On the Sums of Reciprocal Hyperfibonacci Numbers and Hyperlucas Numbers

Rui Liu^{1} and Feng-Zhen Zhao Department of Mathematics Dalian University of Technology Dalian, Liaoning 116024 P. R. China

                     liurui1515@gmail.com
                   fengzhenzhao@yahoo.com.cn

Abstract In this paper, we discuss the properties of hyperfibonacci numbers and hyperlucas numbers. We investigate the sums of reciprocal hyperfibonacci numbers and hyperlucas numbers. In addition, we establish some identities related to reciprocal hyperfibonacci numbers and hyperlucas numbers.

1 Introduction

Fibonacci and Lucas sequences {F_{n}} and {L_{n}} have fascinated both amateurs and professional mathematicians for centuries. They are generalized to many forms. Dil and Mezö [4] introduced the definition of “hyperfibonacci” numbers F_{n}^{(r)} and “hyperlucas” numbers L_{n}^{(r)} :

F_{n}^{(r)} = ∑_{j=0}^{n} F_{j}^{(r−1)}, with F_{n}^{(0)} = F_{n}, F_{0}^{(r)} = 0, F_{1}^{(r)} = 1,
L_{n}^{(r)} = ∑_{j=0}^{n} L_{j}^{(r−1)}, with L_{n}^{(0)} = L_{n}, L_{0}^{(r)} = 2, L_{1}^{(r)} = 2r + 1,

where r is a positive integer. It is well known that the Binet forms of {F_{n}} and {L_{n}} are

F_{n} = \frac{α^{n} − (−1)^{n}α^{−n}}{\sqrt{5}} , L_{n} = α^{n} + (−1)^{n}α^{−n},  (1)

1This work was supported by the Science Research Foundation of Dalian University of Technology (2008)

and the National Natural Science Foundation of China (NSFC Grant # 11061020).

where α = (1 + \sqrt{5})/2. The sequences \{F_{n}\} and \{L_{n}\} satisfy the linear recurrence relation

W_{n} = W_{n−1} + W_{n−2}, n ≥ 2.  (2)

It is clear that

F_{n}^{(1)} = F_{n+2} − 1, L_{n}^{(1)} = L_{n+2} − 1.  (3)

Some values of {F_{n}^{(1)}} and {L_{n}^{(1)}} are given below.

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ F_{n}^{(1)} & 0 & 1 & 2 & 4 & 7 & 12 & 20 & 33 & 54 & 88 & 143 & 232 & 376 & 609 & 986 \\ L_{n}^{(1)} & 2 & 3 & 6 & 10 & 17 & 28 & 46 & 75 & 122 & 198 & 321 & 520 & 842 & 1363 & 2206 \end{matrix}

These are sequences \underline{A000071} and \underline{A001610} in Sloane’s Encylopedia [11]. Some properties of {F_{n}^{(r)}} and {L_{n}^{(r)}} are studied in the paper of Ning-Ning Cao and Feng-Zhen Zhao [3]. In this paper, we investigate the sums of reciprocal hyperfibonacci numbers and hyperlucas numbers.

Now we recall some definitions involved in this paper. The Fibonacci and Lucas zeta functions are defined by

ζ_{F} (s) = ∑_{n=1}^{∞} \frac{1}{F_{n}^{s}} and ζ_{L}(s) = ∑_{n=1}^{∞} \frac{1}{L_{n}^{s}} ,

where {F_{n}} and {L_{n}} are the Fibonacci and Lucas sequences, respectively. Recently, properties of ζ_{F} (s) and ζ_{L}(s) are investigated in several different ways, see for instance [5, 6, 7, 9]. In [5], the partial infinite sums of reciprocal Fibonacci numbers were studied by Ohtsuka and Nakamura, [10]. They proved that

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}} )^{−1}⌋ = \begin{cases}F_{n−2}, & if n is even and n ≥ 2; \\ F_{n−2} − 1, & if n is odd and n ≥ 1, \end{cases}  (4)

where ⌊·⌋ denotes the floor function. In [8], Holliday and Komatsu generalize (4) to the generalized Fibonacci sequence. They showed that

