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Edit. Of course, I would find the mistake after posting. k ranges over the even numbers from 2 to 2h, so the result is not as nice. I will try rescuing the approach. End Edit Each term is close t …

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 …

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 …

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 …

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 …

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 …

I was reading some fascicles of Knuth's volume 4 on combinatorial enumeration. It seems that he was using Binary Decision Diagrams (BDDs) and variations to calculate quickly similar numbers. Perhaps …

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 …

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 …

Here is a partial attempt at an answer. If we're lucky, it will attract someone's attention (Gjergji? Noam?) and they will resolve the question. If there are enough sums of the form $a^ab^b$ the res …

It strikes me that you are going to need to remember either the index i or the value pi(i), so barring "magic memory", you are going to need log n bits as a strict minimum. For small n, you may as we …

I recommend looking at Balanced Incomplete Block Designs. They have some features which resemble (my interpretation of) your scenario. Each block maps to a word (usually a block is a set, so order o …

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 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 …