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

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.

• Just for the record, your function is $N(x)=p_{\pi(x)+1}$. – rtybase May 21 '18 at 20:39
• Upto which $n$ did you check the primality ? – Peter May 21 '18 at 21:05
• Further, from $$\pi(p_n)=n \tag{1}$$ and $$N(x)=p_{\pi(x)+1} \tag{2}$$ we have $$N(N(x))=p_{\pi(N(x))+1}=p_{\pi(p_{\pi(x)+1})+1}=p_{\pi(x)+1+1}=p_{\pi(x)+2}$$ and by induction $$N^{\circ n}(x)=p_{\pi(x)+n} \tag{3}$$ – rtybase May 21 '18 at 21:10
• @rtybase So, the above expression is trivially prime for all $n$ ? – Peter May 21 '18 at 21:15
• Yep ... now, if we can prove that all the primes $\leq \sqrt{N^{\circ n}\left(p_n\#\right)-p_n\#}$ are exactly those $\{p_1,p_2,...,p_n\}$ then we're done. – rtybase May 21 '18 at 21:25