Erdős Problem #626

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.

Erdős Problem #626: independent continuation

Date: 23 September 2026. Source versions read in full: 626.tex in gpt_pro_5.4.

1) OBJECTIVE AND SOURCE AUDIT

The two questions concern existence of asymptotic exponents for extremal girth at fixed chromatic number and extremal chromatic number at fixed girth. The source correctly relates these to inverse threshold functions. Inverting limsup and liminf does not prove that they coincide.

2) A PRECISE SUFFICIENT COMPOSITION PROPERTY

Fix a chromatic requirement $k\ge4$, and let $N_k(m)$ be the least number of vertices in a graph of chromatic number at least $k$ and girth greater than $m$. Suppose, additionally, that there exists $C\ge1$ such that \[ N_k(a+b)\le C N_k(a)N_k(b)\qquad(a,b\ge1). \tag{1} \] Then $u_m=\log(CN_k(m))$ is subadditive. To see its limiting behaviour directly, fix $q$ and write $m=tq+r$, $0\le r<q$. Repeated subadditivity gives $u_m\le t u_q+O_q(1)$, hence $\limsup u_m/m\le u_q/q$. Taking the infimum over $q$ gives existence of $\lim u_m/m$. The usual positive finite exponential bounds then allow the inverse-threshold argument in the source to yield a limit for the girth divided by $\log n$.

3) WHY EXISTING ENVELOPES ARE INSUFFICIENT

Consider the monotone integer function \[ N(m)=\left\lceil\exp(2^{\lceil\log_2m\rceil})\right\rceil. \] It satisfies $e^m\le N(m)\le e^{2m}+1$, but $\log N(m)/m$ tends to $1$ along $m=2^j$ and to $2$ along $m=2^j+1$. Its inverse threshold likewise has two different asymptotic ratios to $\log n$. Thus monotonicity and matching exponential order cannot by themselves establish the desired limit.

4) FAILED CONSTRUCTION AND REMAINING GAP

A disjoint union of two graphs keeps the larger chromatic number but takes the smaller girth. It therefore supplies no inequality of form $1$, whose girth parameter is additive. Identifying vertices has the same obstruction. I have not found a graph operation establishing $1$, or an adequate approximate variant, for these extremal thresholds. The fixed-girth chromatic-exponent question also remains untouched by this conditional result.

7) FINAL

UNRESOLVED. This isolates a concrete missing composition theorem and gives a countermodel to a purely monotone inversion proof; it does not establish a new graph bound. Problem record: https://www.erdosproblems.com/626.

8) COMPLETION ESTIMATE

COMPLETION: 5% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.

Problem 626

1) FORMAL RESTATEMENT

Throughout, graphs are finite, simple, undirected. The girth $\operatorname{girth}(G)$ is the length of the shortest cycle in $G$, with $\operatorname{girth}(G)=\infty$ if $G$ is acyclic. The chromatic number is $\chi(G)$. We take $\log$ to be the natural logarithm.

Fix an integer $k\ge 4$. For each $n\in\mathbb{N}$ define \[ g_k(n)\ :=\ \max\left\{\, m\in\mathbb{N}\ :\ \exists\text{ graph }G\text{ with }|V(G)|=n,\ \chi(G)=k,\ \operatorname{girth}(G)>m\,\right\}. \] Equivalently, $g_k(n)=\max\{\operatorname{girth}(G)-1:\ |V(G)|=n,\ \chi(G)=k\}$, since $k\ge4$ forces $\operatorname{girth}(G)<\infty$.

Question A. Does the limit \[ \lim_{n\to\infty}\frac{g_k(n)}{\log n} \] exist (as a finite real number), and if so what is its value?

Fix an integer $m\ge 3$. For each $n\in\mathbb{N}$ define \[ h^{(m)}(n)\ :=\ \max\left\{\,\chi(G)\ :\ \exists\text{ graph }G\text{ with }|V(G)|=n,\ \operatorname{girth}(G)>m\,\right\}. \]

Question B. Does the limit \[ \lim_{n\to\infty}\frac{\log h^{(m)}(n)}{\log n} \] exist, and if so what is its value?

2) QUICK LITERATURE/CONTEXT CHECK

Web browsing available: YES. Computation available: YES.

The Erdős Problems site lists this problem as open, with the following known bounds (for fixed $k\ge4$ and all sufficiently large $n$): \[ \frac{1}{4\log k}\,\log n\ \le\ g_k(n)\ \le\ \frac{2}{\log(k-2)}\,\log n + 1, \] with the lower bound attributed to Kostochka and the upper bound to Erdős.

For $h^{(m)}(n)$, Erdős proved a general lower bound $\displaystyle \lim_{n\to\infty}\frac{\log h^{(m)}(n)}{\log n}\gg \frac{1}{m}$, and for odd $m$ an upper bound $\displaystyle \lim_{n\to\infty}\frac{\log h^{(m)}(n)}{\log n}\le \frac{2}{m+1}$, and conjectured this upper bound is sharp for odd $m$. For even $m$ Erdős suggested the limit should lie in $[2/(m+2),\,2/m]$ but could not prove this even for $m=4$.

3) ATTACK PLAN

I attempted the following "meta" approaches, aiming either to (i) force the limit to exist by a general regularity principle, or (ii) exhibit mechanisms that could make the limit fail to exist.

