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For a prime $q_k \neq 2$ we can study the corresponding set $\{q_1!+q_k,...,q_{k-1}!+q_k\}$, where $q_1,...q_{k-1}$ are all primes strictly less than the prime $q_k$.

Peter and Mathphile computed that for $q_k=1193$ we have that all the numbers from the corresponding set are composites

I think that $1193$ is rather large as an example of the first prime that generates all composites in the corresponding set and I do not see some reasons, armed with strong principles, of why there shouldn´t be some other primes that generate corresponding sets in which not a single number is prime.

Suppose that $\{r_1,...,r_m,...\}$ is the set of all the prime numbers for which the corresponding sets consist of only the composite numbers.

Are there any reasons and principles for justifying the assertion that $\{r_1,...,r_m,...\}$ is not a finite set?

Edit: This question arrived as the result of thinking about Mathphile´s conjecture.

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Here's a heuristic for the likelihood that $q_k$ generates only composite numbers in this way.

For given $j<k$, the number $q_j!+q_k$ has size about $q_k \sim k\log k$ when $j<\log k$ or so, and size about $q_j!\approx e^{q_j\log q_j} \approx e^{j(\log j)^2}$ when $j$ is larger. The default probabilities that numbers of this size are prime are about $1/k\log k$ and $1/j(\log j)^2$, respectively.

We know a little more about these particular numbers. We know that $q_j!+q_k$ is not divisible by any of the first $j$ primes, which leads to heuristically modifying their probabilities of being prime upwards by a factor of $\prod_{i=1}^j (1-1/p_i)^{-1} \asymp \log j$. (We also know that $q_j!+q_k$ is the sum of two non-multiples of $p$ for every prime $p$ larger than $q_j$, except for $q_k$ itself, which leads to heuristically modifying their probabilities of being prime downwards by a factor of $\prod_{i=j+1}^{\infty} (1-1/p_j)^{-1} (1-1/(p_j-1)) \asymp 1$.) So the working heuristic probabilities that these numbers are prime change into about $(\log j)/k\log k$ for small $j$ and $1/j\log j$ for larger $j$.

The heuristic probability, then, that all these numbers are composite should be about \begin{multline*} \prod_{j=1}^{\log k} \bigg( 1-\frac{\log j}{k\log k} \bigg) \prod_{j=\log k}^{k} \bigg( 1-\frac1{j\log j} \bigg) \approx 1\bigg/\exp\bigg( \sum_{j=1}^{\log k} \frac{\log j}{k\log k} + \sum_{j=\log k}^{k} \frac1{j\log j} \bigg) \\ \approx 1\bigg/\exp\bigg( \frac{\log\log k}{k} + \log\log k - \log\log\log k \bigg) \approx (\log k)^{-C} \end{multline*} for some constant $C$ (because we were ignoring constants along the way).

Since $\sum_{k=1}^\infty (\log k)^{-C}$ diverges, this heuristic leads to the prediction that there should actually be plenty of primes $q_k$ for which all the numbers $q_1!+q_k,\dots,q_{k-1}!+q_k$ are composite.

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