Ordinary two-point multiplicative correlations

Source and verification

This article explains family 007 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Multiplicative functions and two-point averages

Let $\Omega (n)$ count prime factors with multiplicity. The Liouville function is $\lambda (n)=(-1{)}^{\Omega (n)}$; it satisfies $\lambda (mn)=\lambda (m)\lambda (n)$ for all positive integers. For a fixed $h\ge 1$, the average

$$\frac{1}{X}\sum _{n\le X}\lambda (n)\lambda (n+h)$$

measures the difference between the proportions with equal and opposite factor-count parity.

An ordinary average assigns equal weight to integers. A logarithmic average instead weights $n$ by $1/n$ and divides by $\log X$. Partial summation transfers ordinary cancellation to logarithmic cancellation; the converse does not follow, since a logarithmic average mixes many scales.

Claimed quantitative Liouville theorem

There is an absolute $c>0$ such that, for every fixed $a_1,a_2\ge 1$ and $b_1,b_2\ge 0$ with $a_1{b}_2-a_2{b}_1\ne 0$, the manuscript proves

$$\begin{array}{c}\left|\sum _{n\le X}\lambda (a_{1}n+b_1)\lambda (a_{2}n+b_2)\right|\\ {\ll }_{a_1,a_2,b_1,b_2}\frac{X}{(\log X{)}^c}.\end{array}$$

This is asserted for every real $X\ge 3$. The exponent is common to the forms; the implied constants may be ineffective. There is no asserted uniformity for coefficients growing with $X$. (OpenAI, 2026)

The determinant condition excludes proportional forms. Its necessity is visible in the example $\lambda (n)\lambda (2n)=-1$, which has no cancellation.

🏷️ The corrected Elliott hypothesis

A function $f:\mathbb{N}\to \mathbb{C}$ is multiplicative if $f(mn)=f(m)f(n)$ for coprime $m,n$. Assume $|f(n)|\le 1$. A Dirichlet character $\chi$ is a periodic multiplicative phase, extended by zero off the invertible residue classes. The twist $n^{it}=\exp (it\log n)$ is a slowly changing multiplicative phase.

Define

$$\begin{array}{c}D(f,\chi n^{it};X{)}^2\\ =\sum _{p\le X}\frac{1-\mathrm{Re}(f(p)\overline{\chi (p)p^{it}})}{p}.\end{array}$$

Small distance means that $f$ imitates a character times a logarithmic phase on primes. The source’s uniform nonpretentiousness condition is, for every fixed $\chi$,

$$\underset{|t|\le N}{\inf }D(f,\chi n^{it};N)⟶\infty .$$

The infimum matters: the best matching frequency may change with $N$. Divergence for each fixed $t$ is a different hypothesis.

The manuscript claims that if at least one of two bounded multiplicative functions satisfies this condition, then their ordinary correlation at any two fixed distinct nonnegative shifts tends to zero. It also gives the nonproportional affine version. There is no quantitative rate in this general theorem, and the displayed product need not conjugate the second factor. Conjugated versions follow by replacing that factor with its conjugate. (OpenAI, 2026)

🏷️ Dilation graphs and centered divisibility

Multiplicative dilations compare the pair $n,n+h$ with $dn,d(n+h)$. The comparison creates graphs whose edges have displacement $hd$. For a uniformly sampled residue modulo a prime $p$,

$$\mathbb{E}{1}_{p\mid n}=\frac{1}{p},\mathbb{E}\left(1_{p\mid n}-\frac{1}{p}\right)=0.$$

The centered factors are the source of cancellation. To control a graph operator $A$, one estimates a high even trace $\mathrm{tr}(A^{2k})$. Expanding the trace enumerates closed walks and products of edge weights.

A prime occurring once can cancel under residue averaging. A recurring prime creates dependent congruences, so the proof must retain its previous uses. Taking absolute values before averaging would erase precisely the cancellation the graph was designed to exploit.

In the quantitative branch, each edge uses primes from $J$ disjoint supplies of reciprocal mass comparable to a large constant $W$. The normalization supplies roughly $W^J$, whereas the operator estimate costs roughly $(C\sqrt{W}{)}^J$. A sufficiently large fixed $W$, with $J$ proportional to $\log \log X$, converts exponential decay in $J$ into a power of $\log X$.