1. Invert the extremal functions. Define the inverse threshold function $N_k(M):=\min\{n:\exists\,G\text{ on }n\text{ vertices with }\chi(G)=k\text{ and }\operatorname{girth}(G)>M\}$. If $\log N_k(M)$ grows linearly in $M$ with an asymptotic slope, then $g_k(n)$ is asymptotically that slope's reciprocal times $\log n$. This reduces existence of $\lim g_k(n)/\log n$ to existence of $\lim \log N_k(M)/M$.

   Similarly, define $N^{(m)}(t):=\min\{n:\exists\,G\text{ on }n\text{ vertices with }\operatorname{girth}(G)>m\text{ and }\chi(G)\ge t\}$. If $N^{(m)}(t)$ is regularly varying (power law), then $\log h^{(m)}(n)/\log n$ has a limit, and vice versa. So one can try to prove "regular variation" for these inverse functions.

2. Try to prove submultiplicativity/supermultiplicativity for $N_k(M)$. If one could build from two graphs (with chromatic number $k$ and large girth) a new graph whose vertex count is roughly the product while girth adds, one would get a submultiplicativity statement $N_k(M_1+M_2)\ \le\ C\,N_k(M_1)N_k(M_2)$ for some constant $C$, and hence by Fekete's lemma a limit for $\log N_k(M)/M$. I do not know an operation that simultaneously preserves exact chromatic number $k$ and yields additive girth in such a clean way.
3. Oscillation search. To show non-existence, one would want two infinite sequences $n_j,n'_j\to\infty$ such that $g_k(n_j)\sim c_1\log n_j$ but $g_k(n'_j)\sim c_2\log n'_j$ with $c_1\ne c_2$. Since $g_k(n)$ is a maximum over graphs, oscillation would have to come from the availability of much better constructions only at certain sizes $n$, in a way that cannot be "filled in" for intermediate $n$ (e.g. by adding isolated vertices). I did not find a rigorous mechanism producing two different constants.

4) WORK

No complete solution (existence or non-existence of the limits) was found. I record rigorous structural facts that reduce the limit question to an inverse-growth question, plus small-$n$ sanity checks.

4.1. Basic monotonicity

4.2. Inverse function viewpoint for $g_k(n)$

Define \[ N_k(M)\ :=\ \min\left\{\,n\in\mathbb{N}\ :\ \exists\text{ graph }G\text{ on }n\text{ vertices with }\chi(G)=k,\ \operatorname{girth}(G)>M\,\right\}. \] Then by definition, \begin{equation}\label{wsref-0-65-71-3a-69-6e-76-2d-72-65-6c} g_k(n)\ =\ \max\{\,M\in\mathbb{N}: N_k(M)\le n\,\}. \end{equation}

4.3. Inverse function viewpoint for $h^{(m)}(n)$

Fix $m\ge3$ and define \[ N^{(m)}(t)\ :=\ \min\left\{\,n\in\mathbb{N}\ :\ \exists\text{ graph }G\text{ on }n\text{ vertices with }\operatorname{girth}(G)>m,\ \chi(G)\ge t\,\right\}. \] Then \[ h^{(m)}(n)\ =\ \max\{\,t\in\mathbb{N}: N^{(m)}(t)\le n\,\}. \]

4.4. Small-$n$ sanity checks (exhaustive search for $n\le 6$)

I exhaustively enumerated all labeled graphs on $n\le 6$ vertices and computed $\chi(G)$, $\omega(G)$, and $\operatorname{girth}(G)$ by brute force.

• For $k=4$ one finds $g_4(4)=g_4(5)=g_4(6)=2$, i.e. every $4$-chromatic graph on at most $6$ vertices contains a triangle (so $\operatorname{girth}=3$).
• For $m=3$ (triangle-free graphs), one finds $h^{(3)}(5)=h^{(3)}(6)=3$, witnessed by $C_5$ (and $C_5$ plus an isolated vertex).

These checks are consistent with the definitions and with the general phenomenon that forcing larger girth rapidly forces larger $n$ for fixed chromatic number.

5) VERIFICATION

Propositions Proposition and Proposition were checked carefully: they use only the extremal definitions plus monotonicity of the inverse thresholds and basic limit algebra. The exhaustive $n\le 6$ computations serve as a consistency check of the definitions (not as evidence for asymptotics).

6) FINAL

UNRESOLVED.

1. The question asks for existence (and value) of two asymptotic limits. Current known bounds leave a constant-factor gap for $g_k(n)/\log n$ and an exponent gap for $\log h^{(m)}(n)/\log n$ (especially for even $m$).
2. I proved that existence of $\lim g_k(n)/\log n$ would follow from existence of $\lim \log N_k(M)/M$ for the inverse threshold $N_k(M)$, and similarly for $h^{(m)}$ via $N^{(m)}(t)$.
3. I did not find (a) a construction proving the limits exist, nor (b) a mechanism forcing non-existence by oscillation.
4. Any full resolution likely requires either new "product" constructions/subadditivity for $N_k(M)$ or sharp matching upper/lower constructions for $h^{(m)}(n)$.

7) COMPLETION ESTIMATE

COMPLETION: 25%.