scholarly journals PRIME DIVISORS OF LUCAS SEQUENCES AND A CONJECTURE OF SKAŁBA

2005 ◽  
Vol 01 (04) ◽  
pp. 583-591 ◽  
Author(s):  
FLORIAN LUCA ◽  
PANTELIMON STĂNICĂ
Keyword(s):  
Prime Divisors ◽  
Lucas Sequences ◽  
Lucas Sequence ◽  
Almost All ◽  
Positive Integers

In this paper, we give some heuristics suggesting that if (un)n≥0 is the Lucas sequence given by un = (an - 1)/(a - 1), where a > 1 is an integer, then ω(un) ≥ (1 + o(1)) log n log log n holds for almost all positive integers n.

scholarly journals Some arithmetic functions of factorials in Lucas sequences

2021 ◽  
Vol 56 (1) ◽  
pp. 17-28
Author(s):  
Eric F. Bravo ◽  
◽  
Jhon J. Bravo ◽  
Keyword(s):  
General Class ◽  
Catalan Numbers ◽  
Arithmetic Functions ◽  
Integer Sequences ◽  
Lucas Sequences ◽  
Lucas Sequence ◽  
Positive Integers ◽  
Sum Of Divisors

We prove that if {un}n≥ 0 is a nondegenerate Lucas sequence, then there are only finitely many effectively computable positive integers n such that |un|=f(m!), where f is either the sum-of-divisors function, or the sum-of-proper-divisors function, or the Euler phi function. We also give a theorem that holds for a more general class of integer sequences and illustrate our results through a few specific examples. This paper is motivated by a previous work of Iannucci and Luca who addressed the above problem with Catalan numbers and the sum-of-proper-divisors function.

scholarly journals Powers in products of terms of Pell's and Pell–Lucas Sequences

2015 ◽  
Vol 11 (04) ◽  
pp. 1259-1274 ◽  
Author(s):  
Jhon J. Bravo ◽  
Pranabesh Das ◽  
Sergio Guzmán ◽  
Shanta Laishram
Keyword(s):  
General Result ◽  
Diophantine Equation ◽  
Initial Condition ◽  
Lucas Sequences ◽  
Lucas Sequence ◽  
Positive Integers

In this paper, we consider the usual Pell and Pell–Lucas sequences. The Pell sequence [Formula: see text] is given by the recurrence un = 2un-1 + un-2 with initial condition u0 = 0, u1 = 1 and its associated Pell–Lucas sequence [Formula: see text] is given by the recurrence vn = 2vn-1 + vn-2 with initial condition v0 = 2, v1 = 2. Let n, d, k, y, m be positive integers with m ≥ 2, y ≥ 2 and gcd (n, d) = 1. We prove that the only solutions of the Diophantine equation unun+d⋯un+(k-1)d = ym are given by u7 = 132 and u1u7 = 132 and the equation vnvn+d⋯vn+(k-1)d = ym has no solution. In fact, we prove a more general result.

scholarly journals Shifted powers in binary recurrence sequences

2015 ◽  
Vol 158 (2) ◽  
pp. 305-329 ◽  
Author(s):  
MICHAEL A. BENNETT ◽  
SANDER R. DAHMEN ◽  
MAURICE MIGNOTTE ◽  
SAMIR SIKSEK
Keyword(s):  
Standard Technique ◽  
Local Information ◽  
Linear Forms In Logarithms ◽  
Linear Forms ◽  
Lucas Sequences ◽  
Lucas Sequence ◽  
Recurrence Sequences ◽  
Totally Real ◽  
Positive Integers ◽  
Totally Real Fields