⌊( ∑_{k=1000n}^{∞} \frac{1}{G_{k}} )^{−1}⌋ = \begin{cases}G_{n} − G_{n−1}, & if n is even and n ≥ 2; \\ G_{n} − G_{n−1} − 1, & if n is odd and n ≥ 1, \end{cases}

where {G_{n}} is generalized Fibonacci sequence defined by G_{k+2} = aG_{k+1} + G_{k}(k ≥ 0) with G_{0} = 0, G_{1} = 1, and a is a positive integer. In this paper, we discuss the partial infinite sums of reciprocal hyperfibonacci numbers and hyperlucas numbers. In the next section, we investigate the sums of the following forms

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}^{(1)}} )^{−1}⌋, ⌊( ∑_{k=n}^{∞} \frac{1}{L_{k}^{(1)}} )^{−1}⌋.

In addition, we establish some identities related to reciprocal hyperfibonacci numbers and hyperlucas numbers.

2 The partial infinite sums of reciprocal hyperfibonacci

numbers and hyperlucas numbers

In this section, we discuss the partial infinite sums of reciprocal hyperfibonacci numbers and hyperlucas numbers. Lemma 1. For {F_{n}} and {L_{n}}, the following formulas hold:

F_{n+1}F_{n+3} − F_{n}F_{n+4} = 2(−1)^{n},  (5)
L_{n+1}L_{n+3} − L_{n}L_{n+4} = 10(−1)^{n+1},  (6)
F_{n+2}^{2} − F_{n+1}F_{n+3} = (−1)^{n+1},  (7)
L_{n+2}^{2} − L_{n+1}L_{n+3} = 5(−1)^{n}.  (8)

Proof. From (1), we can verify that (5)-(8) hold. Theorem 2. For {F_{n}^{(1)}} and {L_{n}^{(1)}} (n ≥ 3), we have

⌊( ∑_{k=n}^{∞} \frac{1}{F_{k}^{(1)}} )^{−1}⌋ = F_{n} − 1,  (9)
⌊( ∑_{k=n}^{∞} \frac{1}{L_{k}^{(1)}} )^{−1}⌋ = L_{n} − 1, n ≥ 4.  (10)

Proof. By using (2)–(3) and (5), we get

\frac{1}{F_{n}^{(1)} − F_{n−1}^{(1)}} − \frac{1}{F_{n}^{(1)}} − \frac{1}{F_{n+1}^{(1)}} − \frac{1}{F_{n+2}^{(1)} − F_{n+1}^{(1)}} = \frac{1}{F_{n}} − \frac{1}{F_{n}^{(1)}} − \frac{1}{F_{n+1}^{(1)}} − \frac{1}{F_{n+2}}
= \frac{F_{n}^{(1)}(F_{n+1}F_{n+3} − F_{n}F_{n+2} − F_{n+1}) − F_{n}F_{n+2}F_{n+1}^{(1)}}{F_{n}F_{n+2}F_{n}^{(1)}F_{n+1}^{(1)}}
= \frac{F_{n}^{(1)}[2(−1)^{n} + F_{n}F_{n+3} − F_{n+1}] − F_{n}F_{n+2}F_{n+1}^{(1)}}{F_{n}F_{n+2}F_{n}^{(1)}F_{n+1}^{(1)}}
= \frac{F_{n}^{(1)}[2(−1)^{n} − F_{n+1}] + F_{n}[F_{n}^{(1)}F_{n+3} − F_{n+2}F_{n+1}^{(1)}]}{F_{n}F_{n+2}F_{n}^{(1)}F_{n+1}^{(1)}}
= \frac{F_{n}^{(1)}[2(−1)^{n} − F_{n+1}] − F_{n}F_{n+1}}{F_{n}F_{n+2}F_{n}^{(1)}F_{n+1}^{(1)}} , n ≥ 2,

