Related papers: Markoff $m$-triples with $k$-Fibonacci components

We classify all solution triples with Fibonacci components to the equation $a^2+b^2+c^2=3abc+m,$ for positive $m$. We show that for $m=2$ they are precisely $(1,F(b),F(b+2))$, with even $b$; for $m=21$, there exist exactly two Fibonacci…

We study infinite paths of Markoff $m$-triples, that is, solutions to the generalised Markoff equation \[ x^2+y^2+z^2=3xyz+m, \] with $m>0$, with at least two $k$-Fibonacci components. First, we obtain a complete classification of Markoff…

A triple (a, b, c) of positive integers is called a Markoff triple iff it satisfies the Diophantine equation a2+b2+c2=abc . Recasting the Markoff tree, whose vertices are Markoff triples, in the framework of integral upper triangular 3x3…

For a positive integer $m>1$, if the generalized Markoff equation $a^2+b^2+c^2=3abc+m$ has a solution triple, then it has infinitely many solutions. We show that all positive solution triples are generated by a finite set of triples that we…

A triple (a,b,c) of positive integers is called a Markoff triple iff it satisfies the diophantine equation a2 + b2 + c2 = abc . Recasting the Markoff tree, whose vertices are Markoff triples, in the framework of intergral upper triangular…

We show that if $k\ge 2$ is an integer and $(F_n^{(k)})_{n\ge 0}$ is the sequence of $k$-generalized Fibonacci numbers, then there are only finitely many triples of positive integers $1<a<b<c$ such that $ab+1,~ac+1,~bc+1$ are all members of…

An analysis of the Markov tree is presented. Markov triplets, {x,R,z}, are the positive integer solutions to the Diophantine equation x2 + R2 + z2 = 3xRz. Inspired by patterns of the Fibonacci and Pell triplets in Region 1 and Region 2 of…

Let $P(m)$ denote the largest prime factor of an integer $m\geq 2$, and put $P(0)=P(1)=1$. For an integer $k\geq 2$, let $(F_{n}^{(k)})_{n\geq 2-k}$ be the $k-$generalized Fibonacci sequence which starts with $0,...,0,1$ ($k$ terms) and…

The $k$-Markov numbers, introduced by Gyoda and Matsushita, are those which appear in positive integral solutions to $x^2 + y^2 + z^2 + k(xy + xz + yz) = (3+3k)xyz$. When $k =0$, this recovers the ordinary Markov numbers. A long-standing…

Let us denote by $F_n$ the $n$-th Fibonacci number. In this paper we show that for a fixed integer $y$ there exists at most one integer exponent $a>0$ such that the Diophantine equation $F_n+F_m=y^a$ has a solution $(n,m,a)$ in positive…

We study an extension to the uniqueness conjecture for Markov numbers. For any three positive integers $m\geq a$ and $m\geq b$ satisfying $a^2+b^2+m^2=3abm$, this conjecture states that the triple $(a,m,b)$ is uniquely determined by the…

The Markov numbers are the positive integer solutions of the Diophantine equation $x^2 + y^2 + z^2 = 3xyz$. Already in 1880, Markov showed that all these solutions could be generated along a binary tree. So it became quite usual (and…

A generalization of the well-known Fibonacci sequence is the $k$-Fibonacci sequence with some fixed integer $k\ge 2$. The first $k$ terms of this sequence are $0,0, \ldots, 1$, and each term afterwards is the sum of the preceding $k$ terms.…

For an integer $k\geq 2$, let $\{F^{(k)}_{n}\}_{n\geqslant 2-k}$ be the $ k$--generalized Fibonacci sequence which starts with $0, \ldots, 0,1$ (a total of $k$ terms) and for which each term afterwards is the sum of the $k$ preceding terms.…

Markov polynomials are the Laurent-polynomial solutions of the generalised Markov equation $$X^2 + Y^2 + Z^2 = kXYZ, \quad k=\frac{x^2 + y^2 + z^2}{x y z}$$ which are the results of cluster mutations applied to the initial triple $(x, y,…

Let $\left\lbrace F_{k}\right\rbrace_{k\geq0}$ be the Fibonacci sequence defined by $F_{k}=F_{F-1}+F_{k-2}$ for all $ n\geq2$ with initials $F_{0}=0\; F_{1}=1$. Let $\left\lbrace J_{n}\right\rbrace_{n\geq0}$ be the Jacobsthal sequence…

Let $a, b\in \mathbb{N}$ be relatively prime. Previous work showed that exactly one of the two equations $ax + by = (a-1)(b-1)/2$ and $ax + by + 1 = (a-1)(b-1)/2$ has a nonnegative, integral solution; furthermore, the solution is unique.…

Let $F_n$ be the $n$th Fibonacci number. Let $m, n$ be positive integers. Define a sequence $(G(k,n,m))_{k\geq 1}$ by $G(1,n,m) = F^m_n$, and $G(k+1,n,m) = F_{nG(k,n,m)}$ for all $k\geq 1$. We show that $F_n^{k+m-1}\mid G(k,n,m)$ for all…

Let $m(n, k)$ be the maximum number of prime implicants that any $k$-CNF on n variables can have. We show that $3^{n/3} \le m(n,2) \le (1+o(1))3^{n/3}$.

In this paper, we find all the solutions of the Diophantine equation $P_\ell + P_m +P_n=2^a$, in nonnegative integer variables $(n,m,\ell, a)$ where $P_k$ is the $k$-th term of the Pell sequence $\{P_n\}_{n\ge 0}$ given by $P_0=0$, $P_1=1$…