## Another Unsolved Problem- Polignac’s conjecture

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 yet been proven or disproven for a given value of n. In 2013 an important breakthrough was made by Zhang Yitang who proved that there are infinitely many prime gaps of size n for some value of n < 70,000,000.

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

Dickson’s conjecture generalizes Polignac’s conjecture to cover all prime constellations; the Bateman–Horn conjecture gives conjectured asymptotic densities.

#### Conjectured density

Let  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

where Cn is a function of n, and  means that the quotient of two expressions tends to 1 as x approaches infinity.[citation needed]

C2 is the twin prime constant

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

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

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 . 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 , since q divides 1 of the q numbers from a to a + q − 1. Now assume q divides nand consider a potential prime pair (aa + n). q divides a + n if and only if q divides a, and the chance of that is . The chance of (aa + n) being free from the factor q, divided by the chance that (aa + 2) is free from q, then becomes  divided by . This equals  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.