broader meaning of twin prime constant?

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?

• Does “for $p_1 \leq p \leq \sqrt{n}$” mean “for all primes $p$ such that…”? If so please add “all” for clarity. Look up the heuristic probabilistic reasoning behind the Bateman-Horn conjecture. It easily leads to such constants as $2C_2$, but it is pretty hopeless to expect to get actual proofs about (real) prime counts in this way.
– KCd
Commented Dec 30, 2023 at 22:33
• Instead of saying you “drop integers in residue classes…”, I suggest saying “remove integers in residue classes…” because to drop something in something could be misread as dropping into (i.e., putting into) that something, which is the exact opposite of what you want!
– KCd
Commented Dec 30, 2023 at 22:37
• made the edits. Thank you @KCd.
– sku
Commented Dec 30, 2023 at 23:45
• Your idea makes alot of sense !
– mick
Commented Sep 4 at 23:09

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.

• I don't understand how what you wrote is related to what I wrote. Let a $n$-sequence of primes be of the form $p, p+2, p+4,\cdots$ where it is feasible. I am saying the ratio of the ratio of the count of $n$-sequence primes to the count of $n-1$-sequence primes to the ratio of the count of $n-1$-sequence primes to the count of $n-2$-sequence primes is a constant.
– sku
Commented Sep 4 at 5:43
• @sku see the huge edit. I hope it is clear now.
– mick
Commented Sep 4 at 23:06