Some Properties of Hyperfibonacci and Hyperlucas Numbers

Ning-Ning Cao^{1} and Feng-Zhen Zhao^{1} School of Mathematical Sciences Dalian University of Technology Dalian 116024 China

                  caoning0928@163.com
               fengzhenzhao@yahoo.com.cn

Abstract

In this paper, we discuss the properties of hyperfibonacci numbers and hyperlucas numbers. We derive some identities for hyperfibonacci and hyperlucas numbers by the method of coefficients. Furthermore, we give asymptotic expansions of certain sums involving hyperfibonacci and hyperlucas numbers by Darboux’s method.

1 Introduction

Dil and Mező [1] introduced the definitions of “hyperfibonacci” numbers F_{n}^{(r)} and “hyperlucas” numbers L_{n}^{(r)}

F_{n}^{(r)} = ∑_{k=0}^{n} F_{k}^{(r−1)}, with F_{n}^{(0)} = F_{n}, F_{0}^{(r)} = 0, F_{1}^{(r)} = 1,

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

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

where r is a positive integer, and F_{n} and L_{n} are Fibonacci and Lucas numbers, respectively. The generating functions of F_{n}^{(r)} and L_{n}^{(r)} are as follows:

∑_{n=0}^{∞} F_{n}^{(r)}t^{n} = \frac{t}{(1 − t − t^{2})(1 − t)^{r}} ,

∑_{n=0}^{∞} L_{n}^{(r)}t^{n} = \frac{2 − t}{(1 − t − t^{2})(1 − t)^{r}} .

The first few values of F_{n}^{(r)} and L_{n}^{(r)} are as follows:

F_{n}^{(1)} : 0, 1, 2, 4, 7, 12, 20, 33, 54, 88, 143, 232, 376, 609, 986, 1596, 2583, …;
F_{n}^{(2)} : 0, 1, 3, 7, 14, 26, 46, 79, 133, 221, 364, 596, 972, 1581, 2567, 4163, 6746, …;
L_{n}^{(1)} : 2, 3, 6, 10, 17, 28, 46, 75, 122, 198, 321, 520, 842, 1363, 2206, 3570, 5777, …;

L_{n}^{(2)} : 2, 5, 11, 21, 38, 66, 112, 187, 309, 507, 828, 1348, 2190, 3553, 5759, 9329, 15106, ….

There are some elementary identities for F_{n}^{(r)} and L_{n}^{(r)} when r = 1, 2. For example,

F_{n}^{(1)} = F_{n+2} − 1,
F_{n}^{(2)} = F_{n+4} − n − 3
= ∑_{k=0}^{n}(n − k)F_{k},
L_{n}^{(1)} = L_{n+2} − 1
= F_{n} + F_{n+2} − 1,
L_{n}^{(2)} = 4(F_{n+1} − 1) + 3F_{n} − n
= L_{n+3} − (n + 4).

For the above values and elementary identities of F_{n}^{(r)} and L_{n}^{(r)}, see [4] (\underline{A000071}, \underline{A001924}, \underline{A001610}, \underline{A023548}).

Hyperfibonacci numbers and hyperlucas numbers are useful, and Dil and Mező [1] derived some equalities for Fibonacci and Lucas numbers by applying them. Hence, hyperfibonacci numbers and hyperlucas numbers deserve to be investigated. In this paper, we discuss properties of F_{n}^{(r)} and L_{n}^{(r)}. We establish some identities for F_{n}^{(r)} and L_{n}^{(r)}. Furthermore, we give asymptotic expansions of certain sums related to F_{n}^{(r)} and L_{n}^{(r)}.

For convenience, we first recall some notation. Let α = (1 + \sqrt{5})/2. It is well known that

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

and F_{n} and L_{n} satisfy the following recurrence relation

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

As usual, the binomial coefficient \binom{n}{m} is defined by

\binom{n}{m} = \begin{cases}\frac{n!}{m!(n − m)!}, if n ≥ m; \\ 0, if n < m; \end{cases}

where n and m are nonnegative integers. Throughout, [z^{n}]f(z) denotes the coefficient of z^{n} in f(z), where

f(z) = ∑_{n=0}^{∞} f_{n}z^{n}.

