In number theory, Polignac's conjecture was made by Alphonse de Polignac in 1849 and states:[1]
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.[2]
Although the conjecture has not yet been proven or disproven for any given value of n, in 2013 an important breakthrough was made by Yitang Zhang who proved that there are infinitely many prime gaps of size n for some value of n < 70,000,000.[3] [4] Later that year, James Maynard announced a related breakthrough which proved that there are infinitely many prime gaps of some size less than or equal to 600.[5] As of April 14, 2014, one year after Zhang's announcement, according to the Polymath project wiki, n has been reduced to 246.[6] Further, assuming the Elliott–Halberstam conjecture and its generalized form, the Polymath project wiki states that n has been reduced to 12 and 6, respectively.[7]
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.
Dickson's conjecture generalizes Polignac's conjecture to cover all prime constellations.
Let
\pin(x)
The first Hardy–Littlewood conjecture says the asymptotic density is of form
\pin(x)\sim2Cn
| x | |
| (lnx)2 |
\sim2Cn
| x | |
| \int | |
| 2 |
{dt\over(lnt)2}
where Cn is a function of n, and
\sim
C2 is the twin prime constant
C2=\prodp\ge
| p(p-2) | |
| (p-1)2 |
≈ 0.660161815846869573927812110014...
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:
Cn=C2\prodq|n
| 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}
\tfrac{2}{q}
\tfrac{1}{q}
\tfrac{q-1}{q}
\tfrac{q-2}{q}
\tfrac{q-1}{q-2}