 1 or 0.

Long’s identities [16, (4)–(7)] are alternative expressions for special cases of the above proposition. Kronenburg [15, Identity (11.1)] also derived (128).

5.1.8 Generalizations of Hoggatt and Bicknell’s identity (5)

Taking the imaginary part of (51) on page 15, using (89), we have

\begin{aligned}∑_{j=0}^{4n+1} & (−1)^{j−1}\binom{4n + 1}{j}F_{j+k}^{3}ℑ\frac{∂f}{∂k} (j + k) \\ & = 25^{n} (F_{2n+k+1}^{3}ℑ\frac{∂f}{∂k} (2n + k + 1) − F_{2n+k}^{3}ℑ\frac{∂f}{∂k} (2n + k)); \end{aligned}

which, applying (91), gives

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}F_{j+k}^{3}β^{j+k} = 25^{n} (F_{2n+k+1}^{3}β^{2n+k+1} − F_{2n+k}^{3}β^{2n+k}) ,

and hence

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}F_{j+k}^{3}β^{j+r} = 25^{n} (F_{2n+k+1}^{3}β^{2n+r+1} − F_{2n+k}^{3}β^{2n+r});  (129)

where r is an arbitrary integer.

Interchanging α and β in (129), we also have

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}F_{j+k}^{3}α^{j+r} = 25^{n} (F_{2n+k+1}^{3}α^{2n+r+1} − F_{2n+k}^{3}α^{2n+r}) .  (130)

Combining (129) and (130) according to the Binet formula, we have the result stated in the next proposition.

Proposition 31. If n is a non-negative integer and k is an integer, then

∑_{j=0}^{4n+1} (−1)^{j−1}\binom{4n + 1}{j}F_{j+k}^{3}G_{j+r} = 25^{n} (F_{2n+k+1}^{3}G_{2n+r+1} − F_{2n+k}^{3}G_{2n+r}) .  (131)

Observe that (5) and (6) are particular cases of (131).

By applying the method (second component) three more times, with k as the index of interest, the reader is invited to establish the further generalization presented in the next proposition.

Proposition 32. If n is a non-negative integer and r, k, and s are integers, then

\begin{aligned}∑_{j=0}^{4n+1} & (−1)^{j−1}\binom{4n + 1}{j}E_{j+k}G_{j+r}H_{j+s}I_{j+t} \\ & = 25^{n} (E_{2n+k+1}G_{2n+r+1}H_{2n+s+1}I_{2n+t+1} − E_{2n+k}G_{2n+r}H_{2n+s}I_{2n+t}) ; \end{aligned}

where (E)_{j∈ℤ}, (G)_{j∈ℤ}, (H)_{j∈ℤ}, and (I)_{j∈ℤ} are gibonacci sequences with seeds E_{0}, E_{1}; G_{0}, G_{1}; H_{0}, H_{1} and I_{0}, I_{1}.

5.1.9 Generalizations of an identity of Melham

In this section, we present a generalization of Melham’s identity, (58) on page 19, that is

6 (∑_{j=0}^{2n−1} G_{k+j}^{2})^{2} = F_{2n}^{2} (G_{k+n−2}^{4} + 4G_{k+n−1}^{4} + 4G_{k+n}^{4} + G_{k+n+1}^{4});

whose Fibonacci-gibonacci function derivative is

(6 ∑_{j=0}^{2n−1} g(k + j)^{2}) ∑_{j=0}^{2n−1} g(k + j)\frac{∂g}{∂k} (k + j)

= f(2n)^{2} (g(k + n − 2)^{3} \frac{∂g}{∂k} (k + n − 2) + 4g(k + n − 1)^{3} \frac{∂g}{∂k} (k + n − 1)

+4g(k + n)^{3} \frac{∂g}{∂k} (k + n) + g(k + n + 1)^{3} \frac{∂g}{∂k} (k + n + 1)) .

Taking the imaginary part and applying (101) and (102) leads to

\begin{matrix}6 ∑_{j=0}^{2n−1} G_{k+j}^{2} ∑_{j=0}^{2n−1} β^{k+j}G_{k+j} \\ = F_{2n}^{2} (β^{k+n−2}G_{k+n−2}^{3} + 4β^{k+n−1}G_{k+n−1}^{3} + 4β^{k+n}G_{k+n}^{3} + β^{k+n+1}G_{k+n+1}^{3}) , \end{matrix}

which multiplying through by β^{r−k}, r an arbitrary integer, gives

\begin{matrix}6 ∑_{j=0}^{2n−1} G_{k+j}^{2} ∑_{j=0}^{2n−1} β^{r+j}G_{k+j} \\ = F_{2n}^{2} (β^{r+n−2}G_{k+n−2}^{3} + 4β^{r+n−1}G_{k+n−1}^{3} + 4β^{r+n}G_{k+n}^{3} + β^{r+n+1}G_{k+n+1}^{3}) , \end{matrix}  (132)

and also

\begin{matrix}6 ∑_{j=0}^{2n−1} G_{k+j}^{2} ∑_{j=0}^{2n−1} α^{r+j}G_{k+j} \\ = F_{2n}^{2} (α^{r+n−2}G_{k+n−2}^{3} + 4α^{r+n−1}G_{k+n−1}^{3} + 4α^{r+n}G_{k+n}^{3} + α^{r+n+1}G_{k+n+1}^{3}) . \end{matrix} (133)

Combining (132) and (133) according to the Binet formula yields the next result. Proposition 33. If k and r are integers, then

\begin{matrix}6 ∑_{j=0}^{2n−1} G_{k+j}^{2} ∑_{j=0}^{2n−1} H_{r+j}G_{k+j} \\ = F_{2n}^{2} (H_{r+n−2}G_{k+n−2}^{3} + 4H_{r+n−1}G_{k+n−1}^{3} + 4H_{r+n}G_{k+n}^{3} + H_{r+n+1}G_{k+n+1}^{3}) . \end{matrix}

