A Formula for the Generating Functions of Powers of Horadam’s Sequence with Two

Additional Parameters

Emrah Kılıç

TOBB University of Economics and Technology

Mathematics Department

06560 Ankara Turkey

  ekilic@etu.edu.tr

Yücel Türker Ulutaş and Neşe Ömür

Kocaeli University Mathematics Department

41380 İzmit Kocaeli

Turkey

turkery@kocaeli.edu.tr
neseomur@kocaeli.edu.tr

Abstract

In this note, we give a generalization of a formula for the generating function of powers of Horadam’s sequence by adding two parameters. Thus we obtain a generalization of a formula of Mansour.

1 Introduction

Horadam [1, 2] defined the second-order linear recurrence sequence {W_{n} (a, b; p, q)} , or briefly {W_{n}} , as follows:

W_{n+1} = pW_{n} + qW_{n−1,} W_{0} = a, W_{1} = b  (1)

where a, b and p, q are arbitrary real numbers for n > 0 . The Binet formula for the sequence {W_{n}} is

W_{n} = \frac{Aα^{n} − Bβ^{n}}{α − β} ,

where A = b − aβ and B = b − aα. When a = 0, b = 1, and, a = 2, b = 1, denote W_{n} by U_{n} and V_{n}, respectively. If we take p = 1, q = 1, then U_{n} = F_{n} (nth Fibonacci number) and V_{n} = L_{n} (nth Lucas number).

Kılıç and Stanica [8] showed that for r > 0, n > 0, the sequence {W_{n}} satisfies the following recursion

W_{r(n+2)} = V_{r}W_{r(n+1)} − (−q)^{r} W_{rn}.

Riordan [4] found the generating function for powers of Fibonacci numbers. He proved that the generating function S_{k} (x) = ∑_{n≥0} F_{n}^{k}x^{n} satisfies the recurrence relation

(1 − a_{k}x + (−1)^{k} x^{2}) S_{k} (x) = 1 + kx ∑_{j=1}^{[k/2]} (−1)^{j} \frac{a_{kj}}{j} S_{k−2j} ((−1)^{j} x) ,

for k ≥ 1, where a_{1} = 1, a_{2} = 3, a_{s} = a_{s−1} + a_{s−2} for s ≥ 3, and (1 − x − x^{2})^{−j} = ∑_{k≥0} a_{kj}x^{k−2j}. Horadam [2] gave a recurrence relation for H_{k} (x) (see also [5]). Haukkanen [6] studied linear combinations of Horadam’s sequences and the generating function of the ordinary product of two of Horadam’s sequences.

Mansour [3] studied about the generating function for powers of Horadam’s sequence given by H_{k} (x; a, b, p, q) = H_{k} (x) = ∑_{n≥0} W_{n}^{k}x^{n}. Then he showed that the generating function H_{k} (x) can be expressed the ratio of two k by k determinants as well as he gave some applications for the generating function H_{k} (x).

In this study, we consider the generating function for powers of Horadam’s sequence defined by

ℜ_{k,t,r} (x; a, b, p, q) = ℜ_{k,t,r} (x) = ∑_{n≥t} W_{rn}^{k}x^{n}.

We shall derive a ratio to express the generating function ℜ_{k,t,r} (x) by using the method of Mansour. Moreover, we give applications of our results.

2 The Main Result

Firstly, we define two k by k matrices, in order to express the ℜ_{k,t,r} (x) as a ratio of two determinants. Let Δ_{k,r} = (Δ_{k,r} (i, j))_{1≤i,j≤k} = Δ_{k,r}(p, q) be the k × k matrix have the form

Δ_{k,r}(p, q)

