Reused writeup on Math-5817 (gpt pro 5.2)

Erdős Problem 679 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/679

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_5.2/679.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/679.tex.

FORMAL RESTATEMENT. Fix $\varepsilon>0$. Does there exist an infinite set of integers $n$ such that for all integers $k$ with $k<n$ and $k$ sufficiently large (i.e. $k\ge K(\varepsilon)$), one has \[ \omega(n-k)\;<\;(1+\varepsilon)\,\frac{\log k}{\log\log k} \qquad\text{where }\omega(m)=\#\{p:\,p\text{ prime},\ p\mid m\}\,? \]

QUICK LITERATURE/CONTEXT CHECK. The problem statement itself notes (and I will treat as given) that a stronger estimate with an additive $+O(1)$ term, \[\omega(n-k) < \frac{\log k}{\log\log k} + O(1),\] would be false for every large $n$ (by the argument attributed in the statement to DottedCalculator). I do not assume any further external results beyond what is explicitly written in the problem text.

ATTACK PLAN. A natural way to force large values of $\omega(n-k)$ is to make $n-k$ divisible by a primorial $Q:=\prod_{p\le y}p$, because then $\omega(n-k)\ge\omega(Q)=\pi(y)$. One then compares $\pi(y)$ to $\log k/\log\log k$ for $k$ of size $Q$. To prove the original question, one would need to show there are infinitely many $n$ for which every large $k<n$ avoids making $n-k$ have too many distinct prime factors.

WORK. Lemma 1 (Primorial divisor appears with $k$ of comparable size). Let $y\ge2$ and $Q:=\prod_{p\le y}p$. If $n\ge 2Q$, then there exists an integer $k$ with \[Q\le k <2Q\quad\text{and}\quad Q\mid (n-k). \] In particular $\omega(n-k)\ge \omega(Q)=\pi(y)$.

Proof. Write $n\equiv r\pmod Q$ with $0\le r<Q$. Set $k:=Q+r$. Then $Q\le k<2Q$. Also $n-k\equiv r-(Q+r)\equiv -Q\equiv 0\pmod Q$, so $Q\mid(n-k)$. Since $Q$ is squarefree with prime factors exactly the primes $\le y$, we have $\omega(Q)=\pi(y)$, and hence $\omega(n-k)\ge\omega(Q)=\pi(y)$. $\square$

Lemma 2 (Elementary lower bound for $\pi(y)$ in terms of $Q$). With $Q=\prod_{p\le y}p$ as above, \[\pi(y)\ge \frac{\log Q}{\log y}. \]

Proof. Taking logs, \[\log Q=\sum_{p\le y}\log p\le \sum_{p\le y}\log y=\pi(y)\log y. \] Rearranging gives $\pi(y)\ge \log Q/\log y$. $\square$

Lemma 3 (Relating $\log k/\log\log k$ for $k\in[Q,2Q)$ to $Q$). If $Q\ge 3$ and $Q\le k<2Q$, then \[ \frac{\log Q}{\log\log(2Q)}\le \frac{\log k}{\log\log k}\le \frac{\log(2Q)}{\log\log Q}. \]

Proof. Since $Q\le k<2Q$, we have $\log Q\le \log k\le \log(2Q)$. Also $\log\log Q\le \log\log k\le \log\log(2Q)$ because $\log\log$ is increasing for $Q\ge 3$. Combine these monotonicity bounds to obtain the displayed inequalities. $\square$

What these lemmas give (and what they do not). Lemma 1 shows that for every $n$ large enough relative to $Q$ there exists some $k\asymp Q$ with \[\omega(n-k)\ge\pi(y)\ge \frac{\log Q}{\log y}. \] To contradict the desired inequality for a given $\varepsilon>0$, one would like to arrange \[\frac{\log Q}{\log y}\;\gtrsim\;(1+\varepsilon)\frac{\log Q}{\log\log Q}, \] i.e. roughly $\log y\lesssim \frac{1}{1+\varepsilon}\log\log Q$. But $y$ and $Q=\prod_{p\le y}p$ are linked by the distribution of primes (since $\log Q=\sum_{p\le y}\log p$). Without additional input on how $\log Q$ grows with $y$, the above comparison cannot be completed. This is exactly the kind of second-order relationship alluded to in the statement (the disproof of the stronger $+O(1)$ version depends on sharper prime distribution information).

VERIFICATION (FAST REALITY CHECK). I ran a local brute-force sanity check for moderate $n$, computing \[D(n):=\max_{3\le k<n}\left(\omega(n-k)-\frac{\log k}{\log\log k}\right),\] using a sieve for $\omega(\cdot)$. The following maxima were found:

n=5000:    max diff ~ 1.861 at k=170,  n-k=4830 with omega=5, rhs=3.139
n=50000:   max diff ~ 2.270 at k=20,   n-k=49980 with omega=5, rhs=2.730
n=100000:  max diff ~ 2.526 at k=670,  n-k=99330 with omega=6, rhs=3.474
n=100000 with k>=1000: max diff ~ 2.354 at k=1330, n-k=98670 with omega=6, rhs=3.646

These data are only a sanity check; they do not address the quantifier structure in the problem (which asks for $\omega(n-k)$ to be small for all large $k<n$ for infinitely many $n$).

FINAL. UNRESOLVED.

(i) Strongest proved partial result. For any $y$ and its primorial $Q=\prod_{p\le y}p$, every $n\ge 2Q$ admits some $k\in[Q,2Q)$ with $\omega(n-k)\ge\pi(y)\ge \log Q/\log y$ (Lemmas 1--2). This gives a systematic way to build some large $k$ with many prime factors in $n-k$.

(ii) First gap (crisp). One needs a quantitative relationship between $y$ and $Q=\prod_{p\le y}p$ strong enough to compare $\pi(y)$ to $\log Q/\log\log Q$ at the scale $k\asymp Q$. With only the elementary inequalities in Lemma 2 and monotonicity, I cannot decide whether the factor $(1+\varepsilon)$ makes the statement true or false.

(iii) Top 3 next moves. $1$ Prove an explicit bound of the form $\log Q\ge c_1 y$ (or a two-sided comparison) using only tools allowed by the problem context, and then translate it into a comparison between $\pi(y)$ and $\log Q/\log\log Q$. $2$ Numerically, for each $n$ up to a larger cutoff, compute \[\max_{k\ge K} \frac{\omega(n-k)\,\log\log k}{\log k} \] for a fixed moderate $K$, to see whether values exceed $1+\varepsilon$ densely or sparsely. $3$ Try to adapt the construction behind the stated disproof of the $+O(1)$ version to see whether it can also defeat the multiplicative $(1+\varepsilon)$ buffer (or conversely show that it inherently only gives a smaller additive improvement).

(iv) Minimal counterexample structure. A counterexample $n$ to the desired property (for a given $\varepsilon$) would exhibit some large $k<n$ (necessarily with $k\gtrsim K(\varepsilon)$) such that $n-k$ is divisible by a highly squarefree integer (e.g. a primorial-like product) with about $(1+\varepsilon)\log k/\log\log k$ distinct prime factors.

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.