Repeated application of the method (second component) two more times to the identity stated in Proposition 33 with k as the index of interest establishes the next result. Proposition 34. If k, r, s, and t are integers, then

2 ∑_{j=0}^{2n−1} G_{k+j}J_{t+j} ∑_{j=0}^{2n−1} H_{r+j}I_{s+j} + 2 ∑_{j=0}^{2n−1} I_{s+j}J_{t+j} ∑_{j=0}^{2n−1} G_{k+j}H_{r+j}

+ 2 ∑_{j=0}^{2n−1} G_{k+j}I_{s+j} ∑_{j=0}^{2n−1} H_{r+j}J_{t+j}

= F_{2n}^{2} (G_{k+n−2}H_{r+n−2}I_{s+n−2}J_{t+n−2} + 4 G_{k+n−1}H_{r+n−1}I_{s+n−1}J_{t+n−1}

+4 G_{k+n}H_{r+n}I_{s+n}J_{t+n} + G_{k+n+1}H_{r+n+1}I_{s+n+1}J_{t+n+1}) ;

where (G_{j}), (H_{j}), (I_{j}), and (J_{j}) are gibonacci sequences.

5.1.10 Generalizations of an identity of Howard Taking the imaginary part of (44) on page 12 gives

G_{k+r}ℑ\frac{d}{ds}f(s) + (−1)^{r−1}G_{k}ℑ\frac{∂f}{∂s} (s − r) = F_{r}ℑ\frac{∂g}{∂s}(k + s);  (134)

which, on using (91) and (102), yields

β^{s}G_{k+r} + (−1)^{r−1}β^{s−r}G_{k} = β^{k+s−1}G_{0}F_{r} + β^{k+s}G_{1}F_{r}  (135)

and hence, also

α^{s}G_{k+r} + (−1)^{r−1}α^{s−r}G_{k} = α^{k+s−1}G_{0}F_{r} + α^{k+s}G_{1}F_{r}.  (136)

Combining (135) and (136) provides the following generalization of Howard’s identity (11).

Proposition 35. If r, s, and k are integers, then

H_{s}G_{k+r} + (−1)^{r−1}H_{s−r}G_{k} = F_{r} (G_{0}H_{k+s−1} + G_{1}H_{k+s}) .

By taking the imaginary part of (45) on page 12, the reader is invited to establish the result stated in Proposition 36.

Proposition 36. If k, r, s, and t are integers, then

F_{s} (G_{0}H_{k+r+t−1} + G_{1}H_{k+r+t}) − (−1)^{r}F_{s−r}G_{k} (H_{t+1} + H_{t−1}) + (−1)^{r}G_{k}H_{s−r+t} = G_{k+s}H_{r+t}.

5.1.11 Generalizations of Candido’s identity

Taking the imaginary part of (55) on page 18 according to the prescription of (101) and (102), we have

\begin{matrix}2 (β^{k}G_{k}^{3} + β^{k+1}G_{k+1}^{3} + β^{k+2}G_{k+2}^{3}) \\ = (G_{k}^{2} + G_{k+1}^{2} + G_{k+2}^{2}) (β^{k}G_{k} + β^{k+1}G_{k+1} + β^{k+2}G_{k+2}); \end{matrix}

which, upon multiplication by β^{r−k} gives

\begin{matrix}2 (β^{r}G_{k}^{3} + β^{r+1}G_{k+1}^{3} + β^{r+2}G_{k+2}^{3}) \\ = (G_{k}^{2} + G_{k+1}^{2} + G_{k+2}^{2}) (β^{r}G_{k} + β^{r+1}G_{k+1} + β^{r+2}G_{k+2}); \end{matrix}  (137)

which, on account of Proposition 17, also implies

\begin{matrix}2 (α^{r}G_{k}^{3} + α^{r+1}G_{k+1}^{3} + α^{r+2}G_{k+2}^{3}) \\ = (G_{k}^{2} + G_{k+1}^{2} + G_{k+2}^{2}) (α^{r}G_{k} + α^{r+1}G_{k+1} + α^{r+2}G_{k+2}) . \end{matrix}  (138)

Combining according to the Binet formula gives the following generalization of Candido’s identity.

Proposition 37. If r and k are integers, then

\begin{matrix}2 (H_{r}G_{k}^{3} + H_{r+1}G_{k+1}^{3} + H_{r+2}G_{k+2}^{3}) \\ = (G_{k}^{2} + G_{k+1}^{2} + G_{k+2}^{2}) (H_{r}G_{k} + H_{r+1}G_{k+1} + H_{r+2}G_{k+2}) . \end{matrix}  (139)

By differentiating the gibonacci function form of (139) three more times with respect to k, the reader is invited to demonstrate the further generalization of the Candido identity stated in Proposition 38.

Proposition 38. If k, r, s, and t are integers, then

6 (G_{k}H_{r}M_{s}N_{t} + G_{k+1}H_{r+1}M_{s+1}N_{t+1} + G_{k+2}H_{r+2}M_{s+2}N_{t+2})
= (G_{k}M_{s} + G_{k+1}M_{s+1} + G_{k+2}M_{s+2}) (H_{r}N_{t} + H_{r+1}N_{t+1} + H_{r+2}N_{t+2})
+ (G_{k}H_{r} + G_{k+1}H_{r+1} + G_{k+2}H_{r+2}) (M_{s}N_{t} + M_{s+1}N_{t+1} + M_{s+2}N_{t+2})
+ (G_{k}N_{t} + G_{k+1}N_{t+1} + G_{k+2}N_{t+2}) (H_{r}M_{s} + H_{r+1}M_{s+1} + H_{r+2}M_{s+2}) ,

where (M_{j})_{j∈ℤ} and (N_{j})_{j∈ℤ} are gibonacci sequences with seeds M_{0} and M_{1} and N_{0} and N_{1}.