= [ \begin{matrix}1 − xv_{r}^{k} − x^{2} (− (−q)^{r})^{k} & −xv_{r}^{k−1} (− (−q)^{r}) \binom{k}{1} & ... & −xv_{r} (− (−q)^{r})^{k−1} \binom{k}{k−1} \\ −v_{r}^{k−1}x & 1 − xv_{r}^{k−2} (− (−q)^{r}) \binom{k−1}{1} & ... & −x (− (−q)^{r})^{k−1} \binom{k−1}{k−1} \\ −v_{r}^{k−2}x & −xv_{r}^{k−3} (− (−q)^{r}) \binom{k−2}{1} & ... & 0 \\ ... & ... & ... & ... \\ −v_{r}^{2}x & −xv_{r} (− (−q)^{r}) \binom{2}{1} & ... & 0 \\ −v_{r}x & −x (− (−q)^{r}) \binom{1}{1} & ... & 1 \end{matrix} ]

and let δ_{k,t,r} = δ_{k,t,r}(a, b, p, q) be the k × k matrix have the form

δ_{k,t,r}(a, b, p, q)

= [ \begin{matrix}w_{rt}^{k} + x g_{k} & −xv_{r}^{k−1} (− (−q)^{r}) \binom{k}{1} & ... & −xv_{r} (− (−q)^{r})^{k−1} \binom{k}{k−1} \\ x g_{k−1} & 1 − xv_{r}^{k−2} (− (−q)^{r}) \binom{k−1}{1} & ... & −x (− (−q)^{r})^{k−1} \binom{k−1}{k−1} \\ x g_{k−2} & −xv_{r}^{k−3} (− (−q)^{r}) \binom{k−2}{1} & ... & 0 \\ ... & ... & ... & ... \\ x g_{2} & −xv_{r} (− (−q)^{r}) \binom{2}{1} & ... & 0 \\ x g_{1} & −x (− (−q)^{r}) \binom{1}{1} & ... & 1 \end{matrix} ] ,

where g_{j} = (w_{r(t+1)}^{j} − v_{r}^{j}w_{rt}^{j}) w_{rt}^{k−j} for all j = 1, 2, ...k.

Stanica [7] found the generating function of powers of terms of {W_{n}} given by [1], ∑_{n=0}^{∞} W_{n}^{k}x^{n}. Considering Stanica’s result, we give the following result for the generating function

ℜ_{k,t,r} (x) = ∑_{n=t}^{∞} W_{rn}^{k}x^{n}

as the following Lemma 1.

Lemma 1. For odd k,

ℜ_{k,t,r} (x) = \frac{1}{(α − β)^{k}} ∑_{j=0}^{\frac{k−1}{2}} (−AB)^{j} \binom{k}{j}

×\frac{A^{k−2j} − B^{k−2j} + (−q)^{rj} (B^{k−2j}α^{r(k−2j)} − A^{k−2j}β^{r(k−2j)}) x}{1 − (−q)^{rj} V_{r(k−2j)}x − q^{rk}x^{2}}

− ∑_{n=0}^{t−1} W_{rn}^{k}x^{n}

and for even k,

ℜ_{k,t,r} (x) = \frac{1}{(α − β)^{k}} ∑_{j=0}^{\frac{k}{2} −1} (−AB)^{j} \binom{k}{j}

×\frac{A^{k−2j} + B^{k−2j} − (−q)^{rj} (B^{k−2j}α^{r(k−2j)} + A^{k−2j}β^{r(k−2j)}) x}{1 − (−q)^{rj} V_{r(k−2j)}x + q^{rk}x^{2}}

+\binom{k}{k/2}\frac{(−AB)^{k/2}}{1 − (−q)^{k/2} x} − ∑_{n=0}^{t−1} W_{rn}^{k}x^{n}.

Proof. The proof easily follows from [7].

For further use, we define a family {A_{k,d,t,r}}_{d=1}^{k} of generating functions by

A_{k,d,t,r} (x) = ∑_{n=t}^{∞} W_{rn}^{k−d}W_{r(n+1)}^{d}x^{n+1}.  (2)

Now, we give two relations between the generating functions A_{k,d,t,r} (x) and ℜ_{k,t,r} (x). Lemma 2. For k ≥ 1, positive integer r and non-negative integer t,

(1 − V_{r}^{k}x + (− (−q)^{r})^{k} x^{2}) ℜ_{k,t,r} (x) − x ∑_{j=1}^{k−1} \binom{k}{j} (− (−q)^{r})^{j} V_{r}^{k−j}A_{k,k−j,t,r} (x)

= W_{rt}^{k}x^{t} + (W_{r(t+1)}^{k} − V_{r}^{k}W_{rt}^{k}) x^{t+1}.

