The prime number theorem tells us how the primes thin out on average: near a large number x the density is about 1/\ln x. But the primes also come in constellations — twins (11,13), cousins (13,17), sexy pairs (11,17), triplets (11,13,17), and the tight prime quadruplet (11,13,17,19). Zoom in and you find these little clusters recurring seemingly forever.

In their monumental 1923 memoir "Some problems of 'Partitio Numerorum' III," G. H. Hardy and J. E. Littlewood did something audacious: they did not merely conjecture that such constellations occur infinitely often — they wrote down an exact formula for how many there should be up to x, correct down to the leading constant. This page is about that formula: which patterns are even possible, and precisely how common each one is.

Patterns as tuples: the shape of a constellation

Strip a constellation of its position and keep only its shape. A pattern is a set of offsets

and an instance of the pattern is an integer n for which all of n+h_1,\, n+h_2,\, \dots,\, n+h_k are prime. Twin primes are the pattern \mathcal{H} = \{0, 2\}; cousins are \{0, 4\}; the tight quadruplet is \{0, 2, 6, 8\}. The k-tuple conjecture asks: for which shapes \mathcal{H} are there infinitely many instances, and how thickly do they occur?

Some shapes are hopeless — and the reason has nothing to do with the size of the numbers and everything to do with a local arithmetic accident. Meet the gatekeeper.

Admissibility: the local obstruction

Fix a prime p and look at the offsets modulo p. Let

Here is the crux. If \mathcal{H} hits every single residue class mod p — that is, if \nu_{\mathcal{H}}(p) = p — then no matter which n you choose, one of the numbers n + h_i is forced to be divisible by p. Beyond the one lucky case where that number equals p itself, it is a genuine composite. The pattern is strangled by p and can occur only finitely often.

So checking a pattern is quick: run through the primes up to k and make sure none of them is completely covered. That's the whole test.

Worked example: \{0,2,6\} passes, \{0,2,4\} fails

Both are triples, so k = 3 and only the primes p = 2 and p = 3 can obstruct.

Test \{0, 2, 6\}.

• Mod 2: the offsets are 0, 0, 0, so \nu(2) = 1 < 2. The class 1 \bmod 2 is missed. ✓
• Mod 3: the offsets are 0, 2, 0, occupying \{0, 2\}, so \nu(3) = 2 < 3. The class 1 \bmod 3 is missed. ✓

Every prime leaves a gap, so \{0,2,6\} is admissible. And indeed we find instances aplenty: (5,7,11), (11,13,17), (17,19,23), …

Test \{0, 2, 4\}. Mod 3 the offsets are 0, 2, 1 — that is all three classes \{0, 1, 2\}, so \nu(3) = 3 = p. The tuple is not admissible: whatever n is, exactly one of n,\, n+2,\, n+4 is a multiple of 3. The only way that multiple can still be prime is if it is 3, which pins the single instance to (3, 5, 7). After that, three-term arithmetic-style clusters of gap 2 are gone forever — a hard theorem, not a conjecture.

The first Hardy–Littlewood conjecture

Admissibility removes the only obvious obstruction. The daring claim is that it is the only obstruction — and that once it is cleared, the constellation appears with a completely definite frequency.

Read the shape of the formula first. A single "random" number near x is prime with chance about 1/\ln x; if the k events "n+h_i is prime" were independent, all k together would happen with chance 1/(\ln x)^{k}, giving roughly x/(\ln x)^k instances. They are not independent — divisibility by each small prime couples them — and the singular series \mathfrak{S}(\mathcal{H}) is exactly the accumulated correction for that coupling, prime by prime.

Reading the singular series factor by factor

Each prime p contributes the ratio

The denominator (1 - 1/p)^k is what the naive independence guess predicts: each of the k numbers independently avoids the class 0 \bmod p with probability 1 - 1/p. The numerator 1 - \nu_{\mathcal{H}}(p)/p is the true probability that a random n dodges the danger mod p: the tuple forbids n from landing in any of the \nu_{\mathcal{H}}(p) classes that would make some n + h_i \equiv 0. The factor is the ratio of the real world to the naive one.

