# 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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