# Does $\operatorname*{\exists}\limits^{\infty} f(n) :$ prime?

Is it provable that there exist infinitely many primes of the form, $$f(n) = \left\{-1 + (p_{n+2} - 1)\prod_{i=2}^np_i : (p_n)_{n\in\mathbb{N}} = n^{\text{th}}\text{ prime number.}\right\}?$$ This is prime for $n\in\{1,2,3,4,6,7,8,12,13,15,21,48\}$. I have checked for $n\leqslant 54$.

The reason I want to know if there is inf many, is because I conjecture that for every counter-example which we can call $c_1, c_2, c_3,\ldots$ then $\Omega(c_k) \leqslant 4$ such that we denote by $\Omega(c_k)$ the number of prime factors of $c_k$ for which the function does not count the same prime factor twice.

For instance, $\Omega(64) = 1$ because $64 = 2^6 = 2\times 2\times 2\times\cdots \times 2$, and since $2$ is prime, we only count it once, so the function in this case is equal to $1$ (and not $6$).

Is there a way to prove/disprove my conjecture? If there are infinitely many primes $f(n)$ then perhaps I could apply that to proving my conjecture.

If we suppose for a moment that the output of your function is "random" (or perhaps that primes are "random"), then you might expect that the probability that $f(n)$ is prime is approximately $1/\log f(n) \approx 1/\sum_{m \leq n} \log p_m$.
As $p_m \approx m \log m$, we see that $$\sum_{m \leq n} \log p_m \approx \sum_{m \leq n} \log (m \log m) \approx \sum_{m \leq n} \log m + \log \log m \approx n \log n.$$ So we might expect that the number of prime values of $f(n)$ up to $X$ is approximately $$\int_1^X \frac{1}{t \log t} dt \approx \log\log X,$$ which tends to infinity (although very slowly). So under this heuristic argument, we might expect that this function could be prime infinitely often.
• I have to wait an hour before I can vote $(+1)$, but congratulations nonetheless! $$\color{green}{\checkmark}$$ Very nice answer :) – Feeds Feb 12 '18 at 22:59