Let $a_0=0$ and $a_1=1$, and in general define $a_k$ to be the least integer $>a_{k-1}$ for which $(n-a_k,n-a_i)=1$ for all $0\le i<k$. Does$$\sum _{0<a_i<n}\frac{1}{a_i}\to \infty$$as $n\to \infty$? What about if we restrict the sum to those $i$ such that $n-a_j$ is divisible by some prime $\le a_j$, or the complement of such $i$?

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Write [ b_k:=n-a_k\qquad (k\ge 1). ] Then the defining condition is exactly that the (b_k)’s are **pairwise coprime**: $ \gcd(b_k,b_i)=1\quad (1\le i<k). $ And because (a_k) is the *least* integer (>a_{k-1}) with that property, equivalently (b_k) is the **largest** integer (<b_{k-1}) that is coprime to all earlier (b_i…

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.