Is this formula: $81n^2+135n+97$ wealth by prime numbers which $n$ is natural number?

Question

I made some effort to set a wealth quadratic formula for prime, I found this formula:

$A(n)= 81n^2+135n+97$, it gives primes for $n=0 $ to $n=18 $.

I would be like some one to show me if this really a wealth quadratic formula for primes for

large $n$?

Thank you for any replies or any comments.

Answer 1

It gives primes for $26284$ of the integers from $1$ to $10^5$, so it's not too bad. Not quite as good from that point of view as $n^2 + n + 41$, which produces primes for $31984$ of those integers.

EDIT: your $A(n) = (9 n + 7)^2 + (9 n + 7) + 41$, so you just have a minor modification of the Euler polynomial.

Answer 2

It is not currently known whether this polynomial represents an infinite number of primes.

The Bouniakowski Conjecture implies that it does.

No polynomial $P(n)$ of degree $\ge 1$, with integer coefficients, can be prime for all large enough $n$.

Answer 3

If you mean that it produces the most primes for a continuous range of n, no it is not.
The polynomial by Dress and Landreau (2002), Gupta (2006) produces primes for n = 0 to 56 (57 distinct primes)

$A(n) = | \frac 1 4 (n^5-133n^4+6729n^3-158379n^2+1720294n-6823316) |$

Source Prime-Generating Polynomial

An alternate version generates the same numbers for n = 1 to 57

$A(n) = | \frac 1 4 (-8708852 + 2057776 n - 179374 n^2 + 7271 n^3 - 138 n^4 + n^5) |$