Richard Schroeppel <rcs@cs.arizona.edu> wrote:

The next diagonals turn out to be products of three adjacent Fibs divided by 2: 260 = 5*8*13/2. The right way to look at this seems to be 5*8*13/(1*1*2), a kind of binomial coefficent with the regular integers replaced by Fibonacci numbers. The next diagonals will be products of four adjacent Fibs divided by 1*1*2*3, and so on.

Neat! Would you like to call them Fibonomial coefficients, or is that too cute? I wonder if there's an analogue of the role of binomial coefficients in the expansion of (a+b)^n? Dan