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This paper has a reference to a positive answer to the first question.

(Too long for a comment). Note that there are two different ways to define "all the embeddings" of a graph, both completely natural. I'll illustrate by example. Suppose a graph consists of two triangl …

Given that an upper bound is requested, here is one. By symmetry one can replace $\frac1{k_1}+\cdots+\frac1{k_m}$ by $\frac1{k_1}$ and multiply everything by $m$. Now do the sum over $k_1$ (which I'll …

In Ref 1 we called these things semi-Eulerian circular designs or semi-Eulerian quasigroups. They exist for all odd $n\ge 7$. I'll state the problem again. Find an Eulerian circuit in $K_n$ which (co …

If I understand it correctly, there is a counterexample on page 1518 of this paper. Then again, I might not have read it carefully enough.

The complement of a matching generalises. Take $1<k<|X|$ and identify $Y$ with the $k$-subsets of $X$. Let $R(x)$ be the $k$-subsets containing $x$. Note that the automorphism group acts as the symmet …

About 58% of the graphs on 12 vertices have diameter 3, so filtering a complete generation will be as fast as any. On 20 vertices the fraction has dropped to about 31% but the total is so vast that g …

There will be strongly-regular graphs of the same parameters with equal values of all those invariants. Since the parameters determine the eigenvalues, all the invariants determined by the spectrum (s …

All simple non-empty regular graphs of even degree have a two factor, see here . So you are just asking when they have a 1-factor. In addition to having an even number of vertices, the conditions are …

As $k\to\infty$, $\binom{2k-2}{k-1}\sim 2^{2k-2}/\sqrt{\pi k}$. As $N-k\to\infty$, $\binom{2N-2k}{N-1}\sim 2^{2N-2k}/\sqrt{\pi (N-k)}$. Now approximate the sum by an integral: $$\int_0^N \frac{\ln k} …

(A comment rather than an answer.) Here is a plot of $a_n/n^5$ (red) and $b_n/n^5$ (blue). It might not go far enough to show the asymptotic behaviour, but a possibility is that $a_n$ and $b_n$ are as …

(This is a comment, not an answer.) If $f_n(x)$ is your polynomial, starting with $f_0(x)=1$, then $$ \sum_{n=0}^\infty f_n(x) y^n = \frac{1-xy+x^2y^2+x^2y^3}{(1+xy^2)(1-xy-xy^2)} = 1 + \frac{x …

Let $q$ be a prime power and let $P$ be a projective plane of order $q$. It has $q^2+q+1$ points and $q^2+q+1$ lines. Each point lies on $q+1$ lines, and each line has $q+1$ points. Each pair of lines …

As far as I am aware, there is no such program. Also, it needs care to interpret gradpart's claims. Gradpart can make counts greater than one graph per machine instruction, which proves that it doesn' …

(1) In this paper (published J. Combinatorial Designs, 15 (2007) 98-119), in the history section starting page 3, we cite many published errors in counting Latin squares and related objects. Some, but …