Polignac's conjecture

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In number theory, Polignac's conjecture was made by Alphonse de Polignac in 1849 and states:

For any positive even number n, there are infinitely many prime gaps of size n. In other words: There are infinitely many cases of two consecutive prime numbers with difference n.

The conjecture has not been proven or disproven for any value of n.^{[dated info]}

For n = 2, it is the twin prime conjecture. For n = 4, it says there are infinitely many cousin primes (p, p+4). For n = 6, it says there are infinitely many sexy primes (p, p+6) with no prime between p and p+6.

The first Hardy-Littlewood conjecture generalizes Polignac's conjecture to cover all prime constellations, and giving conjectured asymptotic densities.

[edit] Conjectured density

Let $π n (x)$ for even n be the number of prime gaps of size n below x.

The first Hardy-Littlewood conjecture says the asymptotic density is of form

$\pi_n(x) \sim 2 C_n \frac{x}{(\ln x)^2} \sim 2 C_n \int_2^x {dt \over (\ln t)^2}$

where C_n is a function of n, and $˜$ means that the quotient of two expressions tends to 1 as x approaches infinity.

C₂ is the twin prime constant

$C_2 = \prod_{p\ge 3} \frac{p(p-2)}{(p-1)^2} \approx 0.66016 18158 46869 57392 78121 10014\dots$

where the product extends over all prime numbers p ≥ 3.

C_n is C₂ multiplied by a number which depends on the odd prime factors q of n:

$C_n = C_2 \prod_{q|n} \frac{q-1}{q-2}.$

For example, C₄ = C₂ and C₆ = 2C₂. Twin primes have the same conjectured density as cousin primes, and half that of sexy primes.

Note that each odd prime factor q of n increases the conjectured density compared to twin primes by a factor of $\tfrac{q-1}{q-2}$ . A heuristic argument follows. It relies on some unproven assumptions so the conclusion remains a conjecture. The chance of a random odd prime q dividing either a or a + 2 in a random "potential" twin prime pair is $\tfrac{2}{q}$ , since q divides 1 of the q numbers from a to a + q − 1. Now assume q divides n and consider a potential prime pair (a, a + n). q divides a + n if and only if q divides a, and the chance of that is $\tfrac{1}{q}$ . The chance of (a, a + n) being free from the factor q, divided by the chance that (a, a + 2) is free from q, then becomes $\tfrac{q-1}{q}$ divided by $\tfrac{q-2}{q}$ . This equals $\tfrac{q-1}{q-2}$ which transfers to the conjectured prime density. In the case of n = 6, the argument simplifies to: If a is a random number then 3 has chance 2/3 of dividing a or a + 2, but only chance 1/3 of dividing a and a + 6, so the latter pair is conjectured twice as likely to both be prime.