Faster Recursive Fibonacci Numbers

[RJB]

I noticed some discussion of recursion..... the trick is to find a
formula where the arguments are divided, not decremented.
I've had a "divide-and-conquer" recursion for the Fibonacci numbers
for a couple of years in C++ but just for fun rewrote it
in Python.  It was easy.  Enjoy.  And tell me how I can improve it!

def fibo(n):
	"""A Faster recursive Fibonaci function
Use a formula from Knuth Vol 1 page 80, section 1.2.8:
	   If F[n] is the n'th Fibonaci number then
		   F[n+m] = F[m]*F[n+1] + F[m-1]*F[n].
  First set m = n+1
   F[ 2*n+1 ] = F[n+1]**2 + F[n]*2.

  Then put m = n in Knuth's formula,
	   F[ 2*n ] = F[n]*F[n+1] + F[n-1]* F[n],
   and replace F[n+1] by F[n]+F[n-1],
	   F[ 2*n ] = F[n]*(F[n] + 2*F[n-1]).
"""
	if n<=0:
		return 0
	elif n<=2:
		return 1
	elif n%2==0:
		half=n//2
		f1=fibo(half)
		f2=fibo(half-1)
		return f1*(f1+2*f2)
	else:
		nearhalf=(n-1)//2
		f1=fibo(nearhalf+1)
		f2=fibo(nearhalf)
		return f1*f1 + f2*f2


RJB the Lurker
http://www.csci.csusb.edu/dick/cs320/lab/10.html