DTIC ADA548867: Factorizations and Representations of Second Order Linear Recurrences with Indices in Arithmetic Progressions

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FACTORIZATIONS AND REPRESENTATIONS OF SECOND
ORDER LINEAR RECURRENCES WITH INDICES IN
ARITHMETIC PROGRESSIONS
E.KILIC¸ ANDP.STA˘NICA˘
Abstract. Inthispaperweconsidersecondorderrecurrences{Vk}and{Un}.
Wegivesecondorderlinearrecurrencesforthesequences{V }and{U }.
±kn ±kn
Usingtheserecurrencerelations,wederiverelationshipsbetweenthedetermi-
nants of certain matrices and these sequences. Further, as generalizations of
theearlierresults,wegiverepresentationsandtrigonometricfactorizationsof
thesesequencesbymatrixmethodsandmethodsrelyingonChebyshevpoly-
nomials of the first and second kinds. We give the generating functions and
somecombinatorialrepresentationsofthesesequences.
1. Introduction
Let A and B be nonnegative integers such that A2+4B (cid:54)= 0. The generalized
Lucas sequence {V (A,B)} and the generalized Fibonacci sequence {U (A,B)}
n n
are defined by: for n>0
V (A,B) = AV (A,B)+BV (A,B)
n+1 n n−1
U (A,B) = AU (A,B)+U (A,B)
n+1 n n−1
where V (A,B) = 2,V (A,B) = A and U (A,B) = 0,U (A,B) = 1, respectively.
0 1 0 1
WewillfrequentlyusethenotationsV andU insteadofV (A,B)andU (A,B).
n n n n
The Binet formulas of the sequences {U } and {V } are given by
n n
αn−βn
U = and V =αn+βn
n α−β n
where α and β are the roots of the equation t2−At−B =0.
When A = B = 1, then V (1,1) = L (nth Lucas number) and U (1,1) = F
n n n n
(nth Fibonacci number).
Lind (cf. [8, p. 478]) first gave the following trigonometric factorization of
Fibonacci numbers and then, two years later Zeitlin derived a factorization of the
Lucas numbers using trigonometric factorizations of the Chebyshev polynomials of
the first kind [19]:
n
(cid:89)
F = (1−2icos(kπ/n),
n
k=1
n−1
(cid:89)
L = (1−2icos(2k+1)π/2n).
n
k=0
2000 Mathematics Subject Classification. 11B37,11C20,15A15.
Keywordsandphrases. Secondorderrecurrences,factorizations,matrices,determinants,gen-
eratingfunctions.
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1. REPORT DATE 3. DATES COVERED
2009 2. REPORT TYPE 00-00-2009 to 00-00-2009
4. TITLE AND SUBTITLE 5a. CONTRACT NUMBER
Factorizations and Representations of Second Order Linear Recurrences
5b. GRANT NUMBER
with Indices in Arithmetic Progressions
5c. PROGRAM ELEMENT NUMBER
6. AUTHOR(S) 5d. PROJECT NUMBER
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Naval Postgraduate School,Department of Applied REPORT NUMBER
Mathematics,Monterey,CA,93943
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13. SUPPLEMENTARY NOTES
Bulletin Mex. Math. Soc. 15 (2009), 23-36.
14. ABSTRACT
In this paper we consider second order recurrences fVkg and fUng We give second order linear
recurrences for the sequences fVkng and fUkng. Using these recurrence relations, we derive relationships
between the determinants of certain matrices and these sequences. Further, as generalizations of the earlier
results, we give representations and trigonometric factorizations of these sequences by matrix methods and
methods relying on Chebyshev polynomials of the rst and second kinds. We give the generating functions
and some combinatorial representations of these sequences.
