- #1
e2theipi2026
- 8
- 1
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. The prime counting function \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 b.
But, the left side can be both odd and even many times over because f(p_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. Therefore, the two sides cannot be equal for all n.
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 of the twin prime counting function given at the beginning of this question, does that argument hold water?
\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. The prime counting function \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 b.
But, the left side can be both odd and even many times over because f(p_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. Therefore, the two sides cannot be equal for all n.
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 of the twin prime counting function given at the beginning of this question, does that argument hold water?