and

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

= \frac{F_{n+1}F_{n+3} − F_{n+1} − F_{n}F_{n+3} − F_{n+1} + 2F_{n} − F_{n}F_{n+2} + F_{n}^{(1)} + F_{n+1}^{(1)}}{(F_{n} − 1)F_{n}^{(1)}F_{n+1}^{(1)}}
= \frac{F_{n}F_{n+4} + 2(−1)^{n} − F_{n}F_{n+3} − F_{n}F_{n+2} − F_{n+1} + 2F_{n} + F_{n+2} + F_{n+3} − 2}{(F_{n} − 1)F_{n}^{(1)}F_{n+1}^{(1)}}
= \frac{2((−1)^{n} + F_{n} + F_{n+2} − 1)}{(F_{n} − 1)F_{n}^{(1)}F_{n+1}^{(1)}}, n ≥ 3.

By using (2)–(3) and (6), we get

\frac{1}{L_{n}^{(1)} − L_{n−1}^{(1)}} − \frac{1}{L_{n}^{(1)}} − \frac{1}{L_{n+1}^{(1)}} − \frac{1}{L_{n+2}^{(1)} − L_{n+1}^{(1)}} = \frac{(10(−1)^{n+1} − L_{n+1})L_{n}^{(1)} − L_{n}L_{n+1}}{L_{n}L_{n+2}L_{n}^{(1)}L_{n+1}^{(1)}}

for n ≥ 4 and

\frac{1}{L_{n}^{(1)} − L_{n−1}^{(1)} − 1} − \frac{1}{L_{n}^{(1)}} − \frac{1}{L_{n+1}^{(1)}} − \frac{1}{L_{n+2}^{(1)} − L_{n+1}^{(1)} − 1} = \frac{10(−1)^{n+1} + 2(L_{n} + L_{n+2} − 1)}{(L_{n} − 1)L_{n}^{(1)}L_{n+1}^{(1)}},

for n ≥ 2.

From the inequalities

(2(−1)^{n} − F_{n+1})F_{n}^{(1)} − F_{n}F_{n+1} < 0, n ≥ 2,
(10(−1)^{n+1} − L_{n+1})L_{n}^{(1)} − L_{n}L_{n+1} < 0, n ≥ 4,
(−1)^{n} + F_{n} + F_{n+2} − 1 > 0, n ≥ 3,
10(−1)^{n+1} + 2(L_{n} + L_{n+2} − 1) > 0, n ≥ 2,

we obtain

\frac{1}{F_{n}^{(1)} − F_{n−1}^{(1)}} <  ∑_{k=n}^{∞} \frac{1}{F_{k}^{(1)}} < \frac{1}{F_{n}^{(1)} − F_{n−1}^{(1)} − 1}, n ≥ 3,

F_{n} − 1 < ( ∑_{k=n}^{∞} \frac{1}{F_{k}^{(1)}} )^{−1}< F_{n}, n ≥ 3,

\frac{1}{L_{n}^{(1)} − L_{n−1}^{(1)}} <  ∑_{k=n}^{∞} \frac{1}{L_{k}^{(1)}} < \frac{1}{L_{n}^{(1)} − L_{n−1}^{(1)} − 1}, n ≥ 4,

L_{n} − 1 < ( ∑_{k=n}^{∞} \frac{1}{L_{k}^{(1)}} )^{−1} < L_{n}, n ≥ 4.

Hence the relations (9)–(10) hold.

