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I would like to ask you a question regarding the paper "Prime-representing functions" by K. Matomäki.

In particular the author states that, assuming Riemann hypothesis, if $c_i \geq\frac{1+\sqrt5}{2}=\phi\mbox{ }\forall i\in\mathbf{N}$, then it exists a constant $\alpha$ such that the sequence: $$ \lfloor\alpha^{C_n}\rfloor $$ is a prime for all natural numbers $n$. (Here $C_n = \prod_{i=1}^{n}c_i$).

Without assuming Riemann hypothesis it is sufficient that $c_i \geq 2\mbox{ }\forall i\in\mathbf{N}$

An interesting corollary of this fact (which I come up recently) is that it exists a constant $\beta$ such that:

$$ \lfloor \beta^{F_{2n+1}}\rfloor \mbox{ is a prime $\forall n\geq1$}$$ where $F_n$ is the $n$th Fibonacci number.

Consider the sequence defined as follows: $c_1 = F_3 = 2$ and $c_n = \frac{F_{2n}}{F_{2n-1}} + 1$ $\forall n>1$.

First of all note that $c_i \geq 2 \forall i\in\mathbf{N}$ (in fact the sequence is monotone increasing and it converges to $1+\phi \approx 2.618$). Then note how:

$$ C_2 = c_1c_2 = F_3(\frac{F_4}{F_3}+1) = F_4+F_3 = F_5$$ $$ C_3 = c_1c_2c_3 = F_5(\frac{F_6}{F_5}+1) = F_6+F_5 = F_7$$ $$ \vdots $$ $$ C_n = c_1\cdots c_n = F_{2n+1}$$ Finally we can conclude that:

$$ \lfloor \beta^{C_n}\rfloor=\lfloor \beta^{F_{2n+1}}\rfloor \mbox{ is a prime $\forall n\geq1$}$$

I would like to know if I had made some mistakes during the process. Thank you in advance!

CONJECTURE: Playing around with this.. I really think that there must be a constant for which:$$ \lfloor \beta^{F_{n}}\rfloor \mbox{ is a prime $\forall n\in\mathbf{N}$} $$ and that it is around: 2.361452025.

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  • $\begingroup$ This would be a nice prime generator, unfortunately we need extremely many digits to get reasonably many primes. Amazing that the bound of the constant depends on the Riemann hypothesis. Does this mean that , if the Riemann hypothesis is false , the constant must be at least $2$ ? $\endgroup$
    – Peter
    Aug 13, 2022 at 8:57
  • $\begingroup$ Without knowing the concrete proof , I guess that the existence of Mills constant can be shown with the prime number theorem and something similar is probably possible for your conjecture, maybe someone can work it out. $\endgroup$
    – Peter
    Aug 13, 2022 at 8:59
  • $\begingroup$ Hi Peter thank you for your reply. As I said, assuming Riemann hypothesis it is possible to find infinitely many prime generator constants with $c_i\geq\phi$ but, if the Riemann hypothesis is false, then it has been proven that $c_i\geq 2$. But maybe in the future we can prove that $c_i$ must be greater than the golden ratio WITHOUT assuming R.H., and this would be an even nicer result of course. Anyway as you said, computing this constant is very difficult because of the long computations, but anyway it's fun to have a prime generating constant related to FIbonacci numbers! $\endgroup$ Aug 13, 2022 at 9:43
  • $\begingroup$ I agree and upvoted this post. $\endgroup$
    – Peter
    Aug 13, 2022 at 9:45

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