Searcharxiv⌕ Search
Home/Archive/arXiv

arXiv · 2108.13834

A step towards proving de Polignac's Conjecture

John K Sellers
View original source ↗Read source document ↗Export metadata ↓

Abstract

Consider the set of all natural numbers that are co-prime to primes less than or equal to a given prime. Then given a consecutive pair of numbers in that set with an arbitrary even gap, we prove there exists an unbounded number of actual prime pairs with that same gap. This conditional proof of de Polignac's conjecture constitutes a proof for a range of known gaps, but the full conjecture requires additional proof that such number pairs exist for all even gaps.

Explore related subjects

math.GM

Keep this discovery

BibTeXRIS

John K Sellers. 2021-08-27. A step towards proving de Polignac's Conjecture. https://arxiv.org/abs/2108.13834

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

KEEP EXPLORING

Related papers

arXiv2026

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↗
arXiv2026

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↗