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According to Wikipedia the twin prime conjecture is a special case of the first Hardy-Littlewood conjecture: $$\pi_2(n)\sim 2C_2\frac{x}{(\ln x)^2} \sim 2C_2\int_2^n\frac{dt}{(\ln t)^2}$$ where $C_2$ is the twin prime constant. The article states "The conjecture can be justified (but not proven) by assuming that 1 / ln t describes the density function of the prime distribution, an assumption suggested by the prime number theorem and would imply the twin prime conjecture, but remains unresolved." (NB: since Wikipedia does not format "1 / ln t" I have not Mathjaxed it to preserve the quote.)

This language is confusing. I thought that the PNT had now been proved, and that one implication of PNT being true was that "1 / ln t describes the density function of the prime distribution." Moreover, the closing words of the quoted section, "but remains unresolved," seem to refer to the "assumption" about the nature of the density function, although a looser reading of the words might be held to say that it is the twin prime conjecture that remains unresolved.

First question: Why does the Wikipedia language say that this must be assumed? Is this a slight mistatement on Wikipedia's part, or am I missing a subtle distinction between the meaning of the PNT being true and the density function being $\frac{1}{\ln t}$?

Second question: I'm not sure how the twin prime conjecture can be "justified" but "not proven" by a set of circumstances. What specifically is lacking that would be necessary to turn a justification into a proof is not identified in the Wikipedia article. Can someone provide a concise statement of what is missing, and what would suffice to bridge the gap?

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    $\begingroup$ Wikipedia's phrasing is indeed confusing. The justification (but not proof) which the article is alluding to comes from random models, in which a number $n$ is set to be a prime with probability $1/\ln n$ independently of all other numbers, and later accounting for small factors (so that we don't get infinitely many $n,n+1$ prime pairs, for instance). This independence is the crucial assumption. $\endgroup$ – Wojowu May 23 at 21:06
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    $\begingroup$ "Being a prime" is a deterministic property, not really probabilistic (you are not going to have 101 being a prime on Mondays and a composite on Tuesdays). So using a probabilistic model is just heuristics until you do the hard work to justify it. $\endgroup$ – user10354138 May 23 at 21:11
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    $\begingroup$ Let $X_n = 1$ if $n$ is prime, $0$ otherwise, then forget about the primes and consider $X_n$ as a sequence of iid random variables with $P(X_n=1)= 1/\ln n, P(X_n=0) = 1-1/\ln n$, then $E[X_nX_{n+2}] = P(X_nX_{n+2} = 1) \approx \frac{1}{\ln^2 n}, \pi_2(x) \approx E[\sum_{n \le x} X_nX_{n+2} ] \approx \sum_{n \le x} \frac{1}{\ln^2 n}\approx \frac{x}{\ln^2 x}$. As you see this is close to the twin prime conjecture, except that the "twin prime constant" is not the correct one. $\endgroup$ – reuns May 23 at 21:12
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First off, the PNT indeed has been proven (100+ years ago). It discusses a prediction about the distribution of primes, and states that deviations from this prediction go to zero in a very specific sense. What has not been proven (except that is equivalent to the Riemann hypothesis) is a conjecture about the asymptotic form of the difference between the actual and PNT-predicted distributions of primes.

What one often does when faced with a nearly intractable conjecture about numbers having some property is to try to understand its plausibility by pretending that each number has some estimatable probability of having that property, independent of all its neighboring numbers. If the expected count of numbers having the property (in this sense) is much less than 1, then one waves his hands and says the conjecture that no number is has the property is plausible. If that expected count diverges, then one says that unless there is some overlooked reason for the property to be forbidden, there should be some (maybe an infinite set) of such numbers.

Let's look at an example or two:

Conjecture: For all numbers $ 4 < a < b < c $, the property $a^2 + b^2 + c^2$ cannot hold.

Plausibility argument: For any given $n$, the density of perfect squares near $n$ is roughly $\frac1{2\sqrt{n}}$ because the distance between $\sqrt{n}^2$ and $(\sqrt{n}+1)^2$ is about $2\sqrt{n}$. So for any given $b$ the number of such triplets should be something like $$ \sum_{a=4}^b \frac{1}{2\sqrt{a^2+b^2}} \approx \int_{y=4}^b \frac1{2\sqrt{y^2+b^2}}dy=\frac12(\log ((1+\sqrt{2})b)-\frac12\log (4+\sqrt{b^2-16}) $$ And now if we integrate than from $b=5$ to infinity, it diverges, telling us that unless there is some number-theory deep reason to avoid triplets where $a^2+b^2=c^2$ there ought to be an infinite set of such triplets. (In this case, the hand-waving argument leads to the right conclusion.)

Another conjecture:

For all numbers $ 4 < a < b < c $, the property $a^3 + b^3 + c^3$ does not hold.

Plausibility argument: For any given $n$, the density of perfect cubes near $n$ is roughly $\frac1{3n^{2/3}}$ because the distance between $\sqrt[3]{n}^2$ and $(\sqrt[3]{n}+1)^2$ is about $3\sqrt[3]{n}^2$. So for any given $b$ the number of such triplets should be something like $$ \sum_{a=4}^b \frac{1}{3(a^3+b^3)^{2/3}} \approx \int_{y=4}^b \frac1{3(y^3+b^3)^{2/3}}dy $$ The answer here is a hot mess involving a Gauss Hypergeometric function, but we can numerically integrate it from $b=5$ to infinity, and it gives an answer of about $0.261$. So we wave our hands and say there are probably no integer triplets where $a^3+b^3=c^3$. (Again n this case, the hand-waving argument leads to the right conclusion.)

But one has to be careful or lucky when doing such an argument. As an exercise, see if you can say what this sort of reasoning tells you about the number of even integers followed immediately by another even integer.

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  • $\begingroup$ This is informative and helpful. Thanks. $\endgroup$ – Keith Backman May 24 at 0:35
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    $\begingroup$ you put a plus, not an equals, in the original statement. $\endgroup$ – Roddy MacPhee May 24 at 1:10
  • $\begingroup$ Actually we don't need the expected count to be much less than 1. Any finite expected count will suffice even if it is a large number say $1738403.1282$ because our heuristics will still imply that there will be only finitely many such observations. $\endgroup$ – Nilos May 26 at 5:26

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