Branches of Markoff $m$-triples with two $k$-Fibonacci components

Abstract

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 $m$-triples whose last two entries are $k$-Fibonacci numbers and that are not roots of any Markoff trees. We then prove that every such infinite path is contained in a branch, starting at a triple of the form \[ \left(\frac{F_k(4r)}{3F_k(2r)},\,F_k(\ell+2r),\,F_k(\ell+4r)\right), \] where $r$ is an odd integer, $\ell\in\{1,2,\ldots, 2r\}$ and $3\nmid k$. These branches are distributed among exactly $2r$ distinct trees.

Figures (5)

• Figure 1: Beginning of the Markoff $52$-tree, with minimal triple $(4,6,72)$. The bold path shows the principal $(2,4)$-Fibonacci branch formed by the triples $\bigl(6,F_4(n),F_4(n+2)\bigr)$, where $n$ is even.
• Figure 2: Beginning of the Markoff $20$-tree with minimal triple $(1,1,6)$. The bold path shows the principal $(2,4)$-Fibonacci branch formed by the triples $\bigl(6,F_4(n),F_4(n+2)\bigr)$, where $n$ is odd.
• Figure 3: Beginning of the Markoff $100$-tree with minimal triple $(1,6,21)$. The bold path shows the principal $(2,1)$-Fibonacci branch formed by the triples $\bigl(6,F_1(n),F_1(n+6)\bigr)$, where $n$ is even.
• Figure 4: Beginning of the Markoff $100$-tree with minimal triple $(3,6,55)$. The bold path shows the principal $(2,1)$-Fibonacci branch formed by the triples $\bigl(6,F_1(n),F_1(n+6)\bigr)$, where $n$ is even.
• Figure 5: Beginning of the Markoff $100$-tree with minimal triple $(6,8,144)$. The bold path shows the principal $(2,1)$-Fibonacci branch formed by the triples $\bigl(6,F_1(n),F_1(n+6)\bigr)$, where $n$ is even.

Theorems & Definitions (63)

• Theorem 1.1
• Theorem 1.2
• Theorem 1.3
• Remark 2.1
• Lemma 2.2
• proof
• Proposition 2.3
• proof
• Corollary 2.4
• proof