Notice the numerator 1 - \nu_{\mathcal{H}}(p)/p is exactly zero when \nu_{\mathcal{H}}(p) = p — a non-admissible tuple makes \mathfrak{S}(\mathcal{H}) = 0, and the formula honestly predicts zero instances (asymptotically). Admissibility is precisely the condition that keeps the product away from zero. For every prime p > k the ratio is already so close to 1 that the product converges.

The twin-prime case: computing \mathfrak{S}(\{0,2\})

Take \mathcal{H} = \{0, 2\}, so k = 2. Split the product into p = 2 and the odd primes.

• At p = 2: both offsets are even, so \nu(2) = 1. The factor is \dfrac{1 - 1/2}{(1 - 1/2)^{2}} = \dfrac{1/2}{1/4} = 2.
• At an odd prime p: 0 and 2 are distinct mod p, so \nu(p) = 2, and the factor is \dfrac{1 - 2/p}{(1 - 1/p)^{2}} = \dfrac{p(p-2)}{(p-1)^2} = 1 - \dfrac{1}{(p-1)^2}.

Multiplying everything together,

where \Pi_2 = \prod_{p>2}\bigl(1 - (p-1)^{-2}\bigr) \approx 0.6601618 is the twin-prime constant. So Hardy and Littlewood predict

The prediction is uncannily accurate. Below is the true count of twin-prime pairs against 2\,\Pi_2 \int_2^x \frac{dt}{(\ln t)^2} — the same constant, dressed in the smoother logarithmic-integral form that removes the slow bias of the bare x/(\ln x)^2.

A gallery of small admissible tuples

The singular series is a single number attached to each shape — the "clumping factor" by which that constellation beats or trails the naive 1/(\ln x)^k baseline. A few of the smallest patterns:

Two things stand out. First, cousins are exactly as common as twins (\{0,4\} and \{0,2\} share the same singular series) — the extra factor of 2 for sexy primes comes from 6 being divisible by 3, which relaxes the mod-3 constraint. Second, the tight quadruplet \{0,2,6,8\} is the smallest admissible 4-tuple: its rival \{0,2,4,6\} is killed mod 3, exactly like \{0,2,4\}.

Where the formula comes from: the circle method

The precise constant is not guesswork. Hardy and Littlewood derived it heuristically with their circle method, weighing the "major arcs" — the contributions near rational points a/q — where each small prime p throws in exactly the local factor \frac{1 - \nu_{\mathcal{H}}(p)/p}{(1-1/p)^k}. The singular series is the product of these local densities, one per prime, and this "local-to-global" product structure is the same one that appears in Waring's problem and Goldbach's conjecture. Sieve methods approach the same target from below: Brun's sieve cannot prove infinitude, but it does prove the matching upper bound \pi_2(x) \ll x/(\ln x)^2 with a constant of the right order, pinning down the shape even though the exact constant stays conjectural.

The second conjecture — and why it must be false

In the very same paper, Hardy and Littlewood floated a second, innocent-looking guess about primes in intervals:

It sounds obviously true — primes thin out, so surely the richest stretch of length y is the one at the very beginning. Yet in 1974 Douglas Hensley and Ian Richards proved the shock: the two Hardy–Littlewood conjectures cannot both be true. The first conjecture guarantees the existence of very dense admissible constellations — one can build an admissible tuple \mathcal{H} inside an interval of length y that packs in more than \pi(y) offsets. The k-tuple conjecture then says that, infinitely often, some far-out interval (x, x+y] realises all of them as primes — more than \pi(y) \ge \pi(x+y) - \pi(x) allows. Contradiction.

Almost everyone believes the first conjecture (it is the natural, well-tested one), so the second is believed false — even though no explicit counterexample interval has ever been exhibited (the guaranteed dense tuples live absurdly far out). It is a rare and beautiful situation: a plausible statement that is almost certainly wrong, brought down not by a computation but by a different conjecture it quietly contradicts.