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Let $G$ be a multigraph with maximum edge multiplicity $t$ and minimum edge multiplicity $1$ (so that there is at least one 'ordinary' edge). Is there some simple graph $H$ such that the $t$-fold mul …

Let $a,b,c$ be positive integers with gcd$(a,b,c)=1$, and let $\mathbb{N}$ denote the set of nonnegative integers. It is well known that $\mathbb{N} \setminus (a \mathbb{N}+b \mathbb{N} + c \mathbb{N …

For fixed $k \ge 3$, the generic such design is a full orbit with "index" $\lambda = k(k-1)$ and is probably not of much (combinatorial) interest. Short orbits are, of course, very interesting. As a …

In what follows, all graphs $G$ are $K_3$-divisible (all degrees even, number of edges a multiple of three) on $n$ vertices, where $n$ is not too small. The famous Nash-Williams conjecture claims tha …

Here are a few more observations. (1) Recall a consequence of Dirac's Theorem: A simple graph $G$ on $2n$ vertices admits a one-factor if $\delta(G) \ge n$. (This is a sufficient, but not necessary …

I wish I had a real answer for you! You are essentially interested in a tough conjecture of Hartmann, known as the "halving conjecture", which is promoted heavily by Reza Khosrovshahi. Actually, the …

Grids. (for one example only) The grid $\overbrace{P_m \Box P_m \Box \cdots \Box P_m}^k$ has average degree about $2k$ as $m \rightarrow \infty$, but any subgraph seems to have to include a "corner" …

I think your question equivalently asks if there is a universal constant $c>0$ such that every Steiner triple system of order $v$ has a 'cap' (line-free set) of size at least $c v$. The complement of …

I discovered this old question in connection with someone else's similar (and current) question: The Simultaneous Conjugacy Problem in the symmetric group $S_N$ If anyone still cares, here is a sligh …

This is barely of research interest, but I'd classify it as a curiosity with connections to combinatorics. The problem is to place integers in an $n \times n \times n$ array so that all $3n^2$ line s …

Nice question. This is not (any longer) an answer, but a strategy. First, try to construct 25 mutually disjoint squared squares of the same order. Then arrange them according to a 3,4,5 template. …

I guess the answer should be that $(a,b,c)$ fills the line if and only if there is an integral combination of its permutations equaling $(1,1,1)$. We want the elementary divisors of a certain $3 \tim …

Gordon has done a proper search of $(15,3,1)$-designs. I guess my incorrect reasoning does lead to a computer-free proof for (15,3,13)-designs. This is kind of cheating though, because there are rep …

I'll appeal to the linked question's answers for computational details. Intuition only: (as requested here) It is "easier" using your procedure to move from $00110^{n-4}$ to $11000^{n-4}$ than it is …

Personally, I am most interested in design theory with an "asymptotic flavor", and I think there are (edit: were, pre-Keevash) some very interesting open questions in this direction. To cut to the ch …