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 $$\frac{p_n\#}{p_i};\ 1. For example, $$\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);\ 1n$$.

Based on this admittedly extremely small set, I conjecture that it might be the case $$p_n\#=C(p_k-1);\ C\in \{1,p_i\},\ 1n$$ 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?

• 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. – nickgard Jul 9 at 19:40
• @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. – Keith Backman Jul 10 at 0:36
• 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. – Ross Millikan Jul 10 at 0:48
• @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. – Keith Backman Jul 10 at 1:01
• No, you are correct, they are equivalent. – Ross Millikan Jul 10 at 1:04

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 $$1\leq n < 19$$
$$\begin{array}{|c|c|c|c|} n & p_n & \frac{p_{19}\#}{p_n}+1 & \text{smallest divisor of }\frac{p_{19}\#}{p_n}+1\\ \hline 1 & 2 & 3929160775540133527939546 & 2 \\ \hline 2 & 3 & 2619440517026755685293031 & 613 \\ \hline 3 & 5 & 1571664310216053411175819 & 5501 \\ \hline 4 & 7 & 1122617364440038150839871 & 21713 \\ \hline 5 & 11 & 714392868280024277807191 & 389 \\ \hline 6 & 13 & 604486273160020542759931 & 131 \\ \hline 7 & 17 & 462254208887074532698771 & 101 \\ \hline 8 & 19 & 413595871109487739783111 & 136483 \\ \hline 9 & 23 & 341666154394794219820831 & 26801 \\ \hline 10 & 29 & 270976605209664381237211 & 809 \\ \hline 11 & 31 & 253494243583234421157391 & 127 \\ \hline 12 & 37 & 212387068948115325834571 & 3449 \\ \hline 13 & 41 & 191666379294640659899491 & 3593 \\ \hline 14 & 43 & 182751663978610861764631 & 167 \\ \hline 15 & 47 & 167198330874048235231471 & 71 \\ \hline 16 & 53 & 148270217944910699167531 & 2866463 \\ \hline 17 & 59 & 133191890696275712811511 & 283 \\ \hline 18 & 61 & 128824943460332246817691 & 179 \\ \hline \end{array}$$

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.