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Math-88-A1 · revision r1 · on Math-88

gpt 6 astra pro attempt on Math-88

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Contents
  1. Chowla conjecture on correlations of the Liouville function
  2. Definitions and mathematical statement
  3. Short English statement
  4. Research attempt
  5. Current frontier.
  6. Attack route.
  7. Attempt: an exact separation of averaging modes.
  8. Attempt: a finite-energy sufficient lemma.
  9. Attempt: making the missing correlation estimate explicit.
  10. Stress test.
  11. Outcome.
  12. Sources

Chowla conjecture on correlations of the Liouville function

Definitions and mathematical statement

For $n=\prod_p p^{v_p(n)}$, put $\Omega(n)=\sum_pv_p(n)$ and $\lambda(n)=(-1)^{\Omega(n)}$. For the positive-slope linear forms $L_i(n)=a_in+b_i$ specified in the record, require pairwise nonproportionality $a_ib_j-a_jb_i\ne0$. The assertion is \[ \frac1x\sum_{1\le n\le x}\prod_{i=1}^k\lambda(a_in+b_i)\longrightarrow0. \] This is an ordinary, unweighted Cesàro average. Replacing it by a logarithmically weighted average changes the claim. The record's distinct nonnegative offsets and positive natural slopes are retained.

Formulation and concept references: [S1].

Short English statement

Do the parities of prime-factor counts decorrelate along every allowed finite family of distinct linear patterns?

Research attempt

Current frontier.
Tao's two-point theorem is logarithmically averaged [E1]; the logarithmic equivalences in [S1] do not convert it into the ordinary average required here. Chinis's 2026 result assumes Siegel zeros and obtains correlations at infinitely many scales, not all large scales unconditionally [E2]. The dynamical averages in [S4] also retain their averaging qualifications. The attempted upgrade below is a Tauberian one: identify the extra control that would exclude rare macroscopic failures of ordinary cancellation.
Attack route.
Partial summation relates ordinary and logarithmic means, but the inverse implication can fail. A second route would control an average square of ordinary prefix sums on logarithmic scales. The latter is stronger than what the target demands, but has a useful feature: bounded summands prevent a large normalized prefix sum from occupying arbitrarily narrow scale intervals. That persistence can turn an integrable energy bound into pointwise cancellation. A counterexample to the simpler logarithmic route is proved first, rather than concealed.
Attempt: an exact separation of averaging modes.
For any bounded complex sequence $a_n$ put $A(x)=\sum_{n\leq x}a_n$ and $L(x)=\sum_{n\leq x}a_n/n$. Partial summation gives, for $x\geq1$, \begin{equation}\label{wsref-0-65-71-37-37-3a-61-62-65-6c} L(x)=\frac{A(x)}x+\int_1^x\frac{A(t)}{t^2}\,dt. \end{equation} Now take $a_n=(-1)^{\lfloor\log_2n\rfloor}$. At the ends of complete dyadic blocks, \begin{equation}\label{wsref-0-65-71-37-37-3a-63-6f-75-6e-74-65-72} A(2^K-1)=\sum_{j=0}^{K-1}(-2)^j =\frac{1-(-2)^K}{3}. \end{equation} The ratios $A(2^K-1)/(2^K-1)$ have limiting values $1/3$ and $-1/3$ along alternate parities. Nevertheless $L(x)=O(1)$. Indeed the harmonic sum on block $[2^j,2^{j+1})$ is $\log2+O(2^{-j})$; the alternating main terms are bounded and the errors are absolutely summable. A final incomplete block contributes at most $\log2+O(1)$. Thus even bounded logarithmic partial sums, much stronger than $L(x)=o(\log x)$, need not force the ordinary mean to vanish. This is an artificial sequence, not a multiplicative-function counterexample, and precisely rejects an unqualified Tauberian inference.
Attempt: a finite-energy sufficient lemma.
Assume $|a_n|\leq1$ and \begin{equation}\label{wsref-0-65-71-37-37-3a-65-6e-65-72-67-79} \int_2^\infty\frac{|A(t)|^2}{t^3}\,dt<\infty. \end{equation} Then $A(x)=o(x)$. To prove this, suppose instead that $|A(x)|\geq\delta x$ for some fixed $\delta>0$ and arbitrarily large $x$. We may assume $0<\delta\leq1$. On $[x,x+\delta x/4]$, bounded increments imply \[ |A(t)-A(x)|\leq t-x+1\leq\delta x/4+1. \] For sufficiently large $x$ this gives $|A(t)|\geq\delta x/2$ throughout that interval. Since $t\leq5x/4$, its contribution to the integral in (eq77:energy) is at least \[ \frac{\delta x}{4}\, \frac{\delta^2x^2/4}{(5x/4)^3} =\frac{4\delta^3}{125}. \] From an unbounded sequence of such $x$, choose successively disjoint intervals. Their positive contributions contradict finiteness of the integral. This proves the lemma, including the possible complex phase of $A(x)$: only the triangle inequality was used.

