2

I would like to ask a general question that conversely is related to the subject of the post:

Fibonacci pseudoprimes

Has any non-trivial Lucas (or extended Lucas (i.e. Lehmer)) sequence been found to contain no pseudoprimes?

I am interested in an answer to this question because in the theory of the Lucas sequences, if the ``rank of apparition'' of a number $n$ is either $n+1$ or $n-1$, then we can conclude that $n$ is prime. Unfortunately, because of the existence of pseudoprimes in these sequences, the alluded to theorem is not necessary and sufficient.

I caution that if a composite $m$ is a pseudoprime in one Lucas sequence, say $L(a, b)$, it may not be a pseudoprime in another Lucas sequence $L(c,d)$. Surely, except in a degenerate case, $m$ will be a factor of infinitely many terms of $L(c,d)$, but it will divide neither the $m-1st$ or $m+1st$ terms of $L(c,d)$.

The term ``rank of apparition'' refers to the index of the first term in the underlying sequence that contains a given number $n$ as a divisor.

DDS
  • 3,199
  • 1
  • 8
  • 31
  • 1
    There are integer sequences $u_1,u_2,\dots$ satisfying $u_{n+1}=au_n+bu_{n-1}$ with $\gcd(u_1,u_2)=\gcd(a,b)=1$ but with all terms composite. Maybe one of those would be a good place to start. – Gerry Myerson Jul 18 '19 at 05:02

0 Answers0