ON THE LUCAS SEQUENCE EQUATION U1n =P∞

k=1

Uk−1xkSZ. TENGELYAbstract. In 1953 Stancliff noted an interesting property of theFibonacci number F11 = 89. One has that123501
1=+ 2 + 3 + 4 + 5 + 6 + ...,
89
10 1010101010
where in the numerators the elements of the Fibonacci sequenceappear. We provide methods to determine similar identities in caseof Lucas sequences. As an example we prove that∞X
11Uk,==U10416020
647k+1k=0

where U0 = 0, U1 = 1 and Un = 4Un−1 + Un−2 , n ≥ 2.

1. introduction
Stancliff [16] noted without proof an interesting property of the Fibonacci sequence Fn . One has that∞
XFk
1
1== 0.0112358 . . . =.
F1189
10k+1
k=0In 1980 Winans [23] investigated the related sums∞
XFαkk=010k+1

for certain values of α. In 1981 Hudson and Winans [8] provided a complete characterization of all decimal fractions that can be approximatedby sums of the typen1 X Fαk,Fα k=1 10l(k+1)α, l ≥ 1.

2000 Mathematics Subject Classification. Primary 11D25; Secondary 11B39.
Key words and phrases. Lucas sequences, Diophantine equations, Elliptic curves.
1

2SZ. TENGELYLong [12] proved a general identity for binary recurrence sequencesfrom which one obtains e.g.
∞∞XX
1
FkFk
1
==,
.
9899 k=0 102(k+1)
109 k=0 (−10)k+1
In the previous examples decimal fractions were studied, in case ofdifferent bases characterizations were obtained by Jia Sheng Lee [10]and by Köhler [9] and by Jin Zai Lee and Jia Sheng Lee [11]. Here westate a result by Köhler that we will use later in this article.
Theorem A. Let A, B, a0 , a1 be arbitrary complex numbers. Define thesequence {an } by the recursion an+1 = Aan + Ban−1 . Then the formula∞
Xaka0 x − Aa0 + a1
=xk+1x2 − Ax − B
k=0
holds for all complex x such that |x| is larger than the absolute valuesof the zeros of x2 − Ax − B.
Let P and Q be non-zero relatively prime integers. The Lucas sequence {Un (P, Q)} is defined byU0 = 0, U1 = 1 and Un = P Un−1 − QUn−2 , if n ≥ 2.
In this paper we deal with the determination of all integers x ≥ 2 forwhich there exists an n ≥ 0 such that∞
XUk−1
1(1)
=,kUn
x
k=1
where Un is a Lucas sequence with some given P and Q. In case of P =1, Q = −1 one gets the Fibonacci sequence. De Weger [5] computed allx ≥ 2 in case of the Fibonacci sequence, the solutions are as follows∞∞
11
1 X Fk−1
1
1 X Fk−1== =,==,F1
F2
1 k=1 2kF55 k=1 3k∞
X11Fk−1,==F1055 k=1 8k∞
X
11Fk−1.
==F1189 k=1 10k

De Weger applied arguments of algebraic number theory and obtained
two Thue equations, which were solved using Baker’s method (see e.g.
[2, 7, 15]). In the current work, we show how to reduce a search for
P∞ Uk−1
integral x ≥ 2 related to the equation U1n =to elliptic
k=1 xkDiophantine equations or to Thue equations following an elementaryargument by Alekseyev and Tengely [1]. There exists a number of