The full residue period is too large to average directly. The source instead corrects an encoded residue law to exact bounded independence and applies a circuit approximation theorem to the finite tests used in the proof. Repeated-prime patterns are organized through arithmetic constraints and forests, with constants controlled before $W$ is selected.

🏷️ The qualitative order of limits

Uniform nonpretentiousness may diverge arbitrarily slowly. The second branch therefore fixes an auxiliary graph scale $B$, takes the long averaging limit first, and only then sends $B$ to infinity.

A hypothetical biased subsequence produces a graph pairing of size at least a constant times $B^{-1}$. The centering error is $o(B^{-1})$, while the graph bound gains an additional fixed power of $B$. These estimates contradict the bias.

Here some signed factors have the form $1_{p\mid n}-\theta /p$. They are not centered under uniform residues alone. Their cancellation depends on the vertex normalization at both endpoints. This is why the quantitative graph proof cannot simply be copied unchanged.

The affine transfer also requires fixed progression restrictions, including nonunit classes. Complete multiplicativity makes this transfer simpler for Liouville; the general case uses finite local expansions.

🏷️ Key steps and proof outlines

Embed the original correlation in a centered sum

Write $F(X)$ for the normalized Liouville correlation, initially on a fixed progression. The quantitative proof chooses dilation integers $d$, each with one prime from each of $J$ disjoint supplies, and an auxiliary integer $q$ coprime to the progression modulus. It groups $qd$ into short multiplicative intervals. The corresponding terms contain

$$1_{q\mid n}\prod _{p\mid d}\left(1_{p\mid n}-\frac{1}{p}\right)\lambda (n)\lambda (n+hqd).$$

Proof outline. Expand the product. In the term retaining all divisibility indicators, $qd\mid n$, so put $n=qdm$. Complete multiplicativity gives $\lambda (qdm)\lambda (qd(m+h))=\lambda (m)\lambda (m+h)$. If the bin endpoint is $T$, the new cutoff $T/(qd)$ differs from $X$ by a relative amount $O(\eta )$. Boundedness of the summands changes $F$ by only $O(\eta +X^{-1})$. Summing the harmonic dilation weights therefore recovers $S_{0}F(X)$ up to a controlled error, where $S_0$ is their retained total mass. The other expansion terms are handled by the analytic centering estimate. This is the exact bridge from a correlation to a graph pairing; the graph does not estimate an unrelated artificial average. (OpenAI, 2026)

Replace a long residue period by tests of bounded complexity

Independence modulo all selected primes cannot be obtained by averaging over one complete period: that period is too large. The source’s “Comparing finite integer averages with independent residues” instead compares only the residue tests actually used in the proof.

Proof outline. Encode residues by a finite string of bits. The Chinese remainder theorem gives almost uniform marginals on small subsets of these bits. Let $f$ be the density of the encoded integer-origin distribution relative to uniform measure. If the total absolute mass of its nonconstant Fourier coefficients of degree at most $t$ is at most $a$, put

$$H=\sum _{1\le |S|\le t}\hat{f}(S){\chi }_S,g=\frac{f-H+a}{1+a}.$$

Here ${\chi }_S$ is the product of the coordinate signs in $S$. Since $|H|\le a$, $g\ge 0$; its integral is one and all the specified Fourier coefficients vanish. Thus $g$ gives exactly uniform marginals on any $t$ coordinates. Moreover its total variation distance from $f$ is at most $3a/2$. A bounded-independence theorem for circuits then transfers the required finite tests to the fully independent residue model. The correction of the distribution and the circuit theorem have different roles: the first supplies exact independence of small marginals; the second explains why those marginals suffice for the particular large tests.

Nonbacktracking walks control the ordinary edge operator

The trace calculation retains the history of previously used primes and excludes immediate reuse of an edge type. Its output naturally controls a nonbacktracking operator, whereas the desired correlation is a pairing with an ordinary sum of edge operators. The source supplies the following transfer rather than assuming that the two spectra agree.

