# Studying for primes $q_k \neq 2$ the sets $\{q_1!+q_k,…,q_{k-1}!+q_k\}$

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.

Here's a heuristic for the likelihood that $$q_k$$ generates only composite numbers in this way.
For given $$j, the number $$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.