Proof. Using the binomial theorem, we get

W_{r(n+2)}^{k} = (V_{r}W_{r(n+1)} − (−q)^{r} W_{rn})^{k}
= V_{r}^{k}W_{r(n+1)}^{k} + ∑_{j=1}^{k−1} \binom{k}{j} (− (−q)^{r})^{j} V_{r}^{k−j}W_{r(n+1)}^{k−j}W_{rn}^{j} + (− (−q)^{r})^{k} W_{rn}^{k}.

Multiplying by x^{n+2} and summing over all n ≥ t, using definition [2], we get

x^{n+2}W_{r(n+2)}^{k} = x^{n+2}V_{r}^{k}W_{r(n+1)}^{k} + x^{n+2} ∑_{j=1}^{k−1} \binom{k}{j} (− (−q)^{r})^{j} V_{r}^{k−j}W_{r(n+1)}^{k−j}W_{rn}^{j}
+x^{n+2} (− (−q)^{r})^{k} W_{rn}^{k}

and so

ℜ_{k,t,r} (x) − W_{rt}^{k}x^{t} − W_{r(t+1)}^{k}x^{t+1}

= xV_{r}^{k}ℜ_{k,t,r} (x) − x^{t+1}V_{r}^{k}W_{rt}^{k}

+x ∑_{j=1}^{k−1} \binom{k}{j} (− (−q)^{r})^{j} V_{r}^{k−j}A_{k,k−j,r} (x) + x^{2} (− (−q)^{r})^{k} ℜ_{k,t,r} (x) ,

which, by a simple arrangement, completes the proof.

Lemma 3. For any k ≥ 1, positive integer r, non-negative integer t, and d ≥ t + 1,

A_{k,d,t,r} (x) − x^{t+1}W_{rt}^{k−d}W_{r(t+1)}^{d} = xV_{r}^{d} (ℜ_{k,t,r} (x) − x^{t}W_{rt}^{k})

+x ∑_{j=1}^{d} \binom{d}{j} (− (−q)^{r})^{j} V_{r}^{d−j}A_{k,k−j,t,r} (x) .

Proof. Using the binomial theorem, we have

W_{rn}^{k−d}W_{r(n+1)}^{d} = W_{rn}^{k−d} (V_{r}W_{rn} − (−q)^{r} W_{r(n−1)})^{d}
= W_{rn}^{k−d} ∑_{j=0}^{d} \binom{d}{j}V_{r}^{d−j} (− (−q)^{r})^{j} W_{rn}^{d−j}W_{r(n−1)}^{j}.

Multiplying by x^{n+1} and summing over all n ≥ t + 1, we obtain the claimed result:

A_{k,d,t,r} (x) − x^{t+1}W_{rt}^{k−d}W_{r(t+1)}^{d} = xV_{r}^{d} (ℜ_{k,t,r} (x) − x^{t}W_{rt}^{k})

+ x ∑_{j=1}^{d} \binom{d}{j} (− (−q)^{r})^{j} V_{r}^{d−j}A_{k,k−j,t,r} (x) .

