gpt pro 5.4 attempt on Math-879

Problem 626 (Round 2)

1) ROUND-2 OBJECTIVE

Path C: isolate the exact inverse-threshold obstruction left after Round 1.

Round 1 already reduced the problem to the inverse thresholds \[ N_k(M):=\min\{n:\exists\,G,\ |V(G)|=n,\ \chi(G)=k,\ \operatorname{girth}(G)>M\} \] and \[ N^{(m)}(t):=\min\{n:\exists\,G,\ |V(G)|=n,\ \operatorname{girth}(G)>m,\ \chi(G)\ge t\}, \] but only in the one-way form "if the inverse limit exists, then the original limit exists". The natural gap is therefore to prove the exact converse, and more precisely to identify the limsup/liminf of the original quantities in terms of the inverse thresholds.

2) Round-1 FOUNDATION USED

I use the following Round-1 facts without reproving them.

• The monotonicity of $g_k(n)$ and $h^{(m)}(n)$ in $n$.
• The inverse identities \[ g_k(n)=\max\{M:N_k(M)\le n\},\qquad h^{(m)}(n)=\max\{t:N^{(m)}(t)\le n\}. \]
• The Round-1 one-way transfer statements: if $\lim \log N_k(M)/M$ exists then $\lim g_k(n)/\log n$ exists, and similarly for $h^{(m)}$.

3) NEW INSIGHT / TOOL (ROUND 2)

The new tool is an exact generalized-inverse theorem at the two scales relevant here:

1. exponential scale: $N(M)\approx e^{\lambda M}$ versus $g(n)\approx (\log n)/\lambda$;
2. power scale: $N(t)\approx t^{\alpha}$ versus $h(n)\approx n^{1/\alpha}$.

This upgrades the Round-1 one-way implication to an exact two-sided equivalence, and pins down the limsup/liminf constants as reciprocal liminf/limsup growth rates of the inverse thresholds.

4) ATTACK PLAN (ROUND 2)

The exact gaps after Round 1 are:

• for fixed $k\ge 4$, Round 1 did not prove that \[ \lim_{n\to\infty}\frac{g_k(n)}{\log n} \] exists if and only if \[ \lim_{M\to\infty}\frac{\log N_k(M)}{M} \] exists;
• for fixed $m\ge 3$, Round 1 did not prove the analogous equivalence for \[ \frac{\log h^{(m)}(n)}{\log n} \quad\text{and}\quad \frac{\log N^{(m)}(t)}{\log t}. \]

I close exactly these two gaps by proving reciprocal limsup/liminf formulae. This overcomes the main obstacle identified in Round 1: we no longer need any extra regularity assumption beyond monotonicity.

5) WORK (ROUND 2)

Claim 626.1 (generalized inverse on the exponential scale).

Let $N:\mathbb{N}\to\mathbb{N}$ be nondecreasing and satisfy $N(M)\to\infty$. Define \[ g(n):=\max\{M\in\mathbb{N}:N(M)\le n\}. \] Put \[ \lambda_-:=\liminf_{M\to\infty}\frac{\log N(M)}{M}, \qquad \lambda_+:=\limsup_{M\to\infty}\frac{\log N(M)}{M}. \] Assume $0<\lambda_-\le \lambda_+<\infty$. Then \[ \limsup_{n\to\infty}\frac{g(n)}{\log n}=\frac{1}{\lambda_-}, \qquad \liminf_{n\to\infty}\frac{g(n)}{\log n}=\frac{1}{\lambda_+}. \]

Proof. Fix $n$, and let $M:=g(n)$. By definition of the maximum, \[ N(M)\le n < N(M+1). \] For the $\limsup$, let $\varepsilon>0$. By the definition of $\lambda_-$, for all large $M$, \[ \log N(M)\ge (\lambda_--\varepsilon)M. \] Therefore, for all large $n$, \[ \log n\ge \log N(M)\ge (\lambda_--\varepsilon)M, \] so \[ \frac{g(n)}{\log n}=\frac{M}{\log n}\le \frac{1}{\lambda_--\varepsilon}. \] Hence \[ \limsup_{n\to\infty}\frac{g(n)}{\log n}\le \frac{1}{\lambda_-}. \]

