Re: twin-prime counting function: Is this a valid argument?

[e2theipi2026]

Re: twin-prime counting function: Is this a valid argument?

I can prove the twin prime counting function has this form:
$\pi_2(n)=f(n)+\pi(n)+\pi(n+2)-n-1$,
where $\pi_2(n)$ is the twin prime counting function, $f(n)$ is the number of twin composites less than or equal to $n$ and $\pi(n)$ is the prime counting function.
At $n=p_n$, this becomes
$\pi_2(p_n) = f(p_n) + \pi(p_n) + \pi(p_n + 2) - p_n - 1$.
With this form, can I make the following argument?: Assume the twin prime counting function becomes a constant $C$, then I can change the twin prime counting function to C in the equation. $\pi(n)$ at the prime sequence $p_n$ is just $n$ so I can change that to n. Because I'm assuming no more twin primes, $p_n+2$ is not a prime so $\pi(p_n+2)$ will also become $n$, the equation directly above this paragraph can therefore be simplified to:
$C = f(p_n) + 2n - p_n - 1$.
Adding $1$ to both sides of this and rearranging it gives,
$p_n - f(p_n) = 2n - b$, where $b=C+1$
The right side of $p_n - f(p_n) = 2n - b$
has only one possible parity, either odd or even because it is an even number $2n$ minus a constant.
But, the left side can be both odd and even many times over because $f(n)$ can be odd or even and is subtracted from $p_n$ which is odd for $p>2$.
So, the left side will change parity for different values of n, while the right side of the equation will remain one parity, either odd or even.
This seems to show the twin prime counting function cannot become constant and therefore, there are infinite twin primes. Does that argument hold water?

 

[lookagain]

e2theipie2026, in this forum, don't use dollar signs for Latex. Use [ tex ] and [ /tex ]. but without the spaces inside.

I cut-and-pasted and retyped it with the proper code so others could read it:

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I can prove the twin prime counting function has this form:

\(\displaystyle \pi_2(n)=f(n)+\pi(n)+\pi(n+2)-n-1\),

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

At \(\displaystyle n=p_n\), this becomes

\(\displaystyle \pi_2(p_n) = f(p_n) + \pi(p_n) + \pi(p_n + 2) - p_n - 1\).

With this form, can I make the following argument?: Assume the twin prime counting function becomes a constant \(\displaystyle C\),

then I can change the twin prime counting function to C in the equation.

\(\displaystyle \pi(n)\) at the prime sequence \(\displaystyle p_n\) is just \(\displaystyle n\) so I can change that to n. Because I'm assuming no more twin primes, \(\displaystyle p_n+2\) is not a prime so \(\displaystyle \pi(p_n+2)\)

will also become \(\displaystyle n\), the equation directly above this paragraph can therefore be simplified to:

\(\displaystyle C = f(p_n) + 2n - p_n - 1\).

Adding \(\displaystyle 1\) to both sides of this and rearranging it gives,

\(\displaystyle p_n - f(p_n) = 2n - b\), where \(\displaystyle b=C+1\)

The right side of \(\displaystyle p_n - f(p_n) = 2n - b\)

has only one possible parity, either odd or even because it is an even number \(\displaystyle 2n\) minus a constant.

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

So, the left side will change parity for different values of n, while the right side of the equation will remain one parity, either odd or even.

This seems to show the twin prime counting function cannot become constant and therefore, there are infinite twin primes. Does that argument hold water?