In particular, we have 6 (F_{k}F_{r}F_{s}F_{t} + F_{k+1}F_{r+1}F_{s+1}F_{t+1} + F_{k+2}F_{r+2}F_{s+2}F_{t+2}) = (F_{k}F_{s} + F_{k+1}F_{s+1} + F_{k+2}F_{s+2}) (F_{r}F_{t} + F_{r+1}F_{t+1} + F_{r+2}F_{t+2})

+ (F_{k}F_{r} + F_{k+1}F_{r+1} + F_{k+2}F_{r+2}) (F_{s}F_{t} + F_{s+1}F_{t+1} + F_{s+2}F_{t+2}) + (F_{k}F_{t} + F_{k+1}F_{t+1} + F_{k+2}F_{t+2}) (F_{r}F_{s} + F_{r+1}F_{s+1} + F_{r+2}F_{s+2}) .

5.2 Generalizations of a Lucas number identity

Differentiating the Lucas function:

l(2r) + 2(−1)^{r} = l(r)^{2},

of the well-known identity

L_{2r} + 2(−1)^{r} = L_{r}^{2},

gives

\frac{d}{dr}l(2r) + (−1)^{r}iπ = l(r)\frac{d}{dr}l(r),

which, employing (90) and (92), yields

β^{2r} + (−1)^{r} = β^{r}L_{r},

or multiplying through by β^{s−r},

β^{s+r} + (−1)^{r}β^{s−r} = β^{s}L_{r},  (140)

and also

α^{s+r} + (−1)^{r}α^{s−r} = α^{s}L_{r},  (141)

Combining (140) and (141) according to the Binet formula gives the following multiplication formula (also Vajda [23, Formula (10a)]). Proposition 39. If r and s are integers, then

G_{s+r} + (−1)^{r}G_{s−r} = L_{r}G_{s}.

Differentiating the function

g(s + r) + (−1)^{r}g(s − r) = l(r)g(s),

with respect to r gives

\frac{∂g}{∂r} (s + r) + (−1)^{r}πig(s − r) − (−1)^{r} \frac{∂g}{∂r} (s − r) = g(s)\frac{d}{dr}l(r),

and taking the imaginary part, making use of (92), (101) and (102), we find

\begin{matrix}G_{0} (β^{k+r+s−1} − (−1)^{r}β^{k+s−r−1}) + G_{1} (β^{k+s−1+r} − (−1)^{r}β^{k+s−1−r}) \\ = (−1)^{r}β^{k}G_{s−r}\sqrt{5} − β^{r+k}G_{s}\sqrt{5}, \end{matrix}

and also

\begin{matrix}G_{0} (α^{k+r+s−1} − (−1)^{r}α^{k+s−r−1}) + G_{1} (α^{k+s−1+r} − (−1)^{r}α^{k+s−1−r}) \\ = (−1)^{r−1}α^{k}G_{s−r}\sqrt{5} + α^{r+k}G_{s}\sqrt{5}, \end{matrix}

and hence the next result.

Proposition 40. If r, k, and s are integers, then

\begin{matrix}G_{s} (H_{r+k+1} + H_{r+k−1}) + (−1)^{r−1}G_{s−r} (H_{k+1} + H_{k−1}) \\ = G_{0}F_{r} (H_{k+s} + H_{k+s−2}) + G_{1}F_{r} (H_{k+s+1} + H_{k+s−1}) . \end{matrix}

Note that we used

G_{s+r} − (−1)^{r}G_{s−r} = F_{r} (G_{s+1} + G_{s−1}) , [23, Formula (10b)].

Note also that (H_{j}) = (L_{j}) in the proposition gives the Howard identity (11) while (H_{j}) = (F_{j}) produces (12).

Remark 41. It should be noted that in order that the results obtained by applying the method (second component) be valid, it is necessary that the (generalized) Fibonacci function identity obtained from the original (generalized) Fibonacci number identity holds for all real x; not just integers. Also, the derivative of (−1)^{x} is an imaginary number for real x. In order not to lose this value, therefore, (−1)^{2k} must not be set equal to unity in the original Fibonacci identity when converting to the Fibonacci function form. The method (second component) involves taking the imaginary part. This point is taken into consideration in the examples presented in § 5.2.1 to 5.2.3.

5.2.1 A generalization of the Gelin-Cesàro identity

Since

G_{k−1}G_{k+1} = G_{k}^{2} − (−1)^{k}e_{G}, [23, Identity 28],

where e_{G} = G_{0}^{2} − G_{1}^{2} + G_{1}G_{0}, and

G_{k−2}G_{k+2} = (G_{k} − G_{k−1}) (G_{k} + G_{k+1})
= G_{k}^{2} + (G_{k}G_{k+1} − G_{k−1}G_{k}) − G_{k−1}G_{k+1}
= G_{k}^{2} + G_{k}^{2} − G_{k−1}G_{k+1}
= G_{k}^{2} + G_{k}^{2} − (G_{k}^{2} − (−1)^{k}e_{G})
= G_{k}^{2} + (−1)^{k}e_{G};

we have the following generalization of the Gelin-Cesàro identity:

G_{k−2}G_{k−1}G_{k+1}G_{k+2} = G_{k}^{4} − (−1)^{2k}e_{G}^{2},

where we have retained (−1)^{2k} to allow a direct conversion to the gibonacci function form which is required to hold for all real numbers k, namely,

g(k − 2)g(k − 1)g(k + 1)g(k + 2) = g(k)^{4} − (−1)^{2k}e_{G}^{2}.  (142)

Differentiating (142) gives

\frac{d}{dk} g(k − 2)g(k − 1)g(k + 1)g(k + 2)

