GPT 6 Astra Ultra attempt on Math-6257
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Contents
Erdős Problem 1203 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
With $\omega(m)$ equal to the number of different primes dividing $m$, the problem defines \[F(n)=\max_k\left(\omega(n+k)\frac{\log\log k}{\log k}\right)\] and asks whether $F(n)\to\infty$ as $n\to\infty$.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 displayed official definition does not state the range of k or justify attainment of the maximum. At minimum k must avoid the logarithmic singularities. The intended divergence formulation can be stated as: for every M, every sufficiently large n has some integer k>=3 with omega$n+k$ log(log k)/log k > M. No claim about attainment is needed for that formulation.
ATTACK PLAN AND WORK
There is a baseline obtainable by deliberately placing $n+k$ at a primorial. Let $p_j$ be the $j$th prime and $Q_t=p_1\cdots p_t$. For fixed $n$ and large $t$, choose $k_t=Q_t-n\ge3$. Then \[\omega(n+k_t)=t,\qquad \omega(n+k_t)\frac{\log\log k_t}{\log k_t} =t\frac{\log\log(Q_t-n)}{\log(Q_t-n)}.\tag{1}\] The classical prime number theorem gives $\log Q_t\sim t\log t$; hence the right side tends to one as $t\to\infty$. This uses that theorem as an explicit external asymptotic input, not as a new result proved here. Consequently the supremum over admissible $k$ is at least one for every fixed $n$.This argument also respects the source's maximum-attainment issue: it supplies a sequence of values approaching one and does not assert that a maximum is attained. If the maximum in the literal definition exists, it too is at least one. The construction permits astronomically large $k$ relative to the fixed shift $n$; it therefore largely erases the effect of $n$, explaining why it does not make the lower bound grow as $n$ grows.