Theorem 3. For {F_{n}^{(1)}} (n ≥ 2) we have

⌊( ∑_{k=n}^{∞} \frac{1}{(F_{k}^{(1)})^{2}} )⌋ = \begin{cases}F_{n−1}^{(1)}F_{n}^{(1)} + F_{n−1}^{(1)} − 1, & if n is even and n ≥ 2; \\ F_{n−1}^{(1)}F_{n}^{(1)} + F_{n−1}^{(1)}, & if n is odd and n ≥ 1. \end{cases}  (11)

⌊( ∑_{k=n}^{∞} \frac{1}{(L_{k}^{(1)})^{2}} )⌋ = L_{n−1}^{(1)}L_{n}^{(1)} + L_{n−1}^{(1)} − 1,  if n is odd and n > 1.  (12)

Proof. By applying (2)–(3) and (7), we get

\frac{1}{F_{n−1}^{(1)}F_{n}^{(1)} + F_{n−1}^{(1)} − 1} − \frac{1}{(F_{n}^{(1)})^{2}} − \frac{1}{F_{n}^{(1)}F_{n+1}^{(1)} + F_{n}^{(1)} − 1} = \frac{1}{F_{n−1}^{(1)}F_{n+2} − 1} − \frac{1}{(F_{n}^{(1)})^{2}} − \frac{1}{F_{n}^{(1)}F_{n+3} − 1}

= \frac{(F_{n}^{(1)})^{2}(F_{n}^{(1)}F_{n+3} − F_{n−1}^{(1)}F_{n+2}) − (F_{n−1}^{(1)}F_{n+2} − 1)(F_{n}^{(1)}F_{n+3} − 1)}{(F_{n−1}^{(1)}F_{n+2} − 1)(F_{n}^{(1)})^{2}(F_{n}^{(1)}F_{n+3} − 1)}

= \frac{F_{n}^{(1)}^{2}(F_{n+2}^{2} − F_{n+1}) − F_{n−1}^{(1)}F_{n}^{(1)}F_{n+2}F_{n+3} + F_{n−1}^{(1)}F_{n+2} + F_{n}^{(1)}F_{n+3} − 1}{(F_{n−1}^{(1)}F_{n+2} − 1)(F_{n}^{(1)})^{2}(F_{n}^{(1)}F_{n+3} − 1)}

= \frac{(F_{n+3} + (−1)^{n+1}F_{n+2})F_{n}^{(1)} + F_{n}^{(1)}F_{n+1} + F_{n−1}^{(1)}F_{n+2} − 1}{(F_{n−1}^{(1)}F_{n+2} − 1)(F_{n}^{(1)})^{2}(F_{n}^{(1)}F_{n+3} − 1)} > 0.

Thus, we have

∑_{k=n}^{∞} \frac{1}{(F_{k}^{(1)})^{2}} < \frac{1}{F_{n−1}^{(1)}F_{n}^{(1)} + F_{n−1}^{(1)} − 1}.

Similarly, we can prove that

∑_{k=n}^{∞} \frac{1}{(F_{k}^{(1)})^{2}} > \frac{1}{F_{n−1}^{(1)}F_{n}^{(1)} + F_{n−1}^{(1)} + 1}.

On the other hand, we have

\frac{1}{F_{n−1}^{(1)}F_{n}^{(1)} + F_{n−1}^{(1)}} − \frac{1}{(F_{n}^{(1)})^{2}} − \frac{1}{F_{n}^{(1)}F_{n+1}^{(1)} + F_{n}^{(1)}} = \frac{1}{F_{n−1}^{(1)}F_{n+2}} − \frac{1}{(F_{n}^{(1)})^{2}} − \frac{1}{F_{n}^{(1)}F_{n+3}}

= \frac{(F_{n}^{(1)})^{2}F_{n+3} − F_{n−1}^{(1)}F_{n}^{(1)}F_{n+2} − F_{n−1}^{(1)}F_{n+2}F_{n+3}}{F_{n−1}^{(1)}(F_{n}^{(1)})^{2}F_{n+2}F_{n+3}}

= \frac{F_{n}^{(1)}(F_{n}^{(1)}F_{n+3} − F_{n−1}^{(1)}F_{n+2}) − F_{n−1}^{(1)}F_{n+2}F_{n+3}}{F_{n−1}^{(1)}(F_{n}^{(1)})^{2}F_{n+2}F_{n+3}}

= \frac{(−1)^{n+1}F_{n+2} + F_{n+1}}{F_{n−1}^{(1)}(F_{n}^{(1)})^{2}F_{n+2}F_{n+3}} .

