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[tex]\pi_2(n)=f(n)+\pi(n)+\pi(n+2)-n-1,[/tex]

where [itex]\pi_2(n)[/itex] is the twin prime counting function, [itex]f(n)[/itex] is the number of twin composites less than or equal to [itex]n[/itex] and [itex]\pi(n)[/itex] is the prime counting function.

At [itex]n=p_n,[/itex] this becomes

[tex]\pi_2(p_n) = f(p_n) + \pi(p_n) + \pi(p_n + 2) - p_n - 1.[/tex]

With this form, can I make the following argument?: Assume the twin prime counting function becomes a constant [itex]c[/itex], then I can change the twin prime counting function to [itex]c[/itex] in the equation. The prime counting function [itex]\pi(n)[/itex] at the prime sequence [itex]p_n[/itex] is just [itex]n[/itex], so I can change that to [itex]n[/itex]. Because I'm assuming no more twin primes, [itex]p_n+2[/itex] is not a prime so [itex]\pi(p_n+2)[/itex] will also become [itex]n[/itex], the equation directly above this paragraph can therefore be simplified to:

[tex]c = f(p_n) + 2n - p_n - 1.[/tex]

Adding [itex]1[/itex] to both sides of this and rearranging it gives,

[tex]p_n - f(p_n) = 2n - b[/tex], where [itex]b=c+1.[/itex]

The right side of [itex]p_n - f(p_n) = 2n - b[/itex]

has only one possible parity, either odd or even because it is an even number [itex]2n[/itex] minus a constant [itex]b.[/itex]

But, the left side can be both odd and even many times over because [itex]f(p_n)[/itex] can be odd or even and is subtracted from [itex]p_n[/itex] which is odd for [itex]p>2.[/itex]

So, the left side will change parity for different values of [itex]n,[/itex] while the right side of the equation will remain one parity. Therefore, the two sides cannot be equal for all [itex]n.[/itex]

This seems to show the twin prime counting function cannot become constant and therefore, there are infinite twin primes. Now assuming I can prove the form given at the beginning of this question, does that argument hold water?