For any integer $n=\prod p^{k_p}$ let $Q_2(n)$ be the powerful part of $n$, so that$$Q_2(n)=\prod _{\begin{array}{c}p\\ k_p\ge 2\end{array}}{p}^{k_p}.$$Is it true that, for every $ϵ>0$ and $\ell \ge 1$, if $n$ is sufficiently large then$$Q_2(n(n+1)\cdots (n+\ell ))<n^{2+ϵ}?$$If $\ell \ge 2$ then is$$\underset{n\to \infty }{\mathrm{limsup}}\frac{Q_2(n(n+1)\cdots (n+\ell ))}{n^2}$$infinite? If $\ell \ge 2$ then is$$\underset{n\to \infty }{\lim }\frac{Q_2(n(n+1)\cdots (n+\ell ))}{n^{\ell +1}}=0?$$

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

Let [ P_{n,\ell}:=n(n+1)\cdots(n+\ell), \qquad P_{n,\ell}=\prod_p p^{v_p(P_{n,\ell})}, ] and [ Q_2(P_{n,\ell})=\prod_{p:,v_p(P_{n,\ell})\ge 2} p^{v_p(P_{n,\ell})} ] [[nomath]](the “powerful/squarefull part” of $P_{n,\ell}$)[[/nomath]].

llm-hunter · gpt pro 5.2 · unverified

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