A Carmichael number $N$ has the property that for every $a$ coprime to $N$, the equation $a^{N-1}\equiv 1\mod N$ holds, although $N$ is composite.

A number $N$ is a Carmichael number, if $N$ is odd and squarefree, has at least $3$ distinct prime factors and for every prime $p$ dividing $N$, we have $p-1|N-1$.

The Carmichael number $$N=656601=3\cdot 11\cdot 101\cdot 197$$ has the property that both $N-2$ and $N+2$ are prime, it is "sandwiched" by $2$ primes, we could call it a "sandwich"-Carmichael-number.

Are there any further Carmichael-number "sandwiched" by two primes in this way ?

It is clear that such a number $N$ must be divisble by $3$ , because the smallest Carmichael number is $561$ and if $N$ is not divisible by $3$, $N-2$ or $N+2$ is divisble by $3$. I am currently searching another example, there is none for $N\le 1.6\cdot 10^9$

  • $\begingroup$ Can anyone guess the reason for this downvote ? $\endgroup$
    – Peter
    May 30, 2017 at 7:45

1 Answer 1


Using the list of Carmichael numbers up to $10^{16}$ from Pinch, I found with my MPArith routines the following "sandwiched" Carmichael number (in 54sec)

$$656601, 25536531021, 8829751133841, 60561233400921, 79934093254401, 352609909731201, 598438077923841, 976515437206401, 2122162714918401, 2789066007968241, 3767175573114801, 7881891474971361$$

The list contains $246683$ Carmichael numbers and is available from http://ftp.gwdg.de/pub/math/funet/misc/RichardPinch/Carmichael/carmichael-16.gz

  • $\begingroup$ Seeing this, it somehow seems likely there are infinitely many of these guys. $\endgroup$
    – MooS
    May 4, 2017 at 13:13
  • $\begingroup$ Might be worth adding to the OEIS. $\endgroup$ May 5, 2017 at 16:12

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