GPT 6 Astra Ultra attempt on Math-5469

Erdős Problem 164: the weighted sum of a primitive set

1) Statement and scope

A set $A\subseteq\{2,3,\ldots\}$ is primitive if distinct members never divide one another. The assertion is \[ \sum_{a\in A}\frac1{a\log a} \leq \sum_{p\ {\rm prime}}\frac1{p\log p}. \] It was proved by Lichtman. The following reconstruction uses the von Mangoldt-chain approach of Alexeev and collaborators $2026$, with one analytic inequality stated explicitly as an external input. The probability argument and the exceptional prime-power adjustment are proved here. Logarithms are natural.

2) The analytic input and a downward chain

Write $w(n)=1/(n\log n)$ and let $\Lambda$ be the von Mangoldt function. Lemma 3.3(ii) of the cited 2026 paper states, for every real $m\geq2$, with $x=\log m/\log2$, \[ \log m\sum_{q\geq2}\frac{\Lambda(q)}{q\log^2(mq)} \leq\frac{x}{x+1/2}. \tag{1} \] This analytic estimate is not reproved in this note.

For a composite integer $n$ which is not a prime power, set $T(n,n/q)=\Lambda(q)/\log n$ for each divisor $q\geq2$ of $n$. Terms with zero numerator are ignored. The destination is always at least $2$, and $\sum_{q\mid n}\Lambda(q)=\log n$ shows that these probabilities sum to one. For $n=p^k$, $k\geq2$, instead put \[ T(p^k,p)=2/k,\qquad T(p^k,p^j)=1/k\quad(2\leq j\leq k-1). \] The empty second range when $k=2$ is allowed. These probabilities also sum to one. Primes have no downward transitions.

Let $W(m)=\sum_{n>m}w(n)T(n,m)$. If $m$ is composite, direct substitution gives \[ \frac{W(m)}{w(m)}= \log m\sum_{q\geq2}\frac{\Lambda(q)}{q\log^2(mq)}\leq1. \] If $m=p$ is prime, the doubled last transition contributes the additional term $\sum_{k\geq2}1/(k^2p^{k-1})$. Since $x=\log_2p\leq p-1$, \[ \sum_{k\geq2}\frac1{k^2p^{k-1}} \leq\frac1{4(p-1)}\leq\frac1{2p-1} \leq\frac1{2x+1}. \] Together with $1$, this yields $W(p)/w(p)\leq x/(x+1/2)+1/(2x+1)=1$. Thus $W(m)\leq w(m)$ for every $m\geq2$.

3) Reverse the chain and count visits

The prime mass $P=\sum_p w(p)$ is finite. For completeness, primes in $(2^j,2^{j+1}]$ divide $\binom{2^{j+1}}{2^j}\leq4^{2^j}$, so there are at most $2^{j+1}/j$ of them for $j\geq1$. Their total weight is at most $2/(j^2\log2)$; adding the prime $2$ proves convergence.

Start a random chain at a prime $p$, with probability $w(p)/P$. From $m$ move to a proper multiple $n>m$ with probability \[ Q(m,n)=\frac{w(n)T(n,m)}{w(m)}. \] The preceding bound says that the row sum of $Q$ is at most one. With the remaining probability the chain terminates. Thus this defines a countable-state process; its integer states strictly increase and each divides the next.

Let $h(n)$ be $P$ times the probability that the process ever visits $n$. A prime can only be visited initially, so $h(p)=w(p)$. For composite $n$, its possible predecessors are its proper divisors at least $2$. Induction on $n$ and the fact that a state cannot be visited twice give \[ h(n)=\sum_{\substack{m\mid n\\2\leq m<n}}h(m)Q(m,n) =w(n)\sum_m T(n,m)=w(n). \] Every trajectory visits at most one member of a primitive set. Therefore, by the nonnegative sum rule for expectations, \[ \sum_{a\in A}w(a) =P\,\mathbb E\left[\#\{\hbox{visited states in }A\}\right]\leq P. \] This includes infinite primitive sets, without an unproved exchange of signed series. The primes themselves are primitive and attain $P$.

4) Outcome and sources

Outcome: SOLVED using the stated analytic lemma. All steps from $1$ to the desired inequality are proved above. The original resolution and the later chain method belong to the cited authors; this is an exposition, not a novelty or formal-verification claim.

• B. Alexeev, K. Barreto, Y. Li, J. D. Lichtman, L. Price, J. I. Shah, Q. Tang, and T. Tao, Primitive sets and von Mangoldt chains: Erdős Problem #1196 and beyond, Lemma 3.3(ii) and Section 5, https://arxiv.org/abs/2605.00301.
• T. F. Bloom, Problem 164, https://www.erdosproblems.com/164, checked 23 September 2026.

The percentage includes the explicitly cited analytic input.

5) COMPLETION ESTIMATE

COMPLETION: 100%