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On some properties of Chebyshev polynomials

Hacène Belbachir, Farid Bencherif (2008)

Discussiones Mathematicae - General Algebra and Applications

Letting T n (resp. U n ) be the n-th Chebyshev polynomials of the first (resp. second) kind, we prove that the sequences ( X k T n - k ) k and ( X k U n - k ) k for n - 2⎣n/2⎦ ≤ k ≤ n - ⎣n/2⎦ are two basis of the ℚ-vectorial space n [ X ] formed by the polynomials of ℚ[X] having the same parity as n and of degree ≤ n. Also T n and U n admit remarkableness integer coordinates on each of the two basis.

On square classes in generalized Fibonacci sequences

Zafer Şiar, Refik Keskin (2016)

Acta Arithmetica

Let P and Q be nonzero integers. The generalized Fibonacci and Lucas sequences are defined respectively as follows: U₀ = 0, U₁ = 1, V₀ = 2, V₁ = P and U n + 1 = P U + Q U n - 1 , V n + 1 = P V + Q V n - 1 for n ≥ 1. In this paper, when w ∈ 1,2,3,6, for all odd relatively prime values of P and Q such that P ≥ 1 and P² + 4Q > 0, we determine all n and m satisfying the equation Uₙ = wUₘx². In particular, when k|P and k > 1, we solve the equations Uₙ = kx² and Uₙ = 2kx². As a result, we determine all n such that Uₙ = 6x².

On sums of binomial coefficients modulo p²

Zhi-Wei Sun (2012)

Colloquium Mathematicae

Let p be an odd prime and let a be a positive integer. In this paper we investigate the sum k = 0 p a - 1 ( h p a - 1 k ) ( 2 k k ) / m k ( m o d p ² ) , where h and m are p-adic integers with m ≢ 0 (mod p). For example, we show that if h ≢ 0 (mod p) and p a > 3 , then k = 0 p a - 1 ( h p a - 1 k ) ( 2 k k ) ( - h / 2 ) k ( ( 1 - 2 h ) / ( p a ) ) ( 1 + h ( ( 4 - 2 / h ) p - 1 - 1 ) ) ( m o d p ² ) , where (·/·) denotes the Jacobi symbol. Here is another remarkable congruence: If p a > 3 then k = 0 p a - 1 ( p a - 1 k ) ( 2 k k ) ( - 1 ) k 3 p - 1 ( p a / 3 ) ( m o d p ² ) .

On terms of linear recurrence sequences with only one distinct block of digits

Diego Marques, Alain Togbé (2011)

Colloquium Mathematicae

In 2000, Florian Luca proved that F₁₀ = 55 and L₅ = 11 are the largest numbers with only one distinct digit in the Fibonacci and Lucas sequences, respectively. In this paper, we find terms of a linear recurrence sequence with only one block of digits in its expansion in base g ≥ 2. As an application, we generalize Luca's result by finding the Fibonacci and Lucas numbers with only one distinct block of digits of length up to 10 in its decimal expansion.

On the arrowhead-Fibonacci numbers

Inci Gültekin, Ömür Deveci (2016)

Open Mathematics

In this paper, we define the arrowhead-Fibonacci numbers by using the arrowhead matrix of the characteristic polynomial of the k-step Fibonacci sequence and then we give some of their properties. Also, we study the arrowhead-Fibonacci sequence modulo m and we obtain the cyclic groups from the generating matrix of the arrowhead-Fibonacci numbers when read modulo m. Then we derive the relationships between the orders of the cyclic groups obtained and the periods of the arrowhead-Fibonacci sequence...

On the Diophantine equation ( 2 x - 1 ) ( p y - 1 ) = 2 z 2

Ruizhou Tong (2021)

Czechoslovak Mathematical Journal

Let p be an odd prime. By using the elementary methods we prove that: (1) if 2 x , p ± 3 ( mod 8 ) , the Diophantine equation ( 2 x - 1 ) ( p y - 1 ) = 2 z 2 has no positive integer solution except when p = 3 or p is of the form p = 2 a 0 2 + 1 , where a 0 > 1 is an odd positive integer. (2) if 2 x , 2 y , y 2 , 4 , then the Diophantine equation ( 2 x - 1 ) ( p y - 1 ) = 2 z 2 has no positive integer solution.

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

Gökhan Soydan, László Németh, László Szalay (2018)

Archivum Mathematicum

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.

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