Download References - DergiPark

∞
{𝒂𝒌,𝒏 (𝝉)}𝝉 geneleştirilmiş Fibonacci otokorelasyon dizilerinin terimlerini içeren
bazı bağıntılar
ÖZ
Bu çalışmada
a k,n (τ) ≔ a n (Uki , τ)
∞
terimlerine sahip genelleştirilmiş Fibonacci otokorelasyon dizileri {a k,n (τ)}τ ve bu dizilerin terimlerini içeren bazı
toplamlar verildi.
Anahtar Kelimeler: Fibonacci sayıları, genelleştirilmiş Fibonacci otokorelasyon diziler, toplamlar.
Some sums related to the terms of generalized Fibonacci autocorrelation
∞
sequences {𝒂𝒌,𝒏 (𝝉)}𝝉
ABSTRACT
∞
In this paper, we give the terms of the generalized Fibonacci autocorrelation sequences {a k,n (τ)}τ defined as
a k,n (τ) ≔ a n (Uki , τ)
and some interesting sums involving terms of these sequences for odd integer number k and nonnegative integers τ, n.
Keywords: Fibonacci numbers, generalized Fibonacci autocorrelation sequences, sums.
1. GİRİŞ ( INTRODUCTION)
For a, b, p, q ∈ ℤ, the second order sequence {Wn (a, b; p, q)} is defined for n > 0 by
Wn+1 (a, b; p, q) = pWn (a, b; p, q) − qWn−1 (a, b; p, q)
in which W0 (a, b; p, q) = a, W1 (a, b; p, q) = b. Throughout this paper, we will take {Wn } instead of {Wn (a, b; p, q)}. As
some special cases {Wn (a, b; p, q)}, denote Wn (0,1; p, −1), Wn (2, p; p, −1) by Un , Vn , respectively. When 𝑝 = 1, Un =
Fn (nth Fibonacci number) and Vn = Ln (nth Lucas number).
If α and β are the roots of equation x 2 − px − 1 = 0, the Binet formulas of the sequences {Un } and {Vn } have the
Un =
αn −βn
α−β
and Vn = αn + βn ,
1
respectively.
E. Kılıç and P. Stanica [1], derived the following recurrence relations for the sequences {Ukn } and {Vkn } for k ≥ 0, n >
0,
Uk(n+1) = Vk Ukn + (−1)k+1 Uk(n−1)
and
Vk(n+1) = Vk Vkn + (−1)k+1 Vk(n−1) ,
where the initial conditions of the sequences {Ukn } and {Vkn } are 0, Uk and 2, Vk , respectively. The Binet formulas of the
sequences {Ukn } and {Vkn } are given by
Ukn =
αkn −βkn
α−β
and Vkn = αkn + βkn ,
respectively.
P. Filipponi and H.T. Freitag [2] defined the terms a n (Si , τ) of the autocorelation sequences of any sequence {Si }∞0 as
n
a n (Si , τ): = ∑ Si Si+τ
(0 ≤ τ ≤ n),
(1)
i=0
where the subscript i + τ must be considered as reduced modulo n+1 and τ, n are nonnegative integers. It is clearly that
autocorrelation sequences differ from the definition of cyclic autocorrelation function for periodic sequences with period
n+1 [3].
For positive integer number τ, the authors gave
a n (Si , τ) = a n (Si , n − τ + 1)
and
τ−1
n−τ
a n (Si , τ) = ∑ Si Si+τ + ∑ Si+n−τ+1 Si .
i=0
i=0
∞
The terms of the Fibonacci autocorrelation sequences {a k,n (τ)}τ were defined as
a n (τ): = a n (Fi , τ)
and they obtained some sums involving the terms a n (τ) as follows:
𝑛
∑ a n (i) = (Fn+2 − 1)2 ,
𝑖=0
2
n
n
2L3n+2 − 5F2n+2 + Ln+1 ,
10 ∑ ( ) a n (i) = {
2L
i
3n+2 − Ln+1 (5Fn+1 − 1),
i=0
if n is even
.
if n is odd
Inspiring by working in [2], we consider subsequence {Si }∞0 of the autocorrelation sequences of subsequence {Ski }∞0
defined as
n
a n (Ski , τ): = ∑ Ski Sk(i+τ)
(0 ≤ τ ≤ n),
(2)
i=0
where the subscript i + τ must be considered as reduced modulo n+1. It can clearly be seen that
a n (Ski , τ) = a n (Ski , n − τ + 1)
(3)
And
τ−1
n−τ
n
∑ Ski Sk(i+τ) = ∑ Ski Sk(i+τ) + ∑ Sk(i+n−τ+1) Ski ,
i=0
i=0
i=0
where τ is positive integer number.
