Two claims about compositorials

Question

This question is inspired by Fortune's conjecture .

Can you provide proofs or counterexamples for the following two claims :

First claim

If $q$ is the smallest prime greater than $\displaystyle\prod_{i=1}^n C_i+1$ , where $\displaystyle\prod_{i=1}^n C_i$ is the product of the first $n$ composite numbers , then $q-\displaystyle\prod_{i=1}^n C_i$ is prime .

The first few such differences are :

3,5,5,5,11,7,23,11,29,17,31...

Try it Yourself !

Second claim

If $q$ is the greatest prime less than $\displaystyle\prod_{i=1}^n C_i-1$ , where $\displaystyle\prod_{i=1}^n C_i$ is the product of the first $n$ composite numbers , then $\displaystyle\prod_{i=1}^n C_i-q$ is prime .

The first few such differences are :

2,5,11,5,23,17,13,11,13,23,53...

Try it Yourself !

I have tested both claims up to $n=660$ and there were no counterexamples .

Answer 1

here's a list of things we know ( it's a hint or help not an strictly an answer ,partly because I don't know ):

1. q-product for n>1 will have the same modular remainder mod 6 as q.

2. When we take the product up to 2n, we get at least all the primes less than or equal to n up to exponents of 1 or more.

3. all primes less than ${2\over 3}n$ appear at least twice in the product in 2.

4. Generally,all primes less than ${2\over a}n$ will appear at least a-1 times in the product in 2.

   This all means the difference can't divide by any product within the product because then q would not be prime. n# is within this product. so the difference ( again 2n case) has already been trial factored up to n.

Answer 2

For the "first claim" we have...

(perhaps the notation is not correct)

Given

$$1 \le k \le n \textrm{ and } 1 < \ell < q,$$

then

$$q > \prod_{\imath=1}^n C_\imath \Rightarrow (q{ \not\mid}C_k) \wedge (C_k{\not\mid}q) \Rightarrow \ell{\not\mid}\left(q - \prod_{\imath=1}^n C_\imath\right),$$

whence

$$q - \prod_{\imath=1}^n C_\imath$$

is a prime.

I am working on the "second claim" :)

(corrected the typo... thanks to Peđa Terzić)