I noticed that for a Fibonacci sequence starting with seeds $(2,1)$, there is an awful amount of primes in the first $20$ elements of the sequence ($11$ primes), far more than $(0,1)$'s prime density. For $100$ elements, the density is much less than $\frac12 (18)$, but still surprisingly more than the prime density of the first $100$ 'normal-Fibonacci' integers.
Seeing this, I got curious about other seeds that could potentially give better prime density results. I don't know where to start from just guessing, though, and still don't know why the seed $(2,1)$ has a higher prime density. Is it just a coincidence? Can anyone help me out?
Work Until Now
Proving if $L_a$ is prime, $a$ must be a power of $2$, prime, or $0$:
In my proof, I show why every Lucas prime's index (n-value) must be $0$, a power of $2$, or a prime.
A Lucas number, by definition, is a number $L_n$, such that $L_n = A^n+B^n$, where $A$ and $B$ are the roots of $x^2-x-1$.
We want to prove that for any odd $m$, $L_{nm}$ can be divided. By acknowledging this fact, we can show that for $L_a$ to be prime, $a$ must not have odd factors other than $1$. If $a$ does not have any odd factors, it either is a power of $2$, prime, or $0$.
If $m$ is odd, then $X^m+Y^m = (x+y)(x^{m−1}−x^{m−2}y+\dots)$ Plugging in $A^n$ and $B^n$ for $X$ and $Y$, we get, $A^{nm}+B^{nm}=(A^n+B^n)\times S$, where $S$ is the rest of the binomial equation. This means $L_{nm}$ (which equals $A^{nm}+B^{nm}$) can be factored when $m$ is odd. Proving this, we now know that for $L_a$ to be prime, a must be a power of $2$, prime, or $0$.