For example, for n = 6, k = 5 and τ = 3 in (3),
a 6 (S5i , 3) = S0 S15 + S5 S20 + S10 S25 + S15 S30 + S20 S0 + S25 S5 + S30 S10 = a 6 (S5i , 4).
In this study, taking generalized Fibonacci subsequence {Uki }∞0 instead of subsequence {Ski }∞0 in (2), we write the terms
∞
of the generalized Fibonacci autocorrelation sequences {a k,n (τ)}τ as
n
a k,n (τ) = ∑ Uki Uk(i+τ)
i=0
and obtain some sums involving the numbers a k,n (τ), where odd integer k and nonnegative integers 𝜏, n. The following
Fibonacci identities and sums in [4], will be used widely throughout the proofs of Theorems:
Vk(m+n) + Vk(m−n) = {
Vkm Vkn , if n is even
,
ΔUkm Ukn , if n is odd
(4)
ΔU U , if n is even
Vk(m+n)− Vk(m−n) = { km kn
,
Vkm Vkn , if n is odd
(5)
U V , if n is even
Uk(m+n)+ Uk(m−n) = { km kn
,
Vkm Ukn , if n is odd
(6)
V U , if n is even
Uk(m+n)− Uk(m−n) = { km kn
,
Ukm Vkn , if n is odd
(7)
3
n
∑ Wk(ci+d)
(8)
i=r
=
Wk(cr+d) − Wk(c(n+1)+d) − (−1)c Wk(c(r−1)+d) + (−1)c Wk(cn+d)
1 − Vkc + (−1)c
n
∑(−1)i Wk(ci+d)
(9)
i=r
=
(−1)r Wk(cr+d) + (−1)n Wk(c(n+1)+d) + (−1)c+r Wk(c(r−1)+d) + (−1)c+n Wk(cn+d)
1 + Vkc + (−1)c
𝑛
∑ 𝑖𝑊𝑘(𝑐𝑖+𝑑)
(10)
𝑖=𝑟
[nWk(c(n+2)+d) − (n + 1 + 2n(−1)c )Wk(c(n+1)+d) + (n + 2(n + 1)(−1)c )Wk(cn+d)
−(n + 1)Wk(c(n−1)+d) − (r + 1 + 2r(−1)c )Wk(c(r−1)+d) + rWk(c(r−2)+d)
=
+(r + 2(r − 1)(−1)c )Wk(cr+d) − (r − 1)Wk(c(r+1)+d) ]/(1 − Vkc + (−1)c )
and
n
∑(−1)i−1 iWk(ci+d)
(11)
i=r
[n(−1)n+1 Wk(c(n+2)+d) + r(−1)r−1 Wk(c(n−2)+d) − (n + 1)(−1)n Wk(c(n−1)+d)
=
+(2n(−1)c+n+1 − (n + 1)(−1)n )Wk(c(n−1)+d) − ((r − 1)(−1)r − 2r(−1)c+r−1 )Wk(c(r−1)+d)
,
+(r(−1)r−1 − 2(r − 1)(−1)c+r )Wk(cr+d) + (n(−1)n+1 − 2(n + 1)(−1)c+n )Wk(cn+d)
−(r − 1)(−1)r Wk(c(r+1)+d)]/ (1 + Vkc + (−1)c )
where Δ = (Vk2 + 4)⁄Uk2 .
2. SOME IDENTITIES INVOLVING THE TERMS a k,n (τ)
In this section, we will give closed-form expressions for terms of the generalized Fibonacci autocorrelation sequences
∞
{a k,n (τ)}τ . Now, we give auxiliary Lemma before the proof of main Theorems.
Lemma 2.1. Let k be odd integer number. For even τ,
Vk a k,n (τ) = {
Uk(n+1) Uk(n−τ) + Ukn Ukτ ,
Ukn Uk(n−τ+1) + Ukτ ,
if n is even
,
if n is odd
and for odd τ,
Vk a k,n (τ) = {
Ukn Uk(n−τ+1) + Uk(n+1) Uk(τ−1) ,
if n is even
Uk(n+1) (Uk(n−τ) + Uk(τ−1) ),
if n is odd
.
