Is it true for any $n=2p$ where $p$ is prime, that the number of twin primes less than $n$ approaches the number of prime pairs $(p_{1},p_{2})$ such that $p_{1} + p_{2} = n$?

For example, If we choose prime number 499, then $n=998$. For $n = 998$ there are 33 prime pairs and there are 35 twin primes less than 998.

If we choose larger values of $n=2p$, the number of prime pairs will converge to the number of twin primes less than $n$?

  • $\begingroup$ This may be a reasonable conjecture. However proving it will be quite difficult - just proving that the number of prime pairs $(p_1, p_2)$ such that $p_1 + p_2 = n$ is always nonzero is Goldbach's conjecture, a long-standing unsolved problem. $\endgroup$ – Michael Lugo Dec 12 '18 at 19:26

The Hardy-Littlewood conjecture would tell us that the number of primes $p$ less than $x$ such that $p+2$ is also a prime should asymptotically satisfy: $$\pi_2(x)\sim 2C_2\frac x{(\ln x)^2}$$

Where $C_2\approx .66$

On the other hand, standard heuristics tell us that the number of ways to express an even $x$ as the sum of two primes should be asymptotically $$2C_2\times \prod_{p\,|\,x,\;p≥3}\frac {p-1}{p-2}\times \frac x{(\ln x)^2}$$

See, e.g., this.

Thus, standard conjectures would tell us that, for large even $x$:

$$\frac {\text {the number of prime pairs} ≤ x}{\text {the number of ways to write} \;x\;\text{as the sum of two primes}}\sim \prod_{p\,|\,x,\;p≥3}\frac {p-2}{p-1}$$

To be sure, both conjectures are entirely unproven.

  • $\begingroup$ Thank you for your response. I am only looking for values of x that are equal to 2p where p is a prime number. $\endgroup$ – temp watts Dec 13 '18 at 14:21
  • $\begingroup$ Well, my response covers that case. If $x=2p$ then that product is just $\frac {p-2}{p-1}$. $\endgroup$ – lulu Dec 13 '18 at 15:04
  • $\begingroup$ They key point, of course, is that neither of these conjectures is even close to being proven. There are sensible heuristic arguments for both, but so what? Neither Twin Primes nor Goldbach is at all well understood so the heuristic arguments might entirely miss the point. $\endgroup$ – lulu Dec 13 '18 at 15:07

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