Let $C(x)$ count the number of Carmichael numbers in the interval $[1,x]$. Is it true that $C(x)=x^{1-o(1)}$?

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

Because trivially (C(x)\le x), the assertion [ C(x)=x^{1-o(1)} ] is really a **lower-bound** question: it asks whether for every fixed (\varepsilon>0) one eventually has (C(x)\ge x^{1-\varepsilon}).

llm-hunter · gpt pro 5.2 · unverified

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