ON THE LUCAS SEQUENCE EQUATION

3software implementations for determining integral points on elliptic
curves, these procedures are based on a method developed by Stroekerand Tzanakis [18] and independently by Gebel, Pethő and Zimmer [6].
The elliptic logarithm method for determining all integer points on anelliptic curve has been applied to a variety of elliptic equations (see e.g.
[19, 20, 21, 22]).
2. main results
Theorem 1. Let {Un (P, Q)} be a Lucas sequence with Q ∈ {±1}and (P, Q) ∈/ {(−2, 1), (2, 1)}. Then equation (1) possesses only finitely
many solutions in n, x which can be effectively determined.
The proof of Theorem 1 provides an algorithm to determine all solutions of equation (1). Following this algorithm we obtain numericalresults.
Theorem 2. Let {Un (P, Q)} be a Lucas sequence with −10 ≤ P ≤10, Q ∈ {±1} and (P, Q) 6= (−2, 1), (2, 1). Then equation (1) has thefollowing solutions(P, Q, n, x) ∈ {(−3, 1, 5, 6), (−1, −1, 5, 2), (−1, −1, 11, 9), (1, −1, 1, 2),(1, −1, 2, 2), (1, −1, 5, 3), (1, −1, 10, 8), (1, −1, 11, 10), (2, −1, 2, 3),(3, −1, 2, 4), (3, 1, 1, 3), (3, 1, 5, 9), (4, −1, 2, 5), (4, −1, 10, 647),(4, 1, 1, 4), (5, −1, 2, 6), (5, 1, 1, 5), (6, −1, 2, 7), (6, 1, 1, 6),(7, −1, 2, 8), (7, 1, 1, 7), (8, −1, 2, 9), (8, 1, 1, 8), (9, −1, 2, 10),
(9, 1, 1, 9), (10, −1, 2, 11), (10, 1, 1, 10)}.
Remark. We note that the sequences with (P, Q) ∈ {(−2, 1), (2, 1)}
are degenerate sequences, that is the discriminant D = P 2 −4Q is zero.
One has that Un (−2, 1) = (−1)n+1 n and Un (2, 1) = n. In these cases
there are infinitely many solutions of equation (1):∞X
1
1
Uk−1 (−2, 1)==, if x is even,2(x + 1)U(x+1)2 (−2, 1)
xk
k=1
∞
X
1
1
Uk−1 (2, 1)==.
2(x − 1)U(x−1)2 (2, 1)
xkk=1

3. auxiliary results
The Lucas sequences {Un (P, Q)} and associated Lucas sequences{Vn (P, Q)} are defined by the same linear recurrent relation with P, Q ∈4SZ. TENGELY

Z \ {0} but different initial terms:
Un = P Un−1 − QUn−2 , if n ≥ 2,Vn = P Vn−1 − QVn−2 , if n ≥ 2.

U0 = 0, U1 = 1 and
V0 = 2, V1 = P andTerms of Lucas sequences and associated Lucas sequences satisfy thefollowing identity
Vn2 − DUn2 = 4Qn ,(2)where D = P 2 − 4Q. To determine the appropriate Thue equations weuse parametric solutions of ternary quadratic equations. Such parametrization are given in [1, Theorem 1].
Lemma 1. Let A, B, C be non-zero integers and let (x0 , y0 , z0 ) withz0 6= 0 be a particular non-trivial integer solution to the Diophantine
equation Ax2 + By 2 + Cz 2 = 0. Then its general integer solution isgiven byp(x, y, z) = (Px (m, n), Py (m, n), Pz (m, n))qwhere m, n as well as p, q are coprime integers with q > 0 dividing
2 lcm(A, B)Cz02 , andPx (m, n) = x0 Am2 + 2y0 Bmn − x0 Bn2 ,Py (m, n) = −y0 Am2 + 2x0 Amn + y0 Bn2 ,Pz (m, n) = z0 Am2 + z0 Bn2 .
4. proofs
Proof of Theorem 1. Let {Un (P, Q)} be a Lucas sequence with Q ∈{±1}. Theorem A implies that(3)∞X
Uk−1
k=1xk

=1
.
x2 − P x ± 1

Hence from equations (1) and (3) it follows that Un = x2 − P x ± 1.
Finiteness follows from results by Nemes and Pethő [13] and by Pethő
[14]. We provide two approaches to determine a finite set of possible
values of x for which Un = x2 −P x±1. The first one is based on ellipticcurves. It only works if one can determine the ranks of the appropriate
Mordell-Weil groups. The second method is based on an elementary
reduction algorithm which yield finitely many quartic Thue equations
to solve. After computing the finite set of possiblePvalues of x we use
Uk−1
eigenvalues and eigenvectors to compute the sum ∞k=1 xk .

ON THE LUCAS SEQUENCE EQUATION5Substituting Un = x2 − P x ± 1 into the identity (2) yields a genus 1
curve
C(P,Q,n) : y 2 = (P 2 − 4Q)(x2 − P x + Q)2 + 4Qn .
Bruin and Stoll [4] described and algorithmized the so-called two-coverdescent, which can be used to prove that a given hyperelliptic curve
has no rational points. This algorithm is implemented in Magma [3],the procedure is called TwoCoverDescent. If it turns out that there
are no rational points on the curves y 2 = (P 2 − 4)(x2 − P x + 1)2 +4 and y 2 = (P 2 + 4)(x2 − P x − 1)2 ± 4, then equation (1) has nosolution. If TwoCoverDescent yields that rational points may exist,
but the procedure Points fails to find one, then we follow the second
approach, solution via Thue equations that we consider later in theproof. Now we assume that we could determine points on curves forwhich TwoCoverDescent predicts existence of rational points. Thatmeans we are given elliptic curves in quartic model. Tzanakis [21]provided a method to determine all integral points on quartic models,the algorithm is implemented in Magma as IntegralQuarticPoints.
The second approach is based on Lemma 1. There are three ternary
quadratic equations to parametrize
Q1 :
Q2 :
Q3 :

X 2 − (P 2 − 4)Y 2 − 4Z 2 = 0,
X 2 − (P 2 + 4)Y 2 − 4Z 2 = 0,
X 2 − (P 2 + 4)Y 2 + 4Z 2 = 0.

