Generating prime numbers of the form $\lfloor \sqrt{3} \cdot n \rfloor $

by Peđa Terzić   Last Updated July 12, 2019 07:20 AM

How to prove the following claims ?

Let $b_n=b_{n-1}+\operatorname{lcm}(\lfloor \sqrt{3} \cdot n \rfloor , b_{n-1})$ with $b_1=3$ and $n>1$ . Let $a_n=b_{n+1}/b_n-1$ .

  1. Every term of this sequence $a_i$ is either prime or $1$ .

  2. Every odd prime of the form $\left\lfloor \sqrt{3}\cdot n \right\rfloor$ is a term of this sequence.

  3. At the first appearance of each prime of the form $\left\lfloor \sqrt{3}\cdot n \right\rfloor$, it is the next prime after the largest prime that has already appeared.

A few first terms of this sequence can be found at A323388 .

Implementation of this generator in PARI/GP can be found here.

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