If f(t) and g(t) are formal power series, the following relations hold [2]

[t^{n}](af(t) + bg(t)) = a[t^{n}]f(t) + b[t^{n}]g(t),  (2)
[t^{n}]tf(t) = [t^{n−1}]f(t),  (3)
[t^{n}]f(t)g(t) = ∑_{k=0}^{n}[y^{k}]f(y)[t^{n−k}]g(t).  (4)

The above relations (2)–(4) will be used later on.

Now we recall the notation of the binomial transform of a sequence, the inverse binomial transform of a sequence and Euler-Seidel infinite matrix [1, 5]. Let {a_{k}}_{k=0}^{∞} be a sequence. The binomial transform of \{a_{k}\} is given by ∑_{k=0}^{n} \binom{n}{k}a_{k}, the inverse binomial transform of \{a_{k}\} is given by ∑_{k=0}^{n} \binom{n}{k}(−1)^{k}a_{k}, and the Euler-Seidel infinite matrix corresponding to the sequence {a_{k}} is determined by the following formulas

a_{n}^{[0]} = a_{n} (n ≥ 0),
a_{n}^{[k]} = a_{n}^{[k−1]} + a_{n+1}^{[k−1]} (n ≥ 0, k ≥ 0),

where a_{n}^{[k]} is the element at the (k +1)th row and the (n+1)th column. The sequences {a_{n}^{[0]}} and {a_{0}^{[n]}} satisfy [1]

a_{0}^{[n]} = ∑_{k=0}^{n} \binom{n}{k}a_{k}^{[0]},  (5)

a_{n}^{[0]} = ∑_{k=0}^{n} \binom{n}{k}(−1)^{n−k}a_{0}^{[k]}.  (6)

Let a(t) and ā(t) be the generating function of {a_{n}^{[0]}} and {a_{0}^{[n]}}, respectively,

a(t) = ∑_{n=0}^{∞} a_{n}^{[0]}t^{n}, ā(t) = ∑_{n=0}^{∞} a_{0}^{[n]}t^{n}.

The functions a(t) and ā(t) satisfy [1]

ā(t) = \frac{1}{1 − t}a(\frac{t}{1 − t}),  (7)
a(t) = \frac{1}{t + 1}ā(\frac{t}{t + 1}).  (8)

2 Main Results

In this section, we derive some identities for F_{n}^{(r)} and L_{n}^{(r)}. Later, we give asymptotic expansions of certain sums involving F_{n}^{(r)} and L_{n}^{(r)}.

Various identities involving Fibonacci and Lucas numbers were established. The following sums were investigated [3, 7, 8]

∑_{j_{1}+j_{2}+···+j_{k}=n} F_{j_{1}}F_{j_{2}} ⋯ F_{j_{k}}, ∑_{j_{1}+j_{2}+···+j_{k}=n} L_{j_{1}}L_{j_{2}} ⋯ L_{j_{k}}.

For example,

∑_{j_{1}+j_{2}=n} F_{j_{1}}F_{j_{2}} = \frac{(n − 1)L_{n} + 2F_{n−1}}{5} ,
∑_{j_{1}+j_{2}=n} L_{j_{1}}L_{j_{2}} = (n + 1)L_{n} + 2F_{n+1}.

Now we derive some identities for F_{n}^{(r)} and L_{n}^{(r)}. Denote

A_{n,k,r} = ∑_{j_{1}+j_{2}+···+j_{k}=n} F_{j_{1}}^{(r)}F_{j_{2}}^{(r)} ⋯ F_{j_{k}}^{(r)}, B_{n,k,r} = ∑_{j_{1}+j_{2}+···+j_{k}=n} L_{j_{1}}^{(r)}L_{j_{2}}^{(r)} ⋯ L_{j_{k}}^{(r)}.

