SearcharxivSearch

arXiv · 2507.00317

Fixed Points of the Josephus Function via Fractional Base Expansions

Abstract

In this paper, we investigate properties of the fixed point sequence of the Josephus function $J_3$. First, we establish a connection between this sequence and the Chinese Remainder Theorem. Next, we identify a clear numerical pattern for the digits of two consecutive fixed points when they are written in a non-standard fractional number system in base $3/2$. This result enables us to derive a recursive procedure for determining the digits of their base $3/2$ expansions.

Explore related subjects

Keep this discovery

Explore connections, maps & timelines

BibTeXRIS

Yunier Bello-Cruz, Roy Quintero-Contreras. 2025-06-30. Fixed Points of the Josephus Function via Fractional Base Expansions. https://arxiv.org/abs/2507.00317

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

Average Chord Lengths in a Triangle

Let $P$ be a point inside a triangle $T$. We consider the average length of the chords of $T$ through $P$, where the direction of the chord is chosen uniformly. An elementary formula is obtained in terms of the distances from $P$ to the sides and vertices of the triangle. Several classical triangle centers give especially simple specializations. For example, if $I$ is the incenter, then \[ M_T(I)=\frac{2r}{\pi} \log\left(\cot\frac A4\cot\frac B4\cot\frac C4\right). \] Our main result is the sharp inequality \[ M_T(P)\le \frac{p}{\pi\sqrt3}\log(2+\sqrt3), \] valid simultaneously for every triangle of perimeter $p$ and every interior point $P$. Thus, among all such pairs $(T,P)$, the largest possible average chord length occurs only when $T$ is equilateral and $P$ is its center. The proof is an elementary symmetrization argument. We close with brief remarks relating the problem to the radial center of a convex body, the electrostatic potential center of a triangle, and dual quermassintegrals.

math.GM

A Proof of Liu's Conjecture on the Fundamental Triangle Inequality

Let $a,b,c$ be the side lengths of a triangle, and let $R$ and $r$ denote its circumradius and inradius, respectively. We prove a conjecture of Liu stating that \[\sum_{\mathrm{cyc}} \left(\frac{a(b+c-a)}{bc}\right)^k \geq 2+\left(\frac{2r}{R}\right)^k,~~k>1, \] with the reverse inequality for $0<k<1$. The proof reduces the problem to three positive variables with fixed sum and product. We also determine the equality cases.

math.GM