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This is a question related to my recent question Conjecture: “For every prime $k$ there will be at least one prime of the form $n!\pm k$” true?

Using PARI/GP I searched for the number of primes of the form $n!+k$ possible for each prime $k\le 2300$. I observed that for some prime $k$, all numbers less than $24$ were found except $2$.

Note that when $n \ge k$, $n! \pm k$ cannot be prime as $k$ will be a factor of $n! \pm k$. This means that there are a finite number of primes of the form $n! \pm k$ for each $k$.

User Peter and I, searched for some prime $k$ for which $n!+k$ has two primes upto $k\le 10^6$ and found none in the range. Here is the code we used in PARI/GP if anyone wants to extend the search:

forprime(k=2,10^7,s=0;for(n=1,k,if(s<3,if(ispseudoprime(n!+k)==1,s=s+1)));if(s==2,print(k)))

Conjecture:

For any prime $k$ there cannot be exactly two primes of the form $n!+k$.

Can prove/disprove this? Any ideas, heuristics, guesses are welcome.

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    $\begingroup$ Can someone explain the downvotes? $\endgroup$ – Mathphile Jul 15 at 15:02
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    $\begingroup$ Although I don't see a quick proof, I think it extremely unlikely. For $n\ge 3,\ 6\mid n!$. Therefore, $k=2$, the only prime $n!+k=3$ and $k=3$ the only prime is $n!+k=5$. For $k>3,\ k=6m\pm 1$. Thus $n!+k=6N+6m\pm 1=6M\pm 1$. Hard to believe that for $M=\frac{n!}{6}+m$, there is some $m$ for which only two instance of $n$ render $6M\pm1$ prime. $\endgroup$ – Keith Backman Jul 15 at 17:16
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    $\begingroup$ @KeithBackman hence why it's rather interesting that the PARI/GP code may suggest otherwise ;) $\endgroup$ – Mr Pie Jul 15 at 18:03
  • $\begingroup$ Any pair of n,m such that adding k to their factorials is prime, will knock out a value $p=m+n$ such that $p! +k^2{{p}\choose{n}}$ and $p!+k^2{{p}\choose{m}}$ are both 3-almost primes. Taking $k=1$ and choose functions that are primes, we would get to knock out primes. I know it's too stupid. $\endgroup$ – Roddy MacPhee Jul 24 at 15:22

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