For the reverse inequality, choose a sequence $M_j\to\infty$ such that \[ \frac{\log N(M_j)}{M_j}\to \lambda_-. \] Set $n_j:=N(M_j)$. Then $g(n_j)\ge M_j$, so \[ \frac{g(n_j)}{\log n_j}\ge \frac{M_j}{\log N(M_j)}\to \frac{1}{\lambda_-}. \] Thus \[ \limsup_{n\to\infty}\frac{g(n)}{\log n}\ge \frac{1}{\lambda_-}. \] This proves the first identity.

Now turn to the $\liminf$. Let $\varepsilon>0$. By the definition of $\lambda_+$, for all large $M$, \[ \log N(M)\le (\lambda_++\varepsilon)M. \] Using $n<N(M+1)$, we get \[ \log n<\log N(M+1)\le (\lambda_++\varepsilon)(M+1). \] Hence \[ \frac{g(n)}{\log n}=\frac{M}{\log n} > \frac{M}{(\lambda_++\varepsilon)(M+1)}. \] Since $M=g(n)\to\infty$ as $n\to\infty$, it follows that \[ \liminf_{n\to\infty}\frac{g(n)}{\log n}\ge \frac{1}{\lambda_++\varepsilon}. \] Letting $\varepsilon\downarrow 0$ gives \[ \liminf_{n\to\infty}\frac{g(n)}{\log n}\ge \frac{1}{\lambda_+}. \]

For the reverse inequality, choose $M_j\to\infty$ with \[ \frac{\log N(M_j)}{M_j}\to \lambda_+. \] Set $n_j:=N(M_j)-1$. Since $N(M_j)\to\infty$, we have $n_j\to\infty$, and because $n_j<N(M_j)$, we get $g(n_j)\le M_j-1$. Therefore \[ \frac{g(n_j)}{\log n_j}\le \frac{M_j-1}{\log(N(M_j)-1)}. \] Now $\log(N(M_j)-1)\sim \log N(M_j)$, so \[ \frac{M_j-1}{\log(N(M_j)-1)}\to \frac{1}{\lambda_+}. \] Hence \[ \liminf_{n\to\infty}\frac{g(n)}{\log n}\le \frac{1}{\lambda_+}. \] This proves the second identity. $\square$

Claim 626.2 (application to $g_k(n)$).

For fixed $k\ge 4$, define \[ \lambda_k^-:=\liminf_{M\to\infty}\frac{\log N_k(M)}{M}, \qquad \lambda_k^+:=\limsup_{M\to\infty}\frac{\log N_k(M)}{M}. \] Then \[ \limsup_{n\to\infty}\frac{g_k(n)}{\log n}=\frac{1}{\lambda_k^-}, \qquad \liminf_{n\to\infty}\frac{g_k(n)}{\log n}=\frac{1}{\lambda_k^+}. \]

Proof. Round 1 already established the inverse identity \[ g_k(n)=\max\{M:N_k(M)\le n\}. \] Also $N_k(M)\to\infty$: indeed, any graph of girth $>M$ containing a cycle must have at least $M+1$ vertices, and $\chi(G)=k\ge 4$ certainly forces $G$ to contain a cycle. Thus Claim 626.1 applies directly. $\square$

Corollary 626.3 (exact equivalence for the first limit).

For fixed $k\ge 4$, \[ \lim_{n\to\infty}\frac{g_k(n)}{\log n} \quad\text{exists} \quad\Longleftrightarrow\quad \lim_{M\to\infty}\frac{\log N_k(M)}{M} \quad\text{exists}, \] and in that case the two limits are reciprocal.

Proof. The forward and backward implications follow immediately from Claim 626.2, because a real sequence has a limit if and only if its limsup and liminf are equal. $\square$

Claim 626.4 (generalized inverse on the power scale).

