Let \pi_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
Cn is a function of
n, and \sim 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.
Cn 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 = 2
C2. 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 one 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 a probability of 2/3 of dividing
a or
a + 2, but only a probability of 1/3 of dividing
a and
a + 6, so the latter pair is conjectured twice as likely to both be prime. == Notes ==