Date: Tue, 7 Jan 2003 17:12:06 -0800 (PST) From: "R. William Gosper" <rwg@spnet.com> Rich said

This can probably be connected with q-binomial coefficients. Bingo. But to make matters worse, there are q-Fibonacci numbers.

Some other curiosities: (-1 2 2 -1) also works for F*F' It works for Fib(n+a)*Fib(n+b), a, b arbitrary integers.

(-1 -3 6 3 -1) works for F*F'*F'' Likewise. If I load("c:\\rwg\\climax\\share\\differ6.mac")

Warning - you are redefining the MACSYMA function DIFFERENCEQ (which is why it doesn't load automatically) (c17) makerecurrence(f(n),fib(n-12)*fib(n-9)*fib(n+5)) (d17) f(n) - 3 f(n - 1) - 6 f(n - 2) + 3 f(n - 3) + f(n - 4) However, trying to recover fib(n)^3 by imposing initial conditions, (c18) differenceq(%,f(n),f(0)=0,f(1)=1,f(2)=1,f(3)=8) we get instead the alternative form fib(3 n) + 3 fib(- n) (d18) f(n) = --------------------- 5 Without the initial conditions, (c19) differenceq(d17,f(n)) (d19) f(n) = %r15 fib(3 n + 1) + %r16 fib(3 n) + %r14 fib(- n) + %r13 fib(1 - n) I.e., for any integers a,b,c, there are constant values of %r13,...,%r16 to make f(n) = fib(n+a)*fib(n+b)*fib(n+c). Similarly for fib(n)^2: (c21) makerecurrence(g(k),fib(k)^2) (d21) g(k) - 2 g(k - 1) - 2 g(k - 2) + g(k - 3) (c22) differenceq(%,g(k),g(0)=0,g(1)=1,g(2)=1) k 2 fib(2 k + 1) - fib(2 k) - 2 (- 1) (d22) g(k) = ------------------------------------ 5 a peculiar expression for fib(k)^2. To those of you running my Macsyma enhancements, I apologize upon discovering that the MAKERECURRENCE in \rwg\climax\share\differ6.mac returns recurrences of needlessly high order, which I just corrected in differ6 by exhaustively trying all "factors" of the difference operator, in spite of a so-far successful conjecture as to which factor to try. Also, since the even power annihilators contain the difference operator as a factor, there is a lower order operator that renders fib(k)^(2*n) a non-zero constant. --rwg