Let $H (n)$ be the smallest integer $l$ such that there exist $k < l$ with $(k^{n} - 1, l^{n} - 1) = 1$ .Is it true that $H (n) = 3$ infinitely often? (That is, $(2^{n} - 1, 3^{n} - 1) = 1$ infinitely often?)Estimate $H (n)$ . Is it true that there exists some constant $c > 0$ such that, for all $ϵ > 0$ , $H (n) > exp (n^{(c - ϵ) / l o g l o g n})$ for infinitely many $n$ and $H (n) < exp (n^{(c + ϵ) / l o g l o g n})$ for all large enough $n$ ?Does a similar upper bound hold for the smallest $k$ such that $(k^{n} - 1, 2^{n} - 1) = 1$ ?

Worked, still open.

number theory · open · 0 attempts

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vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

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unverified AI candidates (2)

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

[ H(n)=\min\\{\ell\ge 3:\ \exists,k\in{2,\dots,\ell-1}\text{ with }\gcd(k^n-1,\ell^n-1)=1,\\}. ]

llm-hunter · gpt pro 5.2 · unverified

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

h (n)

2^{n} - 1, 3^{n} - 1, \dots, h (n)^{n} - 1

p

δ_{p}

h (n) = p

lim inf h (n) = \infty

p

p - 1 ∣ n

p > n^{ϵ}

h (n) = p

open