+ g(k − 2)\frac{d}{dk} g(k − 1)g(k + 1)g(k + 2)

+ g(k − 2)g(k − 1)\frac{d}{dk} g(k + 1)g(k + 2)

+ g(k − 2)g(k − 1)g(k + 1)\frac{d}{dk} g(k + 2)

= 4g(k)^{3} \frac{d}{dk} g(k) − 2i(−1)^{2k}πe_{G}^{2};

so that taking the imaginary part, we have

B_{G}β^{k−2}G_{k−1}G_{k+1}G_{k+2} + G_{k−2}B_{G}β^{k−1}G_{k+1}G_{k+2}
+ G_{k−2}G_{k−1}B_{G}β^{k+1}G_{k+2} + G_{k−2}G_{k−1}G_{k+1}B_{G}β^{k+2}
= 4G_{k}^{3}B_{G}β^{k} + 2(−1)^{2k}e_{G}^{2}\sqrt{5};

and substituting (G_{1} − G_{0}α) for B_{G} from (102) and multiplying through by β^{r} yields

\begin{matrix}β^{k+r−2}G_{k−1}G_{k+1}G_{k+2} + G_{k−2}β^{k+r−1}G_{k+1}G_{k+2} \\ + G_{k−2}G_{k−1}β^{k+r+1}G_{k+2} + G_{k−2}G_{k−1}G_{k+1}β^{k+r+2} \\ = 4G_{k}^{3}β^{k+r} + 2 (β^{r+1}G_{0} − β^{r}G_{1}) e_{G}\sqrt{5}, \end{matrix}  (143)

which also implies

\begin{matrix}α^{k+r−2}G_{k−1}G_{k+1}G_{k+2} + G_{k−2}α^{k+r−1}G_{k+1}G_{k+2} \\ + G_{k−2}G_{k−1}α^{k+r+1}G_{k+2} + G_{k−2}G_{k−1}G_{k+1}α^{k+r+2} \\ = 4G_{k}^{3}α^{k+r} − 2 (α^{r+1}G_{0} − α^{r}G_{1}) e_{G}\sqrt{5}. \end{matrix}  (144)

Note that we used

\frac{1}{B_{G}} = \frac{1}{G_{1} − G_{0}α} = \frac{G_{0}β − G_{1}}{e_{G}} .

Combining (143) and (144), we arrive at the generalization of the Gelin-Cesàro identity stated in Proposition 42.

Proposition 42. If k and r are integers, then

H_{k+r−2}G_{k−1}G_{k+1}G_{k+2} + G_{k−2}H_{k+r−1}G_{k+1}G_{k+2}
+ G_{k−2}G_{k−1}H_{k+r+1}G_{k+2} + G_{k−2}G_{k−1}G_{k+1}H_{k+r+2}
= 4H_{k+r}G_{k}^{3} − 2e_{G}G_{0} (H_{r+2} + H_{r}) + 2e_{G}G_{1} (H_{r+1} + H_{r−1}) .

5.2.2 Generalizations of Catalan’s identity

Upon differentiating the Fibonacci function form

f(k − r)f(k + r) = f(k)^{2} + (−1)^{k+r+1}f(r)^{2},

of Catalan’s identity

F_{k−r}F_{k+r} = F_{k}^{2} + (−1)^{k+r+1}F_{r}^{2},

with respect to k and applying the prescription (89) and (91), we obtain

F_{k+r}β^{k−r+s} + F_{k−r}β^{k+r+s} = 2F_{k}β^{k+s} + (−1)^{k+r}β^{s}F_{r}^{2}\sqrt{5},  (145)

and hence also

F_{k+r}α^{k−r+s} + F_{k−r}α^{k+r+s} = 2F_{k}α^{k+s} − (−1)^{k+r}α^{s}F_{r}^{2}\sqrt{5}.  (146)

Combining (145) and (146) according to the Binet formula gives the following generalization of Catalan’s identity.

Proposition 43. If k, r, and s are integers, then

F_{k+r}G_{k−r+s} + F_{k−r}G_{k+r+s} = 2F_{k}G_{k+s} + (−1)^{k+r+1}F_{r}^{2} (G_{s+1} + G_{s−1}) .  (147)

Writing

F_{k−s+r}G_{k−r} + F_{k−s−r}G_{k+r} = 2F_{k−s}G_{k} + (−1)^{k−s+r+1}F_{r}^{2} (G_{s+1} + G_{s−1}) ,  (148)

and setting k = s gives the multiplication formula

G_{s+r} − (−1)^{r}G_{s−r} = F_{r} (G_{s+1} + G_{s−1}) ,  (149)

derived also by Vajda [23, Formula (10b)].

Applying the method (second component) to (149) with r as the index of interest gives the next result.

Proposition 44. If r, s, and t are integers, then

L_{r} (G_{0}H_{s+t−1} + G_{1}H_{s+t}) − (−1)^{r}G_{s−r} (H_{t+1} + H_{t−1}) = H_{r+t} (G_{s+1} + G_{s−1}) .

In particular, setting t = s, (H_{j}) = (G_{j}), and using [23, Formula (10a)]:

G_{n+m} + (−1)^{m}G_{n−m} = L_{m}G_{n},

we have

G_{0}G_{2s−1} + G_{1}G_{2s} = G_{s}(G_{s+1} + G_{s−1}).

Differentiating the Fibonacci function form of (148), namely,

\begin{matrix}f(k − s + r)g(k − r) + f(k − s − r)g(k + r) \\ = 2f(k − s)g(k) + (−1)^{k−s+r+1}f(r)^{2} (g(s + 1) + g(s − 1)) , \end{matrix}

with respect to s, taking the imaginary part and making use of (89), (91), (101) and (102) yields the identity stated in Proposition (45).