15. SUBJECT TERMS
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ABSTRACT OF PAGES RESPONSIBLE PERSON
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2 E.KILIC¸ ANDP.STA˘NICA˘
In [17] and [4], the authors gave complex factorization of the Fibonacci numbers
by considering the roots of Fibonacci polynomials. In [10], the author established
the following representations:
F =in−1sin(cid:16)ncos−1(cid:16)−2i(cid:17)(cid:17), L =2incos(cid:0)ncos−1(cid:0)−i(cid:1)(cid:1), n≥1
n sin(cos−1(−i)) n 2
2
Alsoin[3],theauthorsobtainedthesameresultsonthetrigonometricfactoriza-
tionsoftheFibonacciandLucasnumbersbymatrixmethods. Thematrixmethod
was first used by them.
Recently, in [7], the authors consider the backward second order linear recur-
rences and they gave the trigonometric factorizations and representations of these
sequences. Note that this case will be special case with k =1 of the present paper.
The second order linear recurrences have been studied by many authors. For
example, in [1], the author gave the following combinatorial representation:
(cid:98)n/2(cid:99) (cid:18) (cid:19)
(cid:88) n n−k
V = An−2kBk (1.1)
n n−k k
k=0
(cid:98)n/2(cid:99)(cid:18) (cid:19)
(cid:88) n−k
U = An−2kBk (1.2)
n+1 k
k=0
There are many relationships between linear recurrence relations and determi-
nants of certain matrices. For example, the generalized Lucas sequence can be
obtained by the following determinant (see [15, 16, 5, 6]):
(cid:12) (cid:12)
(cid:12) A −2B (cid:12)
(cid:12) (cid:12)
(cid:12) 1 A −B (cid:12)
(cid:12) (cid:12)
(cid:12)(cid:12) 1 A ... (cid:12)(cid:12)=Vn.
(cid:12) (cid:12)
(cid:12)(cid:12) ... ... −B (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) 1 A (cid:12)
Especially the case A = B = 1 can be found in [2]. Furthermore one can find
similar special relationships in [16, 15, 9, 5, 6].
In this paper, we consider the positively and negatively kn subscripted terms of
the sequences {V } and {U }, and we derive relationships between these and the
n n
determinants of certain tridiagonal matrices. Then we give the general trigono-
metric factorizations and representations of terms of these sequences {V } and
∓kn
{U }. We also present generating functions and combinatorial representations
∓kn
of these sequences.
2. Forward and Backward Generalized Lucas Sequences {V },{V }
kn −kn
In this section, for an arbitrary positive integer k, we consider the terms V
kn
andgiveasecondorderlinearrecurrencerelationforthesequence{V }. Westart
kn
with the following useful lemma.
Lemma 1. For k > 0, and n > 1, the terms V satisfy the following recurrence
kn
relation
V =V V +(−1)k+1BkV .
kn k k(n−1) k(n−2)
GENERALIZED LUCAS SEQUENCES 3
Proof. From the Binet formula and since αβ =−B, we can write
V V +(−1)k+1BkV
k k(n−1) k(n−2)
= V V −(−B)kV
k k(n−1) k(n−2)
= (cid:16)(cid:0)αk+βk(cid:1)(cid:16)αk(n−1)+βk(n−1)(cid:17)−(αβ)k(cid:16)αk(n−2)+βk(n−2)(cid:17)(cid:17)
= αkn+βkn
= V ,
kn
which is as desired. (cid:3)
Now we describe a relationship between the terms of sequence {V } and the
kn
determinant of a certain tridiagonal matrix.
Define the n×n tridiagonal matrix T =(t ) by
n ij
 V 2(−B)k/2 
k
 (−B)k/2 V (−B)k/2 
 k 
T = (−B)k/2 V ... .
n  k 
 ... ... (−B)k/2 
 
