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I am hoping to learn enough about additive permutations to help with a number theory problem. These permutations also have connections to difference sets, orthomorphisms, transversals, and other stru …

I will think of $X$ as the set of allowed degree values, with largest value being $b$, and the total number of elements of $M$ to be $m$. If $m=1$ with nonnegative value $b$ as the sole member, then …

I want to share a partial answer to question 1), and raise a few more questions. I found what I think is a neat and likely unoriginal bijection; I'm hoping the combinatorialists can provide a referen …

For squarefree $d$, we can translate this into a design problem. Given an $r$ by $c$ array (which correspond to your $r$ many $k$-tuples, but I use $c$ instead of $k$), you need to divide the $k$ dis …

This is the closest I can come to a positive test, but I don't know how well it will work for you. It is essentially taking the intersection of possible solution sets. Let us cut down on symmetry by …

I would like to maintain my basic position on the other post: that this forum does not do well with questions that frequently change. Since the last version has hit rather close to home, I will remar …

I am afraid the weak version (at this writing) involving density being less than $1/(x-2)$ doesn't work either. Simply pick $d_i$ and $x$ near but less than $\sqrt{N}$, and arrange $K$ and $L$ so tha …

Will Orrick might have a good guess for this one. As far as I know the answer has only been determined for n up to 8. The number of matrices with odd determinant is known: it is $$\prod_{i=0}^{n-1}( …

Since you are looking for finding a feasible solution of a particular instance in a relatively short time, I would combine a couple techniques. I would start by doing some breadth first searches, ess …

For sums like this, usually one term dominates. Taking $p(n)=n^d$ to play with, consider when $\binom{(n-k)^d}{k} \lt \binom{(n-k-1)^d}{k+1}$ stops holding. This is not far from when $(k+1) \lt (\fr …

Your sum is just a few terms off from $\sum_{1 \lt j \lt n} (f(j)-f(j-1))(f(n-j)f(n-j-1))$, where hopefully I did not mess up the indices too much. This in turn "looks like it is majorized" by $\int …

Intrigued by the notion in other posts and comments that there might be solutions to this problem involving Hamiltonian paths, I wrote a program to do breadth-first enumeration of such paths for the 3 …

As I understand it, your question has the answer no. Since you ask for $1-$designs, $\lambda$ is essentially how many times one of the $v$-many points appear in a block, which has size $k=4$ in the d …

This is a list of suggestions, some of which may help. First try smaller cases. For n=3 there is 1 cycle; for n=5 I don't know but I suspect it is a small multiple of 120. The nice thing is that fr …

It may be obvious to the casual observer, but it only just hit me recently that Hamiltonian cycle can be reduced to this problem, so of course the decision and counting problems are hard. I do not kn …