How to explain twice as much primes like $(2n-1)^2-2$ than like $(2n-1)^2+2$?

Question

Are there explanations why there should be about twice as much primes on the form $(2n-1)^2-2$ than on the form $(2n-1)^2+2$?

      n  f(n)-2  f(n)+2
     10       6       4 
    100      33      17 
   1000     208     110 
  10000    1558     817 
 100000   12390    6440 
1000000  102204   52980 

Where f(n)=(2n-1)^2

Example:
for $10$ there are $6$ and $4$ primes of the kind for $1 \le n < 10$.
for $100$ there are $33$ and $17$ primes of the kind for $1 \le n < 100$.

(My test computations was not up to 10, 100 etc)

Answer 1

Just looking modulo $3$, we have that $(2n-1)^2+2 = 4n^2-4n+3 \equiv n(n-1)$, so it will be a multiple of 3 whenever in is $0$ or $1$ modulo $3$. So for two-thirds of the $n$-values it cannot be prime.

On the other hand, $(2n-1)^2-2$ is always $1$ or $2$ modulo $3$, so none of them are immediately proven to be non-prime.