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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 …

If $k$ is allowed to be much, much larger than $n$, then no. A consequence of the assumption is that $a$ and $b$ each have fixed points. Let's take a toy example and see for what $n$ the example wor …

I have given some detail in a comment to another answer. I have a proof that the number of determinants is greater than 4 times the nth Fibonacci number for (n+1)x(n+1) (0,1) matrices, and I conjectu …

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 …

Others have posted that a resolvable block design will help answer the question. If you want to chew up some computer cycles, consider the following approach. There are 122 ways to divide a set of 6 …

This is a transform which tightens the problem and asks the reader to consult the bracelet literature. Consider first the question as specified with the additional proviso that the number of beads is …

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 …

The cheapest way of finding the value of a polynomial, given unlimited preprocessing resources, is to look up the precalculated value in the table. However, if you know you are going to need several …

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 …

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 …

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 …

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}( …

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 …

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 …