GPT 6 Astra Ultra attempt on Math-6200
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Contents
Erdős Problem 1146 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
For $E\subseteq\mathbb N$, put \[ d_s(E)=\inf_{N\geq1}\frac{|E\cap\{1,\ldots,N\}|}{N}. \] Let $A=\{2^m3^n:m,n\geq0\}$. Is $A$ an essential component: for every $B\subseteq\mathbb N$ with $0<d_s(B)<1$, does \[ d_s\left((A\cup\{0\})+(B\cup\{0\})\right)>d_s(B) \] hold?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.The adjoined zero terms are the sumset convention explicitly used by the formalization for Schnirelmann density.
ATTACK PLAN AND WORK
Since $1\in A$, the target sumset contains $B\cup(B+1)$, as well as zero. This proves the essential-component inequality for a concrete infinite family of test sets. For $q\ge2$ let $B_q$ be the positive integers not divisible by $q$. Then \[d_s(B_q)=\inf_{N\ge1}\left(1-\frac{\lfloor N/q\rfloor}{N}\right)=1-1/q.\] Every positive integer outside $B_q$ is a positive multiple of $q$; its predecessor lies in $B_q$. Therefore $B_q\cup(B_q+1)=\mathbb N_{\ge1}$ and \[d_s((A\cup\{0\})+(B_q\cup\{0\}))=1>d_s(B_q).\tag{1}\] More generally $1$ holds for any $B$ of density strictly below one whose complement contains no two consecutive positive integers, provided $1\in B$. Positive Schnirelmann density itself forces $1\in B$ by considering $N=1$.This also explains a limitation of the elementary shift argument. For a general $B$, long consecutive missing blocks can remain after adding the single element 1. The full question requires the extra shifts $2^m3^n$, with a strict improvement uniform over all initial intervals in the infimum, not just an improvement in asymptotic density along selected intervals.