How is the Twin Primes Constant useful? What value does it provide over Brun's Constant?

Question

The Twin Primes Constant is:

$$\prod_{p > 2 \text{ and a prime }}\left(1 - \frac{1}{(p-1)^2}\right) = 0.6601618158\ldots$$

It appears that in this case $p$ does not have to be a prime. But if that's true, why is it called the twin primes constant?

Brun's constant is:

$$\frac{1}{3} + \sum_{p>3 \text{ and } p,p+2 \text{ are primes }}\left(\frac{1}{p} + \frac{1}{p+2}\right) = 1.902160540\ldots$$

Brun's constant seems straight forward to me. The sum of the reciprocal of twin primes converges so there is no proof in the sum of reciprocals for the infinitude of twin primes.

I am not clear what to make of the twin primes constant. What value does it give? How did Merten use it when he came up with it? Why is it interesting?

Answer 1

It appears to be called the "twin prime constant" because it comes up a lot in theorems and conjectures about twin primes. For example, from the relevant MathWorld page:

Answer 2

This constant appears in the conjectured asymptotic estimate on the number of twin primes below a given size, see here. It's also noteworthy that this constant wasn't just taken out of thin air, this can be actually derieved heuristically via Cramer's random model.

Answer 3

@LarryFreeman I think you will better understand the meaning of the twin primes constant, if you write it in the form $$ C _2 = \frac {{ \prod_\limits{p \geqslant 3 }} \left( 1 -\frac {2}{ p} \right)} {{ \prod_\limits{p \geqslant 3 }} \left( 1 - \frac 1 p \right)^2} = { \prod \limits_{p \geqslant 3 }} \, \frac{p(p-2)}{(p-1)^2}. $$