Reused writeup on Math-819 (gpt pro 5.2)

Erdős Problem 452 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/452

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/452.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/452.tex.

FORMAL RESTATEMENT

Let $\omega(n)$ denote the number of distinct prime divisors of the positive integer $n$. Fix a real parameter $x\ge 3$. Define \[ L(x):=\max\Big\{\,\ell\in\mathbb N:\ \exists N\in\mathbb Z\ \text{s.t.}\ x\le N\le 2x-\ell+1\ \text{and}\ \omega(N+j)>\log\log(N+j)\ \forall j\in\{0,\dots,\ell-1\}\Big\}. \] (So $L(x)$ is the length of the longest interval of consecutive integers contained in $[x,2x]$ on which $\omega(n)>\log\log n$ holds throughout.) The problem asks for the order of magnitude / asymptotics of $L(x)$ as $x\to\infty$.

QUICK LITERATURE/CONTEXT CHECK

Only what is in the problem text is used: Erdős $1937$ proved that the natural density of integers with $\omega(n)>\log\log n$ is $1/2$. The problem text also states that CRT gives an interval with \[ |I|\ge (1+o(1))\frac{\log x}{(\log\log x)^2}. \] Per the integrity constraint, I do not import additional results about primes/smooth numbers beyond elementary lemmas proved below.

ATTACK PLAN

• Reduce "consecutive integers all have many distinct prime factors" to a CRT divisibility construction: enforce $N+j$ divisible by a prescribed modulus $M_j$ with many distinct prime factors.
• Isolate the purely CRT part (which is elementary and provable here) from the number-theoretic part (constructing many such $M_j$ with total product $\le x$).
• FAST REALITY CHECK: compute $L(x)$ for modest $x$ by sieving $\omega$.

WORK

Fast reality check (exact computation for small/moderate $x$). By a direct sieve computation of $\omega(n)$ for $n\le 2x$, I found the longest run length $L(x)$ in $[x,2x]$ for several $x$:

These are sanity checks only; they substantially exceed the very small value of $(\log x)/(\log\log x)^2$ at these $x$.

Lemma 452.1 (integer threshold inside $[x,2x]$). If $n\in[x,2x]$ and $\omega(n)>\log\log n$, then \[ \omega(n)\ge \big\lfloor \log\log x\big\rfloor+1. \]

Proof. Since $n\ge x$, we have $\log\log n\ge \log\log x$. If $\omega(n)>\log\log n\ge \log\log x$ and $\omega(n)$ is an integer, then $\omega(n)\ge \lfloor \log\log x\rfloor+1$. $\square$

Lemma 452.2 (CRT block lemma). Fix integers $L\ge 1$ and $r\ge 1$, and a real $x\ge 3$. Suppose there exist integers $M_0,\dots,M_{L-1}\ge 2$ such that

1. the $M_j$ are pairwise coprime,
2. $\omega(M_j)\ge r$ for every $j$, and
3. $M:=\prod_{j=0}^{L-1} M_j\le x-L+2$.

Then there exists an integer $N\in[x,2x-L+1]$ such that for every $j\in\{0,\dots,L-1\}$, \[ M_j\mid (N+j),\quad\text{hence}\quad \omega(N+j)\ge r. \] In particular, if additionally $r>\log\log(2x)$, then $\omega(N+j)>\log\log(N+j)$ for all $j$ (since $N+j\le 2x$), so $L(x)\ge L$.

Proof. For each $j$ impose the congruence \[ N\equiv -j\pmod{M_j}. \] Because the $M_j$ are pairwise coprime, CRT yields an integer $a$ (unique modulo $M=\prod_j M_j$) satisfying all these congruences simultaneously. Then for that residue class, each $a+j$ is divisible by $M_j$, so $\omega(a+j)\ge \omega(M_j)\ge r$.

We now choose a representative of the class $a\pmod M$ inside $[x,2x-L+1]$. The interval \[ [x, x+M-1] \] has length $M$ and contains exactly one representative of each residue class modulo $M$. By the size condition $M\le x-L+2$ we have \[ x+M-1\le x+(x-L+2)-1=2x-L+1, \] so $[x,x+M-1]\subseteq [x,2x-L+1]$. Therefore there exists (indeed, uniquely) an integer $N\in[x,x+M-1]$ with $N\equiv a\pmod M$, and this $N$ lies in $[x,2x-L+1]$. For each $j$, since $N\equiv a\pmod{M_j}$ we have $N+j\equiv a+j\equiv 0\pmod{M_j}$, hence $M_j\mid(N+j)$ and $\omega(N+j)\ge r$. Finally, if $r>\log\log(2x)$ then for all $j$ we have $N+j\le 2x$, hence $\log\log(N+j)\le \log\log(2x)< r\le \omega(N+j)$. $\square$

VERIFICATION

• Lemma 452.1 is purely monotonicity + integrality; no hidden assumptions.
• In Lemma 452.2, the extra size condition $M\le x-L+2$ guarantees that $[x,x+M-1]\subseteq[x,2x-L+1]$, so the required residue class modulo $M$ has a representative $N$ with $N+L-1\le 2x$.
• The computational sanity checks match the definition of $L(x)$ (longest consecutive run in $[x,2x]$).

FINAL. UNRESOLVED

(i) Strongest proved partial result. The existence of long intervals is reduced to constructing pairwise coprime $M_0,\dots,M_{L-1}$ with $\omega(M_j)$ large and $\prod_j M_j\le x-L+2$ (Lemma 452.2). This isolates the CRT mechanism cleanly.

(ii) First gap (crisp). Construct, for arbitrarily large $x$, explicit pairwise coprime integers $M_0,\dots,M_{L-1}$ with $\omega(M_j)\ge \lceil\log\log(2x)\rceil+1$ and $\prod_j M_j\le x-L+2$ for $L$ as large as $\gg \frac{\log x}{(\log\log x)^2}$ (or even $L\gg (\log x)^k$).

(iii) Top 3 next moves.

1. Attempt a prime-allocation construction inside Lemma 452.2: set $M_j$ to be products of $r\approx\log\log x$ distinct primes, arranged so that all primes used are distinct across all $j$, and then prove (with fully justified prime-counting/size bounds) that one can choose $L\approx \log x/(\log\log x)^2$ while keeping $\prod_j M_j\le x-L+2$.
2. Explore a "reuse small primes" variant: allow the same small prime to divide many $N+j$ when indices are congruent modulo that prime, aiming to reduce the total modulus size.
3. Computation: for $x$ up to (say) $10^8$, compute empirical $L(x)$ and the factorization pattern of the maximizing runs; use this to guess a plausible true order (e.g. polylog power).

(iv) Minimal counterexample structure. A minimal counterexample to a conjectured lower bound $L(x)\ge G(x)$ would be an $x$ for which every block of $G(x)$ consecutive integers in $[x,2x]$ contains at least one $n$ with $\omega(n)\le \log\log n$; such an $n$ would likely be either prime, a prime power, or a product of unusually few primes.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.