(−B)k/2 V
k
Theorem 1. For n>1
detT =V ,
n kn
where detT =V .
1 k
Proof. We will use the principle of mathematical induction to show that detT =
n
V . If n=2, then, by Lemma 1, we obtain
kn
(cid:12) (cid:12)
(cid:12) V 2(−B)k/2 (cid:12)
detT =(cid:12) k (cid:12)=V .
2 (cid:12)(cid:12) (−B)k/2 Vk (cid:12)(cid:12) 2k
Suppose that the equation holds for n−1. Then we show that the equation holds
for n. Expanding detT by the Laplace expansion of a determinant according to
n
the last row, we obtain
detT =V detT −(−B)kdetT .
n k n−1 n−2
By our assumption and the result of Lemma 1, we have the required conclusion:
detT =V V −(−B)kV =V .
n k k(n−1) k(n−2) kn
(cid:3)
In the remaining of the section we consider the terms of the backward Lucas
sequence {V }, and we give a second order linear recurrence relation for these,
−kn
similar to the positively subscripted terms. Then we determine a certain matrix
whose successive determinants equal the terms V .
−kn
Lemma 2. For k ≥1 and n>1,
V =(−B)−k(cid:0)V V −V (cid:1).
−kn k −k(n−1) −k(n−2)
Proof. From the Binet formulas of sequence {V }, we have that αβ =−B and so
−n
V =V (−B)−n. The proof follows from Lemma 1. (cid:3)
−n n
4 E.KILIC¸ ANDP.STA˘NICA˘
Now we give a relationship between the determinant of a certain tridiagonal
matrix and the terms of the backward general Lucas sequence.
Define the n×n tridiagonal matrix H by
n
 V 2(−B)−k/2 
−k
 (−B)−k/2 V (−B)−k/2 
 −k 
H = 0 (−B)−k/2 V ... .
n  −k 
 ... ... (−B)−k/2 
 
