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These elementary author-recorded deductions by Plasma AI test specific routes. Source results retain the qualifications stated below; no full catalog resolution or independent verification is claimed.

Spread does not remove the random-hitting logarithm

For Problem 20, a constant base in a sunflower bound cannot be obtained just by deleting the logarithm in the spread hypothesis of Bell–Chueluecha–Warnke's Theorem 3. Their Lemma 4 uses the following product family, credited there to Alweiss, Lovett, Wu and Zhang. Partition a universe into $n$ classes of $s$ points and take every transversal. There are $s^n$ members. Every fixed $t$-set belongs to zero or $s^{n-t}$ members, so the family is $s$-spread. A random subset containing each point independently with probability $\delta$ contains a transversal exactly when it meets every class. The probability is

$$\bigl(1-(1-\delta)^s\bigr)^n\longrightarrow0 \quad(n\longrightarrow\infty,\ s,\delta\text{ fixed},\ 0<\delta<1).$$

This is an obstruction to a constant-spread random-hitting lemma. It is not an obstruction to a constant-base sunflower bound: the family itself has $s$ disjoint transversals. A successful route must change that intermediate lemma or use a different way to obtain disjoint petals.

A strict cube-counting threshold

Use the unrestricted map in Problem 624, without imposing that the image of a subset lie outside it. Suppose $n=2^h$, $h\ge2$, and the map works at threshold $h$. On the subsets of every $h$-set it is a surjection between sets of size $n$, hence injective. The empty set and every pair of distinct singletons lie in some such cube. Their $n+1$ images must therefore all differ, a contradiction. Thus $H(2^h)\ge h+1$. At nonpowers, $2^{H(n)}\ge n$ and integrality give $H(n)>\log_2 n$, so the strict inequality holds for every $n\ge3$.

This proves only a unit gap at powers of two. The requested divergent gap would require control when each cube has a fixed multiplicative surplus of inputs over colors, where surjectivity no longer implies injectivity. It also does not identify the extremal function for every extension of the outside-the-input convention in the original Erdős–Hajnal source.

Exactly three colors and failure of two colors

For uniformity $n\ge2$, the minimum number of edges of an exactly 3-chromatic hypergraph equals the minimum number of edges of a non-2-colorable one, as used in Problem 901. To see this, choose an edge-minimal non-2-colorable hypergraph. Deleting an edge $E$ gives a proper 2-coloring. The removed edge must be monochromatic; give one of its vertices a new third color. Every other edge either keeps its proper coloring or contains the unique third-colored vertex and at least one other vertex, so remains nonmonochromatic. The resulting hypergraph is exactly 3-chromatic. The opposite inequality is immediate. This reconciles the extremal definitions; it improves no asymptotic bound.

What maximal clique packing misses

For Problem 719, let $G$ have $e$ edges, and let $p$ edge-disjoint copies of $K_{r+1}^r$ be maximal. Deleting their $(r+1)p$ edges leaves no such clique, whence

$$p\ge\frac{e-\operatorname{ex}_r(n;K_{r+1}^r)}{r+1}.$$

But the decomposition into those cliques and leftover single edges has $e-rp$ pieces. The conjectured number of pieces requires the stronger $p\ge(e-\operatorname{ex}_r(n;K_{r+1}^r))/r$. Maximality alone loses the necessary factor. A new packing improvement is needed; the maximal-packing argument is not a proof of the conjecture.

A Latin-rectangle normalization boundary

Let $A_{k,n}$ denote the Godsil–McKay expression recorded on Problem 725, for labeled rows, columns and symbols. Taking logarithms gives

$$\log\frac{A_{k,n}}{(n!)^k} =n\sum_{i=0}^{k-1}\log(1-i/n)-\frac n2\log(1-k/n)-\frac k2.$$

Taylor expansion, uniformly when $k/n\to0$, yields

$$\log\frac{A_{k,n}}{(n!)^k} =-\frac{k(k-1)}2 +\frac{-2k^3+6k^2-k}{12n} +O\!\left(\frac{k^4+k^3}{n^2}\right).$$

Indeed the sum of the cubic remainders is $O(k^4/n^2)$ after multiplication by $n$, and the second logarithm contributes $O(k^3/n^2)$. Thus, for $k\sim cn^{1/3}$ with fixed $c>0$,

$$\frac{A_{k,n}}{(n!)^k e^{-\binom k2}}\longrightarrow e^{-c^3/6}.$$

This compares formulas. Identifying $A_{k,n}$ with the actual count at that boundary imports Godsil–McKay's theorem; its primary proof is not inspected here. Even the pending extension recorded on the problem page has $k=o(n)$ and leaves $k$ proportional to $n$ outside its scope.

Uniformity reduction keeps the unbounded edge parameter

In Problem 1178, put $k=2,d=1$ in Conlon–Gishboliner–Levanzov–Shapira's Proposition 1.2. Writing $f_r(n,v,e)$ for the largest number of edges of an $r$-graph without $e$ edges on at most $v$ vertices, it gives

$$f_r(n,(r-2)e+3,e) \le\left(\binom r3(e-1)+1\right)f_3(n,e+3,e),\qquad r,e\ge3.$$

For each fixed $r,e$ the factor is constant. Hence the 3-uniform conjectured upper bound implies the upper bound in every uniformity; the converse is the case $r=3$. Isolated vertices reconcile “at most” with “exactly” the forbidden vertex count in a sufficiently large host. The separately recorded Brown–Erdős–Sós lower bound is still needed for equality of the threshold. This is a specialization of the inspected source statement, not an independent audit of its proof or the lower bound.

The reduction leaves all $e$ unbounded. The held upper theorem permits $e+\lceil26\log e/\log\log e\rceil$ vertices, an excess that cannot be rounded down to three. Neither a fixed-$e$ computation nor this uniformity transfer settles the remaining universal parameter.