AbstractLet {uk} be a Lucas sequence. A standard technique for determining the perfect powers in the sequence {uk} combines bounds coming from linear forms in logarithms with local information obtained via Frey curves and modularity. The key to this approach is the fact that the equation uk = xn can be translated into a ternary equation of the form ay2 = bx2n + c (with a, b, c ∈ ℤ) for which Frey curves are available. In this paper we consider shifted powers in Lucas sequences, and consequently equations of the form uk = xn+c which do not typically correspond to ternary equations with rational unknowns. However, they do, under certain hypotheses, lead to ternary equations with unknowns in totally real fields, allowing us to employ Frey curves over those fields instead of Frey curves defined over ℚ. We illustrate this approach by showing that the quaternary Diophantine equation x2n±6xn + 1 = 8y2 has no solutions in positive integers x, y, n with x, n > 1.

scholarly journals On the Sum of Powers of Two k-Fibonacci Numbers which Belongs to the Sequence of k-Lucas Numbers

2016 ◽  
Vol 67 (1) ◽  
pp. 41-46
Author(s):  
Pavel Trojovský
Keyword(s):  
Recurrence Relation ◽  
Initial Conditions ◽  
Fibonacci Numbers ◽  
Fibonacci Sequence ◽  
Lucas Numbers ◽  
Initial Values ◽  
Lucas Sequence ◽  
Positive Integers

Abstract Let k ≥ 1 and denote (Fk,n)n≥0, the k-Fibonacci sequence whose terms satisfy the recurrence relation Fk,n = kFk,n−1 +Fk,n−2, with initial conditions Fk,0 = 0 and Fk,1 = 1. In the same way, the k-Lucas sequence (Lk,n)n≥0 is defined by satisfying the same recurrence relation with initial values Lk,0 = 2 and Lk,1 = k. These sequences were introduced by Falcon and Plaza, who showed many of their properties, too. In particular, they proved that Fk,n+1 + Fk,n−1 = Lk,n, for all k ≥ 1 and n ≥ 0. In this paper, we shall prove that if k ≥ 1 and $F_{k,n + 1}^s + F_{k,n - 1}^s \in \left( {L_{k,m} } \right)_{m \ge 1} $ for infinitely many positive integers n, then s =1.

scholarly journals Solution of certain Pell equations

2018 ◽  
Vol 11 (04) ◽  
pp. 1850056 ◽  
Author(s):  
Zahid Raza ◽  
Hafsa Masood Malik
Keyword(s):  
Positive Integer ◽  
Fundamental Solution ◽  
Positive Solutions ◽  
Continued Fraction ◽  
Pell Equation ◽  
Lucas Sequences ◽  
Pell Equations ◽  
Integer Solutions ◽  
Positive Integers

Let [Formula: see text] be any positive integers such that [Formula: see text] and [Formula: see text] is a square free positive integer of the form [Formula: see text] where [Formula: see text] and [Formula: see text] The main focus of this paper is to find the fundamental solution of the equation [Formula: see text] with the help of the continued fraction of [Formula: see text] We also obtain all the positive solutions of the equations [Formula: see text] and [Formula: see text] by means of the Fibonacci and Lucas sequences.Furthermore, in this work, we derive some algebraic relations on the Pell form [Formula: see text] including cycle, proper cycle, reduction and proper automorphism of it. We also determine the integer solutions of the Pell equation [Formula: see text] in terms of [Formula: see text] We extend all the results of the papers [3, 10, 27, 37].

scholarly journals Prime divisors of some shifted products

2005 ◽  
Vol 2005 (19) ◽  
pp. 3057-3073
Author(s):  
Eric Levieil ◽  
Florian Luca ◽  
Igor E. Shparlinski
Keyword(s):  
Prime Divisors ◽  
Positive Integers

We study prime divisors of various sequences of positive integersA(n)+1,n=1,…,N, such that the ratiosa(n)=A(n)/A(n−1)have some number-theoretic or combinatorial meaning. In the casea(n)=n, we obviously haveA(n)=n!, for which several new results about prime divisors ofn!+1have recently been obtained.

scholarly journals On some new results for the generalised Lucas sequences

2021 ◽  
Vol 29 (1) ◽  
pp. 17-36
Author(s):  
Dorin Andrica ◽  
Ovidiu Bagdasar ◽  
George Cătălin Ţurcaş
Keyword(s):  
Integer Sequences ◽  
Lucas Sequences ◽  
Lucas Sequence ◽  
Necessary And Sufficient