These sums are interesting because they can help us to find some new convolution properties. For A_{n,k,r} and B_{n,k,r}, we have

Theorem 1. Let k, n ≥ 1 and r ≥ 1 be positive integers. For A_{n,k,r} and B_{n,k,r}, we have

A_{n,2,1} = n + 5 − 2F_{n+4} + \frac{(n + 1)L_{n+4} − 2F_{n+1}}{5} ,  (9)
B_{n,2,1} = n + 9 − 10F_{n+4} − 2F_{n+1} + \frac{(5n + 9)L_{n+4} + 4L_{n+6}}{5} ,  (10)
A_{n,k+1,r} = ∑_{j=0}^{n} A_{n,k,r}F_{j}^{(r)},  (11)

B_{n,k+1,r} = ∑_{j=0}^{n} B_{n,k,r}L_{j}^{(r)}.  (12)

Proof. Let

F_{r}(t) = \frac{t}{(1 − t − t^{2})(1 − t)^{r}} , L_{r}(t) = \frac{2 − t}{(1 − t − t^{2})(1 − t)^{r}} .

Clearly,

F_{1}(t) = (\frac{α^{2} − α^{−2}}{t − 1} − \frac{α}{t − α^{−1}} − \frac{α^{−1}}{t + α})\frac{1}{\sqrt{5}}
= ( ∑_{n=0}^{∞} F_{n+2}t^{n} − ∑_{n=0}^{∞} t^{n})t,
A_{n,2,1} = [t^{n}]F_{1}^{2}(t),
B_{n,2,1} = [t^{n}]L_{1}^{2}(t)
= [t^{n}]F_{1}^{2}(t) − 4[t^{n+1}]F_{1}^{2}(t) + 4[t^{n+2}]F_{1}^{2}(t)
= A_{n,2,1} − 4A_{n+1,2,1} + 4A_{n+2,2,1}.

Then we get

∑_{j_{1}+j_{2}=n} F_{j_{1}}^{(1)}F_{j_{2}}^{(1)} = [t^{n}]F_{1}^{2}(t)

= [t^{n}]\frac{1}{5}[\frac{(α^{2} − α^{−2})^{2}}{(t − 1)^{2}} + \frac{α^{2}}{(t − α^{−1})^{2}} + \frac{α^{−2}}{(t + α)^{2}} − \frac{2α(α^{2} − α^{−2})}{(t − 1)(t − α^{−1})}

−\frac{2α^{−1}(α^{2} − α^{−2})}{(t − 1)(t + α)} + \frac{2}{(t − α^{−1})(t + α)}]

= [t^{n}]\frac{1}{5}\{\frac{(α^{2} − α^{−2})^{2}}{(t − 1)^{2}} + \frac{α^{4}}{(1 − αt)^{2}} + \frac{α^{−4}}{(1 + α^{−1}t)^{2}}

+\frac{2(α^{2} − α^{−2})(α^{3} + α^{−3})}{1 − t} − 2[α^{4}(α^{2} − α^{−2}) + \frac{α}{α + α^{−1}} ]\frac{1}{1 − αt}

+2[α^{−4}(α^{2} − α^{−2}) − \frac{α^{−1}}{α + α^{−1}} ]\frac{1}{1 + α^{−1}t}\}

= [t^{n}]\frac{1}{5}\{(α^{2} − α^{−2})^{2} ∑_{n=0}^{∞}(n + 1)t^{n} + ∑_{n=0}^{∞}[α^{n+4} + (−1)^{n}α^{−n−4}](n + 1)t^{n}

