This is just idle curiosity; I noticed something and I’m trying to track down a simple way to express it/prove it.
Take a rectangular ‘chunk’ of Pascal’s triangle in the following way: Count off k diagonals (starting from the upper left), up to n numbers in the diagonal. Now at the lower edges, copy the values from the last diagonal edges down one. Do this to both the n-length diagonal chosen last, and for the k-length rightward diagonal at the bottom.
You should have a n+1 x k+1 tilted rectangle with the point on the bottom corner missing. For example, for n=5 and k=3, the copied values on the left are 1,3,6,10,15; on the right bottom are 1,5,15.
It turns out for what I was looking at, the values on the lower right are ‘successes’, which I’ll call ‘s’. (In the 5x3 case, we have s = 1+5+15 = 21) . What I discovered is the relation between the sum of all the values in this near-rectangle and the total successes s:
Total = 2*(n/k + 1)*s - 1
That looks so simple that it seems there must be a fairly easy way to derive it. I’m just not sure how, and I’m looking for any insight into how this comes about.