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Hi all

Next to solving serious and trivial problems of Erdos,
I commented on some problems in the forum.
It would be nice to hear from you for the following ones (can we do more by discussing? Do others think it is basically unsolvable?):

$\textbf{#278}$
#386
#690
#694
$\textbf{#700}$
$\textbf{#770}$
#854
#891

The last upper bound is actually given in
Ramsey number of a cycle versus a graph of a given size - Cambie, Freschi, Morawski, Petrova, and Pokrovskiy,
which also implies the $2+o(1)$ result.

It is indeed assumed $d$ is large/ at least 3, since the d=2 case is a trivial case,
since the only connected graphs with $\Delta \le 2$ are paths and cycles, and thence $h_t(2)=2t+2.$

For $d \ge 3$, let G have at least $d+1$ edges, then all other $d$ edges are incident to $uv$ and thus $u$ or $v$.
They cannot all be adjacent to a fixed one.
Now any edge incident with $u$ has to be incident with any edge incident to $v$, thus all of them are of the form $uz, vz$ for the same vertex $z$, which cannot happen.

The paper appeared, see [Arxiv: Cambie, Decadt, Dong, Hu & Tang '26]

The second part is definitely a trivial statement.
For any prime $p$, $v_p(n!)$ is bounded by $n/(p-1)$.
Now $\sqrt[p-1]{p} \le 2$ and $n!\gg 4^n$ for $n$ large.

The first part is not trivial, but again easy. It is again since $n!$ is much larger than the square of a divisor which is $k$-free.

Zach, you will have to read the question again.
All odd cycles have to be large.
The clique has the opposite condition that all odd cycles are small.

So the statement is true for $n=1$, since a graph on $r=2k+1$ vertices whose odd cycles has size $r$ is either bipartite, or $C_r$ (adding a chord results in a smaller odd cycle).