(−B)−k/2 V
−k
As a consequence of Theorem 1, we have the following result.
Corollary 1. For n>1,
detH =V
n −kn
where detH =V .
1 −k
Proof. From the definitions of the matrices H and T , using the identity V =
n n −n
(−B)−nV , it is seen that H =(−B)−kT , and so
n n n
detH =(−B)−kndetT . (2.1)
n n
By Theorem 1 and equation 2.1, we obtain
detH =(−B)−knV =V .
n kn −kn
The proof is complete. (cid:3)
3. Trigonometric factorizations of the General Lucas sequences
{V } and {V }
kn −kn
In this section we give the trigonometric factorizations and representations of
the generalized Lucas sequences {V } and {V } by matrix methods.
kn −kn
Define the n×n tridiagonal matrix Q as below:
 
0 2
 1 0 ... 
Q= .
 ... 1 
1 0
The characteristic equation of the matrix Q satisfies the following equation
t (λ)=−λt (λ)−t (λ), n>0,
n+1 n n−1
where t (λ)=−λ and t (λ)=λ2−2.
0 1
TheChebyshevpolynomialsofthefirstkindaredefinedbythefollowingequation
T (x)=2xT (x)−T (x), n>1,
n n−1 n−2
where T (x)=1, T (x)=x.
0 1
The zeros of the Chebyshev polynomials of the first kind are given by (for more
details see [11, 12, 14])
x =cos(2k−1)π, k =1,2,...,n.
k 2n
GENERALIZED LUCAS SEQUENCES 5
If we take λ ≡ −2x, then the sequence {t (λ)} is reduced to the sequence of
n
Chebyshev polynomials of the first kind, {2T (x)}. Therefore the zeros of the
n
characteristic equation of matrix Q are given by
λ =−2cos(2k−1)π, for k =1,2,...,n (3.1)
k 2n
From the definitions of Q and T , we can write T = V I +(−B)k/2Q where I
n n k n n
is the n×n unit matrix.
Theorem 2. For n>1,
n
V = (cid:89)(cid:104)V −2(−B)k/2cos(cid:16)(2r−1)π(cid:17)(cid:105).
kn k 2n
r=1
Proof. Let λ , r =1,2,...,n, be the eigenvalues of Q with respect to eigenvectors
r
x . Then, for all r
r
(cid:104) (cid:105) (cid:104) (cid:105)
T x = V I +(−B)k/2Q x =V I x +(−B)k/2Qx = V +(−B)k/2λ x .
n k k n r k n r r k r r
Thus µ = V +(−B)k/2λ , r = 1,2,...,n, are the eigenvalues of T . Thus by
r k r n
(3.1)
n n
detT = (cid:89)(cid:104)V +(−B)k/2λ (cid:105)= (cid:89)(cid:104)V −2(−B)k/2cos(cid:16)(2r−1)π(cid:17)(cid:105),
n k r k 2n
r=1 r=1
and the proof is complete. (cid:3)
As a corollary, we obtain Lind’s result [8, p. 478].
Corollary 2. When A=B =k =1, then by the above theorem, we obtain
n
L = (cid:89)(cid:104)1−2icos(cid:16)(2r−1)π(cid:17)(cid:105).
n 2n
r=1
As a consequence of Theorem 2, we give the following corollary.
Corollary 3. For n>1,
n
V = (cid:89)(cid:104)V −2(−B)−k/2cos(cid:16)(2r−1)π(cid:17)(cid:105).
−kn −k 2n
r=1
Proof. From Theorem 2, we have
n
V = (cid:89)(cid:104)V −2(−B)k/2cos(cid:16)(2r−1)π(cid:17)(cid:105).
kn k 2n
r=1
Multiplying the above equation by (−B)kn, we have the conclusion since V =
−n
(−B)−nV . (cid:3)
n
Alternatively,onecanconsidertheequationH =V I +(−B)−k/2Q,andthe
n −k n
next result follows.
Theorem 3. For k ≥1 and n>1,
(cid:16) (cid:16) (cid:17)(cid:17)
V =(−B)∓kn/2cos ncos−1 V∓k .
∓kn 2(−B)∓k/2
6 E.KILIC¸ ANDP.STA˘NICA˘
Proof. First, we consider the case V . If the n×n matrix G has the following
kn n
form
 
2x 2 0
 1 2x ... 
Gn(x)= ... ... 1 , (3.2)
0 1 2x
then it is seen that detG (x) = 2T (x) where {T (x)} is the sequence of the
n n n
Chebyshev polynomials of the first kind. Thus
(cid:16) (cid:17) (cid:16) (cid:17)
detTn =(−B)kn/2detGn 2(−VBk)k/2 =(−B)kn/2Tn 2(−VBk)k/2
Defining x = cosθ allows the Chebyshev polynomials of the second kind to be
written as (see [12])
T (x)=cosnθ. (3.3)
n
Then by (3.3) and the value of determinant of matrix T , we obtain
n
(cid:16) (cid:16) (cid:17)(cid:17)
Vkn =(−B)kn/2cos ncos−1 2(−VBk)k/2 .
For the case of V , we consider
−kn
(cid:16) (cid:17) (cid:16) (cid:17)
detH =(−B)−kn/2detG V−k =(−B)−kn/2T V−k ,
n 2(−B)−k/2 n 2(−B)−k/2
from which the proof follows. (cid:3)
Corollary 4. For k ≥1 and n>1 even,
(cid:98)n/2(cid:99)
V = (cid:89) (cid:104)V2 −4(−B)∓kcos2(cid:16)(2r−1)π(cid:17)(cid:105)
∓kn ∓k 2n
r=1
and for n>1 odd
(n−1)/2
V =V (cid:89) (cid:104)V2 −4(−B)∓kcos2(cid:16)(2r−1)π(cid:17)(cid:105).
∓kn ∓k ∓k 2n
k=1
Proof. These are immediate consequences of Theorem 2 and Corollary 3, since, for
1≤k <n/2, cos(kπ/n)=−cos((n−k)π/n). (cid:3)
4. The Generalized Fibonacci Sequence {U (A,B)}
n
In this section, we consider the recurrence {U } and then obtain two recurrence
n
relations for the sequences {U } and {U }. Also we determine certain tridiag-
kn −kn
onal matrices and then we obtain relationships between the determinants of these
matricesandthesequences{U }and{U }.Therefore,weobtaintrigonometric
kn −kn
factorizations and representations of these sequences. We start with the following
useful lemma.
Lemma 3. For k ≥1 and n>1,
U =V U +(−1)k+1BkU .
kn k k(n−1) k(n−2)
GENERALIZED LUCAS SEQUENCES 7
Proof. From the Binet formula of the sequences {U }, {V } and since αβ = −B,
n n
we can write
V U +(−1)k+1BkU
k k(n−1) k(n−2)
= V U −(−B)kU
k k(n−1) k(n−2)
= (cid:16)(cid:0)αk+βk(cid:1)(cid:16)αk(n−1)−βk(n−1)(cid:17)−(αβ)k(cid:16)αk(n−2)−βk(n−2)(cid:17)(cid:17)
α−β α−β
= αkn−βkn
α−β
= U .
kn
The proof is complete. (cid:3)
Define the n×n tridiagonal Toeplitz matrix M by
n
 
