Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Lucas sequences and infinite sums
Szabolcs Tengely
tengely@science.unideb.hu
http://www.math.unideb.hu/~tengely
Numeration and Substitution 2014
University of Debrecen
DebrecenThis research was supported by the European Union 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’.

Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Summary of the talkBackground
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

The Fibonacci number 89
Let Fn be the Fibonacci sequence, that is
F0 = 0,F1 = 1,Fn = Fn−1 + Fn−2 ,n ≥ 2.

The first few values are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.

Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

3 / 21Background
Related results
Lucas sequences
Main results
Auxiliary results
ProofThe Fibonacci number 89
Let Fn be the Fibonacci sequence, that is
F0 = 0,F1 = 1,Fn = Fn−1 + Fn−2 ,n ≥ 2.

The first few values are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.
In 1953 Stancliff noted an interesting property of the Fibonacci
number F11 = 89. One has that
10112
35+++++ ...,=+89
10 102 103 104 105 106
where in the numerators the elements of the Fibonacci sequenceappear.
Szabolcs Tengely2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
ProofDecimal expansion
In 1980 Winans investigated the related sums
∞XFαk10k+1
k=0

for certain values of α. In 1981 Hudson and Winans provided acomplete characterization of all decimal fractions that can beapproximated by sums of the type
n1 X Fαk,Fα10l(k+1)

α, l ≥ 1.

k=1

Long proved a general identity for binary recurrence sequencesfrom which one obtains e.g.
∞

X Fk
1=,9899102(k+1)
k=0

Szabolcs Tengely

2014.07.07∞X1Fk=.
109(−10)k+1
k=0

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Different bases
In case of different bases characterizations were obtained by Jia
Sheng Lee and by Köhler and by Jin Zai Lee and Jia Sheng Lee.
Here we state a result by Köhler that we will use later

Theorem A
Let A, B, a0 , a1 be arbitrary complex numbers. Define the sequence{an } by the recursion an+1 = Aan + Ban−1 . Then the formula∞
Xaka0 x − Aa0 + a1= 2x − Ax − Bx k+1
k=0

holds for all complex x such that |x| is larger than the absolutevalues of the zeros of x 2 − Ax − B.
Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Lucas sequences
Let P and Q be non-zero relatively prime integers. The Lucassequence {Un (P, Q)} is defined by
U0 = 0, U1 = 1 and Un = PUn−1 − QUn−2 , if n ≥ 2.
In this talk we deal with the determination of all integers x ≥ 2 forwhich there exists an n ≥ 0 such that∞

X Uk−1
1=,Unxk(1)k=1

where Un is a Lucas sequence with some given P and Q.
Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
ProofFibonacci caseIn case of P = 1, Q = −1 one gets the Fibonacci sequence. DeWeger computed all x ≥ 2 in case of the Fibonacci sequence, thesolutions are as follows∞

1
1 X Fk−1
1== =,
F1F2
1
2k
k=1

1
1==F1055

Szabolcs Tengely∞XFk−1k=1

2014.07.07

8k,∞

1 X Fk−1
1= =,F553k
k=1∞X

1
1==F11
89

k=1

Fk−1.
10k

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
ProofStatements

TheoremLet {Un (P, Q)} be a Lucas sequence with Q ∈ {±1}. Thenequation (1) possesses only finitely many solutions in n, x which
can be effectively determined.
The proof of the above Theorem provides an algorithm todetermine all solutions of equation (1). Following this algorithm
we obtained numerical results.

Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Numerical result
TheoremLet {Un (P, Q)} be a Lucas sequence with−10 ≤ P ≤ 10, Q ∈ {±1} and (P, Q) 6= (−2, 1), (2, 1). Thenequation (1) has the following 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)}.

Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Associated Lucas sequences
The Lucas sequences {Un (P, Q)} and associated Lucas sequences{Vn (P, Q)} are defined by the same linear recurrent relation withP, Q ∈ Z \ {0} but different initial terms:
U0 = 0, U1 = 1 and

Un = PUn−1 − QUn−2 , if n ≥ 2,V0 = 2, V1 = P and

Vn = PVn−1 − QVn−2 , if n ≥ 2.

Terms of Lucas sequences and associated Lucas sequences satisfy
the following identity
Vn2 − DUn2 = 4Q n ,(2)where D = P 2 − 4Q.
Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
ProofTernary quadratic equations
To determine the appropriate Thue equations we use parametricsolutions of ternary quadratic equations.

Lemma (Alekseyev-Tengely)
Let A, B, C be non-zero integers and let (x0 , y0 , z0 ) with z0 6= 0 bea particular non-trivial integer solution to the Diophantine equation
Ax 2 + By 2 + Cz 2 = 0. Then its general integer solution is given byp(x, y , z) =q(Px (m, n), Py (m, n), Pz (m, n))where m, n as well as p, q are coprime integers with q > 0 dividing
2 lcm(A, B)Cz02 , andPx (m, n)Py (m, n)Pz (m, n)Szabolcs Tengely===

2014.07.07

22x0 Am + 2y0 Bmn − x0 Bn ,22−y0 Am + 2x0 Amn + y0 Bn ,22z0 Am + z0 Bn .
Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
ProofFinitenessLet {Un (P, Q)} be a Lucas sequence with Q ∈ {±1}. Theorem A
implies that∞X1
Uk−1= 2.
(3)kx − Px ± 1x
k=1Hence from equations (1) and (3) it follows that
Un = x 2 − Px ± 1. Finiteness follows from results by Nemes andPethő and by Pethő. We provide two approaches to determine afinite set of possible values of x for which Un = x 2 − Px ± 1.

Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Elliptic curves
The first one is based on elliptic curves. It only works if one candetermine the rank 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.
Substituting Un = x 2 − Px ± 1 into the identity (2) yields a genus
1 curve
C(P,Q,n) :y 2 = (P 2 − 4Q)(x 2 − Px + Q)2 + 4Q n .

Bruin and Stoll described and algorithm the so-called two-coverdescent, which can be used to prove that a given hyperelliptic
curve has no rational points.
Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Elliptic curves
This algorithm is implemented in Magma, the procedure is called
TwoCoverDescent. If it turns out that there are no rational points
on the curves y 2 = (P 2 − 4)(x 2 − Px + 1)2 + 4 andy 2 = (P 2 + 4)(x 2 − Px − 1)2 ± 4, then the 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 the proof. Now we assume that we could determine points
on curves for which TwoCoverDescent predicts existence ofrational points. That means we are given elliptic curves in quartic
model. Tzanakis provided a method to determine all integral
points on quartic models, the algorithm is implemented in Magma
as IntegralQuarticPoints.
Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Elementary reduction
There are three ternary quadratic equations to parametrize

Szabolcs Tengely

Q1 :X 2 − (P 2 − 4)Y 2 − 4Z 2 = 0,Q2 :X 2 − (P 2 + 4)Y 2 − 4Z 2 = 0,Q3 :X 2 − (P 2 + 4)Y 2 + 4Z 2 = 0.

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Elementary reduction
There are three ternary quadratic equations to parametrize
Q1 :X 2 − (P 2 − 4)Y 2 − 4Z 2 = 0,Q2 :X 2 − (P 2 + 4)Y 2 − 4Z 2 = 0,Q3 :X 2 − (P 2 + 4)Y 2 + 4Z 2 = 0.

There are points on these curves:

Szabolcs Tengely

Q1 :(X , Y , Z ) = (2, 0, 1),

Q2 :(X , Y , Z ) = (2, 0, 1),

Q3 :(X , Y , Z ) = (P, 1, 1).

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Elementary reduction
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)(x 2 − Px + 1)2 + 4, the other twocases are similar. We obtain thatx 2 − Px + 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.
Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Eigenvalues and eigenvectors
If we have a possible solution x ∈ N of equation (1), then we havePUk−1
to compute the value of the sum ∞k=1 x k . We define
P/x
T =1/x−Q/x.
0Following standard arguments one has that 1 ∗
1 01
2N−1
T + T + T + ... + T= PN Uk−1 .
0x
k=1 x kUsing eigenvectors and eigenvalues one can determine a formulafor the powers of TP, hence one obtains a formula depending only
Uk−1
on N for the sum N
k=1 x k .
Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
ProofIllustration of the algorithm
We will illustrate how one can use the approaches provided in theproof of Theorem 1 to determine all solutions of equation (1) forgiven P and Q. First we deal with the case P = 4, Q = −1. We
have thaty 2 = 20(x 2 − 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}.
Szabolcs Tengely

2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Illustration of the algorithm
Since x ≥ 2 only 4 values remain. In case of x = 647 the matrix T
is as follows

4/647 1/647
1/647
0and we obtain that √ N √ N√√2− 52+ 5N
5−1)−(129(129 5 + 1)
X
647
647
Uk−1
1
=+.
k832040
416020647
k=1

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2014.07.07

Lucas sequences and infinite sums

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Background
Related results
Lucas sequencesMain results
Auxiliary results
Proof

Illustration of the algorithm
Thuslim

N→∞NX
Uk−1
k=1

647k

=

1
1=.
416020
U10

In a similar way we get thatlim

N→∞

lim

N→∞

lim

N→∞
Szabolcs Tengely

NX
Uk−1
k=1

3k

N
X
Uk−1
k=1
N
X
k=1

2014.07.07

4k
Uk−1
5k

= +∞,

= +∞,

=1
1=.
4U2
Lucas sequences and infinite sums

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Background
Related results
Lucas sequences
Main results
Auxiliary results
Proof

Illustration of the algorithm
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(x 2 − 3x + 1)2 + 4. It can be written asv 2 = 5u 4 − 50u 2 + 189 with v = 4y and u = 2x − 3. The second
approach has been implemented in Sage by Alekseyev and Tengely.
Using their procedure QuarticEq([5,-50,189]) we obtain thatu ∈ {±1, ±3, ±15}, therefore x ∈ {−6, 0, 1, 2, 3, 9}. We have that
lim

N→∞

lim

N→∞limN→∞Szabolcs TengelyNXUk−1
k=12kNX
Uk−1
k=1

3kNX
Uk−1
k=1

2014.07.07

9k=+∞,=

1==155

1U1=,
1U5.

Lucas sequences and infinite sums

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