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 $2$)?



Is there a linear recurrence with infinitely many zeros, conjecturally infinitely many primes and non-zero terms of exponential growth?
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: $0,1,0,1,0,2,0,3,0,5,0,8\ldots$.


