Primes of the form $4n^2-4n–1$

Question

I noticed for the polynomial $4n^2-4n–1$ it produced 42 primes when testing from n=2 to 100 (results below, primes in bold).

7, 23, 47, 79, 119, 167, 223, 287, 359, 439, 527, 623, 727, 839, 959, 1087, 1223, 1367, 1519, 1679, 1847, 2023, 2207, 2399, 2599, 2807, 3023, 3247, 3479, 3719, 3967, 4223, 4487, 4759, 5039, 5327, 5623, 5927, 6239, 6559, 6887, 7223, 7567, 7919, 8279, 8647, 9023, 9407, 9799, 10199, 10607, 11023, 11447, 11879, 12319, 12767, 13223, 13687, 14159, 14639, 15127, 15623, 16127, 16639, 17159, 17687, 18223, 18767, 19319, 19879, 20447, 21023, 21607, 22199, 22799, 23407, 24023, 24647, 25279, 25919, 26567, 27223, 27887, 28559, 29239, 29927, 30623, 31327, 32039, 32759, 33487, 34223, 34967, 35719, 36479, 37247, 38023, 38807, 39599

Does this polynomial have a higher chance of producing primes than random chance?

Answer 1

Hint The chance of hitting a prime by random chance should be close to the prime density function which can be approximated as the derivative (with respect to $N$) of $$\Pi (N) = \frac{N}{\log(N)}$$ Which is proven in the prime number theorem.

Answer 2

It is known that some quadratic functions generate disproportionately many primes, at least initially. This manifests visually as diagonals in the Ulam spiral.

You seem to have stumbled upon one of these. For a more famous example, take $n^2-n+41$, which gives primes for all $0\leq n\leq 40$.

Answer 3

Suppose you choose a random integer $x$. Then the probability that $x$ is prime is of course $1/$$\log(x)$ while the probability for $4x^2-4x-1$ is $1$/$\log(2x)$, roughly twice as much as $x$. So it is estimated that for $x$, there is one integer $k$ in the range $x, x+\log(2x)$ which $4k^2-4k-1$ is prime for. Compare this with finding a prime the size of $x$ where it is expected there is one prime in the range $x, x+\log(x)$. Again this is basic, and least probability, adding in factors such that each prime factor of $4k^2-4k-1$ is congruent to $1$ or $7 \pmod 8$, the probability may increase.