Now, we shall mention our main result:

Theorem 4. For any k ≥ 1, positive integer r, non-negative integer t, the generating function ℜ_{k,t,r} (x) is

\begin{matrix}det & (δ_{k,t,r}) \\ det & (Δ_{k,r}) \end{matrix}.  (3)

Proof. By using Lemma 1 and Lemma 2, we obtain

Δ_{k,r} [ℜ_{k,t,r} (x) , A_{k,k−1,t,r} (x) , A_{k,k−2,t,r} (x) , ...A_{k,1,t,r} (x)]^{T} = υ_{k,t,r},

where υ_{k,t,r} is given by

[W_{rt}^{k}x^{t} + (W_{r(t+1)}^{k} − V_{r}^{k}W_{rt}^{k}) x^{t+1}, (W_{rt}W_{r(t+1)}^{k−1} − V_{r}^{k−1}W_{rt}^{k}) x^{t+1},
(W_{rt}^{2}W_{r(t+1)}^{k−2} − V_{r}^{k−2}W_{rt}^{k}) x^{t+1}, ..., (W_{rt}^{k−1}W_{r(t+1)} − V_{r}W_{rt}^{k}) x^{t+1}] .

Hence the solution of the above equation gives the generating function ℜ_{k,t,r} (x) = (det (δ_{k,t,r})) / (det (Δ_{k,r})) .

3 Applications

We state some applications of our main result by the following tables:

Table 1: The generating function for the powers of Fibonacci numbers

k t r The generating function ℜ_{k,t,r} (x; 0, 1, 1, 1)
1 1 2 \frac{1}{1−3x+x^{2}}
2 1 2 \frac{1+x}{(1−x)(1−7x+x^{2})}
3 1 2 \frac{1+6x+x^{2}}{1−21x+56x^{2}−21x^{3}+x^{4}}
4 1 2 \frac{16+1712x+1712x^{2}+17x^{3}}{(1−x)(1−34x+x^{2})(1−1154x+x^{2})}

Table 2: The generating function for the powers of Lucas numbers

k t r The generating function ℜ_{k,t,r} (x; 2, 1, 1, 1)
1 1 2 \frac{3−2x}{1−3x+x^{2}}
2 1 2 \frac{9−23x+4x^{2}}{(1−x)(1−7x+x^{2})}
3 1 2 \frac{27−224x+141x^{2}−8x^{3}}{1−21x+56x^{2}−21x^{3}+x^{4}}
4 1 2 \frac{81−2054x+452913226x^{2}−78298x^{3}−2864x^{4}}{(1−x)(1−7x+x^{2})(1−47x+x^{2})}

Table 3: The generating function for the powers of Pell numbers

k t r The generating function ℜ_{k,t,r} (x; 0, 1, 2, 1)
1 1 2 \frac{2}{x^{2}−6x+1}
2 1 2 \frac{4+4x}{(1−x)(1−34x+x^{2})}
3 1 2 \frac{8(1+12x+x^{2})}{1−204x+1190x^{2}−204x^{3}+x^{4}}
4 1 2 \frac{16(x+1)(1+106x+x^{2})}{(1−x)(1−34x+x^{2})(1−1154x+x^{2})}

Table 4: The generating function for the powers of Chebyshev polynomials of the second kind

k t r The generating function ℜ_{k,t,r} (x; 1, 2t, 2t, −1)
1 1 2 \frac{−1+4t^{2}−x}{1+(2−4t^{2})x+x^{2}}
2 1 2 \frac{(16t^{4}−8t^{2}+1)+(16t^{2}−16t^{4}−2)x+x^{2}}{(1−x)(1+(−2+12t^{2})x+x^{2})}
3 1 2 \frac{12t^{2}−48t^{4}+64t^{6}+27−(4−24t^{2}+288t^{4}−256t^{6}+576t^{8})x+(40t^{2}−336t^{4}+64t^{6}−3)x^{2}−x^{3}}{1+(−64t^{6}+96t^{4}−40t^{2}+4)x+(256t^{8}−512t^{6}+336t^{4}−80t^{2}+6)x^{2}+(−64t^{6}+96t^{4}−40t^{2}+4)x^{3}+x^{4}}

Fibonacci numbers. If a = 0 and p = q = b = 1, then Theorem 4 for k = 1, 2, 3, 4 yields Table 1.

Lucas numbers. If a = 2 and p = q = b = 1, then Theorem 4 for k = 1, 2, 3, 4 yields Table 2.

Pell numbers. If a = 0 and p = 2, q = b = 1, then Theorem 4 for k = 1, 2, 3, 4 yields Table 3.

