Someone would have to make a much more convincing argument before I would believe there can be no single line simple closed form expression that gives the number at any particular position.

There is a whole field called 'generating functions' which can sometimes come up with the most surprising solutions to problems like this. Mathematica implements this using RSolve. If you could express the value at any position as the sum of the three values at the next lower positions along with the values of the base cases for the bottom layer then RSolve might find such an expression.

There is even a free online course underway right now on the topic of generating functions called Analytic Combinatorics over at coursera.org. The class has been going for a couple of weeks and is only six weeks long, so it might be a challenge, but if it isn't beyond you then it might be an interesting new way of looking at problems like this.

There is also a text called Concrete Mathematics written by Knuth and a couple of other authors that tries to introduce the subject. I had a difficult time trying to adapt to the style of that book, but you might do better. Surprisingly, there were a number of errors in earlier printings of that so if you consider buying an inexpensive used copy then you might try to get the latest printing you can find.

There is a text called Proofs that Really Count which tries to show that many times hard problems in combinatorics are perhaps only hard because we are trying to count up the numbers in an inconvenient way. He tries to show how you look for a different way of counting exactly the same number of items and which makes the problem much easier to do. I'd have to think a while before deciding whether that could help with this particular problem, but I would expect that it would apply.

I tried to give Simplify some assumptions about the domain of r and c and hoped that it might surprise me with a greatly simpler expression, but I was unsuccessful. Perhaps someone else would have more luck.

I did not find this as an example or exercise in Concrete Mathematics, but if you could swim in that text I suppose you could easily solve this problem.

I did not find this as a sequence in the OEIS, but finding multidimensional sequences in that is usually difficult for me.

Using FunctionExpand on things like Binomial[ r+2,c+1 ] / Binomial[ r,c ] I tried expressing coefficients in one layer as a rational times the Binomial of the previous layer. But I just can't see a nice simple factorial or binomial that relates a layer to the previous layer. Fumbling with RSolve didn't seem to help either. Perhaps someone else can do better.

From that you may be able to see the pattern for how to generate all subsequent  layers.