Erdős Problem #1181

Erdős Problem 1181 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

For $n\in\mathbb N$ and $k\geq1$, define $q(n,k)$ as the smallest prime not dividing $\prod_{1\leq i\leq k}(n+i)$, with $i$ integral. Is there an absolute constant $c>0$ such that \[q(n,\log n)<(1-c)(\log n)^2\] for every sufficiently large integer $n$?

CONTEXT AND SCOPE

This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.

The product over integral i interprets the nonintegral argument k=log n using floor(log n).

ATTACK PLAN AND WORK

Let $k=\lfloor\log n\rfloor$ and $M=\prod_{i=1}^k(n+i)$. Every prime at most $k$ divides $M$, since any $k$ consecutive integers contain a multiple of it. Thus $q(n,k)>k$. At the other end, put $\vartheta(y)=\sum_{p\le y}\log p$. If $q(n,k)>y$, then the squarefree product of all primes at most $y$ divides $M$, so \[\vartheta(y)\le\log M\le k\log(n+k).\tag{1}\] This is an exact obstruction, with no assumption of independence among the factors.

Using the classical prime number theorem $\vartheta(y)\sim y$ as an external input, $1$ immediately gives \[q(n,\lfloor\log n\rfloor)\le(1+o(1))(\log n)^2.\tag{2}\] Indeed for each fixed $\delta>0$, the choice $y=(1+\delta)k\log(n+k)$ eventually makes $\vartheta(y)>k\log(n+k)$, forcing a missing prime at or below $y$. The prime number theorem is not reproved in this attempt.

This exposes the exact constant problem: taking logarithms counts each forced prime only once but counts all multiplicities in $M$. A fixed-factor saving would need to show a positive proportion of this logarithmic budget is spent on repetitions or on primes outside the forced initial segment.

VERIFICATION AND REMAINING GAP

Turn the coefficient $1+o(1)$ into $1-c$ with a uniform positive $c$. The product-size obstruction alone cannot produce that saving.

SOURCES

https://www.erdosproblems.com/1181

FINAL

UNRESOLVED. The full source problem remains outside the scope of the proved partial results.

COMPLETION ESTIMATE

$8\%$. This is a conservative subjective estimate toward the whole problem, not a probability that the argument is correct or a claim that a fixed fraction of the problem has been solved.