V (−B)k/2
k
 (−B)k/2 V ... 
M = k .
n  ... ... (−B)k/2 
 
(−B)k/2 V
k
Since U =U V , we have immediately the following result.
2k k k
Theorem 4. For n>1,
U
k(n+1)
detM = ,
n U
k
where detM =U /U .
1 2k k
Proof. It is known that the tridiagonal Toeplitz matrices satisfy
detM =V detM −(−B)kdetM .
n k n−1 n−2
FromtheprincipleofmathematicalinductionandLemma3,theresultfollows. (cid:3)
Lemma 4. For k ≥1 and n>1,
U =(−B)−k(cid:0)V U −U (cid:1)=V U −(−B)−kU
−k(n+1) k −kn −k(n−1) −k −kn −k(n−1)
Proof. By the Binet formulas of {U } and {V }, we can write
−n −n
(−B)−kV U −(−B)−kU
k −kn −k(n−1)
= (−B)−k(cid:0)V U −U (cid:1)
k −kn −k(n−1)
= (−B)−k(cid:16)(cid:0)αk+βk(cid:1)(cid:16)α−kn−β−kn(cid:17)−(cid:16)α−k(n−1)−β−k(n−1)(cid:17)(cid:17)
α−β α−β
(cid:16) (cid:17)
= (−B)−k α−kn+k−β−kn+k−αkβ−kn+βkα−kn−α−kn+k+β−kn+k
α−β
= (cid:0)α−kβ−k(cid:1)(cid:16)−αkβ−kn+βkα−kn(cid:17)
α−β
= α−kn−k−β−kn−k
α−β
= U ,
−k(n+1)
and the conclusion follows. (cid:3)
8 E.KILIC¸ ANDP.STA˘NICA˘
Define the n×n tridiagonal Toeplitz matrix E as shown below:
n
 
V (−B)−k/2
−k
 (−B)−k/2 V ... 
E = −k .
n  ... ... (−B)−k/2 
 
(−B)−k/2 V
−k
Corollary 5. For n>1,
U
−k(n+1)
detE =
n U
−k
where detE =U /U .
1 2k k
Proof. SinceE =(−B)−kM ,detE =(−B)kndetM .ByTheorem4,theproof
n n n n
follows easily. (cid:3)
5. Trigonometric factorization of the General Fibonacci sequences
{U } and {U }
kn −kn
In this section, we give the trigonometric factorizations and representations of
sequences {U } and {U } by matrix methods and the Chebyshev polynomials
kn −kn
of the second kind.
Define the n×n tridiagonal matrix W as shown:
 
