gpt pro 5.4 attempt on Math-880

Problem 627 (Round 2)

1) ROUND-2 OBJECTIVE

Path C: isolate the exact Ramsey-theoretic obstruction and convert the problem to an inverse-threshold statement on the $n/(\log n)^2$ scale.

Round 1 produced a self-contained probabilistic lower bound and a basic Ramsey-number lower bound, but it still did not identify the correct $\limsup$, nor did it isolate the exact place where the open difficulty remains. The most promising Round-2 move is therefore to combine the new 2026 Ramsey-theoretic information with a precise inverse-threshold theorem.

2) Round-1 FOUNDATION USED

I use the following Round-1 facts.

• The probabilistic lower bound \[ f(n)\ge \left(\frac14+o(1)\right)\frac{n}{(\log_2 n)^2}. \]
• The Ramsey lower-bound mechanism: if $R(k,k)\le n < R(k+1,k+1)$, then \[ f(n)\ge \frac{n}{k^2}. \]

Throughout this section, $\log$ means $\log_2$.

3) NEW INSIGHT / TOOL (ROUND 2)

There are two genuinely new ingredients in this round.

1. A new exact theorem of Araujo--Filipe--Miyazaki gives the exact value of \[ \limsup_{n\to\infty}\frac{f(n)(\log n)^2}{n} \] in terms of a Ramsey envelope parameter $M$, and a lower bound on the corresponding $\liminf$ in terms of the diagonal Ramsey lower rate $L$.

2. I prove an inverse-threshold theorem for monotone functions on the $n/(\log n)^2$ scale. Applied to \[ Q(x):=\min\{n\in\mathbb N:f(n)\ge x\}, \] it converts limsup/liminf information about $f(n)(\log n)^2/n$ into exact liminf/limsup information about $Q(x)/(x(\log x)^2)$.

4) ATTACK PLAN (ROUND 2)

The exact gaps after Round 1 are:

• Round 1 did not identify $\limsup f(n)(\log n)^2/n$.
• Round 1 did not isolate the exact obstruction to the existence of the limit.
• Round 1 did not convert the problem to an equivalent inverse-threshold problem.

The plan is:

1. import the checked external theorem of Araujo--Filipe--Miyazaki;
2. prove an inverse theorem for $Q(x)$;
3. combine the two to obtain an exact obstruction statement: the limsup side is already completely understood, and the remaining open gap is the mismatch between the diagonal quantity $L$ and the off-diagonal envelope $M$.

5) WORK (ROUND 2)

Checked external theorem ([AFM25], Theorems 2.1 and 2.2).

Define \[ g(n):=\frac{f(n)(\log n)^2}{n}, \] \[ M:=\limsup_{t\to\infty}\max_{1\le s\le t}\frac{\log R(s,t)}{\sqrt{st}}, \qquad L:=\liminf_{k\to\infty}\frac{\log R(k,k)}{k}. \] Then \[ \limsup_{n\to\infty} g(n)=M^2, \qquad \liminf_{n\to\infty} g(n)\ge L^2. \] Moreover, [AFM25] proves the currently best asymptotic upper bound \[ f(n)\le (3.71943+o(1))\frac{n}{(\log n)^2}. \]

No extra hypotheses are needed for these statements.

Claim 627.1 (monotonicity and logarithmic location of the inverse threshold).

The function $f(n)$ is nondecreasing. If \[ Q(x):=\min\{n\in\mathbb N:f(n)\ge x\}, \] then $Q(x)$ is well-defined for all sufficiently large $x$, and \[ \frac{\log Q(x)}{\log x}\to 1 \qquad (x\to\infty). \]

Proof. Monotonicity follows by adding isolated vertices: if $G$ witnesses $f(n)$, then $G\cup I_1$ has the same values of $\chi$ and $\omega$, so $f(n+1)\ge f(n)$.

By Erdős' theorem, $f(n)=\Theta(n/(\log n)^2)$, so there exist $0<c<C$ and $n_0$ such that for all $n\ge n_0$, \[ c\frac{n}{(\log n)^2}\le f(n)\le C\frac{n}{(\log n)^2}. \] Let $n:=Q(x)$. Then $x\le f(n)\le Cn/(\log n)^2$, hence \[ \log x\le \log n + O(1). \] Also $x>f(n-1)\ge c(n-1)/(\log(n-1))^2$ for large $x$, so \[ \log x\ge \log(n-1)-2\log\log(n-1)+O(1). \] Thus \[ \log n-\log x=O(\log\log n), \] and dividing by $\log x\sim \log n$ gives \[ \frac{\log Q(x)}{\log x}\to 1. \] $\square$

Claim 627.2 (inverse theorem on the $n/(\log n)^2$ scale).