+2(α^{2} − α^{−2})(α^{3} + α^{−3}) ∑_{n=0}^{∞} t^{n}

−2[α^{4}(α^{2} − α^{−2}) + \frac{α}{α + α^{−1}} ] ∑_{n=0}^{∞} α^{n}t^{n}

+2[α^{−4}(α^{2} − α^{−2}) − \frac{α^{−1}}{α + α^{−1}} ] ∑_{n=0}^{∞}(−1)^{n}α^{−n}t^{n}\}

= [t^{n}]\{ ∑_{n=0}^{∞}(n + 1)t^{n} + ∑_{n=0}^{∞} \frac{α^{n+4} + (−1)^{n}α^{−n−4}}{5} (n + 1)t^{n} + 2F_{3} ∑_{n=0}^{∞} t^{n}

−\frac{2}{5}(α^{2} − α^{−2}) ∑_{n=0}^{∞}[α^{n+4} − (−1)^{n}α^{−n−4}]t^{n}

−\frac{2}{5(α + α^{−1})} ∑_{n=0}^{∞}[α^{n+1} + (−1)^{n}α^{−n−1}]t^{n}\}

= [t^{n}] ∑_{n=0}^{∞}(n + 1)t^{n} + [t^{n}] ∑_{n=0}^{∞} \frac{(n + 1)L_{n+4}}{5} t^{n} + [t^{n}]4 ∑_{n=0}^{∞} t^{n}

−[t^{n}]2 ∑_{n=0}^{∞} F_{n+4}t^{n} − [t^{n}]\frac{2}{5} ∑_{n=0}^{∞} F_{n+1}t^{n}.

By applying (3) and the definitions of F_{n} and L_{n}, we have (9). Naturally, we deduce that

B_{n,2,1} = n + 9 − 10F_{n+4} + \frac{4(n + 3)L_{n+6} − 4(n + 2)L_{n+5} + (n + 1)L_{n+4}}{5} − 2F_{n+1}.

By using (1), we prove that (10) holds. By means of (4), we can show that (11) and (12) hold.

It is interesting that we can get congruence relations from (9) and (10)

A_{n,2,1} ≡ (n − 2F_{n+4} + \frac{(n + 1)L_{n+4} − 2F_{n+1}}{5} ) (mod 5),

B_{n,2,1} ≡ (n − 10F_{n+4} − 2F_{n+1} + \frac{(5n + 9)L_{n+4} + 4L_{n+6}}{5} ) (mod 9).

When k or r gets large, it is difficult to compute the closed forms of A_{n,k,r} and B_{n,k,r}. However, we can give their asymptotic values. Now we recall a lemma [6].

Lemma 2. Assume that f(t) = ∑_{n≥0} a_{n}t^{n} is an analytic function for |t| < r and with a finite number of algebraic singularities on the circle |t| = r. α_{1}, α_{2}, ⋯, α_{l} are singularities of order ω, where ω is the highest order of all singularities. Then

a_{n} = (n^{ω−1}/Γ(ω)) × ( ∑_{k=1}^{l} g_{k}(α_{k})α_{k}^{−n} + o(r^{−n})),  (13)

where Γ(ω) is the gamma function, and

g_{k}(α_{k}) = lim_{t→α_{k}}(1 − (t/α_{k}))^{ω}f(t).

Theorem 3. Suppose that k and r are fixed positive integers. For A_{n,k,r} and B_{n,k,r}, when n → ∞,

A_{n,k,r} = \frac{n^{k−1}}{(k − 1)!}[(\frac{α}{α^{2} + 1})^{k}(1 + α)^{kr}α^{n} + o(α^{n}))],  (14)
B_{n,k,r} = \frac{n^{k−1}}{(k − 1)!}[(\frac{2α^{2} − α}{α^{2} + 1})^{k}(1 + α)^{kr}α^{n} + o(α^{n}))].  (15)

Proof. Let

f_{k,r}(t) = \frac{t^{k}}{(1 − t − t^{2})^{k}(1 − t)^{kr}} .