Let $N:\mathbb{N}\to\mathbb{N}$ be nondecreasing and $N(t)\to\infty$. Define \[ h(n):=\max\{t\in\mathbb{N}:N(t)\le n\}. \] Put \[ \alpha_-:=\liminf_{t\to\infty}\frac{\log N(t)}{\log t}, \qquad \alpha_+:=\limsup_{t\to\infty}\frac{\log N(t)}{\log t}. \] Assume $0<\alpha_-\le \alpha_+<\infty$. Then \[ \limsup_{n\to\infty}\frac{\log h(n)}{\log n}=\frac{1}{\alpha_-}, \qquad \liminf_{n\to\infty}\frac{\log h(n)}{\log n}=\frac{1}{\alpha_+}. \]

Proof. Fix $n$, and let $t:=h(n)$. By definition, \[ N(t)\le n < N(t+1). \] For the $\limsup$, let $\varepsilon>0$. By the definition of $\alpha_-$, for all large $t$, \[ \log N(t)\ge (\alpha_--\varepsilon)\log t. \] Since $n\ge N(t)$, we get \[ \log n\ge \log N(t)\ge (\alpha_--\varepsilon)\log t, \] so \[ \frac{\log h(n)}{\log n}=\frac{\log t}{\log n}\le \frac{1}{\alpha_--\varepsilon}. \] Thus \[ \limsup_{n\to\infty}\frac{\log h(n)}{\log n}\le \frac{1}{\alpha_-}. \]

For the reverse inequality, choose $t_j\to\infty$ such that \[ \frac{\log N(t_j)}{\log t_j}\to \alpha_-. \] Set $n_j:=N(t_j)$. Then $h(n_j)\ge t_j$, hence \[ \frac{\log h(n_j)}{\log n_j}\ge \frac{\log t_j}{\log N(t_j)}\to \frac{1}{\alpha_-}. \] Therefore \[ \limsup_{n\to\infty}\frac{\log h(n)}{\log n}\ge \frac{1}{\alpha_-}. \] This proves the first identity.

Now let $\varepsilon>0$. By the definition of $\alpha_+$, for all large $t$, \[ \log N(t)\le (\alpha_++\varepsilon)\log t. \] Using $n<N(t+1)$, we obtain \[ \log n<\log N(t+1)\le (\alpha_++\varepsilon)\log(t+1), \] so \[ \frac{\log h(n)}{\log n}=\frac{\log t}{\log n}> \frac{\log t}{(\alpha_++\varepsilon)\log(t+1)}. \] As $t=h(n)\to\infty$, the right-hand side tends to $1/(\alpha_++\varepsilon)$. Hence \[ \liminf_{n\to\infty}\frac{\log h(n)}{\log n}\ge \frac{1}{\alpha_++\varepsilon}. \] Letting $\varepsilon\downarrow 0$ yields \[ \liminf_{n\to\infty}\frac{\log h(n)}{\log n}\ge \frac{1}{\alpha_+}. \]

For the reverse inequality, choose $t_j\to\infty$ with \[ \frac{\log N(t_j)}{\log t_j}\to \alpha_+, \] and set $n_j:=N(t_j)-1$. Then $h(n_j)\le t_j-1$, so \[ \frac{\log h(n_j)}{\log n_j} \le \frac{\log(t_j-1)}{\log(N(t_j)-1)}. \] Since $\log(t_j-1)\sim \log t_j$ and $\log(N(t_j)-1)\sim \log N(t_j)$, the right-hand side tends to $1/\alpha_+$. Therefore \[ \liminf_{n\to\infty}\frac{\log h(n)}{\log n}\le \frac{1}{\alpha_+}. \] This proves the second identity. $\square$

Claim 626.5 (application to $h^{(m)}(n)$).