Let \[ \gamma_-:=\liminf_{n\to\infty}\frac{f(n)(\log n)^2}{n}, \qquad \gamma_+:=\limsup_{n\to\infty}\frac{f(n)(\log n)^2}{n}. \] Then \[ \liminf_{x\to\infty}\frac{Q(x)}{x(\log x)^2}=\frac{1}{\gamma_+}, \qquad \limsup_{x\to\infty}\frac{Q(x)}{x(\log x)^2}=\frac{1}{\gamma_-}. \]

Proof. We first prove the formula for the liminf.

Fix $\varepsilon>0$. By the definition of $\gamma_+$, for all large $n$, \[ f(n)\le (\gamma_++\varepsilon)\frac{n}{(\log n)^2}. \] Let $n:=Q(x)$. Then $x\le f(n)$, so \[ x\le (\gamma_++\varepsilon)\frac{n}{(\log n)^2}, \] that is, \[ \frac{Q(x)}{x(\log x)^2} = \frac{n}{x(\log x)^2} \ge \frac{1}{\gamma_++\varepsilon}\cdot \frac{(\log n)^2}{(\log x)^2}. \] By Claim 627.1, $\log n/\log x\to 1$. Hence \[ \liminf_{x\to\infty}\frac{Q(x)}{x(\log x)^2}\ge \frac{1}{\gamma_++\varepsilon}. \] Letting $\varepsilon\downarrow 0$ gives \[ \liminf_{x\to\infty}\frac{Q(x)}{x(\log x)^2}\ge \frac{1}{\gamma_+}. \]

For the reverse inequality, choose a sequence $n_j\to\infty$ such that \[ \frac{f(n_j)(\log n_j)^2}{n_j}\to \gamma_+. \] Set $x_j:=f(n_j)$. Then $Q(x_j)\le n_j$, so \[ \frac{Q(x_j)}{x_j(\log x_j)^2} \le \frac{n_j}{f(n_j)(\log x_j)^2} = \frac{1}{\frac{f(n_j)(\log n_j)^2}{n_j}}\cdot \frac{(\log n_j)^2}{(\log x_j)^2}. \] Since $x_j=f(n_j)=\Theta(n_j/(\log n_j)^2)$, we have $\log x_j/\log n_j\to 1$. Therefore \[ \liminf_{x\to\infty}\frac{Q(x)}{x(\log x)^2}\le \frac{1}{\gamma_+}. \] This proves the liminf identity.

Now prove the formula for the limsup.

Fix $\varepsilon>0$. By the definition of $\gamma_-$, for all large $n$, \[ f(n)\ge (\gamma_--\varepsilon)\frac{n}{(\log n)^2}. \] Let $m:=Q(x)-1$. Then $f(m)<x$, so \[ x> (\gamma_--\varepsilon)\frac{m}{(\log m)^2}. \] Hence \[ \frac{Q(x)}{x(\log x)^2} = \frac{m+1}{x(\log x)^2} < \frac{m+1}{(\gamma_--\varepsilon)m}\cdot \frac{(\log m)^2}{(\log x)^2}. \] Since $x>f(m)$ and $x\le f(m+1)$, while $f(n)=\Theta(n/(\log n)^2)$, we also have $\log m/\log x\to 1$. Hence \[ \limsup_{x\to\infty}\frac{Q(x)}{x(\log x)^2}\le \frac{1}{\gamma_--\varepsilon}. \] Let $\varepsilon\downarrow 0$ to get \[ \limsup_{x\to\infty}\frac{Q(x)}{x(\log x)^2}\le \frac{1}{\gamma_-}. \]

For the reverse inequality, choose $m_j\to\infty$ such that \[ \frac{f(m_j)(\log m_j)^2}{m_j}\to \gamma_-. \] Set \[ x_j:=f(m_j)+\frac{1}{j}. \] Then $x_j>f(m_j)$, so $Q(x_j)\ge m_j+1$. Also $x_j/f(m_j)\to 1$, and $f(m_j)=\Theta(m_j/(\log m_j)^2)$, so $\log x_j/\log m_j\to 1$. Therefore \[ \frac{Q(x_j)}{x_j(\log x_j)^2} \ge \frac{m_j+1}{x_j(\log x_j)^2} \to \frac{1}{\gamma_-}. \] Thus \[ \limsup_{x\to\infty}\frac{Q(x)}{x(\log x)^2}\ge \frac{1}{\gamma_-}. \] This proves the second identity. $\square$

Corollary 627.3 (exact inverse reformulation of Problem 627).