4
Proof. Let n and τ be even integers. Using Binet formula of generalized Fibonacci sequence {Ukn }, we write
τ−1
n−τ
a k,n (τ) = ∑ Uki Uk(i+τ) + ∑ Uki Uk(i+n−τ+1)
i=0
i=0
n−τ
=
1
{∑(αk(2i+τ) + βk(2i+τ) − βki αk(i+τ) − αki βk(i+τ) )
Δ
i=0
τ−1
+ ∑(αk(2i+n−τ+1) + βk(2i+n−τ+1) − βki αk(i+n−τ+1) −αki βk(i+n−τ+1) )}
i=0
n−τ
τ−1
i=0
i=0
1
= {∑(Vk(2i+τ) − (−1)ki Vkτ ) + ∑(Vk(2i+n−τ+1) − (−1)ki Vk(n−τ+1) )}.
Δ
From (8) and the sums
τ−1
n−τ
if n, τ are same parities
1,
, ∑(−1)i = {
if n, τ are different parities
0,
1,
∑(−1) = {
0,
i
i=0
if τ is odd
,
if τ is even
i=0
we write
Δa k,n (τ) =
(Vk(2n−τ+1) − Vk(τ−1) + Vk(n+τ) )
⁄
(Vk − Vkr ).
By (5), we have
Vk a k,n (τ) = Uk(n+1) Uk(n−τ) + Ukn Ukr .
The other equalities are obtained similar to the proof. Thus we have the conclusion.
For example, for 𝑎 = 0, 𝑏 = 𝑘 = 𝑝 = 1 and 𝜏 = 0 in Lemma 2.1, it is clearly seen that
a1,n (0) = Fn+1 Fn [1].
Now, we will investigate some sums involving the terms a k,n (τ).
Theorem 2.1. Let k be odd integer number. We have
2
n
(Uk(n+1) − Ukn + Uk ) ⁄
Vk ∑(−1)i a k,n (i) = {
Vk , if n is odd ,
Uk(n+1) Ukn ,
if n is even
i=0
n
i
Vk V3k ∑(−1) a k,n−i (i) =
i=0
Uk(n−1) + (Vk(2n+4) − Vk(2n+1) )⁄Δ
−Vk(n−1) + V3k (Vk + V2k )⁄ΔVk ,
if n is odd
Uk(n+2) + (Vk(2n+4) − Vk(2n+1) )⁄Δ
{ −Vk(n+2) − V3k (Vk − 2)⁄ΔVk ,
if n is even
.
Proof. For even number n, observed that
5
n
∑(−1)i a k,n (i) = a k,n (0) − a k,n (1) + ⋯ + a k,n (n).
i=0
From the equality a k,n (i) = a k,n (n − i + 1) and Lemma 2.1, we get
𝑛
∑(−1)𝑖 a k,n (i) = a k,n (0) − a k,n (1) + a k,n (2) − ⋯ − a k,n (n − 1) + a k,n (n) = a k,n (0) =
𝑖=0
Uk(n+1) Ukn
⁄V .
k
For odd number n, we write
(n−1)⁄2
𝑛
(n−1)⁄2
𝑖
∑(−1) a k,n (i) = a k,n (0) − a k,n (1) + ⋯ + a k,n (n) = a k,n (0) + ∑ a k,n (2τ) − ∑ a k,n (2τ − 1).
τ=1
𝑖=0
τ=1
2
Using the equality Ukn
− Uk(n−1) Uk(n+1) = Uk2 in [5] and (7), (8) and Lemma 2.1, we have the claimed result. The
remaining formulas are similarly proven.
Theorem 2.2. Let k be odd integer number. We have
n
Vk2
(Uk(n+1) + Ukn )(Ukn − Uk ),
∑ i a k,n (i) = (n + 1) {
Uk(n+1) (Ukn + Uk(n−1) − Uk ) − Ukn Uk ,
i=0
𝑛
𝑉𝑘2 ∑(−1)𝑖 ia k,n (i) =
𝑖=0
{
if n is odd
if n is even
(n + 1)(Ukn − Uk(n+1) )(Ukn − Uk ),
if n is odd
(n + 1)(Ukn Uk − Uk(n+1) Uk(n−1) )
+(n − 1)Uk(n+1) (Ukn − Uk )
if n is even
,
(13)
+ 4Ukn (Uk(n+1) − Ukn )⁄Vk ,
Proof. For odd number n, we write
(n−1)⁄2
𝑛
(n−1)⁄2
∑ 𝑖 a k,n (i) = a k,n (1) + 2a k,n (2) + ⋯ + na k,n (n) = ∑ 2ia k,n (2i) + ∑ (2i + 1)a k,n (2i + 1).
