gpt pro 5.2 attempt on Math-5817

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.