Let $d_{n} = p_{n + 1} - p_{n}$ , where $p_{n}$ is the $n$ th prime. Let $r (x)$ be the smallest even integer $t$ such that $d_{n} = t$ has no solutions for $n \leq x$ .Is it true that $r (x) \to \infty$ ? Or even $r (x) / lo g x \to \infty$ ?

Worked, still open.

number theory · open · formalized (Lean) · 0 attempts

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vela reproduce examples/erdos-problems

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unverified AI candidates (2)

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

Write [ d_n=p_{n+1}-p_n\qquad (n\ge 1), ] and for (x\ge 1) let $r(x)$ be the *least even* $t$ such that no (d_n=t) occurs with (n\le x).

llm-hunter · gpt pro 5.2 · unverified

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

AMS 11 · open (literature)

theorem erdos_853.parts.i : atTop.Tendsto r atTop

C \geq 0

n_{i}

i \to \infty lim \frac{p _{n_{i} + 1} - p _{n_{i}}}{lo g n _{i}} = C ?

d_{n} = p_{n + 1} - p_{n}

p_{n}

n

h (x)

n < x

d_{n}, d_{n + 1}, \dots, d_{n + h (x) - 1}

h (x)

h (x) > (lo g x)^{c}

c > 0

h (x) = o (lo g x)?

open