2014/06/03 by Germán Paz, Paz, Germán
Mathematics · #00-XX #00A05 #11-XX #11A41 #Algebraic Geometry and Number Theory #Analytic Number Theory Research #FOS: Mathematics #History and Theory of Mathematics #Mathematics and Applications #Number Theory (math.NT) #math.NT #msc:00-XX #msc:00A05 #msc:11-XX #msc:11A41
paper · pdf · doi:10.48550/arxiv.1406.4801
16 pages, 3 figures (version 2 includes them also as ancillary files, no changes to version 1 have been made), Mathematica code; keywords: Andrica's conjecture, Brocard's conjecture, Legendre's conjecture, Oppermann's conjecture, prime numbers, triangular numbers. arXiv admin note: text overlap with arXiv:1310.1323
openalex publication_date 2014/06/03 · arxiv created 2014/06/19 · arxiv updated 2014/06/20 · openalex created_date 2025/10/10 · openalex updated_date 2026/07/28
Let n∈ℤ+. In [8] we ask the question whether any sequence of n consecutive integers greater than n2 and smaller than (n+1)2 contains at least one prime number, and we show that this is actually the case for every n≤ 1,193,806,023. In addition, we prove that a positive answer to the previous question for all n would imply Legendre's, Brocard's, Andrica's, and Oppermann's conjectures, as well as the assumption that for every n there is always a prime number in the interval [n,n+2\lfloor√(n)\rfloor-1]. Let π[n+g(n),n+f(n)+g(n)] denote the amount of prime numbers in the interval [n+g(n),n+f(n)+g(n)]. Here we show that the conjecture described in [8] is equivalent to the statement that π[n+g(n),n+f(n)+g(n)]≥ 1, ∀ n∈ℤ+, where f(n)=((n-\lfloor√(n)\rfloor2-\lfloor√(n)\rfloor-β)/(|n-\lfloor√(n)\rfloor2-\lfloor√(n)\rfloor-β|))(1-\lfloor√(n)\rfloor), g(n)=\lfloor1-√(n)+\lfloor√(n)\rfloor\rfloor, and β is any real number such that 1<β<2. We also prove that the conjecture in question is equivalent to the statement that π[Sn,Sn+\lfloor√(Sn)\rfloor-1]≥ 1, ∀ n∈ℤ+, where Sn=n+(1)/(2)\lfloor(√(8n+1)-1)/(2)\rfloor2-(1)/(2)\lfloor(√(8n+1)-1)/(2)\rfloor+1. We use this last result in order to create plots of h(n)=π[Sn,Sn+\lfloor√(Sn)\rfloor-1] for many values of n.