Is there a linear recurrence with infinitely many zeros, conjecturally infinitely many primes and non-zero terms of exponential growth?

Question

Let $a_n$ be a linear recurrence with integer constant coefficients and initial values.

Is it possible $a_n$ to satisfy all of these:

1. $a_n = 0$ infinitely often.

2. if $a_n \ne 0$, $ | a_n |$ is of exponential growth (to avoid cases like $\dots 0 , n , 0 ,n+1, \dots$).

3. $ | a_n |$ is conjectured to be prime infinitely often. Since it is not known if there are infinitely many Fibonacci primes, proving primality is hard, but there shouldn't be obvious obstructions like divisibility by a single prime or some polynomial factorization when treating a coefficient as a variable.

If this is possible, what is minimal order of $a_n$ (it can't be $2$)?

Answer 1

Yes. $a_n=a_{n-1}+a_{n-2}-a_{n-3}+a_{n-4}-a_{n-5}$ has a solution containing zeros and Fibonacci numbers: $0,1,0,1,0,2,0,3,0,5,0,8\ldots$.