proportion of primes in a polynomial sequence

Question

It is conjectured (Bunyakovsky) that when $P(x)$ is a polynomial from $\mathbb{Z}[X]$, irreducible, with positive leading coefficient and so that the integers $P(n)$ , $n\gt0$ do not share a common factor different from $1$, then there are infinitely many primes in the sequence $P(n)$ , $n\gt0$ .

But what about the proportion (density) of those primes? It seems that it can vary a lot, even when the degree of the polynomial is not changed.

For instance, for degree $10$, let $$P(x)=1-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+x^{10}$$ and $$Q(x)=-691+2073 x-287 x^2-3285 x^3+1420 x^4+2310 x^5-1190 x^6-1050 x^7+525 x^8+525 x^9+105 x^{10}$$ there are more than $60$ values of $n$ below $1000$, for which $P(n)$ is prime, whereas in the same range, $Q(n)$ is prime only when $n=129$ or $n=539$.

Can somebody find another such polynomial sequence, with an even lesser proportion of primes?

Answer 1

Chinese remainder theorem.

Pick $R(x)$ with positive coefficients so that $$\begin{align} R(x)&\equiv x^9(x-2)\pmod 3\\ R(x)&\equiv x^7(x-1)(x-2)(x-3)\pmod 5\\ R(x)&\equiv x^5(x-1)(x-2)(x-4)(x-5)(x-6)\pmod 7\\ R(x)&\equiv (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)\\&\equiv x^{10}-1\pmod {11} \end{align} $$

Then $R(n)$ will be divisible by one of $3,5,7,11$ unless $n\equiv 1\pmod 3,4\pmod 5,$ etc. This means the only $n$ we need to check is $n\equiv 1144\pmod {1155=3\cdot5\cdot7\cdot 11}$. So in particular, $R(n)$ is not prime for $n=0,1,2,3\dots,2298$.

Note, the above polynomials are not hard to compute, since, for example, $$\begin{align} (x-1)(x-2)(x-4)(x-5)(x-6)&\equiv\frac{x^6-3^6}{x-3} \\&\equiv x^5+3x^4+3^2x^3+3^3x^2+3^4x+3^5\\ &\equiv x^5+3x^4+3x^3+6x^2+4x+5\pmod 7 \end{align}$$

On the other hand, your first $P(x)$ is the polynomial for the primitive $22$nd roots of unity, which means that if $p\mid P(n)$ then $n$ is a primitive $22$nd root of unity, modulo $p$. That can only happen if $p\equiv 1\pmod {22}$, so, in particular, $P(n)\equiv 0\pmod {2,3,5,7}$ is not possible, and the only eliminated value modulo $11$ is $10$. This gives a far larger set of prospective value of $n$.

Answer 2

The single-polynomial case of the Bateman-Horn conjecture says that if $f$ satisfies the Buniakowsky conditions then the number of integers $n$, $1\le n<x$, such that $f(n)$ is prime is asymptotic to $C(f)n/\log n$, where $$C(f)={1\over{\rm deg}\,f}\prod_p\left(1-{1\over p}\right)^{-1}\left(1-{N_f(p)\over p}\right)$$ where the product is over primes $p$, and $N_f(p)$ is the number of solutions of $f(n)\equiv0\pmod p$.