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Maybe , this conept is known , maybe not : If we define $C_n$ to be the $n$-th Carmichael-number and $$F(n):=\prod_{j=1}^n C_j$$ could be called "Carmichael-factorial".

For large $n$ , $F(n)$ has many small factors , so the numbers $F(n)-2$ and $F(n)+2$ have a good chance to be prime ot at least to be squarefree.

In fact, the first impression by checking those numbers is that they are all squarefree , but this is not the case :

We have

  • $71999^2\mid F(93669)-2$
  • $82763^2\mid F(45629)-2$

There can be more examples , but I do not know any more.

Question : Is $F(n)+2$ always squarefree ?

Upto $n\le 246\ 683$ (The product of the Carmichael numbers upto $10^{16}$) , the primes upto $10^6$ are checked. For $n\le 2\ 000$ , the primes upto $22\cdot 10^6$ are checked.

Peter
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1 Answers1

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After searching all Carmichael numbers up to $2^{64}$ together with the primes $p < 1.1 \cdot 10^6$ (and the primes $p<10^7$ up to the $532500$-th Carmichael number), I found that $$\begin{array}{rl} 476423^2 \mid F(2135937)-2 &\hspace{1pt}& C_{2135937} = 3015594892139235121 \\ 62627^2 \mid F(2271719)-2 && C_{2271719} = 3541980836224875241 \\ 464687^2 \mid F(3252745)-2 && C_{3252745} = 9034227090413688241 \\ 113843^2 \mid F(3637866)-2 && C_{3637866} = 12097526583615598945 \\ 219749^2 \mid F(409182)+2 && C_{409182} = 38669635673182801 \\ \end{array}$$ Thus the answer to your question is, quite simply, no.

jorisperrenet
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