2
$\begingroup$

It is not known whether there is a largest prime in the Fibonacci sequence, but of course there are quite a few primes.

Similarly, the Lucas sequence starting with $L_1=1, L_2 = 3$ comes to a prime at $L_4 = 7$. On the other hand, for $x_1=3, x_2 = 11$ all the $x_n$ are composite until $x_7 = 103$.

It seems intuitive that for any arbitrary positive coprime $(x_1, x_2)$ the corresponding sequence eventually arrives at a prime, but I'm wondering whether there is a proof or whether somebody out there can provide a proof. Concretely:

Given $(x_1, x_2) \in \Bbb{Z}+ : \gcd(x_1,x_2) = 1$. For all integer $n > 2$ define $x_n = x_{n-1}+x_{n-2}$. Prove that $\exists n \in \Bbb N, | n> 2 \wedge x_n \in \Bbb{P} $.

$\endgroup$
1
  • 2
    $\begingroup$ There is a pair of coprime numbers which generate a Fibonacci like sequence of composite integers. I don't recall the details, so check out Richard Guy's UPINT. Gerhard "Unsolved Problems In Number Theory" Paseman, 2019.11.14. $\endgroup$ Commented Nov 14, 2019 at 23:19

1 Answer 1

4
$\begingroup$

As Gerhard Paseman said in his comment, there are counterexamples. This is discussed in A3 of UPINT. According to the third edition, the smallest known counterexample as of the printing (in 2003) is $x_1= 8983542533631$ and $x_2 = 248272649660939$ due to John Nicol. This article by Knuth discusses the basic idea behind finding such pairs and may be a good place to start. The relevant section in UPINT has a few references in addition.

$\endgroup$
3

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .