Hardy-Littlewood Prime k-Tuple Conjecture

Hardy-Littlewood Prime k-Tuple Conjecture

Definitions and mathematical statement

For a finite set $H=\{h_1,\ldots,h_k\}\subset{\mathbb{Z}}$ of distinct offsets, let $\nu_p(H)$ be the number of residues it occupies modulo $p$. Admissibility means $\nu_p(H)<p$ for every prime $p$. Define \[ \mathfrak S(H)=\prod_p\frac{1-\nu_p(H)/p}{(1-1/p)^k},\qquad \pi_H(x)=\#\{1\le n\le x:n+h_i\text{ prime for every }i\}. \] For each fixed admissible $H$, the conjecture is $\pi_H(x)\sim\mathfrak S(H)x/(\log x)^k$ as $x\to\infty$. Here $f\sim g$ means $f/g\to1$; this is stronger than infinitude alone.

Formulation and concept references: [S1].

Short English statement

Does every locally admissible fixed pattern of primes occur with exactly the frequency predicted by its divisibility corrections?

Research attempt

Exact target and literature distinction.

The offsets are distinct and fixed before $x$ tends to infinity. Negative offsets merely remove finitely many initial integers: put $n_0=\max(1,2-\min H)$ so all $n+h$ under consideration are at least two. The empty pattern, if permitted, gives $\pi_H(x)=\lfloor x\rfloor$ and is immediate. We assume $k\ge1$ below. Repeated offsets cannot be inserted as extra coordinates: two copies of zero would leave the prime count unchanged while incorrectly changing the proposed logarithmic exponent.

Maynard [E1] proves strong results about many primes in bounded tuples and a positive proportion of patterns satisfying an infinitude statement. Those conclusions do not assert the precise asymptotic for each prescribed $H$ requested here. No primorial-counting claim in catalogue source [S3] is assumed to prove that stronger assertion. The selected attack is to derive everything supplied by local divisibility exactly, then attempt to pass to the sieve level that detects actual primes. This isolates a quantifiable missing term, rather than treating local independence as an unproved probability law.

Admissibility and positivity of the singular series.

If $\nu_p(H)=p$ for some prime $p$, every integer $n$ makes some $n+h_i$ divisible by $p$. If all these integers are prime, that particular coordinate must equal $p$, so $n=p-h_i$ for one of at most $k$ choices. Thus an inadmissible pattern has only finitely many occurrences. This proves the necessity of local admissibility for infinitude, allowing exceptional small tuples such as $(3,5,7)$ for $H=\{0,2,4\}$.

Conversely, assume admissibility. Let $B=\max_{i,j}|h_i-h_j|$, taking $B=0$ for $k=1$, and choose $y>\max(B,2k,2)$. For $p>y$ all offsets are distinct modulo $p$, so $\nu_p(H)=k$. Write $F_p=(1-k/p)/(1-1/p)^k$. Expanding the logarithms absolutely gives \[ \log F_p=-\sum_{m\ge2}\frac{k^m-k}{mp^m},\qquad -\frac{k^2}{p^2}\le\log F_p\le0. \] Indeed $1/m\le1/2$ for $m\ge2$ and $k/p<1/2$, giving $\sum_{m\ge2}(k/p)^m/m\le k^2/p^2$. Hence the infinite product converges to a positive value, and its tail has the explicit enclosure \begin{equation}\label{wsref-0-72-61-3a-31-38-3a-74-61-69-6c} \exp\!\left(-\frac{k^2}{y-1}\right) \le\prod_{p>y}F_p\le1. \end{equation} We used the elementary bound $\sum_{p>y}p^{-2}\le\sum_{n>y}n^{-2}\le1/(y-1)$. All earlier factors are positive by admissibility. This proves that the catalogue's constant is well-defined and nonzero, with a computable tail certificate. It proves no occurrence of the pattern.

Translation of $H$ preserves every $\nu_p$ and the singular series. Multiplying all offsets by an integer need not preserve it: additional collisions arise at primes dividing that integer. The proposed frequency therefore contains arithmetic information beyond the diameter of the pattern. In the special case $k=1$ every local factor is exactly one.

A finite admissibility test is not a finite primality test.

Admissibility itself is decidable by checking only primes $p\le k$: for $p>k$ the elementary inequality $\nu_p(H)\le k<p$ is automatic. For example, $H=\{0,2,6\}$ occupies one residue modulo two and two residues modulo three, so it is admissible without any further local tests. The pattern $\{0,2,4\}$ occupies all three residues modulo three and fails. This finite procedure checks whether a local obstruction exists, not whether the admissible pattern has a single prime occurrence beyond its initial range. There is no analogous cutoff at $p=k$ when testing primality of the values $n+h$: composite values can have their smallest prime factor much larger than the number of coordinates. The distinction matters computationally as well. A program may certify admissibility quickly and compute a positive singular-series enclosure, yet still lack any lower bound for the occurrence count. Those certificates are useful input checks, and the next calculation states exactly how far their counting consequences extend.

Finite wheels: what the Chinese remainder theorem proves.

Let $P(z)=\prod_{p\le z}p$ and \[ T_H(x,z)=\#\{n_0\le n\le x: \gcd(\prod_{h\in H}(n+h),P(z))=1\}. \] For each prime $p\le z$ there are exactly $p-\nu_p(H)$ allowed residues for $n$. The Chinese remainder theorem identifies these choices independently across the distinct prime moduli. Thus a complete block modulo $P(z)$ has exactly \[ R_H(z)=\prod_{p\le z}(p-\nu_p(H)) \] survivors. In an interval of length $L=\lfloor x\rfloor-n_0+1\ge0$, each allowed residue appears either $\lfloor L/P(z)\rfloor$ or $\lceil L/P(z)\rceil$ times. Therefore \begin{equation}\label{wsref-0-72-61-3a-31-38-3a-77-68-65-65-6c} T_H(x,z)=L D_H(z)+O(R_H(z)),\qquad D_H(z)=\prod_{p\le z}(1-\nu_p(H)/p), \end{equation} with absolute error at most $R_H(z)$.

For every fixed $z$, this is an exact density theorem as $x\to\infty$. Its implied error depends on $z$; the statement does not license taking $z=\sqrt x$ while retaining a bounded error. Admissibility ensures that every finite wheel has survivors, but most survivors can still have prime factors beyond the wheel. The infinite sequence of positive finite-wheel densities is not itself a prime-pattern count.

Exact inclusion--exclusion exposes the accumulated error.

For squarefree $d\mid P(z)$ put $\nu_d=\prod_{p\mid d}\nu_p(H)$, with $\nu_1=1$, and let $A_d(x)$ count $n_0\le n\le x$ such that $d\mid\prod_h(n+h)$. CRT now counts forbidden residues, giving $A_d=L\nu_d/d+e_d$ with $|e_d|\le\nu_d$. Inclusion--exclusion gives the exact identities \begin{align} T_H(x,z)&=\sum_{d\mid P(z)}\mu(d)A_d(x) =L D_H(z)+\mathcal R_H(x,z),\label{wsref-0-72-61-3a-31-38-3a-49-45}\\ \mathcal R_H(x,z)&=\sum_{d\mid P(z)}\mu(d)e_d, \qquad |\mathcal R_H(x,z)|\le\prod_{p\le z}(1+\nu_p(H)).\notag \end{align} The expansion is finite, so there is no convergence issue. The difficulty is the size and signed behavior of the remainder when $z$ grows. A bound on each progression separately does not survive summing over the enormous family of divisors at the needed accuracy.

Now set $z=\sqrt{x+\max H}$ for sufficiently large $x$. Every positive integer $n+h\le x+\max H$ which is composite has a prime divisor at most $z$. Hence every survivor is an actual prime tuple, with all its components greater than $z$. Conversely any prime tuple omitted by the wheel has one component at most $z$, hence $n\le z-\min H$. We have proved the unconditional relation \begin{equation}\label{wsref-0-72-61-3a-31-38-3a-70-72-69-6d-65-2d-6c-65-76-65-6c} \pi_H(x)=T_H(x,\sqrt{x+\max H})+O_H(\sqrt x). \end{equation} The error is negligible compared with $x/(\log x)^k$ for fixed $k$. This brings the argument to the exact sieve level where the desired asymptotic would follow, but the remainder in (ra:18:IE) is now uncontrolled.

A sharper falsification of simply discarding that remainder.

Mertens' classical product theorem is $\prod_{p\le z}(1-1/p)\sim e^{-\gamma}/\log z$; see [E2], introduction. Together with the proved convergence of the singular series, it gives \[ D_H(z)\sim\mathfrak S(H)\frac{e^{-k\gamma}}{(\log z)^k}. \] At the prime-detecting level $z=\sqrt{x+\max H}$ this predicts \[ L D_H(z)\sim (2e^{-\gamma})^k\mathfrak S(H) \frac{x}{(\log x)^k}. \] The multiplier is not one. Already for $H=\{0\}$, dropping $\mathcal R_H$ contradicts the prime number theorem [E3]: the actual count is asymptotic to $x/\log x$, whereas $2e^{-\gamma}=1.1229\ldots$. Thus the issue is stronger than a missing small-error proof. At this sieve level the signed correction has a nonzero main term even in the solved one-coordinate case.

For general fixed admissible $H$, the precise estimate that would complete this particular calculation is \begin{equation}\label{wsref-0-72-61-3a-31-38-3a-67-61-70} \mathcal R_H(x,\sqrt{x+\max H}) =\left(1-(2e^{-\gamma})^k+o(1)\right) \mathfrak S(H)\frac{x}{(\log x)^k}. \end{equation} Equations (ra:18:IE) and (ra:18:prime-level) prove its sufficiency. No derivation of (ra:18:gap) is given. It is a specific arithmetic correction problem, not a consequence of multiplying the finite local densities. This elementary obstruction is consistent with the limitations of sieve methods discussed in [E1].

Stress tests for order of limits and pattern choice.

An abstract family makes the order-of-limits issue unmistakable. Let $f_z(x)=1$ when $x\ge z^3$ and zero otherwise. For each fixed $z$, $f_z(x)\to1$ as $x\to\infty$, but $f_{\sqrt x}(x)=0$ for $x>1$. Thus a theorem valid for each fixed primorial does not become a theorem at a primorial whose cutoff grows with the observation range. The actual wheel has additional arithmetic structure, but that structure must enter a uniform estimate; the quantifiers alone do not provide it.

One also cannot choose a new favorable pattern $H_x$ at each scale and claim the required result for every fixed $H$. The bounds (ra:18:tail) and (ra:18:prime-level) explicitly depend on the offsets and their diameter. Conversely, a proof for one prescribed admissible pattern with $k\ge2$ would be genuine progress, even without uniformity in all offsets. The catalogue asks for all such fixed patterns, not a density-one family of them and not merely an unspecified pair of prime positions inside a larger tuple.

Finite verification.

The standard-library script checks CRT and complete inclusion--exclusion for the wheel $P(7)=210$, for $H=\{0\},\{0,2\},\{0,2,6\}, \{0,2,4\},\{-2,0,4\}$ and $x=100,1000,10000$. It also independently sieves the actual prime tuples. Selected results are \[ \begin{array}{c|r|r|r|r} H&R_H(7)&x&\pi_H(x)&T_H(x,7)\\\hline \{0,2\}&15&100&8&6\\ \{0,2\}&15&1000&35&70\\ \{0,2\}&15&10000&205&713\\ \{0,2,6\}&8&10000&55&381\\ \{0,2,4\}&0&10000&1&0 \end{array} \] At small $x$ a wheel can exclude genuine tuples containing its small primes, explaining the first row. At larger $x$ it admits many composites, explaining the later discrepancies. The inadmissible example retains its one exceptional tuple. All counts and identities are exact; no asymptotic constant was fitted to the data. Reproduction files are in _Local/top_problems/continuation_15_25_2026_09_27/analytic/.

Remaining gap and outcome.

The proved results are positivity with a quantitative tail bound for the singular series, exact finite-wheel counts, and the prime-level sieve identity. They isolate the missing correction (ra:18:gap) and show why setting it to zero gives the wrong answer even for one prime. Neither the required signed remainder nor a counterexample pattern has been obtained. The next mathematical advance would need uniform control of that remainder or a different method detecting joint primality, while retaining the precise leading constant.

Outcome: UNRESOLVED. The local computations and obstruction proofs are rigorous partial analysis, not a solution of the asymptotic conjecture. Direct derivation, primary references, and exact finite checks were used; no independent formal certification is claimed.

Sources

• S1 k-Tuple Conjecture (MathWorld). https://mathworld.wolfram.com/k-TupleConjecture.html. Formulation source.
• S2 Correlations of almost primes — Institut Henri Poincaré seminar abstract. https://www.ihp.fr/en/node/3096. Further reference; not a proof endorsement.
• S3 Proof of the Hardy-Littlewood K-tuple Conjecture in the Distribution of Numbers Coprime with the Primorial — Tim Samshuijzen, May 2026 revision. https://rxiv.org/pdf/2501.0157v4.pdf. Further reference; not a proof endorsement.
• S4 Higher uniformity of arithmetic functions in short intervals II. Almost all intervals — Matomäki, Radziwiłł, Shao, Tao and Teräväinen $2026$. https://link.springer.com/article/10.1007/s00222-026-01408-6. Further reference; not a proof endorsement.

• E1 James Maynard, Small gaps between primes, Annals of Mathematics 181 $2015$, 383--413. https://annals.math.princeton.edu/wp-content/uploads/annals-v181-n1-p07-p.pdf. Primary research context: bounded tuples and infinitude for a positive proportion of patterns, not the asymptotic for every specified pattern. Consulted 27 September 2026.
• E2 Youness Lamzouri, A bias in Mertens' product formula, arXiv:1410.3777v2 (6 February 2015), introduction. https://arxiv.org/pdf/1410.3777v2. Research source recording the classical unconditional Mertens product theorem used in the cutoff stress test. The paper's conditional bias results are not invoked. Consulted 27 September 2026.
• E3 Andrew V. Sutherland, 18.785 Number Theory I, Lecture 16: Riemann's zeta function and the prime number theorem, MIT, fall 2021. https://math.mit.edu/classes/18.785/2021fa/LectureNotes16.pdf. Author lecture notes for the one-coordinate prime number theorem, used to falsify the naive sieve asymptotic. Consulted 27 September 2026.

Hardy-Littlewood Prime k-Tuple Conjecture

Definitions and mathematical statement

For a finite set $H=\{h_1,\ldots,h_k\}\subset{\mathbb{Z}}$ of distinct offsets, let $\nu_p(H)$ be the number of residues it occupies modulo $p$. Admissibility means $\nu_p(H)<p$ for every prime $p$. Define \[ \mathfrak S(H)=\prod_p\frac{1-\nu_p(H)/p}{(1-1/p)^k},\qquad \pi_H(x)=\#\{1\le n\le x:n+h_i\text{ prime for every }i\}. \] For each fixed admissible $H$, the conjecture is $\pi_H(x)\sim\mathfrak S(H)x/(\log x)^k$ as $x\to\infty$. Here $f\sim g$ means $f/g\to1$; this is stronger than infinitude alone.

Formulation and concept references: [S1].

Short English statement

Does every locally admissible fixed pattern of primes occur with exactly the frequency predicted by its divisibility corrections?

Research attempt

Current frontier.

Research check: 27 September 2026. Maynard's multidimensional sieve proves bounded intervals containing many primes, not the asymptotic for each prescribed tuple. Green--Tao's linear-equations framework excludes the parallel linear parts in the single-variable forms $n+h_i$. [E1], [E2]. The 2026 short-interval uniformity theorem in the supplied sources gives Hardy--Littlewood consequences with an additional short average; it does not remove that average for every fixed $H$. [S4]. The primorial proof claim in [S3] is retained as an original source, not accepted as a verified proof. The following argument tests the precise local-to-global passage on which such an approach would have to improve.

Attack route.

We compare three possible routes: direct simultaneous sieving, cancellation in von Mangoldt correlations, and averaging over offsets followed by removal of the average. They respectively require control of a growing sieve modulus, a genuine fixed-shift correlation estimate, or a uniform bound on exceptional offsets. We pursue the first two together, so that one can see exactly which information survives from the local factors.

Attempt: the exact local calculation and its range.

Assume first $k\geq1$ and fix $H$. Write \[ W(z)=\prod_{p\leq z}p,\qquad R_z=\{a\bmod W(z):\gcd(\prod_i(a+h_i),W(z))=1\}. \] The Chinese remainder theorem gives \[ |R_z|=\prod_{p\leq z}(p-\nu_p(H)),\qquad \#\{n\leq x:n\bmod W(z)\in R_z\} =x\prod_{p\leq z}(1-\nu_p(H)/p)+O(|R_z|). \tag{18.1} \] The error follows by counting each residue class as $x/W(z)+O(1)$; its implied constant is absolute. No independence assumption is used. Normalizing the main term by the density for $k$ individually rough integers gives exactly \[ \mathfrak S_z(H)= \prod_{p\leq z}\frac{1-\nu_p(H)/p}{(1-1/p)^k}. \tag{18.2} \] For every sufficiently large prime the offsets are distinct modulo $p$, so $\nu_p(H)=k$. Taylor expansion at zero yields \[ \log\frac{1-k/p}{(1-1/p)^k} =-\frac{k(k-1)}{2p^2}+O_k(p^{-3}). \] The tail converges absolutely; admissibility makes each of the finitely many remaining factors positive. This proves existence and positivity of the singular series, rather than just treating its product formally.

For fixed $z$, $18.1$ is excellent as $x\to\infty$, but it counts composite survivors too. To make roughness equivalent to primality for positive $n+h_i\leq x+\max_i|h_i|$, one would sieve through its square root (apart from the small primes themselves). Formula $18.1$ supplies no useful relative error at that growing modulus. The problem is not the CRT identity: it is the unproved cancellation in its errors when the cutoff is increased with $x$. Taking $x\to\infty$ first at each fixed $z$ does not justify evaluating a diagonal $z=z(x)$.

Attempt: an exact weighted reduction.

Extend $\Lambda(t)$ by zero for integers $t\leq0$, and put \[ C_H(x)=\sum_{1\leq n\leq x}\prod_{i=1}^k\Lambda(n+h_i). \tag{18.3} \] A precise sufficient statement is $C_H(x)=\mathfrak S(H)x+o_H(x)$ for every fixed admissible $H$. Here is the removal of both prime powers and logarithmic weights. The number of $n\leq x$ for which some $n+h_i=p^a$ with $a\geq2$ is $O_H(\sqrt{x}\log x)$, by summing the elementary bounds $(x+O_H(1))^{1/a}$ over $2\leq a\leq\log_2(x+O_H(1))$. Every term in $18.3$ is $O_H((\log x)^k)$. Therefore the total contribution from such powers is $o_H(x)$.

Let $A_H(t)$ count genuine prime tuples up to $t$ and let $w(t)=\prod_i\log(t+h_i)$ for $t$ larger than a fixed constant $t_0(H)$. The weighted count $B_H(x)=\int_{t_0}^xw(t)\,dA_H(t)$ consequently satisfies $B_H(x)=\mathfrak S(H)x+o_H(x)$. For large $t$, \[ w(t)\sim(\log t)^k,\qquad \frac{w'(t)}{w(t)^2}=O_H\!\left(\frac1{t(\log t)^{k+1}}\right). \] Stieltjes integration by parts gives \[ A_H(x)=\frac{B_H(x)}{w(x)} +\int_{t_0}^x B_H(t)\frac{w'(t)}{w(t)^2}\,dt+O_H(1). \tag{18.4} \] Since $B_H(t)=O_H(t)$, the integral is $O_H(x/(\log x)^{k+1})$; one can bound it by splitting at $\sqrt{x}$. Thus $18.3$'s proposed estimate implies exactly the stated unweighted asymptotic, with no extra uniformity in $H$ demanded. Conversely the unweighted asymptotic and integration by parts recover $18.3$, after the same prime-power estimate. This is an equivalent analytic target, not a new proof of it. Negative offsets affect only finitely many initial terms. If the empty tuple is admitted, its assertion is simply $\lfloor x\rfloor \sim x$ and needs no argument.