The limit \[ \lim_{n\to\infty}\frac{f(n)}{n/(\log n)^2} \] exists if and only if the limit \[ \lim_{x\to\infty}\frac{Q(x)}{x(\log x)^2} \] exists, and in that case the two limits are reciprocal.

Proof. This is immediate from Claim 627.2. $\square$

Corollary 627.4 (the exact Ramsey-theoretic obstruction).

With $Q(x)$, $L$, and $M$ as above, \[ \liminf_{x\to\infty}\frac{Q(x)}{x(\log x)^2}=\frac{1}{M^2}, \qquad \limsup_{x\to\infty}\frac{Q(x)}{x(\log x)^2}\le \frac{1}{L^2}. \]

Proof. By the checked theorem of Araujo--Filipe--Miyazaki, \[ \gamma_+=M^2,\qquad \gamma_-\ge L^2. \] Apply Claim 627.2: \[ \liminf \frac{Q(x)}{x(\log x)^2}=\frac{1}{\gamma_+}=\frac{1}{M^2}, \] and \[ \limsup \frac{Q(x)}{x(\log x)^2}=\frac{1}{\gamma_-}\le \frac{1}{L^2}. \] $\square$

Corollary 627.5 (a clean sufficient condition for the original limit).

If $L=M$, i.e. if \[ \liminf_{k\to\infty}\frac{\log R(k,k)}{k} = \limsup_{t\to\infty}\max_{1\le s\le t}\frac{\log R(s,t)}{\sqrt{st}}, \] then \[ \lim_{n\to\infty}\frac{f(n)}{n/(\log n)^2} \] exists and equals $L^2=M^2$.

Proof. The checked theorem gives \[ \liminf g(n)\ge L^2,\qquad \limsup g(n)=M^2. \] If $L=M$, then \[ M^2\le \liminf g(n)\le \limsup g(n)=M^2, \] so $g(n)\to M^2$. $\square$

Corollary 627.6 (updated interval for any putative limit).

If the limit \[ \lim_{n\to\infty}\frac{f(n)}{n/(\log n)^2} \] exists, then it lies in the interval $[1/4,\,3.71943]$.

Proof. The lower bound $1/4$ is exactly the Round-1 probabilistic bound. The upper bound $3.71943$ is the checked asymptotic upper bound from [AFM25]. $\square$

6) ADVERSARIAL VERIFICATION

I checked the new material against the main hidden pitfalls.

• Plateaux of $f(n)$. Because $f(n)$ is only nondecreasing, not strictly increasing, the choice $x_j=f(m_j)+1/j$ in Claim 627.2 is essential. It guarantees $Q(x_j)\ge m_j+1$ even if $f$ stays constant for long intervals.

• Replacement of $\log Q(x)$ by $\log x$. This was justified separately in Claim 627.1 using only the known theorem $f(n)=\Theta(n/(\log n)^2)$.

• Use of external results. The Araujo--Filipe--Miyazaki theorem was stated exactly as checked; no extra hypothesis was inserted into Corollary 627.4. Only Corollary 627.5 adds the extra condition $L=M$, and it is explicitly isolated there.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advances over Round 1 are:

1. the limsup side of Problem 627 is now exact via the checked theorem \[ \limsup_{n\to\infty}\frac{f(n)(\log n)^2}{n}=M^2; \]
2. the problem has been converted exactly into an inverse-threshold statement for $Q(x)$;
3. the remaining obstruction is sharply isolated: the original limit would follow from the collapse of the diagonal lower Ramsey rate $L$ and the off-diagonal envelope $M$.

8) COMPLETION ESTIMATE

COMPLETION: 58%

9) REFERENCES

• [AFM25 ] I. Araujo, R. Filipe, and R. Miyazaki, A note on the maximum ratio between chromatic number and clique number, arXiv:2512.16062v2, 2026.
• [Bloom627 ] T. F. Bloom, Erdős Problem #627, accessed 2026-03-08.