Is $N^n(p_n\#)-p_n\#$ always prime for all $n\in\mathbb{N}_{>1}$?

91 Views Asked by At

I have made a conjecture, and it seems like this can generate prime numbers very very well and in their oder too.

Conjecture:

Consider the following primorial for $n\in\mathbb{N}$: $$p_n\#_c=\prod_{i=1}^n(p_i+c)\tag{$p_n\#_0=p_n\#$}$$ such that we denote by $p_n$ the $n^{\text{th}}$ prime number. Consider a function $$N:\mathbb{Z}^+\rightarrowtail \mathbb{Z}^+:x\mapsto \min\big\{p_n > x\big\}$$ then, by denoting $N^m(x) = \underbrace{N(N(\ldots N}_{m \text{ times}}(x)$, I conjecture the following:

$$N^n\left(p_n\#\right)-p_n\#\text{ is always prime for $n > 1$}.$$

I did a little bit of research, and this is very similar to Fortune's Conjecture, but slightly extended. In fact, it appears that if $m>1$ is natural, then $N^m(p_n\#)-p_n\#$ is prime, but this is much too broad.

Unfortunately, though, I do not know how to even attempt at proving any of these claims. Also, it ap-pears (to me at least) that for a fixed $m > 1$, running through $n=1,2,3,\ldots$ generates the primes in their order, though it sometimes skips a prime number or equals a prime number squared for some natural $n$; for example, try $m=2$.

You can use what I used to run the tests, namely, the Alpertron $-$ Integer Factorization Calculator. The same function $N$ exists. (According to the calculator, the function $N(x)$ calculates the next prob-able prime after $x$, so it is not entirely the same as the way I have defined $N$; but, it is very very accur-ate for the first few hundred digits of $x$.)

But could someone please help me? This looks very close to finding a pattern in the sequence of prime numbers.


Thank you in advance.