Let $B_i$ be self-adjoint edge matrices and $P$ a projection deleting vertices of excessive weighted degree. Suppose $\Vert B_i\Vert \le a$ and ${\sum }_i\Vert B_{i}v{\Vert }^2\le b^2\Vert v{\Vert }^2$ for $v=Pv$. Define the block matrix ${\mathcal{B}}_{ij}=1_{i\ne j}{B}_j$. Then

$$\Vert P\left(\sum _i{B}_i\right)P\Vert \le 3\max \{a,b,\rho (\mathcal{B})\},$$

where $\rho$ denotes spectral radius.

Proof outline. Put $m=\max (a,b,\rho )$ and, when $m>0$, consider

$$F(u)=I-\sum _{i}u{B}_i(I+u{B}_i{)}^{-1},|u|\le (2m{)}^{-1}.$$

A vector in the kernel of $F(u)$ would produce, by setting $z_i=(I+u{B}_i{)}^{-1}v$, a vector in the kernel of $I-u\mathcal{B}$. Thus $F(u)$ is invertible. It is self-adjoint and starts at $I$, so continuity makes it positive throughout the interval. The resolvent identity writes it as $I-u{\sum }_i{B}_i+Q(u)$ with $0\le ⟨Q(u)v,v⟩\le 2u^2{b}^2\Vert v{\Vert }^2$ on the range of $P$. Evaluate at both endpoints of the interval to bound the absolute quadratic form of ${\sum }_i{B}_i$ by $3m\Vert v{\Vert }^2$. No two distinct edge matrices were commuted. High moments of the nonbacktracking operator, followed by a tail estimate and this transfer, can now bound the original graph pairing.

Keep the losses when converting a graph gain into a logarithmic saving

Each prime supply has reciprocal mass comparable to $W$. The retained correlation mass is at least a constant times $W^J$, while the spectral argument has square-root cost per supply. But binning and degree truncation introduce extra parameters: the source obtains a spectral contribution

$$\frac{K}{\eta }{\left(\frac{C_3}{\sqrt{W}}\right)}^J.$$

Proof outline. Choose $\eta =e^{-J}$ and $K=e^{4J}$. Their combined cost is $e^{5J}$, so select the fixed constant $W$ only after $C_3$ is known, with $e^5{C}_3/\sqrt{W}\le e^{-1}$. The displayed term is then at most $e^{-J}$. The source checks its centering, deletion, and binning errors against the same budget. Finally $J$ is proportional to $\log \log X$, giving $e^{-J}\ll (\log X{)}^{-c}$. Leaving out the $K/\eta$ factor would miss a necessary part of the parameter choice.

Qualitative cancellation uses endpoint weights and an ordered limit

For general nonpretentious functions, no fixed logarithmic rate is available. The proof holds the auxiliary graph scale $B$ fixed while sending the integer averaging length to infinity. A hypothetical correlation bias supplies a weighted graph pairing of size at least $c{L}_0/B$, where $L_0$ is the graph’s normalization. Centering changes it by $o(L_0/B)$, whereas the graph estimate bounds it by $L_0{B}^{-1-c_{\ast }/2}$. Sending $B$ to infinity afterwards gives the contradiction.

Proof outline for the signed prime gain. On a good edge a prime has one occurrence, and its active residue normalizes both endpoints by ${\beta }_Z$. The coefficient is chosen as $\theta ={\beta }_Z^2$. The active residue contributes $\theta (1-\theta /p)/p$ to the residue average; inactive residues contribute negative terms of size $\theta {\beta }_Z^{\nu (a)}/p^2$, where $\nu (a)$ counts the visits active at residue $a$. Their order-$1/p$ terms cancel. Only the residues visited by the walk differ from the baseline, leaving $O(\ell /p^2)$ for a walk of length $\ell$. Repeated primes require the source’s memory and forest bookkeeping, since their residue conditions cannot be resampled independently. This calculation explains both the extra saving and why taking absolute values before the prime average destroys it.

🏷️ Connections

Consecutive largest prime factors also distinguishes ordinary from logarithmic averages, but uses functions whose prime bins depend on the scale. Prime predecessors develops related marked dilation graphs for prime extraction.

🏷️ Manuscripts and verification scope

• Ordinary two-point correlations of multiplicative functions — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The page selects quantitative affine Liouville cancellation and qualitative corrected Elliott cancellation under uniform nonpretentiousness. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. Ordinary two-point correlations of multiplicative functions.