We know that f_{k,r}(t) is analytic for |t| < 1/α, and with one algebraic singularity on the circle |t| = 1/α. The order of 1/α is k. One can compute that

lim_{t→1/α}(1 − αt)^{k}f_{k,r}(t) = (\frac{α}{α^{2} + 1})^{k}(1 + α)^{kr}.

By using Lemma 2, we can prove that (14) holds. Using the same method, we can prove that (15) holds.

In addition, we give asymptotic expansions of other sums for F_{n}^{(r)} and L_{n}^{(r)}.

Theorem 4. Let n be a positive integer. When n → ∞,

∑_{k=0}^{n} \binom{n}{k}F_{k}^{(r)} = \frac{α(1 + α)^{r}}{(α^{2} + 1)(2 − α)^{n}} + o((2 − α)^{−n}),  (16)
∑_{k=0}^{n} \binom{n}{k}L_{k}^{(r)} = \frac{(1 + α)^{r}}{(2 − α)^{n}} + o((2 − α)^{−n}),  (17)
∑_{k=0}^{n} \binom{n}{k}(−1)^{k}F_{k}^{(r)} = \frac{−α^{n+1}(2 − α)^{r}}{α^{2} + 1} + o((−α)^{n}),  (18)
∑_{k=0}^{n} \binom{n}{k}(−1)^{k}L_{k}^{(r)} = (2 − α)^{r}α^{n} + o((−α)^{n}).  (19)

Proof. We only give the proofs of (16) and (18). The proofs of (17) and (19) are similar to those of (16) and (18), respectively, and they are omitted here. Let F_{k}^{(r)} be the first row, and we get the Euler-Seidel infinite matrix A. Then let F_{k}^{(r)} be the first column, and we get Euler-Seidel infinite matrix B. The elements of A and B are denoted by a_{n}^{[k]} and b_{n}^{[k]}. By using (5) and (6), we obtain

a_{0}^{[n]} = ∑_{k=0}^{n} \binom{n}{k}F_{k}^{(r)},
b_{n}^{[0]} = ∑_{k=0}^{n} \binom{n}{k}(−1)^{n−k}F_{k}^{(r)}.

By means of (7) and (8), we get

∑_{k=0}^{n} \binom{n}{k}F_{k}^{(r)} = a_{0}^{[n]}
= [t^{n}]\frac{t(1 − t)^{r}}{(t^{2} − 3t + 1)(1 − 2t)^{r}} ,
∑_{k=0}^{n} \binom{n}{k}(−1)^{n−k}F_{k}^{(r)} = b_{n}^{[0]}
= [t^{n}]\frac{−t(1 + t)^{r}}{t^{2} − t − 1} .

We know that the function ā(t) = \frac{t(1 − t)^{r}}{(t^{2} − 3t + 1)(1 − 2t)^{r}} is analytic for |t| < 2 − α with one

algebraic singularity α_{1} = 2 − α on the circle |t| = 2 − α, and a(t) = \frac{−t(1 + t)^{r}}{t^{2} − t − 1} is analytic for |t| < α^{−1} with one algebraic singularity β_{1} = −α^{−1} on the circle |t| = α^{−1}. The orders of α_{1} and β_{1} are 1. We can compute that

lim_{t→2−α}(1 − \frac{t}{2 − α})ā(t) = \frac{α(1 + α)^{r}}{α^{2} + 1} ,
lim_{t→−α^{−1}} (1 + \frac{t}{α^{−1}} )a(t) = −\frac{α(2 − α)^{r}}{α^{2} + 1} .

By using Lemma 2, we prove that (16) and (18) hold.

