Generating prime numbers of the form $\lfloor \sqrt{3} \cdot n \rfloor $ 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 $a_n=b_{n+1}/b_n-1$ .


*

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

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

*At the first appearance of each prime of the form $\left\lfloor \sqrt{3}\cdot n \right\rfloor$ greater than $5$, it is the next prime of the given form 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.

We prove that the second and the third claims are true.
The second claim is true.
If $a=d\alpha$, $b=d\beta$ and $(a,b)=d$, we have $\mathrm{lcm}(a,b) / b = \alpha = a / d.$
We may rewrite the sequence $a_n$ using above. 
$$
a_1=1,$$
$$
a_n=\frac{\lfloor(n+1)\sqrt 3\rfloor}{\left(\lfloor(n+1)\sqrt 3\rfloor, (a_{n-1}+1)\cdots (a_1+1) 3 \right)}, \ \ n\geq 2.
$$
Thus, if $n\geq 2$ and $\lfloor(n+1)\sqrt 3\rfloor=p>3$ is prime, then $p$ cannot divide $a_i+1$ for all $1\leq i\leq n-1$.
It is clear that $p$ cannot divide $a_i+1$ for $1\leq i\leq n-2$. Otherwise, $p | a_i+1 \leq \lfloor (n-1)\sqrt 3 \rfloor +1<\lfloor(n+1)\sqrt 3\rfloor=p$ is a contradiction. 
To see that $p$ cannot divide $a_{n-1}+1$, assume otherwise. Then $p|a_{n-1}+1 \leq \lfloor n\sqrt 3\rfloor +1\leq \lfloor(n+1)\sqrt 3\rfloor=p$. This gives the equality 
$$
p=a_{n-1}+1=\lfloor n\sqrt 3\rfloor +1=\lfloor(n+1)\sqrt 3\rfloor.  
$$
Then $\lfloor n\sqrt 3 \rfloor = p-1$ is an even number. Moreover, $\lfloor \sqrt 3\rfloor = 1$, $\lfloor 2\sqrt 3\rfloor=5$ gives $n\geq 3$. But, $2=a_1+1$ gives
$$
a_{n-1}=\frac{\lfloor n\sqrt 3 \rfloor}{\left(\lfloor n\sqrt 3\rfloor, (a_{n-2}+1)\cdots (a_1+1) 3 \right)}\leq \frac{p-1}2 <p-1.
$$
This is also a contradiction. 
Therefore, we have $a_n=p$ in such case. 
The third claim is true.
Recall that
$$
a_1=1,$$
$$
a_n=\frac{\lfloor(n+1)\sqrt 3\rfloor}{\left(\lfloor(n+1)\sqrt 3\rfloor, (a_{n-1}+1)\cdots (a_1+1) 3 \right)}, \ \ n\geq 2.
$$
Suppose that $n\geq 2$ and $\lfloor(n+1)\sqrt 3\rfloor = p>3$ is prime. Then we have for any $1\leq i\leq n-1$, 
$$
a_i\leq \lfloor (i+1)\sqrt 3 \rfloor \leq \lfloor n\sqrt 3\rfloor < \lfloor (n+1)\sqrt 3 \rfloor =p, 
$$
Therefore, $p$ is the next prime of the given form after the largest prime that has already appeared.