Proposition 45. If k, r, s, and t are integers, then

G_{k+s−r}H_{k+r+t} + G_{k+s+r}H_{k−r+t}
= 2G_{k+s}H_{k+t}
+ (−1)^{k+r}F_{r}^{2} (G_{0} (H_{s+t} + H_{s+t−2}) + G_{1} (H_{s+t+1} + H_{s+t−1}))
− (−1)^{k+r}F_{r}^{2} (G_{s+1} + G_{s−1}) (H_{t+1} + H_{t−1}) .

In particular,

G_{k−r}G_{k+r} − G_{k}^{2} = (−1)^{k+r}F_{r}^{2}e_{G}.

5.2.3 Generalization of an identity obtained from an inverse tangent relation

The method (second component) cannot be applied to (10) because (54) on page 16 is valid only for integers k. In order to redeem the situation, we proceed as follows:

tan^{−1} \frac{1}{F_{2k}} − tan^{−1} \frac{1}{F_{2k+2}} = tan^{−1} \frac{F_{2k+2} − F_{2k}}{F_{2k}F_{2k+2} + 1}

= tan^{−1} \frac{F_{2k+1}}{F_{2k+1}^{2} + (−1)^{2k+1} + 1},

where we refrained from setting (−1)^{2k+1} = −1 to ensure that the Fibonacci function form

tan^{−1} \frac{1}{f(2k)} − tan^{−1} \frac{1}{f(2k + 2)} = tan^{−1} \frac{f(2k + 1)}{f(2k + 1)^{2} + (−1)^{2k+1} + 1},  (150)

holds for all real numbers k.

Differentiating (150) with respect to k and taking the imaginary part, we find

\begin{matrix}\frac{1}{F_{2k}^{2} + 1}ℑ\frac{df}{dk} (2k) − \frac{1}{F_{2k+2}^{2} + 1}ℑ\frac{df}{dk} (2k + 2) \\ = \frac{1}{F_{2k+1}^{2} + 1}ℑ\frac{df}{dk} (2k + 1) − \frac{π}{F_{2k+1} (F_{2k+1}^{2} + 1)}, \end{matrix}

which, upon use of (89) and (91), gives

\frac{β^{2k}}{F_{2k}^{2} + 1} − \frac{β^{2k+2}}{F_{2k+2}^{2} + 1} = \frac{β^{2k+1}}{F_{2k+1}^{2} + 1} + \frac{\sqrt{5}}{F_{2k+1} (F_{2k+1}^{2} + 1)},

that is

\frac{α^{r+2}}{F_{2k}^{2} + 1} − \frac{α^{r}}{F_{2k+2}^{2} + 1} = \frac{α^{2k+r+2}\sqrt{5}}{F_{2k+1} (F_{2k+1}^{2} + 1)} − \frac{α^{r+1}}{F_{2k+1}^{2} + 1},  (151)

where r is an arbitrary integer and also

\frac{β^{r+2}}{F_{2k}^{2} + 1} − \frac{β^{r}}{F_{2k+2}^{2} + 1} = −\frac{β^{2k+r+2}\sqrt{5}}{F_{2k+1} (F_{2k+1}^{2} + 1)} − \frac{β^{r+1}}{F_{2k+1}^{2} + 1}.  (152)

By combining (151) and (152), we have the next result.

Proposition 46. If r and k are integers, then

\frac{G_{r+2}}{F_{2k}^{2} + 1} − \frac{G_{r}}{F_{2k+2}^{2} + 1} = \frac{G_{2k+r+3} + G_{2k+r+1}}{F_{2k+1}(F_{2k+1}^{2} + 1)} − \frac{G_{r+1}}{F_{2k+1}^{2} + 1}.

6 Extension of the method to the Horadam sequence

In this section we extend the method to a general non-degenerate second order sequence (Horadam sequence).

The Horadam sequence (W_{j}) = (W_{j}(W_{0}, W_{1}; P, Q)) is defined, for all integers and arbitrary real numbers W_{0}, W_{1}, P ≠ 0, and Q ≠ 0, by the recurrence relation

W_{j} = P W_{j−1} − QW_{j−2}, j ≥ 2,  (153)

with W_{−j} = (P W_{−j+1} − W_{−j+2}) /Q.

Associated with (W_{j}) are the Lucas sequences of the first kind, (U_{j}(P, Q)) = (W_{j}(0, 1; P, Q)), and the second kind, (V_{j}(P, Q)) = (W_{j}(2, P; P, Q)); that is

U_{0} = 0, U_{1} = 1, U_{j} = P U_{j−1} − QU_{j−2}, j ≥ 2,  (154)

and

V_{0} = 2, V_{1} = P, V_{j} = P V_{j−1} − QV_{j−2}, j ≥ 2,  (155)

with U_{−j} = (P U_{−j+1} − U_{−j+2}) /Q and V_{−j} = (P V_{−j+1} − V_{−j+2}) /Q.

Note that, for convenience and since no confusion can arise, we have retained the notation (W_{j}) = (W_{j}(W_{0}, W_{1}; P, Q)) for the Horadam sequence and (U_{j}) = (U_{j}(P, Q)) and (V_{j}) = (V_{j}(P, Q)) for the Lucas sequences.

The closed formula for W_{j}(W_{0}, W_{1}; P, Q) in the non-degenerate case, P^{2}−4Q > 0, remains

W_{j} = \frac{Aσ^{j} − Bτ^{j}}{σ − τ} = \frac{Aσ^{j} − Bτ^{j}}{δ} ,  (156)

where

A = W_{1} − W_{0}τ, B = W_{1} − W_{0}σ,  (157)

with σ and τ now given by

σ = \frac{P + \sqrt{P^{2} − 4Q}}{2},  τ = \frac{P − \sqrt{P^{2} − 4Q}}{2};  (158)

so that

σ + τ = P, σ − τ = \sqrt{P^{2} − 4Q} = δ, and στ = Q.  (159)

In particular,

U_{j} = \frac{σ^{j} − τ^{j}}{σ − τ} , V_{j} = σ^{j} + τ^{j}.  (160)

The following identities, of which (70) to (73) are particular cases, are easy to derive:

U_{j+1} − QU_{j−1} = V_{j},  (161)
U_{j+1} + QU_{j−1} = P U_{j},  (162)
V_{j+1} − QV_{j−1} = U_{j}δ^{2},  (163)

and

V_{j+1} + QV_{j−1} = P V_{j}.