For fixed $m\ge 3$, define \[ \alpha_m^-:=\liminf_{t\to\infty}\frac{\log N^{(m)}(t)}{\log t}, \qquad \alpha_m^+:=\limsup_{t\to\infty}\frac{\log N^{(m)}(t)}{\log t}. \] Then \[ \limsup_{n\to\infty}\frac{\log h^{(m)}(n)}{\log n}=\frac{1}{\alpha_m^-}, \qquad \liminf_{n\to\infty}\frac{\log h^{(m)}(n)}{\log n}=\frac{1}{\alpha_m^+}. \]

Proof. Round 1 already established \[ h^{(m)}(n)=\max\{t:N^{(m)}(t)\le n\}. \] Also $N^{(m)}(t)\to\infty$, since any graph with chromatic number at least $t$ has at least $t$ vertices. So Claim 626.4 applies directly. $\square$

Corollary 626.6 (exact equivalence for the second limit).

For fixed $m\ge 3$, \[ \lim_{n\to\infty}\frac{\log h^{(m)}(n)}{\log n} \quad\text{exists} \quad\Longleftrightarrow\quad \lim_{t\to\infty}\frac{\log N^{(m)}(t)}{\log t} \quad\text{exists}, \] and in that case the two limits are reciprocal.

Proof. Again this is immediate from Claim 626.5. $\square$

Corollary 626.7 (translation of the currently listed bounds to the inverse scale).

Using the current bounds recorded on the problem page, \[ \frac{1}{4\log k}\log n\le g_k(n)\le \frac{2}{\log(k-2)}\log n+1, \] one gets \[ \frac{\log(k-2)}{2} \le \liminf_{M\to\infty}\frac{\log N_k(M)}{M} \le \limsup_{M\to\infty}\frac{\log N_k(M)}{M} \le 4\log k. \]

Proof. Let $a:=1/(4\log k)$ and $b:=2/\log(k-2)$.

From $g_k(n)\ge a\log n$, choose $n=\lceil e^{M/a}\rceil$. Then $g_k(n)\ge M$, hence $N_k(M)\le n$, so \[ \limsup_{M\to\infty}\frac{\log N_k(M)}{M}\le \frac{1}{a}=4\log k. \]

From $g_k(n)\le b\log n+1$, if $n<e^{(M-1)/b}$ then $g_k(n)<M$, so $N_k(M)\ge e^{(M-1)/b}$. Therefore \[ \liminf_{M\to\infty}\frac{\log N_k(M)}{M}\ge \frac{1}{b}=\frac{\log(k-2)}{2}. \] This proves the claim. $\square$

6) ADVERSARIAL VERIFICATION

The new work was checked against the following possible failure modes.

• Non-strict growth of $N_k(M)$ and $N^{(m)}(t)$. The proofs only use monotonicity, not strict monotonicity. The identities \[ N(g(n))\le n < N(g(n)+1) \] and \[ N(h(n))\le n < N(h(n)+1) \] remain valid for generalized inverses.

• Use of $N(M)-1$. This requires $N(M)\ge 2$, which holds for all sufficiently large $M$ because $N(M)\to\infty$.

• Could $g_k(n)$ or $h^{(m)}(n)$ stay bounded? No. If $g_k(n)$ were bounded along an unbounded sequence of $n$, then $N_k(M)$ would be bounded for some arbitrarily large $M$, impossible. The same argument works for $h^{(m)}$.

• Reciprocal-limsup/lower-limsup interchange. Each direction was proved separately; no hidden continuity assumption was used.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advance over Round 1 is the exact reciprocal limsup/liminf theory: the two Erdős limits in Problem 626 exist if and only if the corresponding inverse-threshold limits exist, and the limsup/liminf constants are exactly reciprocals of the inverse liminf/limsup. Thus the problem is now reduced with no gap to determining the asymptotic growth of $N_k(M)$ and $N^{(m)}(t)$.

8) COMPLETION ESTIMATE

COMPLETION: 40%

9) REFERENCES

• [Bloom626 ] T. F. Bloom, Erdős Problem #626, accessed 2026-03-08.