There are points on these curves:
Q1 :
Q2 :
Q3 :

(X, Y, Z) = (2, 0, 1),(X, Y, Z) = (2, 0, 1),(X, Y, Z) = (P, 1, 1).

It follows from Lemma 1 that ±1 = Pz (m, n) = pq (m2 − (P 2 ± 4)n2 ).
Therefore p = ±1. We deal with the curve y 2 = (P 2 − 4)(x2 − P x +1)2 + 4, the other two cases are similar. We obtain thatx2 − P x + 1 =±4mn.
q

Hence we have thatq(2x − P )2 ± (4 − P 2 )(m2 − (P 2 − 4)n2 ) ∓ 16mn = 0,
where q > 0 divides 8(P 2 − 4). Applying Lemma 1 again we obtainthat m = fm (u, v) and n = fn (u, v), where fm , fn are homogeneous
quadratic polynomials. It remains to compute integral solutions of

6SZ. TENGELY

finitely many quartic Thue equations (it may happen that the defining quartic polynomials are reducible, in those cases the resolution issimpler)±q = fm (u, v)2 − (P 2 − 4)fn (u, v)2 .
If we have a possible solution x ∈ N of equation (1), then we havePUk−1
to compute the value of the sum ∞k=1 xk . We define
P/x −Q/xT =.
1/x0
Following standard arguments one has that   11 0∗12N −1= PN Uk−1 .
T + T + T + ... + T0x
k=1 xkUsing eigenvectors and eigenvalues one can determine a formula forthe powers
hence one obtains a formula depending only on N for
PNof T,Uk−1
the sum k=1 xk . To find solutions of the equation (1) it remains tocomputeNX
Uk−1
lim.
kN →∞xk=1
Proof of Theorem 2. We will illustrate how one can use the approaches
provided in the proof of Theorem 1 to determine all solutions of equation (1) for given P and Q. First we deal with the case P = 4, Q = −1.
We have thaty 2 = 20(x2 − 4x − 1)2 ± 4.
To determine all integral solutions we use the Magma commands
IntegralQuarticPoints([20, -160, 280, 160, 16]) and
IntegralQuarticPoints([20, -160, 280, 160, 24],[-1,-18]).
One obtains that x ∈ {−643, −1, 0, 1, 3, 4, 5, 647}. Since x ≥ 2 only 4values remain. In case of x = 647 the matrix T is as follows
4/647 1/647
1/647
0
and we obtain that √ N √ N√√
2− 52+ 5N(1295−1)−(129 5 + 1)
X
647
647
Uk−1
1=+.
k647
832040416020k=1

ON THE LUCAS SEQUENCE EQUATION

7Thus
NX
Uk−1

1
1=.
=k
N →∞
647
416020U10k=1lim

In a similar way we get that
NX
Uk−1

lim

N →∞

3kNX
Uk−1

lim

N →∞

lim

k=1

k=1

4k

N
X
Uk−1

N →∞

k=1

5k= +∞,

= +∞,

=1
1=.
4U2

We apply the second method to completely solve equation (1) with
P = 3, Q = 1. The curve CP,Q,n has the form y 2 = 5(x2 − 3x + 1)2 + 4.
It can be written as v 2 = 5u4 − 50u2 + 189 with v = 4y and u = 2x − 3.
The second approach has been implemented in Sage [17] by Alekseyev
and Tengely [1]. Using their procedure QuarticEq([5,-50,189]) we
obtain that u ∈ {±1, ±3, ±15}, therefore x ∈ {−6, 0, 1, 2, 3, 9}. We
have thatlim

N →∞

lim

N →∞lim

N →∞

NX
Uk−1
k=12k

NX
Uk−1
k=13kNX
Uk−1
k=19k= +∞,= 1==1,U1

11.
=55
U5


Acknowledgement. The research was supported by the EuropeanUnion and the State of Hungary, co-financed by the European Social
Fund in the framework of TÁMOP 4.2.4. A/2-11-1-2012-0001 ”National Excellence Program”.
References[1] M. A. Alekseyev and Sz. Tengely. On integral points on biquadratic curves
and near-multiples of squares in lucas sequences. J. Integer Seq., 17(6):Article
14.6.6, 15, 2014.

