Doesn't the first Hardy-Littlewood conjecture imply the finiteness of prime constellations?

Question

The first Hardy-Littlewood conjecture says, in essence, that if all numbers within a prime $k$-tuple do not form a complete residue class with respect to any prime, then they are infinite in number and their asymptotic density is

$$ \pi_k(n) \sim C_k \int_2^n \frac{dt}{\log^kt}, $$

where $\pi_k(n)$ denotes the amount of primes in the constellation less than or equal to $n$ and $C_k$ is a constant calculated using the different residues.

But this conjecture implies that the primes within the constellation will eventually thin out in intervals of the same size, i.e., there will not be the same amount of primes in the constellation between $1\times10^{20}$ and $2\times10^{20}$ than between $100\times10^{20}$ and $101\times10^{20}$. In fact, the further one gets down the number line, the smaller number the integral will evaluate to and eventually that evaluation will be below $1$.

It is therefore reasonable to assume that there will eventually be a positive integer $m$ such that the expected amount of twin primes between $m\times10^{20}$ and $(m+1)\times10^{20}$ will be less than $1$, and there will be no more twin primes.

Is this reasoning correct? If not, what am I getting wrong?

Answer 1

It basically shows that, for any given span, no matter how large, there will eventually be cases where no primes constellation are expected to occur within the span. However, that's not the same thing as stating this will happen for all future spans of that size. It just means that to expect to have at least one primes constellation in a given span, the size of the span will need to increase as you get to larger & larger values. However, at any given point, the required span size to expect having at least one primes constellation in it will always be finite.

Another way to see it is that, for any non-$0$ density, you'll get a result $\gt 1$ just by multiplying by a large enough value, but as the density is decreasing with $n$, to ensure you get a lower bound, you should use the density at the end of the bound instead of the beginning.

As for a proven counter example to your reasoning, note the Prime Number Theorem gives the asymptotic density of primes as being $\frac{1}{\log{n}}$, which goes to $0$ as $n \to \infty$. Thus, for any given span, at a large enough $n$, the expected # of primes within that span will be less than $1$ and, in fact, will be as close to $0$ as you wish by just choosing $n$ large enough. Nonetheless, there are an infinite number of primes, as proven originally by Euclid and with many other proofs available since then.

Answer 2

That’s already true for the primes. If you take M > exp(10^20) then only one in 10^20 consecutive integers > M are prime on average.