GPT 6 Astra Ultra attempt on Math-6264
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Contents
Erdős Problem 1210 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
Does an absolute constant $C$ exist such that for every integer $n\geq2$ and every set $A\subseteq\{1,\ldots,n-1\}$ whose distinct elements are pairwise coprime, \[\sum_{a\in A}\frac1{n-a}\leq\sum_{\substack{p<n\\p\text{ prime}}}\frac1p+C?\]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.ATTACK PLAN AND WORK
Change variables to $D=\{n-a:a\in A\}\subseteq[1,n-1]$. Pairwise coprimality has the exact consequence \[|D\cap(n+p\mathbb Z)|\le1\qquad\text{for every prime }p.\tag{1}\] Indeed two members of that residue class would correspond to two distinct elements of $A$ divisible by $p$. Thus the small denominators cannot be chosen arbitrarily: for each prime, one residue class has occupancy at most one.For example, if $n$ is even, at most one denominator is even. Bounding its contribution by $1/2$ and allowing all odd denominators gives \[\sum_{a\in A}\frac1{n-a} \le H_{n-1}-\tfrac12H_{\lfloor(n-1)/2\rfloor}+\tfrac12 =\tfrac12\log n+O(1).\tag{2}\] More generally, for a fixed squarefree $Q\mid n$, at most one denominator is divisible by each prime of $Q$. After deleting at most $\omega(Q)$ exceptional denominators, the rest are coprime to $Q$; summing by residue classes yields \[\sum_{a\in A}\frac1{n-a} \le\frac{\varphi(Q)}Q\log n+O_Q(1)+\omega(Q).\tag{3}\] The residue-class estimate in $3$ follows by comparing each progression's harmonic sum with its integral. The dependence of the error on $Q$ cannot be ignored when allowing $Q$ to grow with $n$.