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Taking my notation from a previous question

Define a function $P_6$ as $$P_6(n)=\begin{cases} 0, \ \ 6n-1 \not\in \mathbb P \wedge 6n+1 \not\in \mathbb P \\ 1, \ \ (6n-1 \not\in \mathbb P \wedge 6n+1 \in \mathbb P) \vee (6n-1 \not\in \mathbb P \wedge 6n+1 \in \mathbb P)\\ 2, \ \ 6n-1 \in \mathbb P \wedge 6n+1 \in \mathbb P \end{cases}$$

So $P_6(n)$ has value $0$ if neither of the numbers around $6n$ is a prime, $1$ if either but not both are primes and $2$ is both are.

Let sets $P_6^0,P_6^1,P_6^2$ be the corresponding sets of indexes where $P_6 = 0,1 \vee 2$, so, for example, $\forall n\in P_6^1, P_6(n) = 1$

Define three new functions using the indicator functions of above sets:

\begin{cases} \pi_{6\bullet}^0 (n) = \sum_{i=1}^n 1_{P_6^0}(i) \\ \pi_{6\bullet}^1 (n) = \sum_{i=1}^n 1_{P_6^1}(i) \\ \pi_{6\bullet}^2 (n) = \sum_{i=1}^n 1_{P_6^2}(i) \end{cases}

So these functions tell how many such indexes $1 \leq s \leq n$ there are for whom the number of primes surrounding $6s$ is $i$, $i \in \{0,1,2\}$.

These functions have following relations: \begin{equation} \pi_{6\bullet}^0 (n)+\pi_{6\bullet}^1 (n)+\pi_{6\bullet}^2 (n) = n \ \ \ \ \ (1) \end{equation} and \begin{equation} \pi(6n+1)-2 = \pi_{6\bullet}^1 (n)+2 \pi_{6\bullet}^2 (n) \ \ \ \ \ (2). \end{equation}

Here $\pi(n)$ is the prime counting function and it has argument $6n+1$ because the biggest number we test is indeed $6n+1$ and we have to remove $2$ because the first two primes are not reachable via number six.

Now we get from (1): $$\pi_{6\bullet}^1 (n) = n-\pi_{6\bullet}^0 (n)-\pi_{6\bullet}^2 (n)$$

and inserting this to (2), we get: $$\pi(6n+1)+\pi_{6\bullet}^0 (n)=n+2+\pi_{6\bullet}^2 (n) \ \ \ (3).$$

Assume there is finite number of twin prime pairs, let this number be $M$ and let the last of these be of the form $6S \pm 1$.

Now (3) can, after index $S$, be written as $$\pi(6(k+S)+1)+\pi_{6\bullet}^0(k+S)=k+S+2+M \ \ \ \ (4)$$

ie. for all numbers $k\geq 0$ the sum of all primes smaller or equal to $6k+6S+1$ plus the number of numbers divisible by six between $6$ and $6k+6S$ which have no primes within distance 1 of them, can be expressed as linear function of k.

Now, an upper bound for $\pi_{6\bullet}^0 (n)$ is $n$. This is simply assumption that there are no primes. Inserting this into (4) gets us $M < \pi(6(k+S)+1)-2.$

Then a lower bound for $\pi_{6\bullet}^0 (n)$. The number of numbers between $5$ and $6n+1$ divisible by 6 is $n$. So we could estimate $\pi_{6\bullet}^0 (n) > n - \pi(6n+1)$, ie. there are no twin primes. This is always true, but it is too small, for inserting this to (4) gives us $M>-2$, which is useless.

Much better would be to assume that every prime is a part of a twin prime, so we could estimate $\pi_{6\bullet}^0 (n) > n - \frac{\pi(6n+1)}{2}.$ This is, however, not true, especially after $S$.

Assume that there exists a number $\alpha,$ such that $\pi_{6\bullet}^0 (n) > n - \frac{\pi(6n+1)}{\alpha}.$ Inserting this into (4) gives us $M > (1-\frac{1}{\alpha}) \pi(6(k+S)+1)-2$. This means that as long as $\alpha > 1$, there is no such constant $M$.

Is there some way to justify that there can be such an $\alpha>1$?

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up vote 1 down vote accepted

By Brun's theorem, $$ \pi_{6\bullet}^2(n)=O\left(\frac{n(\log\log n)^2}{(\log n)^2}\right) $$ so together with the Prime Number Theorem in the form $$ \pi(n)=\frac{n}{\log n}+O\left(\frac{n}{(\log n)^2}\right) $$ we can conclude that $$ \pi_{6\bullet}^1(n)=\frac{6n}{\log 6n}+O\left(\frac{n(\log\log n)^2}{(\log n)^2}\right) $$ and $$ \pi_{6\bullet}^0(n)=n-\frac{6n}{\log 6n}+O\left(\frac{n(\log\log n)^2}{(\log n)^2}\right). $$

Therefore, $$ \pi_{6\bullet}^0(n)=n-\frac{\pi(6n+1)}{1}+O\left(\frac{n(\log\log n)^2}{(\log n)^2}\right) $$ and any choice of $\alpha>1$ gives the wrong asymptotic. So no, this approach cannot be used to prove the Twin Prime Theorem.

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