Abstract In this paper we introduce the functions which count the number of generalized Lucas and Pell-Lucas sequence terms not exceeding a given value x and, under certain conditions, we derive exact formulae (Theorems 3 and 4) and establish asymptotic limits for them (Theorem 6). We formulate necessary and sufficient arithmetic conditions which can identify the terms of a-Fibonacci and a-Lucas sequences. Finally, using a deep theorem of Siegel, we show that the aforementioned sequences contain only finitely many perfect powers. During the process we also discover some novel integer sequences.

On the Normal Growth of Prime Factors of Integers

1992 ◽  
Vol 44 (6) ◽  
pp. 1121-1154 ◽  
Author(s):  
J. M. De Koninck ◽  
I. Kátai ◽  
A. Mercier
Keyword(s):  
Prime Factor ◽  
Continuous Distribution ◽  
Normal Growth ◽  
Distribution Functions ◽  
Prime Divisors ◽  
Simple Argument ◽  
Prime Factors ◽  
Almost All ◽  
Number Of Prime Divisors ◽  
Class Of Functions

AbstractLet h: [0,1] → R be such that and define .In 1966, Erdős [8] proved that holds for almost all n, which by using a simple argument implies that in the case h(u) = u, for almost all n, He further obtained that, for every z > 0 and almost all n, and that where ϕ, ψ, are continuous distribution functions. Several other results concerning the normal growth of prime factors of integers were obtained by Galambos [10], [11] and by De Koninck and Galambos [6].Let χ = ﹛xm : w ∈ N﹜ be a sequence of real numbers such that limm→∞ xm = +∞. For each x ∈ χ let be a set of primes p ≤x. Denote by p(n) the smallest prime factor of n. In this paper, we investigate the number of prime divisors p of n, belonging to for which Th(n,p) > z. Given Δ < 1, we study the behaviour of the function We also investigate the two functions , where, in each case, h belongs to a large class of functions.

scholarly journals Some Diophantine Problems Related to k-Fibonacci Numbers

Mathematics ◽  
2020 ◽  
Vol 8 (7) ◽  
pp. 1047
Author(s):  
Pavel Trojovský ◽  
Štěpán Hubálovský
Keyword(s):  
Recurrence Relation ◽  
Diophantine Equations ◽  
Initial Conditions ◽  
Fibonacci Numbers ◽  
Fibonacci Sequence ◽  
Recursive Relation ◽  
Initial Values ◽  
Lucas Sequence ◽  
Diophantine Problems ◽  
Positive Integers

Let k ≥ 1 be an integer and denote ( F k , n ) n as the k-Fibonacci sequence whose terms satisfy the recurrence relation F k , n = k F k , n − 1 + F k , n − 2 , with initial conditions F k , 0 = 0 and F k , 1 = 1 . In the same way, the k-Lucas sequence ( L k , n ) n is defined by satisfying the same recursive relation with initial values L k , 0 = 2 and L k , 1 = k . The sequences ( F k , n ) n ≥ 0 and ( L k , n ) n ≥ 0 were introduced by Falcon and Plaza, who derived many of their properties. In particular, they proved that F k , n 2 + F k , n + 1 2 = F k , 2 n + 1 and F k , n + 1 2 − F k , n − 1 2 = k F k , 2 n , for all k ≥ 1 and n ≥ 0 . In this paper, we shall prove that if k > 1 and F k , n s + F k , n + 1 s ∈ ( F k , m ) m ≥ 1 for infinitely many positive integers n, then s = 2 . Similarly, that if F k , n + 1 s − F k , n − 1 s ∈ ( k F k , m ) m ≥ 1 holds for infinitely many positive integers n, then s = 1 or s = 2 . This generalizes a Marques and Togbé result related to the case k = 1 . Furthermore, we shall solve the Diophantine equations F k , n = L k , m , F k , n = F n , k and L k , n = L n , k .

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