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Assume that $x,y > 2$, and that $x<y$. Then the second Hardy–Littlewood conjecture states that

$$\pi(x + y) - \pi(y) \leq \pi(x).$$

We can easily justify this heuristically, since

$$ \textrm{li}(x+y)-\textrm{li}(y) < \textrm{li}(x), $$ where we have applied the prime number theorem on the form

$$\pi(x) \sim \textrm{li}(x):=\int_2^x \frac{1}{\log t}dt.$$

As stated on Wikipedia, the conjecture is believed to be false, as it would be inconsistent with the more confident first Hardy–Littlewood conjecture on prime k-tuples, but it is expected that the first violation will only occur at very large $x$.

Now, as $x$ grows large, we have that $$ \lim_{x\rightarrow \infty} \textrm{li}(x) - [\textrm{li}(x+y)-\textrm{li}(y)] \rightarrow \infty, $$ while asymptotically $$ \lim_{x\rightarrow \infty} \frac{\textrm{li}(x)}{\textrm{li}(x+y)-\textrm{li}(y)} \rightarrow 1. $$

From the first of these expressions it appears naively that the larger $x$ is, the less probability for the conjecture to fail; even more so if $y$ also becomes large relative to $x$. I would very much like to know whether there are other heuristic arguments for understanding why the conjecture likely is false, besides the clash with the first Hardy–Littlewood conjecture.

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You may find this and this interesting. – user170039 Oct 15 '14 at 4:47

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