Theorem 5. Suppose that m and r are fixed positive integers. When n → ∞,

∑_{j=0}^{n} F_{n−j}^{(r)}\binom{j + m − 1}{j} = \frac{α^{n+1}(1 + α)^{r+m}}{α^{2} + 1} + o(α^{n}),  (20)

∑_{j=0}^{n} L_{n−j}^{(r)}\binom{j + m − 1}{j} = (1 + α)^{r+m}α^{n} + o(α^{n}).  (21)

Proof. We can verify that

∑_{j=0}^{n} F_{n−j}^{(r)}\binom{j + m − 1}{j} = ∑_{j=0}^{n}[t^{n−j}]\frac{t}{(1 − t − t^{2})(1 − t)^{r}} [t^{j}]\frac{1}{(1 − t)^{m}}
= [t^{n}]\frac{t}{(1 − t − t^{2})(1 − t)^{m+r}} ,
∑_{j=0}^{n} L_{n−j}^{(r)}\binom{j + m − 1}{j} = [t^{n}]\frac{2 − t}{(1 − t − t^{2})(1 − t)^{m+r}} .

By using Lemma 2, we have (20) and (21).

In the final of this section, we compare the accurate values with the asymptotic values. In Theorem 4, let

X_{n} = ∑_{k=0}^{n} \binom{n}{k}F_{k}^{(r)}, Y_{n} = \frac{α(1 + α)^{r}}{(α^{2} + 1)(2 − α)^{n}} .

\begin{matrix}n & X_{n} & Y_{n} & |X_{n} − Y_{n}|/X_{n} \\ 50 & 9.2737156629317196 × 10^{20} & 9.2737269219308562 × 10^{20} & 1.2141 × 10^{−6} \\ 100 & 7.345448671565505 × 10^{41} & 7.3454486715782857 × 10^{41} & 1.7399 × 10^{−12} \\ 150 & 5.818115698360039 × 10^{62} & 5.8181156983601641 × 10^{62} & 2.1352 × 10^{−14} \end{matrix}

Table 1: some values of X_{n} and Y_{n}

From the above table, we find that the value of |X_{n} − Y_{n}|/X_{n} gets smaller and smaller with the increasing of n.

3 Remarks

Consider the sequences {u_{n}^{(r)}} and {v_{n}^{(r)}} defined by

∑_{n=0}^{∞} u_{n}^{(r)}z^{n} = \frac{z}{(1 − pz − z^{2})(1 − z)^{r}} ,

∑_{n=0}^{∞} v_{n}^{(r)}z^{n} = \frac{2 − pz}{(1 − pz − z^{2})(1 − z)^{r}} ,

where p > 0. It is clear that u_{n}^{(r)} = F_{n}^{(r)} and v_{n}^{(r)} = L_{n}^{(r)} when p = 1. The conclusions of F_{n}^{(r)} and L_{n}^{(r)} can be generalized to the cases of {u_{n}^{(r)}} and {v_{n}^{(r)}}. For example, put

U_{n,k,r} = ∑_{j_{1}+j_{2}+···+j_{k}=n} u_{j_{1}}^{(r)}u_{j_{2}}^{(r)} ⋯ u_{j_{k}}^{(r)}, V_{n,k,r} = ∑_{j_{1}+j_{2}+···+j_{k}=n} v_{j_{1}}^{(r)}v_{j_{2}}^{(r)} ⋯ v_{j_{k}}^{(r)}.

Then we have

U_{n,2,1} = \frac{n − 1}{p^{2}(p^{2} + 4)}(v_{n}^{(0)} + 2v_{n+1}^{(0)} + v_{n+2}^{(0)}) − \frac{2}{p^{3}} (u_{n+1}^{(0)} + 2u_{n}^{(0)} + u_{n−1}^{(0)})
+\frac{2}{p(p^{2} + 4)}u_{n−1}^{(0)} + \frac{np + p + 4}{p^{3}},  (22)
U_{n,k+1,r} = ∑_{j=0}^{n} U_{n,k,r}u_{j}^{(r)}.  (23)

We can verify that

U_{n,2,1} = [t^{n}]\frac{t^{2}}{(1 − pt − t^{2})^{2}(1 − t)^{2}}
= [t^{n}]\{\frac{1}{Δ(1 + τ)^{2}} \frac{t^{2}}{(t + τ)^{2}} + \frac{1}{Δ}(\frac{1}{1 − τ^{−1}} − \frac{1}{1 + τ} )^{2} \frac{t^{2}}{(1 − t)^{2}}

