Let $f: \mathbb{\mathbb{Z}^+} \to \mathbb{Z^+}$ be a function and suppose


$(\star)$ For all integers $x \geq 3$, if $f(x)$ is prime, then $x$ is prime.


A trivial example of such a function is the identity $f(x) = x$. However, a possible non-trivial example which I have come across is
\begin{align\*}
f(x) = \left\lfloor \frac{\cosh(x\ln(2 + \sqrt{3}))}{2}\right\rfloor.
\end{align\*}


This function seems to satisfies $(\star)$ (see [OEIS A198196](https://oeis.org/search?q=3%2C%205%2C%207%2C%2011%2C%2013%2C%2017%2C%2019%2C%2079%2C%20151%2C%20199%2C%20233%2C%20251%2C%20317%2C%20863%2C%20971&language=english&go=Search)). I have two questions:


1. How might one go about proving $f$ satisfies $(\star)$? I'm not sure where to begin with this, and it seems like a difficult task.
2. Have functions with property $(\star)$ been studied before?


Thanks for any information you can give me.




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**Edits:** I've composed the floor function with $f$, and added the condition to $(\star)$ that $x$ must be an integer.



Question about functions $f: \mathbb{Z}^+ \to \mathbb{Z}^+$ such that $x$ is prime whenever $f(x)$ is prime
As observed in comments, we have $f(n) = \lfloor g(n) \rfloor$ where $g(n) = \frac{\alpha^n + \alpha^{-n}}{4}$ and $\alpha = 2 + \sqrt{3}$. From the recurrence $g(n+1) = 4 g(n) - g(n-1)$ we see that $g(n)$ is a half-integer when $n$ is even and an integer when $n$ is odd. In fact we see from induction that for even $n$ we have $g(n) = \frac{1}{2} \hbox{ mod } 3$ and for odd $n$ we have $g(n) = 1 \hbox{ mod } 3$. Hence for even $n$, $f(n)$ is divisible by $3$ and thus not prime except when $n=2$. For odd $n$, we have $f(n)=g(n)$, and for odd $n,m$ we then have
$$ f(nm) = f(n) (\alpha^{n(m-1)} + \alpha^{n(m-3)} + \dots + \alpha^{n(1-m)})$$
thanks to the formula $a^m+b^m = (a+b)(a^{m-1} + a^{m-2} b + \dots + b^{m-1})$. From the Galois group action interchanging $\alpha$ and $\alpha^{-1}$ we see that $\alpha^{n(m-1)} + \alpha^{n(m-3)} + \dots + \alpha^{n(1-m)}$ is an integer, and for $n,m \geq 3$ this integer is larger than $1$. Thus $f(nm)$ is composite when $n,m \geq 3$ are odd, so the only remaining possible values of $n$ for which $f(n)$ can be prime are the primes.


In the language of [divisibility sequences](https://en.wikipedia.org/wiki/Divisibility_sequence), $f(n)$ is a divisibility sequence on the odd natural numbers, though not on the even ones. In retrospect this is not so surprising given that $f$ is so similar to the Fibonacci sequence $F\_n = \frac{\phi^n - \phi^{-n}}{\sqrt{5}}$, which is well known to be a divisibility sequence.