𝑖=0
i=0
i=1
By Lemma 2.1, we have
𝑛
(n−1)⁄2
(n−1)⁄2
𝑖=0
i=1
i=1
1
∑ 𝑖 a k,n (i) = {Ukn ∑ 2i(Uk(n−2i+1) + Uki ) + Uk(n+1) ∑ (2i + 1)(Uk(n−2i−1) + Uki )}.
Vk
From (5), (7), (8) and (10), we get
n
Vk2
∑ i a k,n (i) = (n + 1)(Uk(n+1) + Ukn )(Ukn − Uk )
i=0
6
as claimed. Similarly, for even n, the proof is clearly obtained. With the help of (11), the proof of the other result is
given. Thus the proof is completed.
For example, taking a=0, b= k=1 in (13), it is clearly seen that
𝑛
2
𝑝 ∑(−1)𝑖 𝑖 a1,n (i) = {
𝑖=0
(n + 1)(Un+1 − Un )(1 − Un ),
(n + 1)(Un − Un+1 Un−1 )
+(n − 1)Un+1 (Un − 1) + 4Un (Un+1 − Un ),
if n is odd
if n is even
.
Theorem 2.3. Let k be odd integer number. We have
ΔUk Ukn Uk(n+1) ,
Uk(n+1) × (Vk(n+1) + Vk(n−1) + 2(Vk − Vkn )) ,
if n ≡ 0(mod4)
if n ≡ 1(mod4)
(Vk − 2)Uk(2n+1) + (Vk + 2)Uk + 2Vk (Uk(n+1) − Ukn ),
if n ≡ 2(mod 4)
Vk Ukn (Vk(n+1) − 2),
if n ≡ 3(mod4)
n
(i+1
2 )
ΔU2k ∑(−1)
a k,n (i) =
i=0
{
n+1
2 )
Uk(2n+4) − Uk(2n+1) + U3k (Vk + 1) − (−1)(
n
ΔVk U3k ∑(−1)
(i+1
2 )
a k,n−i (i) =
n+1
2 )
Uk(2n+4) − Uk(2n+1) + U3k − (−1)(
i=0
(Vk + V2k )Uk(n−1) ,
(Vk + V2k )Uk(n+2) ,
{
,
if n is odd
.
if n is even
Proof.For the second sum, the proof can be given. Let 𝑛 ≡ 0(𝑚𝑜𝑑4). Observed that
𝑛
∑(−1)(
𝑖+1
2 )
1
2
n+1
2 )
a k,n−i (i) = (−1)(2) a k,n (0) + (−1)(2) a k,n−1 (1) + ⋯ + (−1)(
a k,0 (n)
𝑖=0
= a k,n (0) − a k,n−1 (1) + ⋯ + a k,1 (n − 1) + a k,0 (n)
𝑛⁄4
𝑛⁄4
𝑛⁄4
𝑛⁄4
= ∑ a k,n−4i (4i) + ∑ a k,n−4i+1 (4i − 1) − ∑ a k,n−4i+2 (4i − 2) − ∑ a k,n−4i+3 (4i − 3) + a k,n (0).
𝑖=1
𝑖=1
𝑖=1
𝑖=1
By (5) and Lemma 2.1, we get
𝑛
𝑉𝑘 ∑(−1)(
𝑖+1
2 )
a k,n−i (i)
𝑖=0
𝑛⁄4
1
= 𝑈𝑘(𝑛+1) 𝑈𝑘𝑛 + ∑(−𝑉3𝑘 (Vk(2n−12i+4) + Vk(2n−12i+7) ) − Vk (V4kτ + Vk(4τ−1) ) + ΔUk(n−8τ+4) U4k ).
𝛥
𝑖=1
From (4)-(6) and (8), we write
𝑛
𝑉𝑘 ∑(−1)(
𝑖+1
2 )
a k,n−i (i)
𝑖=0
= Uk(n+1) Ukn +
1
1
(U
− Uk(2n+1) )−Uk(2n−2) + Uk(−n−2) +
(U − Uk(n+2) )+Uk − Uk(n+1)
ΔU3k k(1−n)
ΔUk 2k
7
1
(U
− Uk(2n+1) − (𝑉𝑘 + 𝑉2𝑘 )𝑈𝑘(𝑛+2) + 𝑈3𝑘 )
ΔU3k k(2n+4)
=
(14)
as claimed. For n ≡ 2(mod4),
𝑛
𝑉𝑘 ∑(−1)(
𝑖+1
2 )
a k,n−i (i) =
𝑖=0
1
(U
− Uk(2n+1) + (𝑉𝑘 + 𝑉2𝑘 )𝑈𝑘(𝑛+2) + U3k ).