+\frac{1}{Δ} \frac{1}{(1 − τ^{−1})^{2}} \frac{t^{2}}{(t − τ^{−1})^{2}} − \frac{2}{Δ(1 + τ)}(\frac{1}{1 − τ^{−1}} − \frac{1}{1 + τ} )\frac{t^{2}}{τ + 1}(\frac{1}{t + τ} + \frac{1}{1 − t})

−\frac{2t^{2}}{Δ(1 + τ)(1 − τ^{−1})}(\frac{−1}{t + τ} + \frac{1}{t − τ^{−1}} )\frac{1}{\sqrt{Δ}}

+\frac{2t^{2}}{Δ(1 − τ^{−1})}(\frac{1}{1 − τ^{−1}} − \frac{1}{1 + τ} )(\frac{1}{t − τ^{−1}} + \frac{1}{1 − t})\frac{1}{1 − τ^{−1}} \}

= [t^{n}]\{\frac{1}{p^{2}Δ} ∑_{n=0}^{∞}(n + 1)t^{n+2}(v_{n+2}^{(0)} + 2v_{n+3}^{(0)} + v_{n+4}^{(0)}) + \frac{1}{p^{2}} ∑_{n=0}^{∞}(n + 1)t^{n+2}

−\frac{2}{p^{3}} ∑_{n=0}^{∞} t^{n+2}(u_{n+3}^{(0)} + 2u_{n+2}^{(0)} + u_{n−1}^{(0)}) + \frac{2}{pΔ} ∑_{n=0}^{∞} t^{n+2}u_{n+1}^{(0)} + \frac{4 + 2p}{p^{3}} ∑_{n=0}^{∞} t^{n+2}\},

where Δ = p^{2} + 4, τ = \frac{p+\sqrt{Δ}}{2}. Hence (22) holds. The proof of (23) is similar to that of (11), and it is omitted here. The identities about {v_{n}^{(0)}} can be found in the same way.

4 Acknowledgments

The authors would like to thank the anonymous referees for their criticism and useful suggestions.

References

[1] A. Dil and I. Mező, A symmetric algorithm for hyperharmonic and Fibonacci numbers, Appl. Math. Comput. 206 (2008), 942–951.

[2] D. Merlini, R. Sprugnoli and M. C. Verri, The method of coefficients, Amer. Math. Monthly 114 (2007), 40–57.

[3] N. Robbins, Some convolution-type and combinatorial identities pertaining to binary linear recurrences, Fibonacci Quart. 29 (1991), 249–255.

[4] N. J. A. Sloane, The On–Line Encyclopedia of Integer Sequences. Published electroni-

   cally at http://www.research.att.com/~njas/sequences, 2010.

[5] M. Z. Spivey, Combinatorial sums and finite difference, Discrete Math. 307 (2007), 3130–3146.

[6] G. Szegő, Orthogonal Polynomials, Amer. Math. Soc. Colloq. Publ., Vol. 23, rev. ed., 1959.

[7] Wenpeng Zhang, Some identities involving the Fibonacci numbers, Fibonacci Quart. 35 (1997), 225–228.

[8] Fengzhen Zhao and Tianming Wang, Generalizations of some identities involving the Fibonacci numbers, Fibonacci Quart. 39 (2001), 165–167.

2000 Mathematics Subject Classification: Primary 11B37, 11B39; Secondary 05A16, 05A19. Keywords: Fibonacci numbers, Lucas numbers, generating function, asymptotic expansion, Darboux’s method.

(Concerned with sequences \underline{A000048}, \underline{A000071}, \underline{A001610}, \underline{A001924}, and \underline{A023548}.)

Received June 6 2010; revised version received September 8 2010. Published in Journal of Integer Sequences, September 10 2010. Corrected, March 26 2011.

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