Generating prime numbers of the form $\lfloor \sqrt{n^3} \rfloor $ How to prove the following claim ?
Let $b_n=b_{n-1}+\operatorname{lcm}(\lfloor \sqrt{n^3} \rfloor , b_{n-1})$ with $b_1=2$ 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 $\lfloor \sqrt{n^3} \rfloor $ is member of this sequence .

*At the first appearance of each prime of the form $\lfloor \sqrt{n^3} \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 A323359 .
Maxima implementation of generator :
/* Enter number n */
load(functs);
n:200;
b1:2;max:2;k:2;i:1;
while max<=floor(sqrt(n^3)) do
(if i=1 then(print(max),i:0),b2:b1+lcm(floor(sqrt(k^3)),b1),a:b2/b1-1,k:k+1,b1:b2,if max<a then (max:a,i:1));

Implementation of this generator in PARI/GP can be found here.
 A: Since $lcm(n,m)=(n\cdot m) / \gcd(n,m)$ for all $n,m\in\Bbb{N}$, we can rewrite the formulae for $b_n$ and $a_n$:
$$b_n=b_{n-1}\left(1+\frac{\lfloor\sqrt{n^3}\rfloor}{\gcd(b_{n-1},\lfloor\sqrt{n^3}\rfloor)}\right)$$
$$a_n=\frac{b_{n+1}}{b_n}-1 = \frac{\lfloor\sqrt{(n+1)^3}\rfloor}{\gcd(b_{n},\lfloor\sqrt{(n+1)^3}\rfloor)}$$

Let $p_n=p$ denote the $n$'th prime that can be written as $\lfloor\sqrt{k^3}\rfloor$ for some $k\in\Bbb{N}$. In fact, let's call the smallest possible value of $k$ $q_n$. Let $f(p_n)$ be the smallest positive integer $k$ such that $p_n\mid b_k$. Since $b_{n}\mid b_{n+1}$ for all $n\in\mathbb{N}$, we have:
$$p_n \mid \frac{b_{f(p_n)}}{b_{f(p_n)-1}}\Longleftrightarrow p\mid 1+\frac{\lfloor\sqrt{f(p)^3}\rfloor}{\gcd(b_{f(p)-1},\lfloor\sqrt{f(p)^3}\rfloor)}\implies p\le 1+\frac{\lfloor\sqrt{f(p)^3}\rfloor}{\gcd(b_{f(p)-1},\lfloor\sqrt{f(p)^3}\rfloor)}$$
So $\lfloor\sqrt{f(p)^3}\rfloor\ge p-1$,but if $\lfloor\sqrt{f(p)^3}\rfloor=p-1$, then $\gcd(b_{f(p)-1},\lfloor\sqrt{f(p)^3}\rfloor)\ge 2$, since $b_m$ is even for all $m\in\Bbb{N}$. Hence:
$$\lfloor\sqrt{f(p)^3}\rfloor\ge p\implies f(p)\ge \lceil\sqrt[3]{p^2}\rceil$$
So $p\nmid b_{q_n-1}$, which means $\gcd(b_{q_n-1},p_n)=1$ for all $n\in\Bbb{N}$ and thus:
$$a_{q_n-1}=\frac{\lfloor\sqrt{q_n^3}\rfloor}{\gcd(b_{q_n-1},\lfloor\sqrt{q_n^3}\rfloor)}=\frac{p_n}{\gcd(b_{q_n-1},p_n)}=p_n$$
This proves your second and third statement (because $a_{q_n-1}$ is also the first element in the sequence that is even divisible by $p_n$).