8SZ. TENGELY[2] A. Baker. Contributions to the theory of Diophantine equations. I. On therepresentation of integers by binary forms. Philos. Trans. Roy. Soc. London
Ser. A, 263:173–191, 1967/1968.
[3] W. Bosma, J. Cannon, and C. Playoust. The Magma algebra system. I. The
user language. J. Symbolic Comput., 24(3-4):235–265, 1997. Computational
algebra and number theory (London, 1993).
[4] N. Bruin and M. Stoll. Two-cover descent on hyperelliptic curves. Math.
Comp., 78(268):2347–2370, 2009.
[5] B. M. M. de Weger. A curious property of the eleventh Fibonacci number.
Rocky Mountain J. Math., 25(3):977–994, 1995.
[6] J. Gebel, A. Pethő, and H. G. Zimmer. Computing integral points on ellipticcurves. Acta Arith., 68(2):171–192, 1994.
[7] K. Győry. Solving Diophantine equations by Baker’s theory. In A panorama ofnumber theory or the view from Baker’s garden (Zürich, 1999), pages 38–72.
Cambridge Univ. Press, Cambridge, 2002.
[8] R. H. Hudson and C. F. Winans. A completecharacterization of the decimal
P −k(i+1)
fractions that can be represented as
10Fαi , where Fαi is the αith
Fibonacci number. Fibonacci Quart., 19(5):414–421, 1981.
[9] G. Köhler. Generating functions of Fibonacci-like sequences and decimal expansions of some fractions. Fibonacci Quart., 23(1):29–35, 1985.
[10] Jia Sheng Lee. A complete characterization of B-power fractions that can berepresented as series of general Fibonacci numbers or of general Tribonaccinumbers. Tamkang J. Management Sci., 6(1):41–52, 1985.
[11] Jin Zai Lee and Jia Sheng Lee. A complete characterization of B-power fractions that can be represented as series of general n-bonacci numbers. FibonacciQuart., 25(1):72–75, 1987.
[12] C. T. Long. The decimal expansion of 1/89 and related results. FibonacciQuart., 19(1):53–55, 1981.
[13] I. Nemes and A. Pethő. Polynomial values in linear recurrences. II. J. NumberTheory, 24(1):47–53, 1986.
[14] A. Pethő. On the solution of the equation Gn = P (x). In Fibonacci numbers
and their applications (Patras, 1984), volume 28 of Math. Appl., pages 193–201. Reidel, Dordrecht, 1986.
[15] N. P. Smart. The algorithmic resolution of Diophantine equations, volume 41of London Mathematical Society Student Texts. Cambridge University Press,Cambridge, 1998.
[16] F. Stancliff. A curious property of aii . Scripta Math., 19:126, 1953.
[17] W. A. Stein et al. Sage Mathematics Software (Version 6.2). The Sage Development Team, 2014. http://www.sagemath.org.
[18] R. J. Stroeker and N. Tzanakis. Solving elliptic diophantine equations by estimating linear forms in elliptic logarithms. Acta Arith., 67(2):177–196, 1994.
[19] R. J. Stroeker and N. Tzanakis. Computing all integer solutions of a generalelliptic equation. In Algorithmic number theory (Leiden, 2000), volume 1838of Lecture Notes in Comput. Sci., pages 551–561. Springer, Berlin, 2000.
[20] R. J. Stroeker and N. Tzanakis. Computing all integer solutions of a genus 1equation. Math. Comp., 72(244):1917–1933, 2003.

ON THE LUCAS SEQUENCE EQUATION

9[21] N. Tzanakis. Solving elliptic Diophantine equations by estimating linear forms
in elliptic logarithms. The case of quartic equations. Acta Arith., 75(2):165–190, 1996.
[22] N. Tzanakis. Effective solution of two simultaneous Pell equations by the elliptic logarithm method. Acta Arith., 103(2):119–135, 2002.
[23] C. F. Winans. The Fibonacci series in the decimal equivalents of fractions.
In A collection of manuscripts related to the Fibonacci sequence, pages 78–81.
Fibonacci Assoc., Santa Clara, Calif., 1980.
Mathematical Institute
University of Derecen
P.O.Box 124010 Debrecen
Hungary
E-mail address: tengely@science.unideb.hu