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.

Thanks in advance.


These numbers are incredibly sparse --- far too sparse for current techniques to show that there are infinitely many primes of this form. We are currently very bad at determining whether sparse sets (i.e. sets with relatively low density in the integers) contain infinitely many primes.

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.

But actually proving it is very unlikely.

  • $\begingroup$ I have to wait an hour before I can vote $(+1)$, but congratulations nonetheless! $$\color{green}{\checkmark}$$ Very nice answer :) $\endgroup$ – Feeds Feb 12 '18 at 22:59

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