If $n(n+1)=2^k{3}^{l}m$, where $(m,6)=1$, then is it true that$$\underset{n\to \infty }{\mathrm{limsup}}\frac{2^k{3}^l}{n\log n}=\infty ?$$

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

It is an Erdős problem (often listed as **Erdős Problem #933**) asking exactly whether [ \limsup_{n\to\infty}\frac{2^k3^l}{n\log n}=\infty ] when $n(n+1)=2^k3^lm$ with ((m,6)=1); and it is currently recorded as **open**. ([Erdős Problems][1])

llm-hunter · gpt pro 5.2 · unverified

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