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Let $(L_n^{(k)})_{n\geq 2-k}$ be the sequence of $k$--generalized Lucas numbers for some fixed integer $k\ge 2$ whose first $k$ terms are $0,\ldots,0,2,1$ and each term afterwards is the sum of the preceding $k$ terms. In this paper, we…

Number Theory · Mathematics 2024-12-18 Herbert Batte , Florian Luca

For an integer $k\geq 2$, let $(L_{n}^{(k)})_{n}$ be the $k-$generalized Lucas sequence which starts with $0,\ldots,0,2,1$ ($k$ terms) and each term afterwards is the sum of the $k$ preceding terms. In this paper, we find all the integers…

Number Theory · Mathematics 2014-02-18 Eric F. Bravo , Jhon J. Bravo , Florian Luca

Let $ k \geq 2 $ and let $ ( L_{n}^{(k)} )_{n \geq 2-k} $ be the $k-$generalized Lucas sequence with certain initial $ k $ terms and each term afterward is the sum of the $ k $ preceding terms. In this paper, we find all repdigits which are…

General Mathematics · Mathematics 2023-07-20 Alaa Altassan , Murat Alan

Here, we find all positive integer solutions of the Diophantine equation in the title, where $(\mathcal{U}_n)_{n\geqslant 0}$ is the generalized Lucas sequence $\mathcal{U}_0=0, \ \mathcal{U}_1=1$ and $\mathcal{U}_{n+1}=r \mathcal{U}_n +s…

Number Theory · Mathematics 2023-06-08 P. Tiebekabe , I. Diouf , A. Tall , K. R. Kakanou

A Wagstaff prime is a prime number of the form $(2^{\mathfrak{p}}+1)/3$, where $\mathfrak{p}$ is an odd prime. Let $(L_n^{(k)})_{n\geq 2-k}$ be the $k$-Lucas number sequence defined by the recurrence relation $ L_n^{(k)} = L_{n-1}^{(k)} +…

Number Theory · Mathematics 2026-02-25 Herbert Batte

Let $T_{k}$ be the $k^{\textrm{th}}$ Tribonacci number and $L_{n}$ be the $n^{\textrm{th}}$ Lucas number defined by their respective recurrence relation $T_{k}=T_{k-1}+T_{k-2}+T_{k-3}$ and $L_{n}=L_{n-1}+L_{n-2}$. In this study, we solve…

Number Theory · Mathematics 2026-02-17 Ama Ahenfoa Quansah

For an integer \( k \geq 2 \), the sequence of \( k \)-generalized Lucas numbers is defined by the recurrence relation \( L_n^{(k)} = L_{n-1}^{(k)} + \cdots + L_{n-k}^{(k)} \) for all \( n \geq 2 \), with initial conditions \( L_0^{(k)} = 2…

Number Theory · Mathematics 2025-09-30 Alex Behakanira Tumwesigye , Mahadi Ddamulira , Prosper Kaggwa

Let L_t denote the t-th Lucas number. We prove that the Diophantine equation L_m^{n+k} + L_m^n = L_r has no solutions in positive integers r, m, n, and k with m >= 2. In the case n = 1, the proof is based on a precise factorization formula…

Number Theory · Mathematics 2026-02-19 Seyran S. Ibrahimov , Nazim I. Mahmudov

Let $(L_n^{(k)})_{n\geq 2-k}$ be the sequence of $k$-generalized Lucas numbers for some fixed integer $k\ge 2$, whose first $k$ terms are $0,\;\ldots\;,\;0,\;2,\;1$ and each term afterward is the sum of the preceding $k$ terms. In this…

Number Theory · Mathematics 2023-11-27 Herbert Batte , Mahadi Ddamulira , Juma Kasozi , Florian Luca

In this paper, we consider the Diophantine equation $(-1)^{\alpha}p^x+(-1)^{\beta}(2^k(2p+1))^y=z^2$ for Sophie Germain prime $p$ with $\alpha, \beta \in\{0,1\}$, $\alpha\beta=0$ and $k\geq 0$. First, for $p=2$, we solve three Diophantine…

