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Displaying similar documents to “Bartz-Marlewski equation with generalized Lucas components”

Padovan and Perrin numbers as products of two generalized Lucas numbers

Kouèssi Norbert Adédji, Japhet Odjoumani, Alain Togbé (2023)

Archivum Mathematicum

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Let P m and E m be the m -th Padovan and Perrin numbers respectively. Let r , s be non-zero integers with r 1 and s { - 1 , 1 } , let { U n } n 0 be the generalized Lucas sequence given by U n + 2 = r U n + 1 + s U n , with U 0 = 0 and U 1 = 1 . In this paper, we give effective bounds for the solutions of the following Diophantine equations P m = U n U k and E m = U n U k , where m , n and k are non-negative integers. Then, we explicitly solve the above Diophantine equations for the Fibonacci, Pell and balancing sequences.

On perfect powers in k -generalized Pell sequence

Zafer Şiar, Refik Keskin, Elif Segah Öztaş (2023)

Mathematica Bohemica

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Let k 2 and let ( P n ( k ) ) n 2 - k be the k -generalized Pell sequence defined by P n ( k ) = 2 P n - 1 ( k ) + P n - 2 ( k ) + + P n - k ( k ) for n 2 with initial conditions P - ( k - 2 ) ( k ) = P - ( k - 3 ) ( k ) = = P - 1 ( k ) = P 0 ( k ) = 0 , P 1 ( k ) = 1 . In this study, we handle the equation P n ( k ) = y m in positive integers n , m , y , k such that k , y 2 , and give an upper bound on n . Also, we will show that the equation P n ( k ) = y m with 2 y 1000 has only one solution given by P 7 ( 2 ) = 13 2 .

Repdigits in generalized Pell sequences

Jhon J. Bravo, Jose L. Herrera (2020)

Archivum Mathematicum

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For an integer k 2 , let ( n ) n be the k - generalized Pell sequence which starts with 0 , ... , 0 , 1 ( k terms) and each term afterwards is given by the linear recurrence n = 2 n - 1 + n - 2 + + n - k . In this paper, we find all k -generalized Pell numbers with only one distinct digit (the so-called repdigits). Some interesting estimations involving generalized Pell numbers, that we believe are of independent interest, are also deduced. This paper continues a previous work that searched for repdigits in the usual Pell sequence ( P n ( 2 ) ) n . ...

Mersenne numbers as a difference of two Lucas numbers

Murat Alan (2022)

Commentationes Mathematicae Universitatis Carolinae

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Let ( L n ) n 0 be the Lucas sequence. We show that the Diophantine equation L n - L m = M k has only the nonnegative integer solutions ( n , m , k ) = ( 2 , 0 , 1 ) , ( 3 , 1 , 2 ) , ( 3 , 2 , 1 ) , ( 4 , 3 , 2 ) , ( 5 , 3 , 3 ) , ( 6 , 2 , 4 ) , ( 6 , 5 , 3 ) where M k = 2 k - 1 is the k th Mersenne number and n > m .

The number of solutions to the generalized Pillai equation ± r a x ± s b y = c .

Reese Scott, Robert Styer (2013)

Journal de Théorie des Nombres de Bordeaux

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We consider N , the number of solutions ( x , y , u , v ) to the equation ( - 1 ) u r a x + ( - 1 ) v s b y = c in nonnegative integers x , y and integers u , v { 0 , 1 } , for given integers a > 1 , b > 1 , c > 0 , r > 0 and s > 0 . When gcd ( r a , s b ) = 1 , we show that N 3 except for a finite number of cases all of which satisfy max ( a , b , r , s , x , y ) < 2 · 10 15 for each solution; when gcd ( a , b ) > 1 , we show that N 3 except for three infinite families of exceptional cases. We find several different ways to generate an infinite number of cases giving N = 3 solutions.

Lucas sequences and repdigits

Hayder Raheem Hashim, Szabolcs Tengely (2022)

Mathematica Bohemica

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Let ( G n ) n 1 be a binary linear recurrence sequence that is represented by the Lucas sequences of the first and second kind, which are { U n } and { V n } , respectively. We show that the Diophantine equation G n = B · ( g l m - 1 ) / ( g l - 1 ) has only finitely many solutions in n , m + , where g 2 , l is even and 1 B g l - 1 . Furthermore, these solutions can be effectively determined by reducing such equation to biquadratic elliptic curves. Then, by a result of Baker (and its best improvement due to Hajdu and Herendi) related to the bounds of the integral...