When n ≥ 2 is even, we can verify that

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

and when n ≥ 1 is odd

∑_{k=n}^{∞} \frac{1}{(F_{k}^{(1)})^{2}} < \frac{1}{F_{n−1}^{(1)}F_{n}^{(1)} + F_{n−1}^{(1)}} .

Hence when n is even, we obtain

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

and when n is odd, we have

F_{n−1}^{(1)}F_{n}^{(1)} + F_{n−1}^{(1)} < ( ∑_{k=n}^{∞} \frac{1}{(F_{k}^{(1)})^{2}} )^{−1} < F_{n−1}^{(1)}F_{n}^{(1)} + F_{n−1}^{(1)} + 1.

Then (11) holds.

By applying (2)–(3) and (8), we get

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

∑_{k=n}^{∞} \frac{1}{(L_{k}^{(1)})^{2}} > \frac{1}{L_{n−1}^{(1)}L_{n}^{(1)} + L_{n−1}^{(1))}} , n is odd.

Then (12) holds.

In the final part of this section, we consider the generalized hyperfibonacci numbers {U_{n}^{(r)}}:

U_{n}^{(r)} = ∑_{j=0}^{n} U_{j}^{(r−1)}, with U_{n}^{(0)} = U_{n}, U_{0}^{(r)} = 0, U_{1}^{(r)} = 1.

where

U_{n} = \frac{τ^{n} − (−1)^{n}τ^{−n}}{\sqrt{Δ}} , τ = (p + \sqrt{Δ})/2, Δ = p^{2} + 4,

and p is a positive integer. It is evident that

U_{n}^{(1)} = \frac{U_{n} + U_{n+1} − 1}{p}.  (13)

And {U_{n}} satisfy that

W_{n} = pW_{n−1} + W_{n−2}, n ≥ 2.  (14)

When p = 1, U_{n}^{(1)} = F_{n}^{(1)}.

Now we discuss the partial infinite sum of reciprocal generalized hyperfibonacci numbers.

Lemma 4. For {U_{n}}, the following formulas hold:

U_{n+1}^{2} − U_{n}U_{n+2} = (−1)^{n},  (15)

From the definition of {U_{n}} , we can prove that (15) holds. Theorem 5. When n ≥ 2,

⌊( ∑_{k=n}^{∞} \frac{1}{U_{k}^{(1)}} )^{−1}⌋ = U_{n} − 1,  (16)

Proof. It follows from (13) and (14)–(15) that

\frac{1}{U_{n}^{(1)} − U_{n−1}^{(1)}} − \frac{1}{U_{n}^{(1)}} − \frac{1}{U_{n+1}^{(1)}} − \frac{1}{U_{n+2}^{(1)} − U_{n+1}^{(1)}} = \frac{pU_{n+1}}{U_{n}U_{n+2}} − \frac{U_{n}^{(1)} + U_{n+1}^{(1)}}{U_{n}^{(1)}U_{n+1}^{(1)}}
= \frac{U_{n+1}(U_{n} + U_{n+1} − 1)(U_{n+1} + U_{n+2} − 1) − U_{n}U_{n+2}(U_{n} + 2U_{n+1} + U_{n+2} − 2)}{pU_{n}U_{n+2}U_{n}^{(1)}U_{n+1}^{(1)}}
= \frac{(U_{n} + U_{n+1} − 1)((−1)^{n} + U_{n+1}U_{n+2} − U_{n+1}) − U_{n}U_{n+2}(U_{n+1} + U_{n+2} − 1)}{pU_{n}U_{n+2}U_{n}^{(1)}U_{n+1}^{(1)}}
= \frac{(U_{n} + U_{n+1} − 1)((−1)^{n} − U_{n+1}) + U_{n+2}(U_{n+1}^{2} − U_{n+1} − U_{n}U_{n+2} + U_{n})}{pU_{n}U_{n+2}U_{n}^{(1)}U_{n+1}^{(1)}}
= \frac{(U_{n} + U_{n+1} − 1)((−1)^{n} − U_{n+1}) + U_{n+2}((−1)^{n} − (p − 1)U_{n} − U_{n−1})}{pU_{n}U_{n+2}U_{n}^{(1)}U_{n+1}^{(1)}}
< 0,

