(Everything that follows is unproven conjecture. It's probably old news to Fibonacci aficionados.) It occurred to me that squares of Fibonacci numbers should have a simple recurrence relation. After a bit of futzing, I found 2 2 2 2 F' = 2 F + 2 'F - ''F (I use ' to mean "next term" and prefix ' to mean "previous term".) So 169 = 2*64 + 2*25 - 9. Another way to say this is that the sequence of Fibonacci squares can be dotted with the vector (-1 2 2 -1) to produce 0. The vector for the plain Fibonacci sequence is (1 1 -1). Experimenting showed the vector for Fibonacci cubes to be (-1 -3 6 3 -1). I worked out fourth and fifth powers, and developed the Fibonacci-Pascal triangle, built from the vectors. 1 -1 1 1 -1 -1 2 2 -1 -1 -3 6 3 -1 1 -5 -15 15 5 -1 1 8 -40 -60 40 8 -1 -1 13 104 -260 -260 104 13 -1 The rule for the signs seems clear enough. The magnitudes turn out to be Fibonacci-Binomials. The outer diagonals are 1 1 1 1 ...; the next diagonals are just the plain Fibs; and the next diagonals are products of adjacent Fibs: 104 = 8*13, etc. 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. This can probably be connected with q-binomial coefficients. Some other curiosities: (-1 2 2 -1) also works for F*F' (-1 -3 6 3 -1) works for F*F'*F'' (1 -5 -15 15 5 -1) works for F*F'*F''*F''' Fib(2N+1) = (1 1) dotted with F^2; Fib(2N) = (-1 0 1) dot F^2. Fib(3N) = (-1 1 1) dotted with F^3. Since (-1 2 2 -1) works with both F^2 and F*F', it will also work for squares of the Lucas sequence. Continuing this idea a bit further, the vectors work with powers of generalized Fibonacci sequences, where the starting pair of numbers is A,B, and the terms continue A+B, A+2B, 2A+3B, ... Rich rcs@cs.arizona.edu