Chebyshev polynomials of the second kind. If a = 1, b = p = 2t and q = −1, then Theorem 4 for k = 1, 2, 3 yields Table 4.

Applying Theorem 4 for k = 1, 2, 3, then we give the following corollary.

Corollary 5. Let k = 1, 2, 3. Then the generating function ℜ_{k,t,r} (x; a, b, p, q) is given by Â_{k,t,r} (x) /Ê_{k,t,r} (x) , where

Â_{1,1,2} (x) = aq + bp − aq^{2}x,
Â_{2,1,2} (x) = a^{2}q^{2} + b^{2}p^{2} + 2abpq + q^{2} (−2a^{2}q^{2} + b^{2}p^{2} − 2abp^{3} − 2a^{2}p^{2}q − 2abpq) x
+a^{2}q^{6}x^{2},
Â_{3,1,2} (x) = b^{3}p^{3} + 3ab^{2}p^{2}q + 3a^{2}bpq^{2} + a^{3}q^{3} − (3a^{3}p^{4}q^{4} + 7a^{3}p^{2}q^{5}
+3a^{3}q^{6} + 6a^{2}bp^{5}q^{3} + 15a^{2}bp^{3}q^{4} + 6a^{2}bpq^{5} + 6ab^{2}p^{4}q^{3}
−2b^{3}p^{5}q^{2} − 4b^{3}p^{3}q^{3} + 3ab^{2}p^{6}q^{2})x + (3a^{3}p^{4}q^{7} + 7a^{3}p^{2}q^{8} + 3a^{3}q^{9}
+3a^{2}bp^{5}q^{6} + 6a^{2}bp^{3}q^{7} + 3a^{2}bpq^{8} − 3ab^{2}p^{4}q^{6}
−3ab^{2}p^{2}q^{7} + b^{3}p^{3}q^{6})x^{2} − a^{3}q^{12}x^{3}

and

Ê_{1,1,2} (x) = 1 − (p^{2} + 2q) x + q^{2}x^{2},
Ê_{2,1,2} (x) = (q^{2}x − 1) (−1 + (p^{4} + 4p^{2}q + 2q^{2}) x − q^{4}x^{2}) ,
Ê_{3,1,2} (x) = (−1 + q^{2} (12q + p^{2}) x − q^{6}x^{2})
× (−1 + (2q + p^{2}) (4p^{2}q + p^{4} + q^{2}x − q^{6}x^{2}) .

References

[1] A. F. Horadam, Basic properties of a certain generalized sequence of numbers,Fibonacci Quart. 3 (1965), 161–176.

[2] A.F. Horadam, Generating functions for powers of a certain generalized sequence of numbers, Duke Math. J. 32 (1965), 437–446.

[3] T. Mansour, A formula for the generating functions of powers of Horadam’s sequence, Australasian J. Combinatorics 30 (2004), 207–212.

[4] J. Riordan, Generating function for powers of Fibonacci numbers, Duke Math. J. 29 (1962), 5–12.

[5] P. Haukkanen and J. Rutkowski, On generating functions for powers of recurrence sequences, Fibonacci Quart. 29 (1991), 329–332.

[6] P. Haukkanen, A note on Horadam’s sequence, Fibonacci Quart., 40 (2002), 358–361.

[7] P. Stanica, Generating functions, weighted and non-weighted sums for powers of secondorder recurrence sequences, Fibonacci Quart. 41 (2003), 321–333.

[8] E. Kılıç and P. Stanica, Factorizations of binary polynomial recurrences by matrix methods, Rocky Mount. J. Math., to appear.

2010 Mathematics Subject Classification: Primary 11B37; Secondary 11B39, 05A15. Keywords: second-order linear recurrence, generating function.

(Concerned with sequences \underline{A000032}, \underline{A000045}, and \underline{A000129}.)

Received December 6 2010; revised version received January 27 2011; May 3 2011. Published in Journal of Integer Sequences, May 3 2011.

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