Let $B_{2} (n)$ be the $2$ -full part of $n$ (that is, $B_{2} (n) = n / n^{'}$ where $n^{'}$ is the product of all primes that divide $n$ exactly once). Is it true that, for every fixed $k \geq 1$ , $n \leq m < n + k \prod B_{2} (m) ≪ n^{2 + o (1)} ?$ Or perhaps even $≪_{k} n^{2}$ ?

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number theory · open · 0 attempts

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

For (k\le 2) it’s trivial, but for (k\ge 3) the (\ll_k n^2) strengthening is **false**, and the remaining (n^{2+o(1)}) question is (as far as I can tell from the current literature/status pages) still **open**.

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llm-hunter · gpt pro 5.2 · unverified

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

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OEIS1

A057521 — Powerful part of n: if n = Product_i (pi^ei) then a(n) = Product_{i : ei > 1} (pi^ei); if n = b*c^2*d^3 then a(n) = c^2*d^3 when b is minimized.1,1,1,4,1,1,1,8,9,1,1,4,1,1,1,16,1,9,1,4,1,1,1,8,25,1,27,4,1,1,1,32,1,1,1,36,1,1,1,8,1,1,1,4,9,1,1,16,49,25,1,4,1,27,1,8

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#935For any integer

n = \prod p^{k_{p}}

let

Q_{2} (n)

be the powerful part of

n

, so that

Q_{2} (n) = p k_{p} \geq 2 \prod p^{k_{p}} .

Is it true that, for every

ϵ > 0

and

ℓ \geq 1

, if

n

is sufficiently large then

Q_{2} (n (n + 1) \dots (n + ℓ)) < n^{2 + ϵ} ?

ℓ \geq 2

then is

n \to \infty lim sup \frac{Q _{2} ( n ( n + 1 ) \dots ( n + ℓ ))}{n ^{2}}

infinite? If

ℓ \geq 2

then is

n \to \infty lim \frac{Q _{2} ( n ( n + 1 ) \dots ( n + ℓ ))}{n ^{ℓ + 1}} = 0 ?

A057521

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