To show the 5n/2 bound, replace (c!) in a decomposition by (c-4)! 2!2!3! or better, and similarly for 4! and 5!. This gives that a decomposition with optimal sum has only 2's and 3's, from which 5n/2 as an upper bound on the sum follows quickly.

Actually it doesn't,as evidenced by 12! being a multiple of 12^5 giving a sum of 25; it looks like the upper bound will be more like 5n/2.

Indeed it does Will.

I need more care. Not only do I want an upper bound on a+b+...+c when (a!b!...c!) divides n!, I need that the summands are all greater than 1.

Further, 3!5!7! = 10!; I suspect a bound for the general problem is arbitraily close to 2n. (Actually I can get 2n-2 as a lower bound on the upper bound.)

Peter Jipsen did (and posted online) a short course on Universal Algebra for a BLAST conference a few years back. It isn't a royal road to the subject, but as a travelogue and summary it serves well. You might find something similar for Model Theory, which tells you what places to visit and what to look for when you are there.

I don't know if n + O(log(n)) is best possible. In view of (k! - 1)!k! |(k!)!, I would say it is pretty darn good though.

I thought Chris Godsil was (at the time of writing of the referenced paper) of male gender. I don't know about now; if Chris prefers "she" and "her", that memo has yet to reach my desk.

I imagine doing a web search on "partial Hadamard" will go far towards answering your question. Further checking resources such as the handbook of combinatorial designs should provide decent references.

For that matter, one can cite a number of (non- or limited circulating) holdings of institutions, such as MSRI and UC Berkeley's Logic department.

In the previous comment, I should have suggested looking at a cycle of length at least 6.

I don't know if girth is defined for trees. If you look at a cycle and consider two diametrically opposite points, they must share a neighbor if the graph has diameter 2. This means these same two points are on a cycle of length less than the cycle being considered, and eventually on a cycle of length at most 5.

Also, if the graph has diameter 2 and no cycles of length 5 or smaller, then the graph is acyclic and easily characterized. You probably want to look at diameter 2 graphs before speculating on $k$.

$k=4$ suffices only with the condition that the vertices are in the same bipartition, not in general. In general, the best you can say is that among two distinct vertices in a graph not joined by an edge, a sufficient condition is that the sum of their degrees must be at least n-1 for them to have a common neighbor, from which you can infer a sufficient condition on $\delta$. It is not clear if you want to characterize subclasses of graphs of diameter two or if you want to know more about the role of $\delta$ with respect to certain graph properties.