For the original fixed tuple set $a_n=\prod_{i=1}^k\lambda(a_i n+b_i)$; the slopes and offsets guarantee positive arguments, and $|a_n|=1$. Thus establishing (eq77:energy) for every allowed tuple would prove exactly its ordinary Chowla correlation assertion. No uniformity in the tuple is required by this sufficient condition; the integral and its bound may depend on all the fixed coefficients. The method still asks for stronger quantitative summability than $A(x)=o(x)$ itself, so it is not asserted to be equivalent to Chowla.

Attempt: making the missing correlation estimate explicit.
For a finite real $Y>2$, expansion of the finite sum gives \begin{equation}\label{wsref-0-65-71-37-37-3a-6b-65-72-6e-65-6c} \int_2^Y\frac{|A(t)|^2}{t^3}\,dt =\sum_{m,n\leq Y}a_m\overline{a_n} \left(\frac{1}{2\max(2,m,n)^2}-\frac{1}{2Y^2}\right). \end{equation} For each pair the kernel is the integral from $\max(2,m,n)$ to $Y$; terms at the upper endpoint have zero contribution. Finite summation justifies the identity without any rearrangement of a conditionally convergent infinite double series.

The desired arithmetic input is boundedness of the right side as $Y\to\infty$, using cancellation between $m$ and $n$. The diagonal terms are summable, but estimating all terms in absolute value gives only an $O(\log Y)$ bound. Indeed there are $O(j)$ pairs with $\max(m,n)=j$, each of weight $O(j^{-2})$. For the constant sequence $a_n=1$, that logarithmic divergence actually occurs, since $A(t)\asymp t$. Therefore positivity of the quadratic kernel and bounded summands alone do not remove it.

One possible continuation is a bilinear estimate on separated blocks of $m,n$ that is summable over scales. It must apply to the product of Liouville values on the original linear patterns, not to independently resampled values. Replacing the off-diagonal terms by zero because "Liouville is random" assumes the intended cancellation. Applying a one-variable multiplicative mean-value theorem is also unjustified: a shifted product $\prod_i\lambda(a_i n+b_i)$ is generally not a multiplicative function of $n$.

Stress test.
The energy condition excludes repeated large spikes by paying a fixed amount for each spike, but it may fail for sequences whose normalized sums tend slowly to zero. Consequently failure of this sufficient condition would not disprove Chowla. A logarithmic average controls a signed first moment on scales; it may cancel between large positive and negative blocks, as (eq77:counter) shows. It is not an estimate for the nonnegative energy in (eq77:energy).

A bound established only on some scales under an extra zero hypothesis cannot be applied as a bound for every finite $Y$ in (eq77:kernel). Nor can a statement averaged over coefficients or offsets supply a conclusion for each fixed tuple. The companion script verifies (eq77:counter) with exact integers and the finite kernel identity with rational arithmetic on small sequences; it does not test the infinite energy of any unresolved Liouville pattern.

Outcome.
Outcome: Conditional reduction.

A fully proved finite-energy Tauberian lemma would upgrade suitable quantitative correlation control to the ordinary target, while an explicit bounded sequence disproves the simpler logarithmic upgrade. The exact finite kernel identifies the missing two-variable cancellation and avoids an invalid infinite-sum interchange.

The energy estimate is not proved for the required Liouville products, so no new Chowla case or counterexample is claimed. A complete continuation must establish summable scale control or a weaker Tauberian condition actually available arithmetically. The elementary lemmas carry no claim of independent novelty.

Sources