Identity (75) of Lemma 13 on page 24 now reads

Aσ^{j} + Bτ^{j} = W_{j+1} − QW_{j−1}.  (164)

We define the Horadam function w(x) by

w(x) = \frac{Aσ^{x} − Bτ^{x}}{σ − τ} = \frac{Aσ^{x} − Bτ^{x}}{δ} , x ∈ ℝ,  (165)

where A and B are as defined in (157) and σ and τ are as given in (158).

We now discuss the extension of the method to the Horadam sequence. We distinguish the following three cases:

1. Q < 0;

2. P > 0 and Q > 0;

3. P < 0 and Q > 0.

6.1 Case 1: Q < 0

If the Horadam parameter, Q, is negative, then we see from (158) that σ is positive and τ is negative for all real numbers P; and equation (87) on page 27 in the proof of the theorem now reads

\frac{d}{dx}w(x)|_{x=j∈ℤ} = \frac{1}{δ} ((W_{j+1} − QW_{j−1}) ln σ − Bτ^{j} ln (−Q) − Bτ^{j}πi(2m + 1)) , by (164).

Taking the real and imaginary parts, we have

ℜ ( \frac{d}{dx}w(x)|_{x=j∈ℤ}) = \frac{1}{δ} ((W_{j+1} − QW_{j−1}) ln σ − B_{W} τ^{j} ln (−Q))  (166)

and

ℑ ( \frac{d}{dx}w(x)|_{x=j∈ℤ}) = −\frac{B_{W}}{δ} τ^{j}π(2m + 1),  (167)

where m is some integer and

B_{W} = B = W_{1} − W_{0}σ.

Equation (166) reduces to (85) when Q = −1. Equation (167) is the same as (88) on page 27 in Section 5, except that the values of σ and τ now depend on P and Q.

A description of how the method (first component) for obtaining new identities from existing ones works for the general second order (Horadam) sequence (W_{j}(W_{0}, W_{1}; P, Q)) with Q < 0 now follows.

1. Let k be a free index in the known identity. Replace each Horadam number, say W_{h(k,...)}, with a certain differentiable function of k, namely, w(h(k, …)), with k now considered a variable.

2. By applying the usual rules of calculus, differentiate, with respect to k, through the identity obtained in step 1.

3. Simplify the equation obtained in step 2 and take the real part, using also the following prescription:

w(h(k, …)) → W_{h(k,...)},  (168)
ℜ (\frac{∂w}{∂k} (h(k, …))) → \frac{1}{δ} ((W_{h(k+1,...)} − QW_{h(k−1,...)}) ln σ − Bτ^{h(k,...)} ln (−Q)); (169)

where σ = (P + δ)/2, τ = (P − δ)/2, and δ = \sqrt{P^{2} − 4Q}.

In particular, for the Lucas sequences, we have

u(h(k, …)) → U_{h(k,...)},  (170)
ℜ\frac{∂u}{∂k}(h(k, …)) → \frac{V_{h(k,...)}}{δ} ln σ − \frac{τ^{h(k,...)}}{δ} ln (−Q) ;  (171)

and

v(h(k, …)) → V_{h(k,...)},  (172)
ℜ\frac{∂v}{∂k} (h(k, …)) → U_{h(k,...)}δ ln σ + τ^{h(k,...)} ln (−Q) ;  (173)

of which the generalized Fibonacci relations (35)–(40) on page 11 are particular cases.

Next, we describe how the method (second component) works for the general second order (Horadam) sequence (W_{j}(W_{0}, W_{1}; P, Q)) when Q < 0. The scheme is the following.

1. Let k be a free index in the known identity. Replace each Horadam number, say W_{h(k,...)}, with a certain differentiable function of k, namely, w(h(k, …)), with k now considered a variable.

2. By applying the usual rules of calculus, differentiate, with respect to k, through the identity obtained in step 1.

3. Simplify the equation obtained in step 2 and take the imaginary part, using also the following prescription:

w(h(k, …)) → W_{h(k,...)},  (174)
ℑ\frac{∂w}{∂k}(h(k, …)) → −\frac{B_{W}}{δ} πτ^{h(k,...)} = \frac{W_{0}σ − W_{1}}{δ} πτ^{h(k,...)};  (175)

where σ = (P + δ)/2, τ = (P − δ)/2 and δ = \sqrt{P^{2} − 4Q}.

In particular, for the Lucas sequences, we have

u(h(k, …)) → U_{h(k,...)},  (176)
ℑ\frac{∂u}{∂k}(h(k, …)) → −\frac{B_{U}}{δ} πτ^{h(k,...)} = −\frac{πτ^{h(k,...)}}{δ};  (177)

and

v(h(k, …)) → V_{h(k,...)},  (178)
ℑ\frac{∂v}{∂k} (h(k, …)) → −\frac{B_{V}}{δ} πτ^{h(k,...)} = πτ^{h(k,...)};  (179)

of which the Fibonacci and Lucas relations (89)–(92) on page 28 are particular cases.

Note that Proposition 17 on page 29 on the interchangeability of σ and τ, remains valid; and that the method (second component) is applicable provided the expression obtained after substituting Fibonacci, Lucas, gibonacci and Horadam functions f(x), l(x), u(x), v(x), g(x) and g(x) in the given identity holds for all real numbers, as noted in the remark on page 43 .

