Math-6198 · mathematics · Mathematics

Math-6198

Erdős Problem #1144

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Statement

Independently assign each prime $p$ a sign $f(p)\in\{-1,1\}$ with equal probabilities, and extend $f$ completely multiplicatively to positive integers. Does \[ \limsup_{N\to\infty}\frac{1}{\sqrt N}\sum_{m\leq N}f(m)=+\infty \] hold with probability one?

Restatement from Math-6198-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-6198-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6198

Erdős Problem 1144 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

Independently assign each prime $p$ a sign $f(p)\in\{-1,1\}$ with equal probabilities, and extend $f$ completely multiplicatively to positive integers. Does \[ \limsup_{N\to\infty}\frac{1}{\sqrt N}\sum_{m\leq N}f(m)=+\infty \] hold with probability one?

CONTEXT AND SCOPE

This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.

Complete multiplicativity is essential: this is not the squarefree-supported random multiplicative model.

ATTACK PLAN AND WORK

Write $S_N=\sum_{n\le N}f(n)$. Independence of the prime signs implies $\mathbb E f(n)=1$ for a square and zero otherwise, and $\mathbb E[f(m)f(n)]=1$ precisely when $mn$ is a square. Grouping integers by their squarefree parts gives the exact identities \[ \mathbb E S_N=\lfloor\sqrt N\rfloor,\qquad \mathbb E S_N^2=\sum_{\substack{a\le N\\a\ { m squarefree}}} \lfloor\sqrt{N/a}\rfloor^2.\tag{1} \] Here a squarefree part $a$ contributes pairs $(au^2,av^2)$.

These identities already force logarithmically growing second moments after normalization. Let $Q(t)$ count squarefree positive integers at most $t$. The union bound over all possible square divisors gives, for integral $t$, $Q(t)\ge t-t\sum_{j\ge2}j^{-2}\ge t/4$, since $\sum_{j\ge2}j^{-2}\le1/4+\int_2^\infty x^{-2}\,dx=3/4$. Partial summation consequently gives $\sum_{a\le N,\ a\ { m squarefree}}1/a\ge H_N/4$. Using $\lfloor y\rfloor\ge y/2$ for $y\ge1$ in $1$, \[NH_N/16\le\mathbb E S_N^2\le NH_N.\tag{2}\] In particular the variance is unbounded relative to $N$.

This is not an almost-sure positive-limsup argument. Large second moments can come from rare events, and a signed random variable need not have a large positive tail merely because its variance is large. Temporal dependence also prevents applying independent-event Borel--Cantelli to a sequence of $S_N$.

FINITE COMPUTATIONAL CHECK

Every assignment of prime signs was enumerated for each $1\le N\le16$. Exact rational averages agree with both moment identities and satisfy the harmonic-number bounds.

VERIFICATION AND REMAINING GAP

Obtain quantitative positive-tail events at infinitely many scales with enough control of their dependence. Moment growth alone does not establish the almost-sure one-sided assertion.

SOURCES

https://www.erdosproblems.com/1144

https://www.erdosproblems.com/history/1144

FINAL

UNRESOLVED. The full source problem remains outside the scope of the proved partial results.

COMPLETION ESTIMATE

$8\%$. This is a conservative subjective estimate toward the whole problem, not a probability that the argument is correct or a claim that a fixed fraction of the problem has been solved.

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