and

\frac{1}{U_{n}^{(1)} − U_{n−1}^{(1)} − 1} − \frac{1}{U_{n}^{(1)}} − \frac{1}{U_{n+1}^{(1)}} − \frac{1}{U_{n+2}^{(1)} − U_{n+1}^{(1)} − 1} = \frac{pU_{n+1}}{(U_{n} − 1)(U_{n+2} − 1)} − \frac{1}{U_{n}^{(1)}} − \frac{1}{U_{n+1}^{(1)}}
= \frac{U_{n+1}^{(1)}(U_{n+2} + U_{n} − U_{n+1} − 1 + (−1)^{n}) + U_{n}^{(1)}(U_{n} + U_{n+2} − 1)}{(U_{n} − 1)(U_{n+2} − 1)U_{n}^{(1)}U_{n+1}^{(1)}} + \frac{U_{n}(U_{n+1}^{(1)}U_{n+1} − U_{n}^{(1)}U_{n+2})}{(U_{n} − 1)(U_{n+2} − 1)U_{n}^{(1)}U_{n+1}^{(1)}}
= \frac{U_{n+1}^{(1)}(U_{n+2} + U_{n} − U_{n+1} − 1 + (−1)^{n}) + U_{n}^{(1)}(U_{n} + U_{n+2} − 1)}{(U_{n} − 1)(U_{n+2} − 1)U_{n}^{(1)}U_{n+1}^{(1)}} + \frac{U_{n}((−1)^{n} − U_{n+1} + U_{n+2})}{p(U_{n} − 1)(U_{n+2} − 1)U_{n}^{(1)}U_{n+1}^{(1)}}
> 0.

Then we obtain

\frac{1}{U_{n}^{(1)} − U_{n−1}^{(1)}} < ∑_{k=n}^{∞} \frac{1}{U_{k}^{(1)}} < \frac{1}{U_{n}^{(1)} − U_{n−1}^{(1)} − 1}.

Hence (16) holds.

3 Some identities related to reciprocal hyperfibonacci numbers and hyperlucas numbers

In this section, we give some identities related to inverse of hyperfibonacci and hyperlucas numbers. There are some identities containing the reciprocals of Fibonacci and Lucas numbers (see [1, 2]):

∑_{n=0}^{∞} \frac{1}{F_{2n+1} + F_{s}} = \frac{\sqrt{5}s}{2L_{s}}, s odd,

∑_{n=0}^{∞} \frac{1}{F_{2n+1} + L_{s}/\sqrt{5}} = \frac{s}{2F_{s}} , s > 0 is even.

For hyperfibonacci and hyperlucas numbers F_{n}^{(1)} L_{n}^{(1)}, we have

Theorem 6. Let m be a positive integer. For F_{n}^{(1)} and L_{n}^{(1)}, we have

∑_{n=1}^{∞} \frac{L_{n+2m+2}}{F_{n}^{(1)}F_{n+4m}^{(1)}} = \frac{1}{F_{2m}} ∑_{k=1}^{4m} \frac{1}{F_{k}^{(1)}} ,  (17)

∑_{n=1}^{∞} \frac{F_{n+2m+2}}{L_{n}^{(1)}L_{n+4m}^{(1)}} = \frac{1}{5F_{2m}} ∑_{n=1}^{4m} \frac{1}{L_{k}^{(1)}} ,  (18)

