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A primorial, denoted $p_n\#$, is the product of the first $n$ prime numbers ($p_1=2,\ p_2=3$ etc.). The magnitude of primorials grows rapidly beyond the range of convenient arithmetic manipulation. The number $(p_n\#+1)$ is not divisible by any of the first $n$ primes, and so is frequently a prime number itself. For $n=1,2,3,4,5,11$, $(p_n\#+1) \in \mathbb P$.

I noticed (for primorials accessible to calculation) that when $(p_n\#+1) \not \in \mathbb P$, that a 'near primorial' number plus $1$ could be identified that was a prime. By near primorial number, I mean the product of all but one of the first $n$ primes, or $\frac{p_n\#}{p_i};\ 1<i<n$. For example, $\frac{p_8\#}{3}+1,\ \frac{p_{10}\#}{3}+1,\ \frac{p_6\#}{5}+1,\ \frac{p_7\#}{5}+1,\ \frac{p_{12}\#}{7}+1,\ \frac{p_{13}\#}{11}+1,\ \frac{p_{9}\#}{13}+1$ are all primes. Examples of this kind can be rewritten in the form $p_n\#=p_i(p_k-1);\ 1<i<n,\ k>n$.

Based on this admittedly extremely small set, I conjecture that it might be the case $$p_n\#=C(p_k-1);\ C\in \{1,p_i\},\ 1<i<n,\ k>n$$ The signal feature of $C$ is that it is not composite. A single counterexample arrived at by computation would disprove the conjecture, but for $p_{14}\#$ and greater, the numbers are beyond my ability to conveniently calculate.

My questions are: Has this conjecture been previously considered and settled? If not, is there an analytic approach to prove or disprove the conjecture?

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  • $\begingroup$ I think you have some mistakes. Possibly they are just typos. For example: $p_9=23$. $(p_9\#/5)+1=44618575$, which is clearly not prime. I think your last three example primes are, in fact, composite. Their smallest prime factors are $5; 2861; 103$ respectively. $\endgroup$ – nickgard Jul 9 at 19:40
  • $\begingroup$ @nickgard Thanks for the check. Not a simple typo, but a slipped index in the calculations. I've rechecked and edited the post. I think the examples are all correct now, but in any event the conjecture wasn't invalidated up to this point. $\endgroup$ – Keith Backman Jul 10 at 0:36
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    $\begingroup$ Your statement of the conjecture is not correct. As I read the text, the conjecture would be that if $p_n\#+1$ is not prime, there is some $p_i$ with $i \lt n$ such that $\frac {p_n\#}{p_i}+1$ is prime. $\endgroup$ – Ross Millikan Jul 10 at 0:48
  • $\begingroup$ @Ross Millikan Your statement of the conjecture is correct, but I fail to see how it differs from my statement. If $\frac{p_n\#}{C}+1=p_k$, then either $C=1$ and $p_n\# = 1\cdot (p_k-1)$, or $C=p_i$ and $p_n\# = p_i\cdot (p_k-1)$. Unless I messed up notation somehow (always a possibility), I think that's what I said. $\endgroup$ – Keith Backman Jul 10 at 1:01
  • $\begingroup$ No, you are correct, they are equivalent. $\endgroup$ – Ross Millikan Jul 10 at 1:04
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A few lines of Mathematica shows that $p_{19}\#$ is the first counterexample. $$p_{19}\# = \bigg(\prod_{i=1}^{19}p_i\bigg)+1=7858321551080267055879091=54730729297\cdot 143581524529603,$$ so it is composite. The following table shows that $\frac{p_{19}\#}{p_n}+1$ is composite for all $n$ satisfying $1\leq n < 19$
$$\begin{array}{|c|c|c|c|} n & p_n & \frac{p_{19}\#}{p_n}+1 & \text{smallest divisor of }\frac{p_{19}\#}{p_n}+1\\ \hline 1 & 2 & 3929160775540133527939546 & 2 \\ \hline 2 & 3 & 2619440517026755685293031 & 613 \\ \hline 3 & 5 & 1571664310216053411175819 & 5501 \\ \hline 4 & 7 & 1122617364440038150839871 & 21713 \\ \hline 5 & 11 & 714392868280024277807191 & 389 \\ \hline 6 & 13 & 604486273160020542759931 & 131 \\ \hline 7 & 17 & 462254208887074532698771 & 101 \\ \hline 8 & 19 & 413595871109487739783111 & 136483 \\ \hline 9 & 23 & 341666154394794219820831 & 26801 \\ \hline 10 & 29 & 270976605209664381237211 & 809 \\ \hline 11 & 31 & 253494243583234421157391 & 127 \\ \hline 12 & 37 & 212387068948115325834571 & 3449 \\ \hline 13 & 41 & 191666379294640659899491 & 3593 \\ \hline 14 & 43 & 182751663978610861764631 & 167 \\ \hline 15 & 47 & 167198330874048235231471 & 71 \\ \hline 16 & 53 & 148270217944910699167531 & 2866463 \\ \hline 17 & 59 & 133191890696275712811511 & 283 \\ \hline 18 & 61 & 128824943460332246817691 & 179 \\ \hline \end{array}$$

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What says the random model :

Let $j\le n$ and $$f(j,n) = 1+\prod_{i=1, i \ne j}^n p_i$$

By Mertens theorem $\log f(j,n) \approx \sum_{i=1}^n \log p_i \approx n$

Assuming independence of the congruences $\bmod$ different primes $$Pr(f(j,n) \text{ is prime}] \approx \frac{\prod_{i \ne j} (1-p_i)^{-1}}{\ln N} \ge C\exp(\sum_{i=1}^{n-1}\frac{1}{p_i} - \ln \ln N)\\ \approx C \exp( \ln \ln (n-1) - \ln n) \approx C\frac{\ln n}{n}$$

Taking $j $ uniformly in $1\ldots n$, assuming the random variables "$f(j,n)$ is prime" are independent,

the probability that none of the $f(j,n)$ is prime is $$\approx \prod_{j=1}^n (1-C\frac{\ln n}{n})= (1-C\frac{\ln n}{n})^n = \exp(n\log (1-C\frac{\ln n}{n})) \approx \exp(-C \ln n)) = n^{-C}$$

If you redo it replacing $j$ by a subset $J \subset 1 \ldots n$ with $4$ elements and $f(j,n)$ by $f(J,n) = 1+\prod_{i=1, i \not \in J} p_i$ you'll get $C > 1$ so that the probability that for some $n \ge N$, none of the $f(J,n)$ is prime is $\le \sum_{n=N}^\infty n^{-C}$ which $\to 0$ as $N \to \infty$,

ie. it is almost surely true that for every $n$ large enough $p_n\# = a_n (p_{k_n}-1)$ with $a_n$ a product of at most $4$ primes.

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