GPT 6 Astra Ultra attempt on Math-6193
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Erdős Problem 1139 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
List in increasing order all positive integers having at most two prime factors, counted with multiplicity, as $u_1<u_2<\cdots$. Is \[ \limsup_{k\to\infty}\frac{u_{k+1}-u_k}{\log k}=+\infty? \]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.ATTACK PLAN AND WORK
Arbitrarily long intervals missing all the $u_i$ can be constructed directly. Fix $L$ and choose $3L$ different primes, grouped into triples $(p_j,q_j,r_j)$ for $1\le j\le L$. The Chinese remainder theorem supplies $T$ with \[T\equiv-j\pmod{p_jq_jr_j}\qquad(1\le j\le L).\] Choose a positive representative, adding the product $Q$ of the $3L$ primes if necessary. Each $T+j$ is divisible by three different primes, so it has more than two prime factors even under the multiplicity convention. Thus no $u_i$ belongs to $[T+1,T+L]$.Since primes alone give infinitely many $u_i$, there are consecutive members $u_k\le T<T+L<u_{k+1}$, and $u_{k+1}-u_k\ge L+1$. This proves unbounded gaps, but does not prove the requested logarithmically normalized assertion. The construction gives a representative $T\le Q$ (or $T=Q$), hence $k\le T\le Q$ and a certified ratio at least $(L+1)/\log Q$ when $k>1$. Its modulus is costly: $\log Q$ is the sum of $3L$ positive prime logarithms. Using disjoint triples therefore gives no lower bound tending to infinity from this certificate. One must cover much longer intervals with substantially more economical congruence constraints, or use a different mechanism.