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

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?

• If we assume that a large number $x$ is prime with probability $\frac{1}{\ln(x)}$, we can show that under this assumption we can expect infinite many prime constellations , in particular infinite many twin primes. This is of course no proof. The number of primes decreases but not dramatically. A $100$ digit number still is prime with probability about $\frac{1}{230}$ Mar 18, 2019 at 17:28
• Can you apply your same reasoning to the sequence of perfect squares? Mar 18, 2019 at 17:31

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.