broader meaning of twin prime constant?

Question

It appears that the twin prime constant has meaning outside of the strict twin prime constant. I attempted to keep this post as short as possible.

Definitions:

1. Let $p,q$ represent primes and let $n$ be a positive integer.
2. $\pi(n)$ has its usual meaning.
3. Let $p_k$ be the $k^{th}$ prime. ($p_1 = 2$).

From the first $n$ positive integers, consider integers that are $\not\equiv a_1 \bmod p$ for all $p_1 \leq p \leq \sqrt{n}$. Let us call these integers {$a_1$}-primes and let $\pi^{\text{\{$a_1$\}}}(n)$ represent their count. So we remove integers in residue classes $a_1 \bmod p$ where $p_1 \leq p \leq \sqrt{n}$. Slight deviation from definition but we will assume {$0$}-primes as the usual primes.

Let {$a_1,a_2$}-primes represent {$a_1$}-primes that are $\not\equiv a_2 \bmod p$ (for all $p_2 \leq p \leq \sqrt{n}$) and let $\pi^{\text{\{$a_1, a_2$\}}}(n)$ represent their count. So we remove integers in residue classes $a_1 \bmod p$ (for all $p_1 \leq p \leq \sqrt{n}$) and $a_2 \bmod p$ (for all $p_2 \leq p \leq \sqrt{n}$).

and so on.

We can state the most generic claim for removing integers from two residue classes.

\begin{align} \lim\limits_{n\to\infty} \frac{\left(\frac{\pi^{\text{\{$a_1,a_2$\}}}(n)}{\pi^{\text{\{$a_1$\}}}(n)}\right)}{\left(\frac{\pi^{\text{\{$a_1$\}}}(n)}{n}\right)}= 2C_2 \tag{1} \end{align}

We can read the above as saying that the proportion of count of {$a_1,a_2$}-primes to {$a_1$}-primes is $2C_2$ times the proportion of count of {$a_1$}-primes to $n$.

We have to be sure that we are observing the definition of {$a_1, a_2$}-primes. Equation (1) is true for {$a_1,a_2$} = {$0,1$} or {$0,2$} or {$1,2$} or {$0,-2$} or {$0,-4$} but not {$0,-6$} as the last one doesn't strictly follow the {$a_1,a_2$}-primes definition.

I find this result fascinating assuming it is true. I don't know if I have a proof but I do have an argument.

Let $\alpha_2$ be an estimate for the proportion of count of {$a_1,a_2$}-primes to {$a_1$}-primes and let $\alpha_1$ be an estimate for the proportion of count of {$a_1$}-primes to $n$. We can represent $\alpha_1$ and $\alpha_2$ by: \begin{align} \alpha_1 &= \prod\limits_{\substack{p \text{ $prime$ } \\ 2 \leq p \leq \sqrt{n}}} \frac{p-1}{p} \\\\ \alpha_2 &= \prod\limits_{\substack{p \text{ $prime$ } \\ 3 \leq p \leq \sqrt{n}}} \frac{p-2}{p-1} \\\\ \end{align}

as $\frac{p-1}{p}$ represents the number of residue classes of interest to the total and similarly $\frac{p-2}{p-1}$ represents the number of residue classes of interest to the total.(there is implicit use of Dirichlet's theorem for primes in AP and Chebotarev's density theorem). Although $\alpha_1$ and $\alpha_2$ are not perfect estimators (Mertens' theorems), it is easy to see that as $n\to\infty$, their ratio $\frac{\alpha_2}{\alpha_1}$ tends to $2C_2$. In other words, the error in the estimators $\alpha_2$ and $\alpha_1$ cancels out when we take their ratio.

Question: Does any of this make sense? Is it correct?

Answer 1

If $a_2 - a_1 = 2$ you are essentially counting some prime twins in a given interval.

Consider $i > \sqrt n $

If $i = a_1 \mod p$ for all $p< \sqrt n$ with $a_1 \neq 0$ then $i$ is a prime.
Then $i+2 = a_2 \mod p$ for all $p< \sqrt n$ (with $a_2 \neq 0$) then $i+2$ is ALSO a prime.

So you are counting a subset of prime twins.

But I am skeptical about the possibility.

Consider the integers up to $r = 10 002$.

The smallest prime less than the square root of $r$ is $97$.

But I can not think of a prime twin that has its first prime of the form $2 \mod p$ for all primes $p$ up to $97$ , and yet is smaller than $r = 10002$

A similar situation if $a_2 - a_1 = 4$, then you are counting some prime cousins in a given interval.

EDIT :

The above might be a bit confusing, so I will explain a bit further.

If $a_2 - a_1 = 2$ you are essentially counting some prime twins in a given interval.

Consider $i > \sqrt n $

If $i = a_1 \mod p$ for some $p< \sqrt n$ with $a_1 \neq 0$ then $i - a_1 $ is not a prime.
it then also follows;

$i+2 = a_2 \mod p$ for that same $p< \sqrt n$ (with $a_2 \neq 0$) implies $i+2 - a_2$ is ALSO not a prime.

So we have a pair $(q,q+2)$ of composites in the interval near $i$. If we set $\neq$ it follows we do not have a pair of composites in the interval near $i$. Basically we were sieving out composites pairs $(x,x+2)$ with the $=$ case.

So you are counting a subset of prime twins.

Now the tricky question is, why do we not count the primes like this ? Because sieving out pairs of composites seems to leave room for non-twin primes.

However if $(x-2,x)$ is sieved out because $x-2$ is not a prime and $(x,x+2)$ is sieved out because $x+2$ is not a prime, then it seems $x$ is not counted, even if $x$ is a prime.

Ok back to the start. Because this logic is mainly based on mod this OR that prime , but does not explain the for all prime part much.

So we need to show that it is more or less the same situation.

consider first the simple case :

$v = 0 \mod p $ for some prime then $v$ is not a prime > $p$.

Similar if $v$ is a prime $> p$ then

$v \neq 0 \mod p$ for all prime $p$.

This is the basics of logics and quantifiers : the connections between "not","if" etc.

And the basis for sieve theory.

Now notice that in an interval of length $[v - 2p,v+2p]$ you might as well have used $a_1$ instead of $0$.

Indeed

$v = a_1 \mod p$ is the same as $v - a_1 = 0 \mod p$.

So the $a_1$ does not change the count of primes or composites, it just tells exactly which is the prime or composite and such.

So saying $v \neq a_1 \mod p$ (for all $p$) is by the reasoning above, just counting primes.

A similar thing happens for the twins.

Not including those integers $i$ that are $a_1 \mod p$ is counting primes.

Hence if $i$ is such an integer that is not $a_1 \mod p$ then it counts a prime. If $a_2 = 2 + a_1 $ we get :

That same $i$ is also not an integer of the form $a_2 \mod p$, hence $i$ is not an integer of the form $a_1 + 2 \mod p$.

Lets say $i$ counts the prime $x$.

This implies that $i+2$ is also not an integer of the form $a_1 \mod p$ and therefore $i+2$ also counts a prime.

Since $i,i+2$ are both counting a prime they are counting the primes $x,x+2$, or at least $x,x+2 \mod p*$ for some $p* < \sqrt n$.

In most cases this is a prime twin.

I think we can drop the case $x,x+2$ mod p*, because I THINK it must hold for all p*. What reduces to $x,x+2$ because...

Afterall

$$x + p + 2 \neq x + 2 \mod r$$

for prime $p$ and prime $r$ since $r \neq 0 \mod p$.

So yeah, I guess we are counting prime twins.

Similarly if $a_2 - a_1 = 4$ you would count prime cousins.