Searcharxiv⌕ Search
Home/Archive/arXiv

arXiv · 1810.02188

6m Theorem for Prime numbers

Gandarawatta R. W. M. P. I. S. B.·Perera S. P. C.·Rathnayake R. M. L. S.
View original source ↗Read source document ↗Export metadata ↓

Abstract

We show that for any $P= 6^{m+1}.N -1 $ is a prime number for any $1 < N \le 13$ , $N \ne 8$ and $N \ne i^{m+1}Mod(6i+1) $ where $ i \in Z^+ $ and $ m \in $ $odd$ $Z^+ $ for $1 < N \le 13$ and $N \ne 8$ and also we further discussed that $P= 6^{m+1}.N -1 $ is a prime number for $ N >13 $ if and only if , $N \ne i^{m+1}Mod(6i+1) +(6i +1)a $ $ ; i,a \le Z^+ $

Explore related subjects

math.GM

Keep this discovery

BibTeXRIS

Gandarawatta R. W. M. P. I. S. B., Perera S. P. C., Rathnayake R. M. L. S.. 2018-09-19. 6m Theorem for Prime numbers. https://arxiv.org/abs/1810.02188

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↗