We give an example; but first we state a needed Lemma.

Lemma 47. If k is an integer, then

\frac{W_{k+1} − σW_{k}}{W_{1} − σW_{0}} = τ^{k}, \frac{W_{k+1} − τW_{k}}{W_{1} − τW_{0}} = σ^{k}.

Proof. We prove, by induction, the first identity for a non-negative integer k; and invoke a theorem of Bruckman and Rabinowitz [3] that if an identity involving generalized Fibonacci numbers is true for all positive subscripts, it is true for all non-positive subscripts as well.

The identity is obviously true for the base case k = 0. We assume the identity holds for k = 1, 2, … n. We have

\frac{W_{(n+1)+1} − σW_{n+1}}{W_{1} − σW_{0}} = \frac{P W_{n+1} − QW_{n} − σW_{n+1}}{W_{1} − σW_{0}}

= \frac{(σ + τ) W_{n+1} − στW_{n} − σW_{n+1}}{W_{1} − σW_{0}}

= \frac{τW_{n+1} − στW_{n}}{W_{1} − σW_{0}} = \frac{W_{n+1} − σW_{n}}{W_{1} − σW_{0}} τ = τ^{n}τ = τ^{n+1}.

Thus, the identity holds for n + 1 whenever it holds for n.

Consider the following identity ([9, Equation (3.14)]):

U_{r}W_{k+1} − QU_{r−1}W_{k} = W_{k+r}.

We write

u(r)w(k + 1) − Qu(r − 1)w(k) = w(k + r);

and differentiate with respect to r, obtaining

\frac{d}{dr}u(r) · w(k + 1) − \frac{d}{dr}u(r − 1) · Qw(k) = \frac{∂w}{∂r}(k + r).  (180)

Taking the real part, we get

ℜ\frac{d}{dr}u(r) · W_{k+1} − ℜ\frac{d}{dr}u(r − 1) · QW_{k} = ℜ\frac{∂w}{∂r}(k + r);

which, using (171) and (169), after some rearrangement, gives:

\begin{matrix}((V_{r}W_{k+1} − QV_{r−1}W_{k}) − (W_{k+r+1} − QW_{k+r−1})) ln σ \\ = ((W_{k+1} − σW_{k}) − (W_{1} − σW_{0}) τ^{k}) τ^{r} ln (−Q) . \end{matrix}  (181)

On account of Lemma 47, the right hand side of (181) vanishes and we obtain

V_{r}W_{k+1} − QV_{r−1}W_{k} = W_{k+r+1} − QW_{k+r−1}, Q < 0.  (182)

If Q = −1, then (182) reduces to identity (19) derived in Section 2.4.2.

Next, taking the imaginary part of (180), we find

ℑ\frac{d}{dr}u(r) · W_{k+1} − ℑ\frac{d}{dr}u(r − 1) · QW_{k} = ℑ\frac{∂w}{∂r}(k + r);

which, using (177) and (175), gives

−τ^{r}W_{k+1} + Qτ^{r−1}W_{k} = (W_{0}σ − W_{1}) τ^{k+r},

that is

−τ^{r}W_{k+1} + Qτ^{r−1}W_{k} = W_{0}Qτ^{k+r−1} − W_{1}τ^{k+r}  (183)

and also

−σ^{r}W_{k+1} + Qσ^{r−1}W_{k} = W_{0}Qσ^{k+r−1} − W_{1}σ^{k+r}.  (184)

Combining (183) and (184), using the Binet formula, we have the result stated in Proposition 48.

Proposition 48. If r and k are integers, then

Z_{r}W_{k+1} − QZ_{r−1}W_{k} = W_{1}Z_{k+r} − QW_{0}Z_{k+r−1}, Q < 0;  (185)

where W_{j} = W_{j} (W_{0}, W_{1}; P, Q) and Z_{j} = Z_{j} (Z_{0}, Z_{1}; P, Q) are two Horadam sequences.

6.2 Case 2: P > 0 and Q > 0

If the Horadam parameters P and Q are both positive, then it is clear from (158) that σ and τ are both positive. In this case we have

\frac{d}{dx}w(x) = \frac{1}{δ} (A\frac{d}{dx}σ^{x} − B \frac{d}{dx}τ^{x}) = \frac{1}{δ} ((Aσ^{x} + Bτ^{x}) ln σ − Bτ^{x} ln Q) ;

so that

\frac{d}{dx}w(x)|_{x=j∈ℤ} = \frac{1}{δ} ((W_{j+1} − QW_{j−1}) ln σ − Bτ^{j} ln Q) .

A description of how the method (first component) for obtaining new identities from existing ones works for the general second order (Horadam) sequence (W_{j}(W_{0}, W_{1}; P, Q)), P > 0, and Q > 0 now follows.

1. Let k be a free index in the known identity. Replace each Horadam number, say W_{h(k,...)}, with a certain differentiable function of k, namely, w(h(k, …)), with k now considered a variable.

2. By applying the usual rules of calculus, differentiate, with respect to k, through the identity obtained in step 1.

3. Simplify the equation obtained in step 2, using also the following prescription:

w(h(k, …)) → W_{h(k,...)},  (186)
\frac{∂w}{∂k} (h(k, …)) → \frac{1}{δ} ((W_{h(k+1,...)} − QW_{h(k−1,...)}) ln σ − Bτ^{h(k,...)} ln Q);  (187)

where, as usual, σ = (P + δ)/2, τ = (P − δ)/2, and δ = \sqrt{P^{2} − 4Q}.