0 1
 1 0 ... 
W = .
 ... 1 
1 0
The characteristic equation of the matrix W satisfies the following recurrence
f (y)=−yf (y)−f (y), n>0,
n+1 n n−1
where f (y)=−y and f (y)=y2−1.
0 1
The Chebyshev polynomials of the second kind are defined by the recurrence
relation for n>1
U (x)=2xU (x)−U (x)
n n−1 n−2
where U (x)=1, U (x)=2x.
0 1
The zeros of the Chebyshev polynomials of the second kind is given by (see
[11, 12, 14]):
x =cos kπ , k =1,2,...,n,
k n+1
Taking λ≡−2x, the sequence {f (y)} is reduced to the sequence {U (x)}. Then
n n
the zeros of the characteristic equation of matrix W are given by
y =−2coskπ, for k =1,2,...,n. (5.1)
k n
By the definitions of W, M and E , we write M = V I + (−B)k/2W and
n n n k n
E =V I +(−B)−k/2W where I is the n×n unit matrix.
n −k n n
Theorem 5. Then for n>1,
n
U =U (cid:89)(cid:104)V −2(−B)∓k/2cos(cid:16) rπ (cid:17)(cid:105).
∓k(n+1) ∓k ∓k n+1
r=1
GENERALIZED LUCAS SEQUENCES 9
Proof. Let y , r = 1,2,...,n, be the eigenvalues of matrix W with respect to the
r
eigenvectors x . Then, for all r =12,...,n
r
(cid:104) (cid:105) (cid:104) (cid:105)
M x = V I +(−B)k/2W x =V I x +(−B)k/2Wx = V +(−B)k/2y x
n k k n r k n r r k r r
Thus ω = V +(−B)k/2y , r = 1,2,...,n, are the eigenvalues of M . Thus by
r k r n
(5.1) and Theorem 4,
n n
detM =U (cid:89)(cid:104)V +(−B)k/2y (cid:105)=U (cid:89)(cid:104)V −2(−B)k/2cos(cid:0)rπ(cid:1)(cid:105).
n k k r k k n
r=1 r=1
Similarly, one can obtain that c = V +(−B)−k/2y , r = 1,2,...,n, are the
r −k r
eigenvalues of the matrix E . Thus we obtain
n
n n
detE =U (cid:89)(cid:104)V +(−B)−k/2y (cid:105)=U (cid:89)(cid:104)V −2(−B)−k/2cos(cid:16) rπ (cid:17)(cid:105).
n −k −k r −k −k n+1
r=1 r=1
Considering the value of detE , the proof is complete. (cid:3)
n
For example, when k =5, A=1, B =1 in sequence {U (A,B)}, then
n
n
(cid:89)(cid:104) (cid:16) (cid:17)(cid:105)
F =5 11−2icos rπ .
5(n+1) n+1
r=1
Corollary 6. For an arbitrary positive integer k and n>1 even,
(cid:98)n/2(cid:99)
U =U (cid:89) (cid:104)V2 −4(−B)∓kcos2(cid:16) rπ (cid:17)(cid:105)
∓k(n+1) ∓k ∓k n+1
r=1
and for n>1 odd
(cid:98)n/2(cid:99)
U =U (cid:89) (cid:104)V2 −4(−B)∓kcos2(cid:16) rπ (cid:17)(cid:105).
∓k(n+1) ∓2k ∓k n+1
k=1
Proof. The proof follows from Theorem 5, since, for 1 ≤ k < n/2, cos(kπ/n) =
−cos((n−k)π/n) and U V =U . (cid:3)
n n 2n
Theorem 6. For k ≥1 and n>1,
(−B)∓kn/2sin(cid:20)(n+1)cos−1(cid:18) V∓k (cid:19)(cid:21)
U =U 2(−B)∓k/2 .
∓k(n+1) ∓k sin(cid:18)cos−1(cid:18) V∓k (cid:19)(cid:19)
2(−B)∓k/2
Proof. Let the matrix K be defined by
n
 
2x 1 0
 1 2x ... 
Kn(x)= ... ... 1  . (5.2)
0 1 2x
n×n
ItisknownthatdetK (x)=U (x), where{U (x)}isthesequenceoftheCheby-
n n n
shev polynomials of the second kind. Thus we obtain
(cid:16) (cid:17) (cid:16) (cid:17)
detM =U (A,B)(−B)kn/2detK Vk(A,B) =(−B)kn/2U Vk(A,B)
n k n 2(−B)k/2 n 2(−B)k/2