Number Theory · Mathematics 2024-07-17 Yuan Li , Jing Zhang , Baoxing Liu

For an integer $k \geq 2$, let $\{ P_{n}^{(k)} \}_{n}$ be the $k$-generalized Pell sequence which starts with $0, \dots,0,1$($k$ terms) and each term afterwards is the sum of $k$ preceding terms. In this paper, we find all the solutions of…

Number Theory · Mathematics 2023-07-11 Bibhu Prasad Tripathy , Bijan Kumar Patel

Let $k\geq 2$ and let $(P_{n}^{(k)})_{n\geq 2-k}$ be $k$-generalized Pell sequence defined by \begin{equation*}P_{n}^{(k)}=2P_{n-1}^{(k)}+P_{n-2}^{(k)}+...+P_{n-k}^{(k)}\end{equation*} for $n\geq 2$ with initial conditions…

Number Theory · Mathematics 2020-09-29 Zafer Şiar , Refik Keskin

For any integer $k \geq 2$, let $\{Q_{n}^{(k)} \}_{n \geq -(k-2)}$ denote the $k$-generalized Pell-Lucas sequence which starts with $0, \dots ,2,2$($k$ terms) where each next term is the sum of the $k$ preceding terms. In this paper, we…

Number Theory · Mathematics 2022-11-15 Bibhu Prasad Tripathy , Bijan Kumar Patel

Let (F_n)_{n} be the classical Fibonacci sequence. It is well-known that it satisfies F_{n}^2 + F_{n+1}^2 = F_{2n+1}. In this study, we explore generalizations of this Diophantine equation in several directions. First, we solve the…

Number Theory · Mathematics 2025-01-16 Roberto Alvarenga , Ana Paula Chaves , Maria Eduarda Ramos , Matheus Silva , Marcos Sosa

For any integer $k \geq 2$, let $\{Q_{n}^{(k)} \}_{n \geq -(k-2)}$ denote the $k$-generalized Pell-Lucas sequence which starts with $0, \dots ,2,2$($k$ terms) where each next term is the sum of the $k$ preceding terms. In this paper, we…

Number Theory · Mathematics 2023-03-10 Bibhu Prasad Tripathy , Bijan Kumar Patel

Let $(L_n)_{n\geq 0}$ be the Lucas sequence given by $L_0 = 2, L_1 = 1$ and $L_{n+2} = L_{n+1}+L_n$ for $n \geq 0$. In this paper, we are interested in finding all powers of three which are sums of two Lucas numbers, i.e., we study the…

Number Theory · Mathematics 2022-03-01 Pagdame Tiebekabe , Ismaila Diouf

Let k>=2 and let (Q_{n}^{(k)})_{n>=2-k} be the k-generalized Pell sequence defined by Q_{n}^{(k)}=2Q_{n-1}^{(k)}+Q_{n-2}^{(k)}+...+Q_{n-k}^{(k)} for n>=2 with initial conditions Q_{-(k-2)}^{(k)}=Q_{-(k-3)}^{(k)}=...=Q_{-1}^{(k)}=0,…

Number Theory · Mathematics 2022-09-12 Zafer Şiar , Refik Keskin

This paper finds all Lucas numbers which are the sum of two Jacobsthal numbers. It also finds all Jacobsthal numbers which are the sum of two Lucas numbers. In general, we find all non-negative integer solutions $(n, m, k)$ of the two…

Number Theory · Mathematics 2024-10-17 Osama Salah , A. Elsonbaty , Mohammed Abdul Azim Seoud , Mohamed Anwar

We give a shorter simpler proof of a result of Szalay on the equation $2^a + 2^b + 1 = z^2$. We give an elementary proof of a result of Luca on the equation of the title for prime $p > 2$. The elementary treatment is made possible by a…

Number Theory · Mathematics 2023-05-23 Reese Scott

In [1] it is shown that the Diophantine equation $(k!)^n+k^n=(n!)^k+n^k$ only has the trivial solution $n=k$, and $(k!)^n-k^n=(n!)^k-n^k$ only has the solutions $n=k$, $(n, k)=(1, 2),$ and $(2, 1)$. In this article we find all solutions of…

Number Theory · Mathematics 2021-05-25 Addea Gupta
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