The given example can at least be extended
to 14 terms:
7, 13, 31, 61, 181, 541, 2161, 187921,
3570481, 7140961, 31418238813121, 7728886748027521,
And can probably be extended even further
[using a different ‘tail’].
I used Korselt’s criterion: n is a
Carmichael number iff p|n implies p-1|n-1 and n is squarefree.
Start with a Carmicahel number n, try to find m=n*p, p unequal
to the prime factors of n, such that p-1|m-1 which amounts to
p-1|n-1, construct m and test the prime divisors p[i] of n for
p[i]-1|m-1 [or p[i]-1|m/p[i]-1].
I also investigated some emerging patterns
and found the following article by Jack
Chernick (april 1939) in which he calls these universal
forms. In particular:
which is a Carmichael number if all factors are prime, for
instance if k=950560 we have n=9 factors. His theorem 4 is
directly concerned with this puzzle.
However his method of finding these
patterns is slightly different from mine:
Let m=p.q.q..q[n-1], with p and the
q[i] prime. If we choose q[i]=1 mod p-1 it’s easy to see that
m-1=0 mod p-1. Let’s choose q[i]=a[i](p-1)+1 as a nontrivial
simple way to achieve this.
We also must have q[i]-1|m-1 but q[i]-1=a(p-1)
and m-1=(p-1)P(p) where P(p) is some polynomial in p. So a
necessary and sufficient condition for the a[i] is that P(p)=0
This condition translates to:
ap+1=0 mod a
ap+1=0 mod a
a (and the other 2 can be found by cycling the a[i]).
And will give the same results as mentioned
in the article (choose appropriate values for a[i] and a
condition on p will follow..).
A pratical guide to find 3-term Carmichael
numbers (if I may say so) is found in Jameson.
It might also be interesting to read the
article by Renaud
Lifschitz who describes a set of Carmichael numbers C1, C2,
..Cn where each product of any number of these is also
Carmichael, which he calls Nested Carmichael numbers.