∑_{n=1}^{∞} \frac{F_{n+2m+3}}{F_{n}^{(1)}F_{n+4m+2}^{(1)}} = \frac{1}{L_{2m+1}} ∑_{k=1}^{4m+2} \frac{1}{F_{k}^{(1)}} ,  (19)

∑_{n=1}^{∞} \frac{L_{n+2m+3}}{L_{n}^{(1)}L_{n+4m+2}^{(1)}} = \frac{1}{L_{2m+1}} ∑_{k=1}^{4m+2} \frac{1}{L_{k}^{(1)}} .  (20)

Proof. It follows from (2) that

\frac{1}{F_{n}^{(1)}} − \frac{1}{F_{n+4m}^{(1)}} = \frac{F_{n+4m}^{(1)} − F_{n}^{(1)}}{F_{n}^{(1)}F_{n+4m}^{(1)}} ,

\frac{1}{L_{n}^{(1)}} − \frac{1}{L_{n+4m}^{(1)}} = \frac{L_{n+4m}^{(1)} − L_{n}^{(1)}}{L_{n}^{(1)} − L_{n+4m}^{(1)}} ,
\frac{1}{F_{n}^{(1)}} − \frac{1}{F_{n+4m+2}^{(1)}} = \frac{F_{n+4m+4} − F_{n+2}}{F_{n}^{(1)}F_{n+4m+2}^{(1)}} ,
\frac{1}{L_{n}^{(1)}} − \frac{1}{L_{n+4m+2}^{(1)}} = \frac{L_{n+4m+4} − L_{n+2}}{L_{n}^{(1)}L_{n+4m+2}^{(1)}} .

From

F_{n+4m}^{(1)} − F_{n}^{(1)} = F_{2m}L_{n+2m+2},
L_{n+4m}^{(1)} − L_{n}^{(1)} = 5F_{2m}F_{n+2m+2},
F_{n+4m+4} − F_{n+2} = F_{n+2m+3}L_{2m+1},
L_{n+4m+4} − L_{n+2} = L_{n+2m+3}L_{2m+1}.

we obtain the formula (17)–(20).

We can give other identities for F_{n}^{(1)} and L_{n}^{(1)}. The following lemma will be used (see [12]).

Lemma 7. Let t be a real number with |t| > 1, s and a be positive integers, and b be a nonnegative integer. Then one has that

∑_{n=0}^{∞} \frac{1}{t^{2an+b} + t^{−2an−b} − (t^{as} + t^{−as})} = \frac{1}{t^{as} − t^{−as}} ∑_{n=0}^{s−1} \frac{1}{1 − t^{2an+b−as}}  (21)

Theorem 8. Suppose that a, b and s are positive integers with b > as. For F_{n}^{(1)} and L_{n}^{(1)}, we have:

(i) when a, b and s are odd,

∑_{n=0}^{∞} \frac{1}{F_{2an+b−2}^{(1)} − F_{as−2}^{(1)}} = \frac{\sqrt{5}}{L_{as}} ∑_{n=0}^{s−1} \frac{1}{1 − α^{2an+b−as}} ,  (22)

(ii) when b and s are both even,

∑_{n=0}^{∞} \frac{1}{L_{2an+b−2}^{(1)} − L_{as−2}^{(1)}} = \frac{1}{\sqrt{5}F_{as}} ∑_{n=0}^{s−1} \frac{1}{1 − α^{2an+b−as}} .  (23)

Proof. By means of (22) and F_{n}^{(1)} = F_{n+2} − 1, L_{n}^{(1)} = L_{n+2} − 1, we can easily prove that (22) and (23) hold.

4 Acknowledgment

The authors are grateful to the anonymous referee for his/her helpful comments.

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2010 Mathematics Subject Classification: Primary 11B39, 11B37; Secondary 05A19. Keywords: Fibonacci numbers, Lucas numbers.

(Concerned with sequences \underline{A000071} and \underline{A001610}.)

Received December 29 2011; revised version received March 26 2012. Published in Journal of Integer Sequences, April 9 2012.

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