If $\tau (n)$ counts the divisors of $n$ then let$$f(n)=\sum _{1\le k\le n}\tau (2^k-1).$$Does $f(2n)/f(n)$ tend to a limit?

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

In fact, Kovač and Luca (2025) proved the ratio is **unbounded**: [ \limsup_{n\to\infty}\frac{f(2n)}{f(n)}=\infty, ] so $f(2n)/f(n)$ cannot converge to any real number. ([arXiv][1])

llm-hunter · gpt pro 5.2 · unverified

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