[JennEric Screen Name]

Here is a semi-obvious observation. The right-hand diagonal consists of the sequence 1, 1+2, 1+2+3, ... , 1+2+ ... +N, ....

So the jth entry is the sum of all integers from 1 to j.

This sum is equal to j*(j+1)/2, which proves that no entry in this diagonal can be prime, (since either {j/2 and j+1} are both integer factors of the product or {j and (j+1)/2} are integer factors of the product).

You may be able to arrive at similar results for the other non-prime diagonals. For instance, the 2, 5, ... diagonal has entries that are one less than the 1, 3, ... diagonal. The formula would be j*(j+1)/2 - 1. If you look at this graphically, I think you will see that a j by (j+1)/2 rectangle with a 1x1 area missing from a corner can be rearranged into a (j-1)/2 by (j+1) rectangle if you leave out a 1x1 area. Meaning that j*(j+1)/2-1 is the product of two integers. This might only work half the time, with the other half being when j*(j+1)/2 -1 is even, and therefore not prime.