GPT 6 Astra Ultra attempt on Math-6238

Erdős Problem 1184 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

Put $P(m)=\max\{p:p\text{ prime},\ p\mid m\}$, and set \[f(n,k)=\#\{i\in\{1,\ldots,k\}:P(n+i)>k\}.\] For every fixed $\alpha>1$ and every integer sequence $n=n(k)$ satisfying $n=k^{\alpha+o(1)}$, must \[f(n,k)=(1-\rho(\alpha)+o(1))k\quad(k\to\infty)\] hold? Here $\rho$ is the Dickman function.

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The current statement uses rho(alpha); a superseded revision incorrectly used rho(1/alpha).

ATTACK PLAN AND WORK

The large-prime contributions have an exact separation property: a prime $p>k$ divides at most one of $n+1,\ldots,n+k$, since it would otherwise divide a nonzero difference smaller than $p$. Let $R_i$ be the part of $n+i$ supported on primes greater than $k$, including their full multiplicities, and put $L=\sum_{i=1}^k\log R_i$. The number of indices with $R_i>1$ is exactly $f(n,k)$. For each such index $\log k<\log R_i\le\log(n+k)$, whence \[\frac{L}{\log(n+k)}\le f(n,k)\le\frac{L}{\log k}\qquad(k>1).\tag{1}\]

The weighted quantity $L$ itself can be written without factoring each integer: \[ L=\sum_{i=1}^k\log(n+i)- \sum_{p\le k}\log p\sum_{a\ge1} \left(\left\lfloor\frac{n+k}{p^a}\right\rfloor- \left\lfloor\frac n{p^a}\right\rfloor\right).\tag{2} \] The inner sum is finite because the floor difference vanishes once $p^a>n+k$. Equation $2$ follows by separating prime-power valuations of the full product.

For $n=k^{\alpha+o(1)}$, the two denominators in $1$ differ asymptotically by the factor $\alpha$. Thus even a precise estimate of this weighted logarithmic mass would not by itself determine the unweighted count with Dickman coefficient. The remaining problem is distribution of that mass among individual integers, not an unproved independence of their large prime factors.

VERIFICATION AND REMAINING GAP

Control short-interval smooth-number counts uniformly for every allowed sequence $n(k)$, beyond the weighted bounds $1$--$2$. No Dickman asymptotic in such intervals is established here.

SOURCES

https://www.erdosproblems.com/1184

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.