Attempt: what an averaged theorem cannot by itself remove.

The obstacle can be made quantitative without pretending to build fake primes. For any positive comparison constants $s_h$, let nonnegative model counts satisfy $b_h(X)=s_hX$ except for one fixed $h_0$, where $b_{h_0}(X)=0$. Then \[ \frac1H\sum_{h=1}^H \left|\frac{b_h(X)}{s_hX}-1\right|=\frac1H \quad(H\geq h_0). \tag{18.5} \] Perfectly good mean convergence with $H\to\infty$ coexists with a permanently bad fixed offset. This countermodel invalidates removal of an average solely by nonnegativity. It is not a model of the primes and not a counterexample to Hardy--Littlewood.

One possible strengthening would be a uniform estimate $|C_H(x)-\mathfrak S(H)x|\leq \epsilon(x)x$ for every admissible tuple in a box whose radius tends to infinity, with $\epsilon(x)\to0$ and the number of offsets fixed. Every fixed tuple would eventually be included. But that strengthening already contains the fixed-shift difficulty; the current averaged statements do not supply it. A different route would prove $18.3$ by combining the CRT factors with cancellation beyond the sieve's parity obstruction, rather than identifying survivors with primes.

Stress test.

The convergent singular series is only a local density correction. The error in $18.1$ is not uniform enough for primality detection. No argument here promotes higher uniformity in a finite-complexity system to the parallel-shift system without checking its hypotheses. The models in $18.5$ test an inference, not arithmetic reality. Proving $18.3$ even for $H=\{0,2\}$ would yield the twin-prime assertion of excluded UnsolvedMath problem 11; that dependency is explicitly retained, although no separate file for UnsolvedMath problem 11 is produced.

Outcome.

Outcome: Exploratory approach with unresolved gap. The exact CRT count and weighted/unweighted equivalence isolate the missing fixed-tuple asymptotic $18.3$. Neither the sieve error nor the exceptional-offset error is controlled at the required scale. No novelty is claimed for the classical ingredients.

Sources

• S1 k-Tuple Conjecture (MathWorld). https://mathworld.wolfram.com/k-TupleConjecture.html. Catalogue formulation source.
• S2 Correlations of almost primes — Institut Henri Poincaré seminar abstract. https://www.ihp.fr/en/node/3096. Catalogue research/status reference; not a proof endorsement.
• S3 Proof of the Hardy-Littlewood K-tuple Conjecture in the Distribution of Numbers Coprime with the Primorial — Tim Samshuijzen, May 2026 revision. https://rxiv.org/pdf/2501.0157v4.pdf. Catalogue research/status reference; not a proof endorsement.
• S4 Higher uniformity of arithmetic functions in short intervals II. Almost all intervals — Matomäki, Radziwiłł, Shao, Tao and Teräväinen $2026$. https://link.springer.com/article/10.1007/s00222-026-01408-6. Catalogue research/status reference; not a proof endorsement.
• E1 James Maynard, Small gaps between primes, arXiv:1311.4600 $2013$, Theorem 1.1 and the sieve setup. https://arxiv.org/abs/1311.4600. Added research source. Bounded clusters versus a prescribed simultaneous prime pattern. Consulted 27 September 2026.
• E2 Ben Green and Terence Tao, Linear Equations in Primes, arXiv:math/0606088, introduction and finite-complexity hypotheses. https://arxiv.org/abs/math/0606088. Added research source. Scope of the linear-forms results; parallel shifts are not covered by the finite-complexity case. Consulted 27 September 2026.