ΔU3k k(2n+4)
(15)
By (14) and (15), for even number n, the desired results are obtained. Similarly, for n ≡ 1, 3(mod4), the remaining
results are proven. The proof of the other result is hold. Thus, the proof is completed.
Theorem 2.4. Let k be odd integer number. We have
2
Vk (Uk(n+1) + Ukn )(2 − Vkn ) − ΔUk Ukn
,
n
𝛥U2k ∑(−1)
(i+2
2 )
ΔUk (Uk2
a k,n (i) =
i=0
if n ≡ 0(mod4)
2
Ukn
)
− Uk(n+1) Ukn −
+ Vk Uk(n+1) (2 − Vk(n−1) ),
−ΔUk Ukn Uk(n+1) ,
if n ≡ 1(mod4)
if n ≡ 2(mod4) ,
−Vk Ukn (Vk(n+1) − 2),
{
if n ≡ 3(mod4)
n
ΔVk V2k U3k ∑(−1)(
i+2
2 )
a k,n (i)
i=0
2
V2k (U3k − Uk(2n+4) − Uk(2n+1) )+U2k (Vk(n−3) + ΔUkn U3k ) + Vk2 Uk(n+1) + V2k
Uk(n+2) ,
−V2k (Uk(2n+4) + Uk(2n+1) +U3k (Vk − 1)) −
=
2
Uk(n−1) (ΔU2k
+ V2k (Vk + 2)) ,
V2k (−Uk(2n+4)− Uk(2n+1) + (Vk − V2k )Uk(n+2) + U3k ,
if n ≡ 0(mod4)
if n ≡ 1(mod4)
if n ≡ 2(mod4)
{−V2k (Uk(2n+4) + Uk(2n+1) +U3k (Vk − 1)) + Vk V2k Uk(n−1) + 2Uk(n+1) V4k + U4k Vk(n−1) ,
if n ≡ 3(mod4)
Proof. Let n ≡ 0(mod4). Consider that
𝑛
∑(−1)(
𝑖+2
2 )
a k,n (i) = −a k,n (0) − a k,n (1) + a k,n (2) + ⋯ + a k,n (n − 1) − a k,n (n)
𝑖=0
𝑛⁄4
𝑛⁄4
𝑛 ⁄4
𝑛⁄4
= − ∑ a k,n (4i − 4) − ∑ a k,n (4i − 3) + ∑ a k,n (4i − 2) + ∑ a k,n (4i − 1) + a k,n (n).
𝑖=1
𝑖=1
𝑖=1
𝑖=1
From (7) and Lemma 2.1, we write
𝑛⁄4
𝑛
(𝑖+2
2 )
∑(−1)
𝑖=0
1 2
a k,n (i) = − 𝑈𝑘𝑛
− (Uk(n+1) − Ukn ) ∑(Uk(n−4i+3) − Uk(4i−3) ).
Vk
𝑖=1
By (5), (7) and (8), we have
𝑛
∑(−1)(
𝑖=0
𝑖+2
2 )
a k,n (i) = −
1 2
Vk
1 2
1
𝑈 −
(U
+ Ukn )(Vkn − 2) = − 𝑈𝑘𝑛
−
(U
+ Ukn )(Vkn − 2)
Vk 𝑘𝑛 𝛥𝑈2𝑘 k(n+1)
Vk
𝛥𝑈𝑘 k(n+1)
as claimed. For n ≡ 1, 2, 3 (mod4), the proofs are clearly given. Similarly, the other result is given. Thus, we have the
conclusion.
8
References
[1] E. Kılıç and P. Stanica, "Factorizations and Representations of Second Order Linear Recurrences with Indices in
Arithmetic Progressions," Bol. Soc. Mat. Mex. III. Ser.,, vol. 15, no. 1, pp. 23-25, 2009.
[2] P. Filipponi and H.T. Freitag, "Autocorrelation Sequences," The Fibonacci Quarterly, vol. 32, no. 4, pp. 356-368,
1994.
[3] S. Golomb, Shift Register Sequences, Laguna Hills: CA: Aegean Park Press, 1982.
[4] Y. Türker, N. Ömür, E.Kılıç, "Sums of Products of the Terms of the Generalized Lucas Sequence {V_kn },"
Hacettepe Journal of Mathematics and Statistics, vol. 4, no. 2, pp. 147-161, 2011.
[5] T. Koshy, Fibonacci and Lucas Numbers with Applications, New York: Pure and Applied Mathematics, WileyInterscience, 2001.
[6] S. Vajda, Fibonacci&Lucas Numbers and the Golden Section, New York: John Wiley&Sons, Inc., 1989.
9