A Diophantine inequality with four squares and one k th power of primes

Quanwu Mu, Minhui Zhu, Ping Li (2019)

Czechoslovak Mathematical Journal

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Let k 5 be an odd integer and η be any given real number. We prove that if λ 1 , λ 2 , λ 3 , λ 4 , μ are nonzero real numbers, not all of the same sign, and λ 1 / λ 2 is irrational, then for any real number σ with 0 < σ < 1 / ( 8 ϑ ( k ) ) , the inequality | λ 1 p 1 2 + λ 2 p 2 2 + λ 3 p 3 2 + λ 4 p 4 2 + μ p 5 k + η | < max 1 j 5 p j - σ has infinitely many solutions in prime variables p 1 , p 2 , , p 5 , where ϑ ( k ) = 3 × 2 ( k - 5 ) / 2 for k = 5 , 7 , 9 and ϑ ( k ) = [ ( k 2 + 2 k + 5 ) / 8 ] for odd integer k with k 11 . This improves a recent result in W. Ge, T. Wang (2018).

The exceptional set for Diophantine inequality with unlike powers of prime variables

Wenxu Ge, Feng Zhao (2018)

Czechoslovak Mathematical Journal

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Suppose that λ 1 , λ 2 , λ 3 , λ 4 are nonzero real numbers, not all negative, δ > 0 , 𝒱 is a well-spaced set, and the ratio λ 1 / λ 2 is algebraic and irrational. Denote by E ( 𝒱 , N , δ ) the number of v 𝒱 with v N such that the inequality | λ 1 p 1 2 + λ 2 p 2 3 + λ 3 p 3 4 + λ 4 p 4 5 - v | < v - δ has no solution in primes p 1 , p 2 , p 3 , p 4 . We show that E ( 𝒱 , N , δ ) N 1 + 2 δ - 1 / 72 + ε for any ε > 0 .

On the Diophantine equation j = 1 k j F j p = F n q

Gökhan Soydan, László Németh, László Szalay (2018)

Archivum Mathematicum

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Let F n denote the n t h term of the Fibonacci sequence. In this paper, we investigate the Diophantine equation F 1 p + 2 F 2 p + + k F k p = F n q in the positive integers k and n , where p and q are given positive integers. A complete solution is given if the exponents are included in the set { 1 , 2 } . Based on the specific cases we could solve, and a computer search with p , q , k 100 we conjecture that beside the trivial solutions only F 8 = F 1 + 2 F 2 + 3 F 3 + 4 F 4 , F 4 2 = F 1 + 2 F 2 + 3 F 3 , and F 4 3 = F 1 3 + 2 F 2 3 + 3 F 3 3 satisfy the title equation.

Involutivity degree of a distribution at superdensity points of its tangencies

Silvano Delladio (2021)

Archivum Mathematicum

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Let Φ 1 , ... , Φ k + 1 (with k 1 ) be vector fields of class C k in an open set U N + m , let 𝕄 be a N -dimensional C k submanifold of U and define 𝕋 : = { z 𝕄 : Φ 1 ( z ) , ... , Φ k + 1 ( z ) T z 𝕄 } where T z 𝕄 is the tangent space to 𝕄 at z . Then we expect the following property, which is obvious in the special case when z 0 is an interior point (relative to 𝕄 ) of 𝕋 : If z 0 𝕄 is a ( N + k ) -density point (relative to 𝕄 ) of 𝕋 then all the iterated Lie brackets of order less or equal to k Φ i 1 ( z 0 ) , [ Φ i 1 , Φ i 2 ] ( z 0 ) , [ [ Φ i 1 , Φ i 2 ] , Φ i 3 ] ( z 0 ) , ... ( h , i h k + 1 ) belong to T z 0 𝕄 . Such a property has been proved in [9] for k = 1 and its proof in the...

A note on the weighted Khintchine-Groshev Theorem

Mumtaz Hussain, Tatiana Yusupova (2014)

Journal de Théorie des Nombres de Bordeaux

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Let W ( m , n ; ψ ̲ ) denote the set of ψ 1 , ... , ψ n –approximable points in m n . The classical Khintchine–Groshev theorem assumes a monotonicity condition on the approximating functions ψ ̲ . Removing monotonicity from the Khintchine–Groshev theorem is attributed to different authors for different cases of m and n . It can not be removed for m = n = 1 as Duffin–Schaeffer provided the counter example. We deal with the only remaining case m = 2 and thereby remove all unnecessary conditions from the Khintchine–Groshev theorem. ...

Repdigits in the base b as sums of four balancing numbers

Refik Keskin, Faticko Erduvan (2021)

Mathematica Bohemica

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The sequence of balancing numbers ( B n ) is defined by the recurrence relation B n = 6 B n - 1 - B n - 2 for n 2 with initial conditions B 0 = 0 and B 1 = 1 . B n is called the n th balancing number. In this paper, we find all repdigits in the base b , which are sums of four balancing numbers. As a result of our theorem, we state that if B n is repdigit in the base b and has at least two digits, then ( n , b ) = ( 2 , 5 ) , ( 3 , 6 ) . Namely, B 2 = 6 = ( 11 ) 5 and B 3 = 35 = ( 55 ) 6 .