GPT 6 Astra Ultra attempt on Math-5529
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Erdős Problem 248: uniformly few prime factors in all forward shifts
1) Statement and scope
Must infinitely many integers $n$ satisfy $\omega(n+k)\leq Ck$ for every $k\geq1$, with one absolute constant $C$? Tao and Teräväinen proved the stronger statement with $\Omega$, which counts prime factors with multiplicity. This attempt proves the truncation and moment reductions below, but does not construct the necessary sieve measure. The full problem is unresolved within this attempt.2) Only a logarithmic window needs attention
The elementary inequality $2^{\Omega(m)}\leq m$ gives $\Omega(m)\leq\log_2m$, where $\log_2$ here means logarithm to base two. Let $x$ be large, $x\leq n\leq2x$, and put $L=\lceil\log_2(3x)\rceil$. For $L\leq k\leq x$ we have $\Omega(n+k)\leq\log_2(3x)\leq k$. For $k>x$, we have $n+k<3k$, and $\log_2(3k)\leq k$ for $k\geq4$. Thus all $k\geq L$ are automatic. The challenge is to control $1\leq k<L$ with an absolute constant as $x$ grows.3) Separating primes from higher prime powers
Put $R=x^{1/100}$ and, for $m\leq3x$, define \[ Y(m)=\sum_{p\leq R}1_{p\mid m},\qquad T(m)=\sum_{\substack{p^j\leq R\\j\geq2}}1_{p^j\mid m}. \] The prime factors greater than $R$, counted with multiplicity, contribute at most $\log(3x)/\log R\leq200$ for large $x$. For $p\leq R$, put $r_p=\lfloor\log R/\log p\rfloor\geq1$. The contribution of $p$ to $Y+T$ is $\min(\nu_p(m),r_p)$. If $\nu_p(m)>r_p$, then \[ \frac{\nu_p(m)}{r_p} \leq\frac{\log(3x)/\log p} {\tfrac12\log R/\log p}\leq400. \] The same comparison is immediate if $\nu_p(m)\leq r_p$. Therefore the deterministic bound \[ \Omega(m)\leq400\left(Y(m)+T(m)+1\right) \] holds. This accounts for multiplicities, rather than silently replacing $\Omega$ by $\omega$.4) A sufficient pair of moment estimates
Here is a precise probabilistic input that would finish the proof. Suppose there are absolute constants $A,B$ such that for every sufficiently large $x$ there is a probability measure on integers $n\in[x,2x]$ satisfying, for $1\leq k<L$, \[ \mathbb E\left(Y(n+k)-Ak\right)_+^4\leq Bk^2, \qquad \mathbb E T(n+k)^2\leq Bk^{1/2}. \] The subscript $+$ denotes the positive part. Choose a fixed $d>0$ so large that $2B/d^4+3B/d^2<1$. Markov's inequality and the union bound show that the probability of any violation of $Y(n+k)\leq(A+d)k$ or $T(n+k)\leq dk$ is at most \[ \sum_{k=1}^{L-1} \left(\frac{B}{d^4k^2}+\frac{B}{d^2k^{3/2}}\right) \leq\frac{2B}{d^4}+\frac{3B}{d^2}<1. \] We used the integral bounds $\sum k^{-2}\leq2$ and $\sum k^{-3/2}\leq3$. Some integer in $[x,2x]$ therefore satisfies $\Omega(n+k)\leq400(A+2d+1)k$ throughout the finite window. The preceding truncation handles every larger $k$. The constants are independent of $x$; letting $x\to\infty$ then gives infinitely many such integers.5) Remaining gap and attribution
The moment hypotheses are sufficient conditions proved sufficient here, not estimates established in this attempt. Constructing one measure that supplies them uniformly over the growing shift window is the essential missing step. A different measure for each $k$, or constants growing with $x$, would not suffice. The published solution uses a slowly growing-dimensional Maynard-type sieve and suitable second and fourth moments. We have not independently reconstructed its correlation estimates or claimed that our simplified moment hypotheses are its verbatim statement.Sources: https://www.erdosproblems.com/248; T. Tao and J. Teräväinen, Quantitative correlations and some problems on prime factors of consecutive integers, Theorem 1.1 and Section 2, https://arxiv.org/abs/2512.01739. The known resolution belongs to those authors; this is a partial reconstruction of its probabilistic strategy.