In particular, for the Lucas sequences, we have

u(h(k, …)) → U_{h(k,...)},  (188)
\frac{∂u}{∂k}(h(k, …)) → \frac{V_{h(k,...)}}{δ} ln σ − \frac{τ^{h(k,...)}}{δ} ln Q;  (189)

and

v(h(k, …)) → V_{h(k,...)},  (190)
\frac{∂v}{∂k} (h(k, …)) → U_{h(k,...)}δ ln σ + τ^{h(k,...)} ln Q.  (191)

Returning to the example in the previous section, using (189) and (187) in (180) gives

\begin{matrix}((V_{r}W_{k+1} − QV_{r−1}W_{k}) − (W_{k+r+1} − QW_{k+r−1})) ln σ \\ = ((W_{k+1} − σW_{k}) − (W_{1} − σW_{0}) τ^{k}) τ^{r} ln Q, Q > 0, \end{matrix}

which is the same as (181) with ln (−Q) replaced with ln Q; and which in view of Lemma 47 gives

V_{r}W_{k+1} − QV_{r−1}W_{k} = W_{k+r+1} − QW_{k+r−1}, P > 0, Q > 0.  (192)

6.3 Case 3: P < 0 and Q > 0

If the parameter P is negative and the parameter Q is positive, then it is obvious from (158) that σ and τ are both negative numbers. In this case we have

\frac{d}{dx}w(x) = \frac{1}{δ} (A\frac{d}{dx}σ^{x} − B \frac{d}{dx}τ^{x})

= \frac{1}{δ} (Aσ^{x} (iπ(2m + 1) + ln (−σ)) − Bτ^{x} (iπ(2n + 1) + ln (−τ)))

= \frac{1}{δ} ((Aσ^{x} + Bτ^{x}) ln (−σ) − Bτ^{x} ln Q) + \frac{iπ}{δ} ((2m + 1)Aσ^{x} − (2n + 1)Bτ^{x}) ,

where m and n are integers; so that

ℜ ( \frac{d}{dx}w(x)|_{x=j∈ℤ}) = \frac{1}{δ} ((W_{j+1} − QW_{j−1}) ln (−σ) − B_{W} τ^{j} ln Q)  (193)

and

ℑ ( \frac{d}{dx}w(x)|_{x=j∈ℤ}) = \frac{2π}{δ} (mAσ^{j} − nBτ^{j}) + πW_{j},  (194)

where m and n are integers.

We now describe how the method (first component) works for the general second order (Horadam) sequence (W_{j}(W_{0}, W_{1}; P, Q)) with P < 0 and Q > 0.

1. Let k be a free index in the known identity. Replace each Horadam number, say W_{h(k,...)}, with a certain differentiable function of k, namely, w(h(k, …)), with k now considered a variable.

2. By applying the usual rules of calculus, differentiate, with respect to k, through the identity obtained in step 1.

3. Simplify the equation obtained in step 2 and take the real part, using also the following prescription:

w(h(k, …)) → W_{h(k,...)},  (195)
ℜ (\frac{∂w}{∂k} (h(k, …))) → \frac{1}{δ} ((W_{h(k+1,...)} − QW_{h(k−1,...)}) ln (−σ) − Bτ^{h(k,...)} ln Q); (196)

where σ = (P + δ)/2, τ = (P − δ)/2, and δ = \sqrt{P^{2} − 4Q}.

In particular, for the Lucas sequences, we have

u(h(k, …)) → U_{h(k,...)},  (197)
ℜ\frac{∂u}{∂k}(h(k, …)) → \frac{V_{h(k,...)}}{δ} ln (−σ) − \frac{τ^{h(k,...)}}{δ} ln Q;  (198)

and

v(h(k, …)) → V_{h(k,...)},  (199)
ℜ\frac{∂v}{∂k} (h(k, …)) → U_{h(k,...)}δ ln (−σ) + τ^{h(k,...)} ln Q.  (200)

We see from (194) that the method (second component) will, in general, not generate new identities for the general second order (Horadam) sequence (W_{j}(W_{0}, W_{1}; P, Q)) when P < 0 and Q > 0. Taking the principal value in (194) gives

ℑ\frac{∂w}{∂k}(h(k, …)) → πW_{h(k,...)}  (201)

for a Horadam function w(h(k, …)) and hence does not facilitate a generalization of an identity obtained from the first component.

To illustrate the first component for the case P < 0 and Q > 0, equation (181) on page 51 now reads

\begin{matrix}((V_{r}W_{k+1} − QV_{r−1}W_{k}) − (W_{k+r+1} − QW_{k+r−1})) & ln & (−σ) \\ = ((W_{k+1} − σW_{k}) − (W_{1} − σW_{0}) τ^{k}) τ^{r} ln Q, Q > 0, \end{matrix}

which in view of Lemma 47 gives

V_{r}W_{k+1} − QV_{r−1}W_{k} = W_{k+r+1} − QW_{k+r−1}, P < 0, Q > 0.  (202)

Comparing (182), (192) and (202), we find the result stated in the next proposition.

Proposition 49. If k and r are integers, then

V_{r}W_{k+1} − QV_{r−1}W_{k} = W_{k+r+1} − QW_{k+r−1}, Q ≠ 0.  (203)

In particular, on account of (161) and (163), we have for the Lucas sequences,

V_{r}U_{k+1} − QV_{r−1}U_{k} = V_{k+r},
V_{r}V_{k+1} − QV_{r−1}V_{k} = δ^{2}U_{k+r}.

We note that identity (203) is a particular case of Howard [12, Theorem 3.1].

7 Acknowledgment

We thank the referee whose contributions and critique significantly improved the presentation.

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2020 Mathematics Subject Classification: Primary 11B39; Secondary 11B37. Keywords: Fibonacci number, Lucas number, gibonacci sequence, generalized Fibonacci sequence, Fibonacci function, summation identity, Gelin-Cesàro identity, Candido’s identity, d’Ocagne identity, Horadam sequence.

(Concerned with sequences \underline{A000032} and \underline{A000045}.)

Received December 15 2023; revised version received May 29 2024. Published in Journal of Integer Sequences, June 19 2024.

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